operations research
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OPERATIONS RESEARCH
UNIT 1: APPROXIMATIONS Apply given Technologies in estimating solutions in business environment 1.1 Newton – Raphson iteration method for solving polynomial equations. 1.2 Trapezium Rule for approximating a definite integral 1.3 Simpson’ Rule for approximating a definite integral. 1.4 Maclaurin Series expansion
� The Trapezium Rule
The Trapezium Rule is a method of finding the approximate value of an integral between two limits.
The area involved is divided up into a number of parallel strips of equal width.
Each area is considered to be a trapezium (trapezoid).
If there are n vertical strips then there is n+ 1 vertical line (ordinates) bounding them?
The limits of the integral are between a and b, and each vertical line has length y1 y2 y3... yn+1
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Therefore in terms of the all the vertical strips, the integral is given by:
approx. integral = (strip width) x (average of first and last y-values, plus the sum of all y values between the second and second-last value)
The trapezium rule is a way of estimating the area under a curve. We know that the area under a curve is given by integration, so the trapezium rule gives a method of estimating integrals. This is useful when we come across integrals that we don't know how to evaluate.
The trapezium rule works by splitting the area under a curve into a number of trapeziums, which we know the area of.
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If we want to find the area undinto smaller intervals, each of which has length h (see diagram above).
Then we find that:
Where y0 = f(x0) and y1 = f(x1) etc
If the original interval was split up into n smaller interv
Example
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If we want to find the area under a curve between the points x0 and xn, we divide this interval up into smaller intervals, each of which has length h (see diagram above).
) etc
If the original interval was split up into n smaller intervals, then h is given by: h = (x
3
, we divide this interval up
als, then h is given by: h = (xn - x0)/n
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� Maclaurin Series
The infinite series expansion for f(x) about x = 0 becomes:
f '(0) is the first derivative evaluated at x = 0, f ''(0) is the second derivative evaluated at x = 0, and so on.
[Note: Some textbooks call the series on this page "Taylor Series" (which they are, too), or "series expansion" or "power series".]
Maclaurin’s Series. A series of the form
Such a series is also referred to as the expansion (or development) of the function f(x) in powers of x, or its expansion in the neighborhood of zero. Maclaurin’s series is best suited for finding the value of f(x) for a value of x in the neighborhood of zero. For values of x close to zero the successive terms in the expansion grow small rapidly and the value of f(x) can often be approximated by summing only the first few terms.
A function can be represented by a Maclaurin series only if the function and all its derivatives exist for x = 0. Examples of functions that cannot be represented by a Maclaurin series: 1/x, ln x, cot x.
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Example 1 Expand ex in a Maclaurin Series and determine the interval of convergence.
Solution. f(x) = ex, f '(x) = ex, f ''(x) = ex, f '''(x) = ex, ........ , f(n)(x) = ex
and
f(0) = 1, f '(0) = 1, f ''(0) = 1, f '''(0) = 1, ....... ,f(n)(0) = 1
so
Example 2. Expand sin x in a Maclaurin Series and determine the interval of convergence.
Solution. f(x) = sin x, f'(x) = cos x, f''(x) = - sin x, f'''(x) = - cos x, ......
Since sin 0 = 0 and cos 0 = 1 the expansion is
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Simpson's rule
Simpson's rule can be derived by approximating the integrand interpolant P (x) (in red).
In numerical analysis, Simpson's ruleapproximation of definite integrals
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Simpson's rule can be derived by approximating the integrand f (x) (in blue) by the quadratic
Simpson's rule is a method for numerical integration, the numerical definite integrals. Specifically, it is the following approximation:
.
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) (in blue) by the quadratic
, the numerical . Specifically, it is the following approximation:
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� Newton's method
In numerical analysis, Newton's methodafter Isaac Newton and Joseph Rto the roots (or zeroes) of a realmethods, succeeded by Halley's method
The Newton-Raphson method in one variable:
Given a function ƒ(x) and its derivativeis reasonably well-behaved a better approximation
Geometrically, x1 is the intersection point of the process is repeated until a sufficiently accurate value is reached:
A. Description The function ƒ is shown in blue and the tangent line is in red. We see that approximation than xn for the root
The idea of the method is as follows: one starts with an initial guess which is reasonably close to the true root, then the function is approximated by its tools of calculus), and one computes the elementary algebra). This x-intercept will typically be a better approximatthan the original guess, and the method can be
Suppose ƒ : [a, b] → R is a differentiablereal numbers R. The formula for converging on the root can be easily derived. Suppose we have some current approximation xnreferring to the diagram on the right. We know from the definition othat it is the slope of a tangent at that point.
That is
Here, f ' denotes the derivative
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Newton's method
Newton's method (also known as the Newton–Raphson methodJoseph Raphson, is a method for finding successively better approximations
eal-valued function. The algorithm is first in the class of Halley's method.
Raphson method in one variable:
derivative ƒ '(x), we begin with a first guess x0. Provided the function behaved a better approximation x1 is
is the intersection point of the tangent line to the graph of f, with the xprocess is repeated until a sufficiently accurate value is reached:
is shown in blue and the tangent line is in red. We see that xn+1
for the root x of the function f.
The idea of the method is as follows: one starts with an initial guess which is reasonably close to the true root, then the function is approximated by its tangent line (which can be computed using the
), and one computes the x-intercept of this tangent line (which is easily done with intercept will typically be a better approximation to the function's root
than the original guess, and the method can be iterated.
differentiable function defined on the interval [a, b. The formula for converging on the root can be easily derived. Suppose we have
n. Then we can derive the formula for a better approximation, referring to the diagram on the right. We know from the definition of the derivative at a given point that it is the slope of a tangent at that point.
of the function f. Then by simple algebra we can derive
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Raphson method), named , is a method for finding successively better approximations
. The algorithm is first in the class of Householder's
. Provided the function
graph of f, with the x-axis. The
+1 is a better
The idea of the method is as follows: one starts with an initial guess which is reasonably close to the h can be computed using the
intercept of this tangent line (which is easily done with ion to the function's root
b] with values in the . The formula for converging on the root can be easily derived. Suppose we have
. Then we can derive the formula for a better approximation, xn+1 by f the derivative at a given point
. Then by simple algebra we can derive
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We start the process off with some arbitrary initial value in the absence of any intuition about where the zero might lie, a "guess and check" method might narrow the possibilities to a reasonably small interval by appealing to theorem.) The method will usually converge, provided this initial guess is close enough to the unknown zero, and that ƒ'(x0) ≠least quadratic (see rate of convergencethat the number of correct digits roughly at least doubles in every step. More details can be found in the analysis section below.
B. Examples Square root of a number
Consider the problem of finding the square root of a number. There are many computing square roots, and Newton's method is one.
For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to
The function to use in Newton's method is then,
with derivative,
With an initial guess of 10, the sequence given by Newton's method is
Where the correct digits are underlined. With only a few iterations one can obtain a solution accurate to many decimal places.
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ss off with some arbitrary initial value x0. (The closer to the zero, the better. But, in the absence of any intuition about where the zero might lie, a "guess and check" method might narrow the possibilities to a reasonably small interval by appealing to the intermediate value
.) The method will usually converge, provided this initial guess is close enough to the ≠ 0. Furthermore, for a zero of multiplicity 1, the convergence is at
rate of convergence) in a neighbourhood of the zero, which intuitively means that the number of correct digits roughly at least doubles in every step. More details can be found in
Consider the problem of finding the square root of a number. There are many , and Newton's method is one.
For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to
The function to use in Newton's method is then,
With an initial guess of 10, the sequence given by Newton's method is
Where the correct digits are underlined. With only a few iterations one can obtain a solution accurate
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. (The closer to the zero, the better. But, in the absence of any intuition about where the zero might lie, a "guess and check" method might
intermediate value .) The method will usually converge, provided this initial guess is close enough to the
1, the convergence is at of the zero, which intuitively means
that the number of correct digits roughly at least doubles in every step. More details can be found in
Consider the problem of finding the square root of a number. There are many methods of
For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to
Where the correct digits are underlined. With only a few iterations one can obtain a solution accurate
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Solution of a non-polynomial equation
Consider the problem of finding the positive number finding the zero of f(x) = cos(xx3 > 1 for x > 1, we know that our zero lies between 0 and 1. We try a starting value of (Note that a starting value of 0 will lead to an undefined result, showing the importanstarting point that is close to the zero.)
The correct digits are underlined in the above example. In particular, decimal places given. We see that the number of correct digits after the decimal point increase2 (for x3) to 5 and 10, illustrating the quadratic convergence.
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polynomial equation
sider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(
1, we know that our zero lies between 0 and 1. We try a starting value of (Note that a starting value of 0 will lead to an undefined result, showing the importanstarting point that is close to the zero.)
The correct digits are underlined in the above example. In particular, x6 is correct to the number of decimal places given. We see that the number of correct digits after the decimal point increase
) to 5 and 10, illustrating the quadratic convergence.
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. We can rephrase that as . Since cos(x) ≤ 1 for all x and
1, we know that our zero lies between 0 and 1. We try a starting value of x0 = 0.5. (Note that a starting value of 0 will lead to an undefined result, showing the importance of using a
is correct to the number of decimal places given. We see that the number of correct digits after the decimal point increases from
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Newton-Raphson Method
This uses a tangent to a curve near one of its roots and the fact that where the tangent meets the x-axis gives an approximation to the root.
The iterative formula used is:
Example
Find correct to 3 d.p. a root of the equation
f(x) = 2x2 + x - 6
given that there is a solution near x = 1.4
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UNIT 2: LINEAR PROGRAMMING HOURS: 20 LINEAR PROGRAMMING Is a technique used to determine how best to allocate personnel, equipment, materials, finance, land, transport e.t.c. , So that profit are maximized or cost are minimized or other optimization criterion is achieved. Linear programming is so called because all equations involved are linear. The variables in the problem are Constraints. It is these constraints, which gives rise to linear equations or Inequalities. The expression to the optimized is called the Objective function usually represented by an equation. Question 1: A furniture factory makes two products: Chairs and tables. The products pass through 3 manufacturing stages; Woodworking, Assembly and Finishing. The Woodworking shop can make 12 chairs an hour or 6 tables an hour. The Assembly shops can assembly 8 chairs an hour or 10 tables an hour. The Finishing shop can finish 9 chairs or 7 tables an hour. The workshop operates for 8 hours per day. If the contribution to profit from each Chair is $4 and from each table is $5, determine by Graphical method the number of tables and chairs that should be produced per day to maximize profits. Solution: Let number of chairs be X. Let number of tables be Y. Objective Function is: P = 4X + 5Y. Constraints: WW: X/12 + Y/6 <= 8 X + 2Y <= 96 when X=0 Y= 48 when Y=0 X= 96 AW: X/8 + Y/10 <= 8 5X + 4Y <= 320 when X=0 Y=80 when Y=0 X=64 FNW: X/9 + Y/7 <= 8 7X + 9Y <= 504 when X=0 Y=56 when Y=0 X=72 X>=0 Y>=0
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100 90 80 70 60 50 40 30 20 P------� this gives the maximum point 10 0 10 20 30 40 50 60 70 80 90 100 Question 2: Mr. Chabata is a manager of an office in Guruwe; he decides to buy some new desk and chairs for his staff. He decides that he need at least 5 desk and at least 10 chairs and does not wish to have more than 25 items of furniture altogether. Each desk will cost him $120 and each chair will cost him $80. He has a maximum of $2400 to spend altogether. Using the graphical method, obtain the maximum number of chairs and desk Mr. Chabata can buy. Solution: Let X represents number of Desk. Let Y represents number of Chairs. Objective Function is: P = 120X + 80Y. Constraints: X >= 5 Y>= 10 X + Y <= 25
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30 25 20 P�Point P (10,15) gives the maximum point 15 10 5 0 5 10 15 20 25 30 The Optimum Solution = 120 * 10 + 80 * 15 = 1200 + 1200 = 2400 Question 3: A manufacturer produces two products Salt and Sugar. Salt has a contribution of $30 per unit and Sugar has $40 per unit. The manufacturer wishes to establish the weekly production, which maximize the contribution. The production data are shown below: Production Unit Machine Hours Labour Hours Materials in Kg Salt 4 4 1 Sugar 2 6 1 Total available per unit 100 180 40 Because of the trade agreement sales of Salt are limited to a weekly maximum of 20 units and to honor an agreement with an old established customer, at least 10 units of Sugar must be sold per week. Solution: Let X represents Salt. Let Y represents Sugar.
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Objective Function is: P = 30X + 40Y. Constraints: 4X + 2Y <= 100 {machine hours} 4X + 6Y <= 180 {Labour hours} X + Y <= 40 {material} Y>= 10 X<= 20 X>=0; Y>=0;
SIMPLEX METHOD: The graphical outlined above can only be applied to problems containing 2 variables. When 3 or more variables are involved we use the Simplex method. Simplex comprises of series of algebraic procedures performed to determine the optimum solution. In Simplex method we first convert inequalities to equations by introducing a Slack variable. A Slack variable represents a spare capacity in the limitation.
Simplex Method for Standard Maximization Problem
To solve a standard maximization problem using the simplex method, we take the following steps:
Step 1. Convert to a system of equations by introducing slack variables to turn the constraints into equations, and rewriting the objective function in standard form.
Step 2. Write down the initial tableau.
Step 3. Select the pivot column: Choose the negative number with the largest magnitude in the bottom row (excluding the rightmost entry). Its column is the pivot column. (If there are two candidates, choose either one.) If all the numbers in the bottom row are zero or positive (excluding the rightmost entry), then you are done: the basic solution maximizes the objective function (see below for the basic solution).
Step 4. Select the pivot in the pivot column: The pivot must always be a positive number. For each positive entry b in the pivot column, compute the ratio a/b, where a is the number in the Answer column in that row. Of these test ratios, choose the smallest one. The corresponding number b is the pivot.
Step 5. Use the pivot to clear the column in the normal manner (taking care to follow the exact prescription for formulating the row operations and then relabel the pivot row with the label from the pivot column. The variable originally labeling the pivot row is the departing or exiting variable and the variable labeling the column is the entering variable.
Step 6. Go to Step 3.
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Simplex Method for Minimization Problem
To solve a minimization problem using the simplex method, convert it into a maximization problem. If you need to minimize c, instead maximize p = -c.
Example
The minimization LP problem:
Minimize C = 3x + 4y - 8z subject to the constraints 3x - 4y ≤ 12, x + 2y + z ≥ 4 4x - 2y + 5z ≤ 20 x ≥ 0, y ≥ 0, z ≥ 0
can be replaced by the following maximization problem: Maximize P = -3x - 4y + 8z subject to the constraints
3x - 4y ≤ 12, x + 2y + z ≥ 4 4x - 2y + 5z ≤ 20 x ≥ 0, y ≥ 0, z ≥ 0.
OR STEPS TO FOLLOW IN SIMPLEX METHOD:
i. Obtain the pivot column as the column with the most positive indicator row. ii. Obtain pivot row by dividing elements in the solution column by their corresponding pivot
column entries to get the smallest ratio. Element at the intersection of the pivot column and pivot row is known as pivot elements.
iii. Calculate the new pivot row entries by dividing pivot row by pivot element. This new row is entered in new tableau and labeled with variables of new pivot column.
iv. Transfer other row into the new tableau by adding suitable multiplies of the pivot row (as it appears in the new tableau) to the rows so that the remaining entries in the pivot column becomes zeroes.
v. Determine whether or not this solution is optimum by checking the indicator row entries of the newly completed tableau to see whether or not they are any positive entries. If they are positive numbers in the indicator row, repeat the procedure as from step 1.
vi. If they are no positive numbers in the indicator row, this tableau represents an optimum solution asked for, the values of the variables together with the objective function could then be stated.
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Question 1: Maximize Z = 40X + 32Y Subject to: 40X +20Y <= 600 4X + 10Y <= 100 2X + 3Y <= 38 Using the Simplex method
Solution: To obtain the initial tableau, we rewrite the Objective Function as: Z = 40X + 32Y Introducing Slack variables S1, S2, S3 in the 3 inequalities above we get: 40X + 20Y + S1 = 600 4X + 10Y + S2 = 100 2X + 3Y + S3 = 38 X >= 0 Y >=0.
TABLEAU 1: Pivot Element X Y S1 S2 S3 Solution S1 20 1 0 0 600 S2 4 10 0 1 0 100 S3 2 3 0 0 1 38 Z 40 32 0 0 0 0 Indicator Row
Pivot Column
TABLEAU 2: X Y S1 S2 S3 Solution X 0.5 0.025 0 0 15 S2 0 8 -0.1 1 0 40 S3 0 2 -0.05 0 1 8 Z 0 12 -1 0 0 -600
40
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TABLEAU 3:
X Y S1 S2 S3 Solution X 1 0 0.0375 0 -0.25 13 S2 0 0 0.1 1 -4 8 Y 0 0.025 0 0.5 4 Z 0 0 -0.7 0 -6 -648 Indicator Row
Pivot Column Conclusion: Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X = 13 and Y = 4 giving Z = 648. NB:
a) If when selecting a pivot column we have ties in the indicator row, we then select the pivot column arbitrary.
b) If all the entries in the selected pivot column are negative then the objective function is unbound and the maximum problem has no solution.
c) A minimization problem can be worked as a maximization problem after multiply the objective function and the inequalities by –1.
d) Inequalities change their signs when multiplied by negative number.
Question 2: Maximize Z = 5X1 + 4X2 Subject to: 2X1 +3X2 <= 17 X1 + X2 <= 7 3X1 + 2X2 <= 18 Using the Simplex method
Solution: Max Z = 5X1 + 4X2
Subject to: 2X1 + 3X2 + S1 = 17 X1 + X2 + S2 = 7 3X1 + 2X2 + S3 = 18 X1 >= 0 X2>=0.
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TABLEAU 1: X1 X2 S1 S2 S3 Solution S1 2 3 1 0 0 17 S2 1 1 0 1 0 7 S3 2 0 0 1 18 Z 5 4 0 0 0 0 Indicator Row
Pivot Column
TABLEAU 2: X1 X2 S1 S2 S3 Solution S1 0 5/3 1 0 -2/3 5 S2 0 1/3 0 1 -1/3 1 X1 2/3 0 0 1/3 6 Z 0 2/3 0 0 -5/3 -30
TABLEAU 3: X1 X2 S1 S2 S3 Solution X2 0 3/5 0 -2/5 3 S2 0 0 -1/5 1 -1/5 0 X1 1 0 -2/5 0 9/15 4 Z 0 0 -2/5 0 -7/5 -32
Pivot Column Conclusion: Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 4 and X2 = 3 giving Z = 32.
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Question 3: A company can produce 3 products A, B, C. The products yield a contribution of $8, $5 and $10 respectively. The products use a machine, which has 400 hours capacity in the next period. Each unit of the products uses 2, 3 and 1 hour respectively of the machine’s capacity. There are only 150 units available in the period of a special component, which is used singly in products A and C. 200 kgs only of a special Alloy is available in the period. Product A uses 2 kgs per unit and Product C uses 4kgs per units. There is an agreement with a trade association to produce no more than 50 units of product in the period. The Company wishes to find out the production plan which maximized contribution.
Solution: Maximize Z = 8X1 + 5X2+ 10X3
Subject to: 2X1 + 3X2 + X3 <= 400 {machine hour} X1 + X3 <= 150 {component} 2X1 + 4X3 <= 200 {Alloy} X2 <=50 {Sales} X1 >= 0
X2 >= 0 X3 >= 0.
Introducing slack variables: Maximize Z = 8X1 + 5X2+ 10X3
Subject to: 2X1 + 3X2 + X3 + S1= 400 X1 + X3 + S2 = 150 2X1 + 4X3 + S3 = 200 X2 + S4 =50 X1 >= 0
X2 >= 0 X3 >= 0.
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TABLEAU 1: X1 X2 X3 S1 S2 S3 S4 Solution S1 2 3 1 1 0 0 0 400 S2 1 0 1 0 1 0 0 150 S3 2 0 0 0 1 0 200 S4 0 1 0 0 0 0 1 50 Z 8 5 10 0 0 0 0 0 Indicator Row
Pivot Column NB: Ignore S4 in finding pivot row.
TABLEAU 2: X1 X2 X3 S1 S2 S3 S4 Solution S1 3/2 3 0 1 0 -1/4 0 350 S2 1/2 0 0 0 1 -1/4 0 100 X3 1/2 0 0 0 1/4 0 50 S4 0 1 0 0 0 0 1 50 Z 3 5 0 0 0 -5/2 0 -500
Pivot Column
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TABLEAU 3: X1 X2 X3 S1 S2 S3 S4 Solution S1 3/2 0 0 1 0 -1/4 -3 200 S2 1/2 0 0 0 1 -1/4 0 100 X3 1/2 0 4 0 0 1/4 0 50 X2 0 0 0 0 0 1 50 Z 3 0 0 0 0 -5/2 -5 -750
Pivot Column
TABLEAU 4: X1 X2 X3 S1 S2 S3 S4 Solution S1 0 0 -3 1 0 -1 -3 50 S2 0 0 -1 0 1 -1/2 0 50 X1 0 2 0 0 1/2 0 100 X2 0 1 0 0 0 0 1 50 Z 0 0 -6 0 0 -4 -5 -1050
Pivot Column Conclusion: Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 100 and X2 = 50 giving Z = 1050. Two slack variable S1 =0 and S2 =0. This means that there is no value to be gained by altering the machine hours and component constraints.
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� GENERAL RULE: Constraints only have a valuation when they are fully utilized. These valuations are known as the SHADOW Prices or Shadow Costs or Dual Prices or Simplex Multipliers. A constraint only has a Shadow price when it is binding i.e. fully utilized and the Objective function would be increased if the constraint were increased by 1 unit. When solving Linear Programming problems by Graphical means the Shadow price have to be calculated separately. When using Simplex method they are an automatic by product. MIXED CONSTRAINTS:
This involves constraints containing a mixture of <= and >= varieties. Using Maximization problem we use “Less than or equal to” type. (<=). Faced with a problem which involves a mixture of <= and >= variety. The alternative solution to deal with “Greater than or equal to” (>=) type is to multiply both sides by –1 and change the inequality sign.
Question 4:
Maximize Z = 5X1 + 3X2+ 4X3
Subject to: 3X1 + 12X2 + 6X3 <= 660 6X1 + 6X2 + 3X3 <= 1230 6X1 + 9X2 + 9X3 <= 900 X3 >=10 Solution: The only constraint that need to be changed is X3 >=10 by multiply by –1 both sides and we get: -X3 <= -10 Maximize Z = 5X1 + 3X2+ 4X3
Subject to: 3X1 + 12X2 + 6X3 + S1 = 660 6X1 + 6X2 + 3X3 + S2 = 1230 6X1 + 9X2 + 9X3 <+ S3 = 900 -X3+ S4 = -10
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TABLEAU 1: X1 X2 X3 S1 S2 S3 Solution S1 3 12 6 1 0 0 600 S2 6 6 3 0 1 0 1200 S3 9 9 0 0 1 900 Z 5 3 4 0 0 0 0 Indicator Row
Pivot Column
FINAL TABLEAU: X1 X2 X3 S1 S2 S3 Solution S1 0 15/2 3/2 1 0 -1/2 150 S2 0 -3 -6 0 1 -1 300 X1 3/2 3/2 0 0 1/6 150 Z 0 -9/2 -7/2 0 0 -5/6 -750
Pivot Column Conclusion: Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 150 producing Z = $750. Plus production to satisfy constrain (d) 20 units of X3 producing $ 40 contribution. Therefore Total solution is 150 units of X1 and 10 units of X3 giving $790. NB: Maximize Z = 5(150) + 3(0)+ 4(10) = $790.
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Question 5:
Maximize Z = 3X1 + 4X2
Subject to: 4X1 + 2X2 <= 100 4X1 + 6X2 <= 180 X1 + X2 <= 40 X1 <= 20 X2 >=10 Solution: The only constraint that need to be changed is X2 >=10 by multiply by –1 both sides and we get: -X2 <= -10 Maximize Z = 3X1 + 4X2
Subject to: 4X1 + 2X2 + S1 = 100 {1} 4X1 + 6X2 + S2 = 180 {2} X1 + X2 + S3 = 40 {3} X1 + S4 = 20 {4} -X2 + S5 =-10 {5}
TABLEAU 1: X1 X2 S1 S2 S3 S4 S5 Solution S1 4 1 0 0 0 0 100 S2 4 6 0 1 0 0 0 180 S3 1 1 0 0 1 0 0 40 S4 1 0 0 0 0 1 0 20 S5 0 -1 0 0 0 0 1 -10 Z 3 4 0 0 0 0 0 0 Indicator
Row Pivot Column The problem is then solved by the usual Simplex iterations. Each iteration improves on the one before and the process continues until optimum is reached.
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TABLEAU 2: X1 X2 S1 S2 S3 S4 S5 Solution X2 0 0 0 0 0 -1 10 S2 4 0 1 0 0 0 2 80 S3 4 0 0 1 0 0 6 120 S4 1 0 0 0 1 0 1 30 S5 1 0 0 0 0 1 0 20
Z 4 0 0 0 0 0 4 -40 This shows 10X2 being produced and $40 contribution. The first four constraints have surpluses of 80, 120, 30 and 20 respectively. Not optimums as there are still positive values in Z row.
TABLEAU 3: X1 X2 S1 S2 S3 S4 S5 Solution X2 0.667 1 0 0.167 0 0 0 30 S2 2.667 0 1 -0.333 0 0 0 40 S3 -0.333 0 0 -0.167 0 0 0 10 S4 1 0 0 0 1 1 0 20 S5 0.333 0 0 0.167 0 0 20
Z 0.333 0 0 -0.667 0 0 0 -120 This shows 30X2 being produced and $120 contribution. All constraints have surpluses except Labour hours. Not optimum as there is a positive value in Z row.
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TABLEAU 4: X1 X2 S1 S2 S3 S4 S5 Solution X1 0 0.375 0.125 0 0 0 15 X2 0 1 -0.25 0.250 0 0 2 0 S3 0 0 -0.125 -0.125 1 0 0 5 S4 0 0 -0.375 0.125 0 1 0 5 S5 0 0 -0.25 0.25 0 0 1 10
Z 0 0 -0.125 -0.625 0 0 0 -125 Conclusion: Since the indicator row is negative the solution is optimum with 15X1 and 20X2 giving $125 contribution. Shadow prices are X1 = $0.125 and X2 = $0.625. Non-binding constraints are {3}, {4}, {5} with 5, 5 and 10 spare respectively.
DUALITY: There is a dual or inverse for every Linear Programming problem. Because solving Simplex problem in Maximization is quite simple and straightforward, it is usually to convert a Minimization problem into Maximization problem using dual. The dual or inverse of Linear Programming problem is obtained by making the constraints in the inequalities coefficient of the new objective function. The cofficiences of the original inequalities are combined with the cofficiences of the original objective function as the constraints.
Question 6: Minimize Z = 40X1 + 50X2 Subject to: 3X1 + 5X2 >= 150 5X1 + 5X2 >= 200 3X1 + X2 >= 60 X1, X2 >=0
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Solution: The Dual Linear Programming problem is as follows: Maximize P = 150Y1 + 200Y2 +60 Y3 Subject to: 3Y1 + 5Y2 + 3Y3 <= 40 5Y1 + 5Y2 + Y3 <= 50 Y1>=0, Y2>=0, Y3>=0. Y1 Y2 Y3 S1 S2 Solution S1 3 3 1 0 40
S2 5 5 1 0 1 50
P 150 200 60 0 0 0 Indicator Row Pivot Column Y1 Y2 Y3 S1 S2 Solution Y2 3/5 1 3/5 1/5 0 8
S2 0 -2 -1 1 10
P 30 0 -60 -40 0 -1600 Y1 Y2 Y3 S1 S2 Solution Y2 0 1 1.2 0.5 -0.3 5
Y1 0 -1 -0.5 0.5 5
P 0 0 -30 -25 -15 -1750 Conclusion:
Since the indicator row is negative the solution is optimum with 5Y1 and 5Y2 giving $1750 contribution.
5
2
1
OPERATIONS RESEARCH
Question 7:
Minimize Z = 16X1 + 11X2 Subject to: 2X1 + 3X2 >= 3 5X1 + X2 >= 8 X1, X2 >=0 Using the Dual problem. Solution: The Dual Linear Programming problem is as follows: Maximize P = 3Y1 + 8Y2 Subject to: 2Y1 + 5Y2 <= 16 3Y1 + Y2 <= 11 Y1>=0, Y2>=0. Y1 Y2 S1 S2 Solution S1 2 1 0 16
S2 3 1 0 1 11
Z 3 8 0 0 0 Indicator Row Pivot Column Y1 Y2 S1 S2 Solution Y1 0.4 0.2 0 3.2
S2 2.6 0 -0.2 1 7.8
Z -0.2 0 -1.6 0 -25.6 Conclusion:
Since the indicator row is negative the solution is optimum. Hence P = 3Y1 + 8Y2 is maximum when Y1 = 3.2 and Y2= 0 and P = 25.6 In the primary problem, the solution correspond the slack variable values in the final tableau. i.e. X1= S1 = 1.6 X2= S2 = 0. Hence Z = 16*1.6 + 11*0 => 25.6
5
1
OPERATIONS RESEARCH
TERMS USED WITH LINEAR PROGRAMMING:
� FEASIBLE REGION: Represents all combinations of values of the decision variables that satisfy every restriction simultaneous. The corner point of the feasible region gives what is known as BASIC FEASIBLE SOLUTION i.e. the solution that is given by the coordinates at the intersection of any two binding constraints.
� BINDING CONSTRAINTS: Is an inequality whose graph forms the bounder of the feasible region.
� NON BINDING CONSTRAINTS: Is an inequality, which does not conform to the feasible region. � DUAL PRICE / SHADOW PRICES: It is important that management information to value the scarce resources. These are known as Dual price / Shadow price. Derived from the amount of increase (or decrease) in contribution that would arise if one more (or one less) unit of scare resource was available.
OPERATIONS RESEARCH
ASSIGNMENT PROBLEM: This is the problem of assigning any worker to any job in such a way that only one worker is assigned to each job, every job has one worker assigned to it and the cost of completing all jobs is minimized.
� STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
a. Layout a two way table containing the cost for assigning a worker to a job. b. In each row subtract the smallest cost in the row from every cost in the row. Make a new
table. c. In each column of the new table, subtract the smallest cost from every cost in the column.
Make a new table. d. Draw horizontal and vertical lines only through zeroes in the table in such a way that the
minimum number of lines is used. e. If the minimum number of lines that covers zeroes is equal to the number of rows in the
table the problem is finished. f. If the minimum number of lines that covers zeroes is less than the number of rows in the
table the problem is not finished go to step g. g. Find the smallest number in the table not covered by a line.
i. Subtract that number from every number that is not covered by a line. ii. Add that number to every number that is covered by two lines. iii. Bring other numbers unchanged. Make a new table.
h. Repeat step d through step g until the problem is finished.
OPERATIONS RESEARCH
Question 1: Use the assignment method to find the minimum distance assignment of Sales representative to Customer given the table below: What is the round trip distance of the assignment?
Sales Representative Customer Distance (km) A 1 200 A 2 400 A 3 100 A 4 500 B 1 1000 B 2 800 B 3 300 B 4 400 C 1 100 C 2 50 C 3 600 C 4 200 D 1 700 D 2 300 D 3 100 D 4 250
TABLEAU 1:
1 2 3 4 A 200 400 100 500 B 1000 800 300 400 C 100 50 600 200 D 700 300 100 250
TABLEAU 2:
1 2 3 4 A 100 300 0 400 B 700 500 0 100 C 50 0 550 150 D 600 200 0 150
OPERATIONS RESEARCH
TABLEAU 3:
1 2 3 4 A 50 300 0 300 B 650 500 0 0 C 0 0 550 50 D 550 200 0 50
TABLEAU 4:
1 2 3 4 A 0 250 0 300 B 600 450 0 0 C 0 0 600 100 D 500 150 0 50
Conclusion: Since the number of lines is now equal to number of rows, the problem is finished with the following assignment: SALES REP CUSTOMER DISTANCE A 1 200 B 4 400 C 2 50 D 3 100
750 km Therefore total round Trip distance = 750 km * 2 => 1500 km
Question 2:
A foreman has 4 fitters and has been asked to deal with 5 jobs. The times for each job are estimated as follows. A B C D 1 6 12 20 12 2 22 18 15 20 3 12 16 18 15 4 16 8 12 20 5 18 14 10 17 Allocate the men to the jobs so as to minimize the total time taken.
OPERATIONS RESEARCH
Solution: Insert a Dummy fitter so that number of rows will be equal to number of column.
TABLEAU 1: A B C D DUMMY
1 6 12 20 12 0 2 22 18 15 20 0 3 12 16 18 15 0 4 16 8 12 20 0 5 18 14 10 17 0
TABLEAU 2: A B C D DUMMY
1 0 4 10 0 0 2 16 10 5 8 0 3 6 8 8 3 0 4 10 0 2 8 0 5 12 6 0 5 0
TABLEAU 3: A B C D DUMMY
1 0 4 10 0 3 2 13 7 2 5 0 3 3 5 5 0 0 4 10 0 2 8 3 5 12 6 0 5 3
Conclusion: Since the number of lines is now equal to number of rows, the problem is finished with the following assignment: FITTERS JOBS TOTALS A 1 6 B 4 8 C 5 10 D 2 15 Dummy 2 0 39
OPERATIONS RESEARCH
� THE ASSIGNMENT TECHNIQUE FOR MAXIMIZING PROBLEMS: Maximizing assignment problem typically involves making assignments so as to maximize contributions. � STEPS INVOLVED:
a) Reduce each row by largest figure in that row and ignore the resulting minus signs.
b) The other procedures are the same as applied to minimization problems.
Question 3: A foreman has 4 fitters and has been asked to deal with 4 jobs. The times for each job are estimated as follows. W X Y Z A 25 18 23 14 B 38 15 53 23 C 15 17 41 30 D 26 28 36 29 Allocate the men to the jobs so as to maximize the total time taken. Solution:
TABLEAU 1: W X Y Z
A 0 7 2 7 B 15 38 0 30 C 26 24 0 11 D 10 8 0 7
TABLEAU 2: W X Y Z
A 0 0 2 0 B 15 31 0 23 C 26 17 0 4 D 10 1 0 0
OPERATIONS RESEARCH
TABLEAU 3: W X Y Z
A 0 0 3 1 B 14 30 0 23 C 25 16 0 4 D 9 0 0 0
TABLEAU 4: W X Y Z
A 0 0 7 1 B 10 26 0 19 C 21 12 0 0 D 9 0 4 0
Conclusion: Since the number of lines is now equal to number of rows, the problem is finished with the following assignment: A W 25 B Y 53 C Z 30 D X 28 $136 Question 4: A Company has four salesmen who have to visit four clients. The profit records from previous visits are shown in the table and it is required to Maximize profits by the best assignment.
A B C D 1 6 12 20 12 2 22 18 15 20 3 12 16 18 15 4 16 8 12 20 Solution:
TABLEAU 1: W X Y Z
1 6 12 20 12 2 22 18 15 20 3 12 16 18 15 4 16 8 12 20
OPERATIONS RESEARCH
TABLEAU 2: W X Y Z
1 14 8 0 8 2 0 4 7 2 3 6 2 0 3 4 4 10 8 0
TABLEAU 3: W X Y Z
1 14 6 0 8 2 0 2 7 2 3 6 0 0 3 4 4 10 8 0
Conclusion: Since the number of lines is now equal to number of rows, the problem is finished with the following assignment: 4 D 20 2 A 22 1 C 20 3 B 16 $78
OPERATIONS RESEARCH
TRANSPORTATION PROBLEM: This is the problem of determining routes to minimize the cost of shipping commodities from one point to another. The unit cost of transporting the products from any origin to any destination is given. Further more, the quantity available at each origin and quantity required at each destination is known.
⇒ STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
� Arrange the problem in a table with row requirements on the right and column requirements at the bottom. Each cell should contain the unit cost approximates to the shipment.
� Obtain an initial solution by using the North West Corner rule. By this method one begins at the up left corner cell and works up to the lower right corner. Place the quantity of goods in the first cell equal to the smallest of the rows or column totals in the table. Balance the row and column respectively until you reach the lower right hand cell.
� Find cell values for every empty cell by adding and subtract around the closing loop. � If all empty cell have + values the problem is finished. If not pick the cell with most
– (negative) value. Allocate a quantity of goods to that cell by adding and subtract the small value of the column or row entries in the closed loop. The closed loop techniques involves the following steps:
� Pick an empty cell, which has no quantity of goods in it. � Place a + sign in the empty cell. � Use only occupied cells for the rest of the closed loop. � Find an occupied cell that has occupied values in the same row or same
column and place a – (negative) sign in this cell. � Go to the next occupied cell and place + sign in it. � Continue in this manner until you return to the unoccupied cell in which you
started. � A closed loop exists for every empty cell as long as they are occupied cell
equal to number of rows + number of column – 1.
OPERATIONS RESEARCH
Question 1:
A firm has 3 factory (A, B, C) and 4 warehouses (1, 2, 3, 4). The capacities of the factories and the requirements of the warehouse are in the table below. FACTORY CAPACITY WAREHOUSE REQUIREMENTS
A 220 1 160 B 300 2 260 C 380 3 300 4 180
The cost of shipping one unit from each factory to each warehouse is given below. FACTORY WAREHOUSE COST $
o A 1 3 o A 2 5 o A 3 6 o A 4 5 o B 1 7 o B 2 4 o B 3 9 o B 4 6 o C 1 5 o C 2 12 o C 3 10 o C 4 8
Using the Transportation method, find the least cost shipping schedule and state what it is ?. Solution:
TABLEAU 1: 1 2 3 4 Capacity A 160 60 220 B 200 100 300 C 200 180 380 Req 160 260 300 180 900 This is the initial solution, which costs (160 * 3) + (60 * 5) + (200 * 4) + (100 * 9) + (200 * 10) + (180 * 8) = $5920.00
3 5
10 5 8 12
9
6
6 7 4
5
- 1 5
7 2
- 3 - 4
OPERATIONS RESEARCH
TABLEAU 2:
1 2 3 4 Capacity A 160 60 220 B 260 40 300 C 200 180 380 Req 160 260 300 180 900 The costs = (160 * 3) + (60 * 6) + (260 * 4) + (40 * 9) + (200 * 10) + (180 * 8) = $5680.00
TABLEAU 3:
1 2 3 4 Capacity A 220 220 B 260 40 300 C 160 40 180 380 Req 160 260 300 180 900 The costs = (160 * 5) + (40 * 10) + (260 * 4) + (40 * 9) + (220 * 6) + (180 * 8) = $5360.00
TABLEAU 4: 1 2 3 4 Capacity A 220 220 B 260 40 300 C 160 80 140 380 Req 160 260 300 180 900 The costs = (160 * 5) + (80 * 10) + (260 * 4) + (40 * 6) + (220 * 6) + (140 * 8) = $5320.00
3 5
10 5 8 12
9
6
6 7 4
5
- 1 5
7 -2
1 4
3 5
10 5 8 12
9
6
6 7 4
5
- 1 3
7
2 1 4
3 5
10 5 8 12
9
6
6 7 4
5
1 4
6
2 1 3
OPERATIONS RESEARCH
Conclusion: Since all the cell values are positive the solution is optimum with the following allocations: A Supplies 220 to 3 B Supplies 260 to 2
C Supplies 160 to 1 C Supplies 80 to 3
C Supplies 140 to 4
With a minimum cost of $ 5320.00
QUESTION Below is a transportation problem where costs are in thousand of dollars. SOURCES DESTINATIONS A B C CAPACITIES X 14 13 15 500 Y 16 15 12 400 Z 20 15 16 600 REQUIREMENTS 700 300 500
i. Solve this problem fully indicating the optimum delivery allocations and the corresponding total delivery cost. [6 marks]
ii. There are two optimum solutions. Find the second one [4 marks]. iii. Solve the same problem considering X�A is an infeasible (prohibited / impossible) route
and find the new total transportation cost [7 marks]. iv. If under consideration is a road network in a war zone, what is the simple economic effect of
bombing a bridge between X and A? [3 marks]. Solution: PART (i)
TABLEAU 1: A B C Capacity X 500 500 Y 200 200 400 Z 100 500 600 Req 700 300 500 1500 The initial solution = (500 * 14) + (200 * 16) + (15 * 200) + (100 * 15) + (500 * 16)
3 5
10 5 12
9
6
7 4
0
-4
4
1
OPERATIONS RESEARCH
= $227000000
TABLEAU 2: A B C Capacity X 500 500 Y 200 200 400 400 Z 300 300 600 Req 700 300 500 1500 Hence delivery allocations are:
X Supplies 500 to A Y Supplies 200 to A
Y Supplies 200 to C Z Supplies 300 to B
Z Supplies 300 to C
With a minimum cost of (500 * 14) + (200 * 16) + (15 * 300) + (200 * 12) + (300 * 16) = $21900 0000 PART (ii) The existence of an alternative least cost solution is indicated by a value of zero in an unoccupied cell in the final table. We add and subtract the smallest quantity in the column or row of the zero to get the alternative.
TABLEAU 1: A B C Capacity X 500 500 Y 400 400 400 Z 200 300 100 600 Req 700 300 500 1500
3 5
10 5 12
9
6
7 4
4
4
0
5
3 5
10 5 12
9
6
7 4
4
4 0
5
OPERATIONS RESEARCH
Hence delivery allocations are:
X Supplies 500 to A Y Supplies 400 to C
Z Supplies 200 to A Z Supplies 300 to B
Z Supplies 100 to C
With a minimum cost of (500 * 14) + (200 * 20) + (15 * 300) + (400 * 12) + (100 * 16) = $21900 0000 PART (iii)
TABLEAU 1: A B C Capacity X --- 300 200 500 Y 400 400 400 Z 300 300 600 Req 700 300 500 1500 Hence delivery allocations are:
X Supplies 300 to B X Supplies 200 to C
Y Supplies 400 to A Z Supplies 300 to A
Z Supplies 300 to C
With a minimum cost of (300 * 13) + (200 * 15) + (16 * 400) + (300 * 20) + (300 * 16) = $24100 0000 PART (iv) The simple economic effect of bombing the bridge between X and A = 24100 0000 – 21900 0000 = 2200 000
3 5
10 5 12
9
6
7 4
1
5 0
OPERATIONS RESEARCH
� DUMMIES: This is an extra row or column in a transportation table with zero cost in each cell and with a total equal to the difference between total capacity and total demand. In an unbalance transportation problem a dummy source or destination is introduced.
QUESTION:
The transport manager of a company has 3 factories A, B and C and four warehouses I, II, III and IV is faced with a problem of determining the way in which factories should supply warehouses so as to minimize the total transportation costs. In a given month the supply requirements of each warehouse, the production capacities of the factories and the cost of shipping one unit of product from each factory to each warehouse in $ are shown below.
FACTORY WAREHOUSES I II III IV PRO AVAIL A 12 23 43 3 6 B 63 23 33 53 53 C 33 1 63 13 17 REQUIREMENTS 4 7 6 14 31
You are required to determine the minimum cost transportation plan [20 marks]. Solution:
TABLEAU 1: I II III IV Dummy Capacity A 4 2 6 B 5 6 14 28 53 C 17 17 Req 4 7 6 14 45 76 This is the initial solution, which costs (4 * 12) + (2 * 23) + (5 * 23) + (6 * 33) + (14 * 53) + (28 * 0) + (17 * 0) = $1149.00
12 23
63 33 13 1
33
43
53 63
3 0
51
-22 21
- 51 10 0
0 23
0 30 -40
OPERATIONS RESEARCH
TABLEAU 2: I II III IV Dummy Capacity A 4 2 6 B 7 6 12 28 53 C 17 17 Req 4 7 6 14 45 76 The costs= (4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (12 * 53) + (28 * 0) + (17 * 0) = $1049.00
TABLEAU 3:
I II III IV Dummy Capacity A 4 2 6 B 7 6 40 53 C 12 5 17 Req 4 7 6 14 45 76 The costs= (4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (40 * 0) + (12 * 13) + (5 * 0) = $569.00
TABLEAU4: I II III IV Dummy Capacity A 4 2 6 B 2 6 45 53 C 5 12 17 Req 4 7 6 14 45 76 The costs= (4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1) = $459.00
12 23
63 33 13 1
33
43
53 63
3 50
1
-22 -29
50 60 0
0 23
0 30 -40
12 23
63 33 13 1
33
43
53 63
3 10
41
-22 11
16 20 0
0 23
0 30
40
12 23
63 33 13 1
33
43
53 63
3 32
19
22 11
32 42 0
0 23
0 52
18
OPERATIONS RESEARCH
Hence delivery allocations are:
Factory A Supplies Warehouse I Factory A Supplies Warehouse IV
Factory B Supplies Warehouse II Factory B Supplies Warehouse III
Factory B Supplies Warehouse Dummy Factory C Supplies Warehouse II Factory C Supplies Warehouse IV
With a minimum cost of (4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1) = $459.00
QUESTION:
A well-known organization has 3 warehouse and 4 Shops. It requires transporting its goods from the warehouse to the shops. The cost of transporting a unit item from a warehouse to a shop and the quantity to be supplied are shown below.
DESTINATION I II III IV TOTAL SUPPLY SOURCE A 10 0 20 11 15 SOURCE B 12 7 9 20 25 SOURCE C 0 14 16 18 5 TOTAL DEMAND 5 15 15 10
Use any method to find the optimum transportation schedule and indicate the cost [14marks].
� DEGENERATE SOLUTION: It involves working a transportation problem if the number of used routes is equal to: Number of rows + Number of column – 1. However if the number of used routes can be less than the required figure we pretend that an empty route is really used by allocating a zero quantity to that route.
� MAXIMIZATION PROBLEMS: Transportation algorithm assumes that the objective is to minimize cost. However it is possible to use the method to solve maximization problem by either:
� Multiply all the units’ contribution by – 1. � Or by subtracting each unit contribution from the maximum contribution in the
table.
OPERATIONS RESEARCH
UNIT 3: NON-LINEAR FUNCTIONS: HOURS: 20.
� NON-LINEAR FUNCTIONS:
o MARGINAL DISTRIBUTION:
� PARTIAL INTEGRATION: Partial integration is a function with more than one variable or finding the probability of a function with more than one variables i.e. f(X1, X2, X3, ….Xn) and is just the rate at which the values of a function change as one of the independent variables change and all others are held constant. Question 1: If f(x, y) = 2(x + y –2xy) given the intervals 0<= x<=1, 0<=y<=1. Find the marginal distribution of x = f(x). Find the marginal distribution of y = f(x). Solution: Pr {0<=x<=1} = 0∫
1 2(x + y –2xy)δx = 2 0∫
1 (x + y –2xy)δx = 2 [x2/2 + xy + x2y]0
1
= 2 [½ + y – y] = 2[½] = 1 Pr {0<=y<=1} = 0∫
1 2(x + y –2xy)δy = 2 0∫
1 (x + y –2xy)δy = 2 [xy +y2/2 + xy2]0
1
= 2 [x + ½ – x] = 2[½] = 1 Question 1: If f(X1, X2) =(X2
1X2 + X31X
22 + X1) given the intervals 0<= X1<=2, 1<=X2 <=3.
Find the marginal distribution of x = f(x). Find the marginal distribution of y = f(x). Find the Expected value of X1 (E(X1)). Find the variance of X1 (Var (X1)).
OPERATIONS RESEARCH
Solution: Pr {0<=X1<=2} = 0∫
2 (X21X2 + X3
1X22 + X1)δx
= [X31X2 /3+ X4
1X22 /4+ X2
1/2]02
= [8X2 /3+ 4X22 + 2] – [0]
= 8X2 /3+ 4X22 + 2
Pr {1<=X2<=3} = 1∫
3 (X21X2 + X3
1X22 + X1)δy
= [X21X
22 /2+ X3
1X32 /3+ X1X2]1
3 = [9X2
1 /2+ 27X31 /3+ 3X2]1
3 –[X21 /2+ X3
2 /3+ X1] = 9X2
1 /2+ 27X31 /3+ 3X2 – X2
1 /2- X32 /3 - X1
= 8X21 /2+ 26X3
1 /3+ 2X2 Expected value of E(X1) =0∫
2 X. f(X1)δx = 0∫
2 X(X21X2 + X3
1X22 + X1)δx
= 0∫2 (X3
1X2 + X41X
22 + X2
1)δx = [X4
1X2 /4+ X51X
22 /5+ X3
1/3]02
= [4X2 + 32X22 /5 + 8 /3] – [0]
= 4X2 + 32X22 /5 + 8 /3
Variance of X1 = Var (X1) = 0∫
2 ([X1 - E(X1)]2 . f(X1)δx
= 0∫2 ([X1 - 4X2 + 32X2
2 /5 + 8 /3]2 * (X21X2 + X3
1X22 + X1)δx.
Question 2: A manufacturing company produces two products bicycles and roller skates. Its fixed costs production is: $1200 per week. Its variables costs of production are: $40 for each bicycle produced and $15 for each pair of roller skates. Its total weekly costs in producing x bicycles and y pairs of roller skates are therefore c= cost. C(x, y) = 1200 + 40x + 15y for example; in producing x = 20 bicycles and y = 30 pairs of roller skates/ week. The manufacture experiences total cost of: C(20, 30) = 1200 + 40(20) + 15(30) = 1200 + 800 + 450 = 2450. Question 3: A manufacturing of Automobile tyres produces 3 different types: regular, green and blue tyres. If the regular tyres sell for $60 each, the green tyres for $50 each and the blue tyres for $100 each. Find a function giving the manufacture’s total receipts or revenue from the of x regular tyres and y green tyres and z blue tyres. R(x, y, z) = 60x + 50y +100z.
OPERATIONS RESEARCH
Solution: Since the receipts of the sale of any tyre type is the price per tyre times the number of tyres sold: The total receipts are: R(x, y, z) = 60x + 50y + 100z For example receipts from the sell of 10 tyres of each type would be: R(10,10,10) = 60(10) + 50(10) + 100(10) = 600 + 500 + 1000 = $2100
� PARTIAL DIFFERENTIATION: For a function “f” of a single variable, the derivative f measures the rate at which the values of f(x) change as the independent variable x change. A partial derivative of a function i.e. f(X1, X2, X3.. Xn) of several variables is just the rate at which the values of the function change as one of the independent variable changes and all others are held constant. Question 3: For the function f(x, y) = X3 + 4X2Y3
+ Y2 Find
� δf / δx � δf / δy � f(-2; 3)
Solution:
δf / δx = 3X2 + 8XY3 δf / δy = 12X2Y2
+ 2Y f(-2; 3) = X3 + 4X2Y3
+ Y2
= (-2)3 + 4(-2)2(3)3 + (3)2
= -8 + 16(27) +9 = 433
Question 4: A company produces electronic typewriters and word processors, it sells the electronic typewriters for $100 each and word processors for $300 each. The company has determined that its weekly sales in producing x electronic writers and y word processors are given by the following joint cost function. C(x, y) = 200 + 50x +8y + X2 + 2Y2 Find the numbers of x and y of machines that the company should manufacture and sell weekly in order to maximize profits.
OPERATIONS RESEARCH
Solution: Revenue function is given by: R(x, y) = 100x + 300y Profit = Revenue – Cost. Then Profit function is given by: P(x, y) = R(x, y) – C(x, y) = (100x + 300y) – (200 + 50x +8y + X2 + 2Y2) = 50x + 292y – 200 - X2 - 2Y2 To find the critical points of turning points of x and y. We set the partial derivative = 0. Thus δp / δx => 50 – 2x = 0
� 50 = 2x � x = 25
δp / δy => 292 – 4y =0 � 292 = 4y � y = 73
The production schedule for maximum profit is therefore x = 25 type writers and y = 73 word processors which yields a profit of P = 50(25) + 292(73) – 200 – 625 – 2(73)2 = 1250 + 21316 – 200 – 625 – 1065 = 22566 – 11493 = $11083
� NECESSARY AND SUFFICIENT CONDITIONS FOR EXTREMA:
The necessary condition or the GRADIENT VECTOR of the extrema determines the turning points or critical points of a function. Let X0 be a variable representing the turning point and represented mathematically as: X0 = (A0, B0, … N0). In general form; a necessary condition or gradient vector for X0 to be an extrema point of f(x) is that the gradient (∇) ≡ ∇f (X0) = 0.
OPERATIONS RESEARCH
Question 1: Given f(X1, X2, X3) =(X1 + 2X3 + X2X3 – X2
1 - X22 - X
23)
Find the gradient vector for X0 i.e. ∇f (X0) = 0. Solution: The necessary condition (gradient vector) ∇f (X0) = 0 is given by:
δf / δx1 => 1 - 2X1 = 0. 1 - 2X1 = 0. [1] δf / δx2 => X3 - 2X2 = 0. X3 - 2X2 = 0. [2] δf / δx3 => 2 + X2 - 2X3 = 0. 2 + X2 - 2X3 = 0. [3]
(a) Finding X1 is given by 1 = 2X1
X1 = ½
(b) Equation 2 is given by X3 - 2X2 = 0. X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2.
Thus 2 + X2 - 2X3 = 0. � 2 + X2 – 2(2X2) = 0. � 2 + X2 – 4X2 = 0. � 2 – 3X2 = 0. � X2 = 2/3.
Therefore X3 = 2X2. X3 = 2(2 / 3) X3 = 4/3.
Therefore X0 = (½, 2/3,
4/3)
OPERATIONS RESEARCH
Hessian matrix
In mathematics, the Hessian matrixpartial derivatives of a functionvariables.
Given the real-valued function
if all second partial derivatives
where x = (x1, x2, ..., xn) and Di the Hessian becomes
Some mathematicians define the Hessian as the
Bordered Hessian
A bordered Hessian is used for the secondproblems. Given the function as before:
but adding a constraint function such that:
the bordered Hessian appears as
OPERATIONS RESEARCH
Hessian matrix (or simply the Hessian) is the square matrixfunction; that is, it describes the local curvature of a function of many
of f exist, then the Hessian matrix of f is the matrix
is the differentiation operator with respect to the
Some mathematicians define the Hessian as the determinant of the above matrix.
Bordered Hessian
is used for the second-derivative test in certain constrained optimizaproblems. Given the function as before:
but adding a constraint function such that:
the bordered Hessian appears as
54
square matrix of second-order ; that is, it describes the local curvature of a function of many
is the matrix
is the differentiation operator with respect to the ith argument and
of the above matrix.
derivative test in certain constrained optimization
OPERATIONS RESEARCH
If there are, say, m constraints then the zero in the northand there are m border rows at the top and
The above rules of positive definite and negative definite can not apply here since a bordered Hessian can not be definite: we have followed by zeroes.
The second derivative test consists here of sign restrictions of the determinants of a certain set of m submatrices of the bordered Hessian. Intuitively, think of the problem to one with n - m free variables. (For example, constraint x1 + x2 + x3 = 1 can be reduced to the maximization of constraint.)
A sufficient condition for XHESSIAN matrix (denoted by H) eval
i. Positive definite when Xii. Negative definite when X
The Hessian matrix is achieved by finding the 2each equation with respect to all variables Thus the Hessian matrix is evaluated at the point X H/X0 = δ2f/δX
21,
δ2f/ δX1 δX22,
δ2f/δX2 δX
21,
δ2f/ δX22,
δ2f/δX3 δX
21,
δ2f/ δX3 δX22,
OPERATIONS RESEARCH
constraints then the zero in the north-west corner is an m at the top and m border columns at the left.
The above rules of positive definite and negative definite can not apply here since a bordered Hessian can not be definite: we have z'Hz = 0 if vector z has a non-zero as its first element,
he second derivative test consists here of sign restrictions of the determinants of a certain set of submatrices of the bordered Hessian. Intuitively, think of the m constraints as reducing the
free variables. (For example, the maximization of fcan be reduced to the maximization of f(x1,x2,1 − x
A sufficient condition for X0 a point to be extremism is that the matrix (denoted by H) evaluate at X0 is:
Positive definite when X0 is a Minimum point. Negative definite when X0 is a Maximum point.
The Hessian matrix is achieved by finding the 2nd Partial derivation of the first Partial derivative of each equation with respect to all variables defined.
Thus the Hessian matrix is evaluated at the point X0
, δ2f/ δX1 δX
23
δ2f/ δX2 δX
23
2, δ2f/ δX
23
55
× m block of zeroes,
The above rules of positive definite and negative definite can not apply here since a bordered zero as its first element,
he second derivative test consists here of sign restrictions of the determinants of a certain set of n - constraints as reducing the
f(x1,x2,x3) subject to the x1 − x2) without
a point to be extremism is that the
Partial derivation of the first Partial derivative of
OPERATIONS RESEARCH
δf / δx1 => 1 - 2X1 = 0. δf / δx2 => X3 - 2X2 = 0. δf / δx3 => 2 + X2 - 2X3 = 0. To establish the sufficiency the function has to have: H/X0 = -2 0 0 0 -2 1 0 1 -2 Since the Hessian matrix is 3 by 3 matrix then:
� Find the 1st Principal Minor determinant of 1 by 1 matrix in the Hessian matrix. � Find the 2nd Principal Minor determinant of 2 by 2 matrix in the Hessian matrix. � Find the 3rd Principal Minor determinant of 3 by 3 matrix in the Hessian matrix.
The Positive definite when X0 is a Minimum point is evaluated as:
� When 1st PMD = + ve. � When 2nd PMD = +ve. � When 3rd PMD = +ve.
Or � When 1st PMD = - ve. � When 2nd PMD = +ve. � When 3rd PMD = +ve.
Thus 3 by 3 Hessian matrix the number of positive number should be greater than one. (Should be two or more). The Negative definite when X0 is a Maximum point is evaluated as:
� When 1st PMD = - ve. � When 2nd PMD = - ve. � When 3rd PMD = - ve.
Or � When 1st PMD = + ve. � When 2nd PMD = - ve. � When 3rd PMD = - ve.
Thus 3 by 3 Hessian matrix the number of negative number should be greater than two. (Should be two or more). H/X0 = -2 0 0 0 -2 1 0 1 -2 Thus the 1st PMD of (-2) = -2
OPERATIONS RESEARCH
Thus the 2nd PMD of –2 0
0 -2 = (-2 * -2) – (0 * 0) = 4 The 3rd PMD = -2 0 0 0 -2 1 0 1 -2 = -2 –2 1 - 0 0 1 + 0 0 -2 1 -2 0 -2 0 1 = -2 {(-2 * -2) – (1 * 1)} – 0 (0 – 0) + 0 (0 – 0) = - 2 (3) = - 6 Thus the PMD is equal to –2, 4 and –6 and H/X0 is negative definite and X0 = (½, 2/3,
4/3) represents a Maximum point. Question 2: Given f(X1, X2, X3) =(-X1 + 2X3 - X2X3 + X2
1+ X22 - X
23)
i. Find the gradient vector for X0 i.e. ∇f (X0) = 0. ii. Determine the nature of the turning points using Hessian Matrix.
Solution: The necessary condition (gradient vector) ∇f (X0) = 0 is given by:
δf / δx1 => -1 + 2X1 = 0. -1 + 2X1 = 0. [1] δf / δx2 => -X3 + 2X2 = 0. -X3 + 2X2 = 0. [2] δf / δx3 => 2 - X2 - 2X3 = 0. 2 - X2 - 2X3 = 0. [3]
(b) Finding X1 is given by -1 = 2X1
X1 = ½
(b) Equation 2 is given by -X3 + 2X2 = 0. X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2. Thus 2 - X2 - 2X3 = 0.
OPERATIONS RESEARCH
� 2 - X2 – 2(2X2) = 0. � 2 - X2 – 4X2 = 0. � 2 – 5X2 = 0. � X2 = 2/5.
Therefore X3 = 2X2. X3 = 2(2 / 5) X3 = 4/5.
Therefore X0 = (½, 2/5,
4/5)
H/X0 = δ2f/ δX21,
δ2f/ δX1 δX22,
δ2f/ δX1 δX23
δ2f/ δX2 δX
21,
δ2f/ δX22,
δ2f/ δX2 δX23
δ2f/ δX3 δX
21,
δ2f/ δX3 δX22,
δ2f/ δX23
H/X0 = 2 0 0 0 2 -1 0 -1 -2 Thus the 1st PMD of (2) = 2 Thus the 2nd PMD of 2 0
0 2 = (2 * 2) – (0 * 0) = 4 The 3rd PMD = 2 0 0 0 2 -1 0 -1 -2 = -2 2 -1 - 0 0 -1 + 0 0 2 -1 -2 0 -2 0 -1 = 2 {(2 * -2) – (-1 * -1)} – 0 (0 – 0) + 0 (0 – 0) = 2 (-4) - (1) = 2 (-5) = - 10 Thus the PMD is equal to 2, 4 and –10 and H/X0 is positive definite and X0 = (½, 2/5,
4/5) represents a Minimum point.
OPERATIONS RESEARCH
� NON LINEAR ALGORITHMS: (COMPUTATIONS) 1 THE GRADIENT METHODS:
The general idea is to generate successive iterative points, starting from a given initial point, in the direction of the fast and increase (maximization of the function). The method is based on solving the simultaneous equations representing the necessary conditions for optimality namely ∇f (X0) = 0. Termination of the gradient method occurs at the point where the gradient vector becomes null. This is only a necessary condition for optimality suppose that f(x) is maximized. Let X0 be the initial point from which the procedure starts and define ∇f (Xk) as the gradient of f at Kth point Xk. This result is achieved if successive point Xk and Xk+1 are selected such that Xk+1 = Xk + rk ∇f (Xk) where rk is a parameter called Optimal Step Size. The parameter rk is determined such that Xk+1 results in the largest improvement in f. In other words, if a function h(r) is defined such that h(r) = f(Xk )+ rk ∇f (Xk). This function is then differentiated and equate zero to the differentiatable function to obtain the value of rk. Question 3: Consider maximizing f(X1, X2) =(4X1 + 6X2 - 2X2
1- 2X2X1 - 2X22)
And let the initial point be given by X0(1, 1). Hint in X0(1, 1) X1 =1 and X2 = 1 Solution: Find ∇f (X0) = (δf/δx1,
δf/δx2) = (4 – 4X1 – 2X2; 6 – 2X1 - 4X2)
1st iteration
Step 1: Find ∇f (X0) = (4 – 4 – 2; 6 – 2 - 4) = (-2; 0) Step 2: Find Xk+1 = Xk + rk ∇f (Xk)
� X0+1 = X0 + rk ∇f (X0) � X1 = (1, 1) + r(-2; 0) � (1, 1) + (-2r, 0) � (1 + -2r, 1) � (1 – 2r, 1)
Thus h(r) = f(Xk )+ rk ∇f (Xk) = f(X0 )+ r ∇f (X0) = f(1 – 2r; 1) = 4(1 – 2r) + 6(1) – 2(1 – 2r)2 – 2(1)(1 – 2r) - 2(1)2.
OPERATIONS RESEARCH
= 4(1 – 2r) + 6 – 2(1 – 2r) 2 – 2(1 – 2r) - 2. = 4(1 – 2r) – 2(1 – 2r) + 6 – 2 – 2(1 – 2r)2. = (4 – 2)(1 – 2r) + 4 – 2(1 – 2r)2. = 2(1 – 2r) – 2(1 – 2r)2 + 4. = – 2(1 – 2r)2 +2(1 – 2r) + 4. = – 2(1 – 2r)2 + 2 – 4r + 4. h1(r) = 0
� – 2(1 – 2r)2 + 2 – 4r + 4 = 0. � - 4 * -2(1 – 2r) – 4 = 0. � 8(1 – 2r) + - 4 = 0. � 8 – 16r – 4 = 0 � 4 –16r = 0. � r = ¼
The optimum step size yielding the maximum value of h(r) is h1 = ¼. This gives X1= (1 –2(¼); 1) = (1 - ½; 1) = (½; 1)
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UNIT 4: PROJECT MANAGEMENT WITH
PERT/CPM HOURS: 20
� PROJECT MANAGEMENT: TERMS USED IN PROJECT MANAGEMENT: PROJECT:
Is a combination of interrelated activities that must be executed in a certain order before the entire task can be completed?
ACTIVITY:
Is a job requiring time and resource for its completion? ARROW:
Represents a point in time signifying the completion of some activities and the beginning of others.
NETWORK:
Is a graphic representation of a project’s operation and is composed of activities and nodes.
Benefits of PERT
PERT is useful because it provides the following information:
• Expected project completion time. • Probability of completion before a specified date. • The critical path activities that directly impact the completion time. • The activities that have slack time and that can lend resources to critical path activities. • Activity starts and end dates.
Limitations
The following are some of PERT's weaknesses:
• The activity time estimates are somewhat subjective and depend on judgement. In cases where there is little experience in performing an activity, the numbers may be only a guess. In other cases, if the person or group performing the activity estimates the time there may be bias in the estimate.
• Even if the activity times are well-estimated, PERT assumes a beta distribution for these time estimates, but the actual distribution may be different.
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• Even if the beta distribution assumption holds, PERT assumes that the probability distribution of the project completion time is the same as the that of the critical path. Because other paths can become the critical path if their associated activities are delayed, PERT consistently underestimates the expected project completion time.
Critical Path Analysis CPA (Network Analysis)
Critical Path Analysis (CPA) is a project management tool that:
• Sets out all the individual activities that make up a larger project. • Shows the order in which activities have to be undertaken. • Shows which activities can only taken place once other activities have been completed. • Shows which activities can be undertaken simultaneously, thereby reducing the overall time
taken to complete the whole project. • Shows when certain resources will be needed – for example, a crane to be hired for a
building site.
In order to construct a CPA, it is necessary to estimate the elapsed time for each activity – that is the time taken from commencement to completion.
Then the CPA is drawn up a based on dependencies such as:
• The availability of labour and other resources • Lead times for delivery of materials and other services • Seasonal factors – such as dry weather required in a building project
Once the CPA is drawn up, it is possible to see the CRITICAL PATH itself – this is a route through the CPA, which has no spare time (called ‘FLOAT’ or ‘slack’) in any of the activities. In other words, if there is any delay to any of the activities on the critical path, the whole project will be delayed unless the firm makes other changes to bring the project back on track.
The total time along this critical path is also the minimum time in which the whole project can be completed.
Some branches on the CPA may have FLOAT, which means that there is some spare time available for these activities.
What can a business do if a project is delayed?
• Firstly, the CPA is helpful because it shows the likely impact on the whole project if no action were taken.
• Secondly, if there is float elsewhere, it might be possible to switch staff from another activity to help catch up on the delayed activity.
• As a rule, most projects can be brought back on track by using extra labour – either by hiring additional people or overtime. Note, there will be usually be an extra cost. Alternative suppliers can usually be found – but again, it might cost more to get urgent help.
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The key rules of a CPA
• Nodes are numbered to identify each one and show the Earliest Start Time (EST) of the activities that immediately follow the node, and the Latest Finish Time (LFT) of the immediately preceding activities
• The CPA must begin and end on one ‘node’ – see below • There must be no crossing activities in the CPA • East activity is labelled with its name eg ‘print brochure’, or it may be given a label, such as
‘D’, below. • The activities on the critical path are usually marked with a ‘//’
In the example below
• The Node is number 3 • The EST for the following activities is 14 days • The LFT for the preceding activities is 16 days • There is 2 days’ float in this case (difference between EST and LFT) • The activity that follows the node is labelled ‘D’ and will take 6 days
OR
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A simple example – baking a loaf of bread
Here is a simple example, in which some activities depend on others having been undertaken in order, whereas others can be done independently.
Activity Preceded by Elapsed time (minutes) A weigh ingredients - 1 B mix ingredients A 3 C dough rising time B 60 D prepare tins - 1 E pre-heat oven - 10 F knock back dough and place in tins C&D 2 G 2 nd dough rising time F 15 H cooking time E & G 40
In this example, there is a clear sequence of events that have to happen in the right order. If any of the events on the critical path is delayed, then the bread will not be ready as soon. However, tasks D (prepare tins) and E (heat the oven) can be started at any time as long as they are done by the latest finish time in the following node.
So, we can see that the oven could be switched on as early as time 0, but we can work out that it could be switched on at any time before 71 – any later than this and it won’t be hot enough when the dough is ready for cooking. There is some ‘float’ available for tasks D and E as neither is on the critical path.
This is a fairly simple example, and we can see the LST and LFT are the same in each node. In a more complex CPA, this will not necessarily be the case, and if so, will indicate that there is some ‘float’ in at least one activity leading to the node. However, nodes on the critical path will always have the same EST and LFT.
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A
HOW TO CONSTRUCT A CRITICAL PATH NETWORK DIAGRAM Here is the data:
Activity Preceded by Duration (days) A - 2 B - 3 C A 4 D B 5 E C 8 F E 3 G D,F 4
Here is what to do:
1. Draw the first ‘node’ and number it ‘1’. 2. Draw the line to show any ‘activities’ that are not preceded by any other activities. 3. Tick these activities off to show you have done them
4. Look at the next activity and see which it is preceded by. In this case it is activity C and it is preceded by activity A
5. Draw this on the diagram and again tick off the activity on the list. 6. Do the same for activity D, E & F. Look carefully at which activity they are preceded by
7. Now do activity G. This one is a little trickier, as it is preceded by more than one activity.
However, all you have to do is make them meet at one node – easy! 8. Next, draw a node on the end of the network diagram 9. Now number the nodes, following through the activities. If there are 2 activities starting at
the time, you need to number the shortest activity first.
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The diagram is now ready for you to work out (make sure you understand these terms by reading further).
• Earliest Start Time (EST) – top segment or left segment. • Latest Finish Time (LFT) – bottom segment or right segment. • Float Time • The critical path
There are many ways to do the above, but the method below is the simplest, so learn it and follow it!
i. Work out which ‘route’ takes longest (which is the critical path) ii. In this case : A C E F G takes 21 days and B D G takes 12 days iii. Consequently, A C E F G is the critical path
iv. As these activities take 21 days, you can write ‘21’ days in both the top or left and bottom or right
segment of the end node.
v. Now work backwards through the nodes on the critical path and enter the LFT in the bottom or right segment and the EST in the top or left segment.
vi. Taking off the length of time it takes to complete the activity. So from node 6 would have
‘17’ in both segments. vii. Now work backwards through any other nodes and enter the LFT in the bottom or right
segment. You MUST do this backwards for ALL other ‘routes’. You then do the same for the EST for each route, but go forwards this time!
viii. Note: Backwards for LFT and Forwards for EST
ix. Here it is below!
OPERATIONS RESEARCH
Activity EST LFT Duration Float(LFT-D-EST) A 0 2 2 0 B 0 12 3 9 C 2 6 4 0 D 3 17 5 9 E 6 14 8 0 F 14 17 3 0 G 17 21 4 0
RULES FOR CONSTRUCTING NETWORK DIAGRAM:
� Each activity is represented by one and only one arrow in the network. � No two activities can be identified by the same head and tail events. If activities A and B
can be executed simultaneously, then a dummy activity is introduced either between A and one end event or between B and one end event. Dummy activities do not consume time or resources. Another use of the dummy activity: suppose activities A and B must precede C while activity E is preceded by B only.
� To ensure the correct precedence relationships in the network diagram, the following questions must be answered as every activity is added to the network:
� What activities must be completed immediately before this activity can start. � What activities must follow this activity? � What activities must occur concurrently with this activity?
OPERATIONS RESEARCH
Question 1: ACTIVITY PRECEDED BY DURATION (Weeks)
A Initial activity 10 B A 9 C A 7 D B 6 E B 12 F C 6 G C 8 H F 8 I D 4 J G, H 11 K E 5 L I 7 Find the critical path and the time for completing the project. Solution: D I 6 4 B 9 E 12 L 7 K A 5 10 Dummy J 11 C 7 G 8 F H 6 8
� EARLISET START TIME: Represents all the activities emanating from i. Thus ESi represent the earliest occurrence time of event i. Earliest finish time is given by: EF = Max {ESi + D}
0 0 0
1 10 10
2 19 25
3 17 17
4 25 31
5 31 37
7 25 31
6 23 23
8 31 31
9 29 35
10 42 42
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� LATEST COMPLETION TIME: It initiates the backward pass. Where calculations from the “end” node and moves to the “start” node. Latest start time is given by: LSi = Min {LF – D}
� DETERMINATION OF THE CRITICAL PATH: A Critical path defines a chain of critical that connects the start and end of the arrow diagram. An activity is said to be critical if the delay in its start will cause a delay in the completion date of the entire project. Or it is the longest route, which the project should follow until its completion date of the entire project. The critical path calculations include two phases:
FORWARD PASS: Is where calculations begin from the “start” node and move to the “end” node. At each node a number is computed representing the earliest occurrence time of the corresponding event.
BACKWARD PASS:
Begins calculations from the “end” node and moves to the “start” node. The number computed at each node represents the latest occurrence time of the corresponding event.
� DETERMINITION OF THE FLOATS: A Float or Spare time can only be associated with activities which are non critical. By definition activities on the critical path cannot have floats. There are 3 types of floats.
1 TOTAL FLOAT: This is the amount of time a path of activities could be delayed without affecting the overall project duration. Total Float = Latest Head Time – Earliest Tail time – duration. = LS – ES. = LF – ES – D = LF – EF or EC. 1 FREE FLOAT: This is the amount of time an activity can be delayed without affecting the commencement of a subsequent activity at its earliest start time. Free Float = Earliest Head Time – Earliest Tail Time – Duration. = LF – ES – D = ESj – ESi – D.
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1 INDEPENDENT FLOAT: This is the amount of time an activity can be delayed when all preceding activities are completed as late as possible and all succeeding activities completed as early as possible. Independent Float = EF – LS – D.
NORMAL EARLIEST TIME LATEST TIME TOTAL FLOAT
ACTIVITY TIME ES EF = ES + D LS = LF – D LF =LS – ES A 10 0 10 0 10 0 B 9 10 19 16 25 6 C 7 10 17 10 17 0 D 6 19 25 25 31 6 E 12 19 31 25 37 6 F 6 17 23 17 23 0 G 8 17 25 23 31 6 H 8 23 31 23 31 0 I 4 25 29 31 35 6 J 11 31 42 31 42 0 K 5 31 36 37 42 6 L 7 29 36 35 42 6
Question 2: Draw the network for the data given below then find the critical path as well total float and free float. ACTIVITY (I, J) DURATION (0, 1) 2 (0, 2) 3 (1, 3) 2 (2, 3) 3 (2, 4) 2 (3, 4) 0 (3,5) 3 (3, 6) 2 (4, 5) 7 (4, 6) 5 (5, 6) 6
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0 0 0
Solution: 2 2 2 3 3 6 Dummy 3 7 5 2
ACTIVITY D ES EF=ES+D LS=LF-D LF TOTAL FREE Float Float (0, 1) 2 0 2 2 4 2 2 (0, 2) 3 0 3 0 3 0 0 (1, 3) 2 2 4 4 6 2 2 (2, 3) 3 3 6 3 6 0 0 (2, 4) 2 3 5 4 6 1 1 (3, 4) 0 6 6 6 6 0 0 (3,5) 3 6 9 10 13 4 4 (3, 6) 2 6 8 17 19 11 11 (4, 5) 7 6 13 6 13 0 0 (4, 6) 5 6 11 14 19 8 8 (5, 6) 6 13 19 13 19 0 0
1 2 4
2 3 3
3 6 6
6 19 19
4 6 6
5 13 13
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� PERT ALGORITHM:
� PROBABILISTIC TIME DURATION OF ACTIVITIES. The following are steps involved in the development of probabilistic time duration of activities.
� Make a list of activities that make up the project including immediate predecessors.
� Make use of step 1 sketch the required network. � Denote the Most Likely Time by Tm, the Optimistic Time by To and
Pessimistic time by Tp. � Using beta distribution for the activity duration the Expected Time Te for each
activity is computed by using the formula: Te = (To + 4Tm + Tp) / 6. � Tabulate various times i.e. Expected activity times, Earliest and Latest times and
the EST and LFT on the arrow diagram. � Determine the total float for each activity by taking the difference between EST
and LFT. � Identify the critical activities and the expected date of completion of the project. � Using the values of Tp and To compute the variance (δ2) of each activity’s time
estimates by using the formula: δ2 = {{Tp – To} / 6}2. � Compute the standard normal deviate by: Zo = (Due date – Expected date of Completion) / √Project variance. � Use Standard normal tables to find the probability P (Z <= Zo) of completing
the project within the scheduled time, where Z ~ N(0,1). Question 3:
A project schedule has the following characteristics: Activity Most Likely Time Optimistic Time Pessimistic Time 1 – 2 2 1 3 2 – 3 2 1 3 2 – 4 3 1 5 3 – 5 4 3 5 4 – 5 3 2 4 4 – 6 5 3 7 5 – 7 5 4 6 6 – 7 7 6 8 7 – 8 4 2 6 7 – 9 6 4 8 8 – 10 2 1 3 9 – 10 5 3 7
I. Construct the project network.
II. Find expected duration and variance for each activity.
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III. Find the critical path and expected project length. IV. What is the probability of completing the project in 30 days.
Solution: 4 5 6 2 3 7 4 5 2 3 2 5
Expected job Time Te = (To + 4Tm + Tp) / 6. Variance δ2 = {{Tp – To} / 6}2.
Activity Tm To Tp Te δ2 1 – 2 2 1 3 2 0.111 2 – 3 2 1 3 2 0.111 2 – 4 3 1 5 3 0.445 3 – 5 4 3 5 4 0.111 4 – 5 3 2 4 3 0.111 4 – 6 5 3 7 5 0.445 5 – 7 5 4 6 5 0.111 6 – 7 7 6 8 7 0.111 7 – 8 4 2 6 4 0.445 7 – 9 6 4 8 6 0.445 8 – 10 2 1 3 2 0.111 9 – 10 5 3 7 5 0.445 Critical path (*) comprises of activities (1 –2), (2 - 4), (4 –6), (6 –7), (7 –9) and (9 –10) Expected project length is = 28 days. Variance δ2 = 0.111 + 0.445 + 0.445 + 0.111 + 0.445 + 0.445 = 2.00 (on critical path only)
1 0 0
2 2 2
3 4 8
5 8 12
7 17 17
9 23 23
10 28 28
4 5 5
6 10 10
8 21 26
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(iv) Probability of completing the project in 30 days is obtained by: Zo = (Due date – Expected date of Completion) / √Project variance. = (30 – 28) / √2. = 1.414 (Look this from Normal tables) Now from Standard Normal tables Z= 0.4207. P (t <= 30) = P (Z <= 1.414) = 0.5 + 0.4207 = 0.9207 This shows that the probability of meeting the scheduled time will be 0.9207
� COST CONSIDERATIONS IN PERT / CPM: The cost of a project includes direct costs and indirect costs. The direct costs are associated with the individual activities and the indirect costs are associated with the overhead costs such as administration or supervision cost. The direct cost increase if the job duration is to be reduced whereas the indirect costs increase if the job duration is to be increased.
1 TIME COST OPTIMIZATION PROCEDURE: The process of shortening a project is called Crashing and is usually achieved by adding extra resources to an activity. Project crashing involves the following steps:
� Critical Path: Find the normal critical path and identify the critical activities. � Cost Slope: Calculate the cost slope for the different activities by using the Formula: COST SLOPE = Crash cost – Normal cost.
Normal Time – Crash Time.
� Ranking: Rank the activities in the ascending order of cost slope. � Crashing: Crash the activities in the critical path as per the ranking i.e. activities having
lower cost slope would be crashed first to the maximum extent possible. Calculate the new direct cost by cumulatively adding the cost of crashing to the normal cost.
� Parallel Crashing: As the critical path duration is reduced by the crash in step 3 other paths become critical i.e. we get parallel critical paths. This means that project duration can be reduced by simultaneous crashing of activities in the parallel critical paths.
� Optimal Duration. Crashing as per Step 3 and step 4 an optimal project is determined. It would be the time duration corresponding to which the total cost (i.e. Direct cost plus Indirect cost) is a minimum.
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Question 4:
For the network given below find the optimum cost schedule for the completion of the project: JOB NORMAL CRASH TIME COST $ TIME COST $ 1 – 2 10 60 8 120 2 – 3 9 75 6 150 2 – 4 7 90 4 150 3 – 4 6 100 5 140 3 – 5 9 50 7 80 3 – 6 10 40 8 70 4 – 5 6 50 4 70 5 – 6 7 70 5 110 Solution: JOB COST SLOPE *1 – 2 30 ---> (4) = (120 –60) / (10 – 8) *2 – 3 25 ---> (3)
2 – 4 20 ⇒ 3 – 4 40 --->(5)
3 – 5 15 3 – 6 15
� 4 – 5 10 ---> (1) � 5 – 6 20 ---> (2)
10 9 6 9 7 10 7 6 The critical path = 1 –2, 2 –3, 3 –4, 4 –5 and 5 –7. Expected project Length = 38 days. Associated with 38 days the minimum direct project cost = 60 + 75 + 90 + 100 + 50 + 40 + 50 + 70 = $535
1 0 0
3 19 19
5 31 31
6 38 38
4 25 25
2 10 10
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In order to reduce the project duration we have to crash at least one of the jobs on the critical path. This is being done because crashing of the job not on the critical path does not reduce the project length. 1st Crashing: On critical path the minimum cost slope is job 4 –5 and is to be crashed at extra cost of $10 per day. 10 9 6 9 7 10 7 4 Duration of project = 36 days and Total cost = $535 + $10 * 2 = $555. 2nd Crashing: Now crash job 5 –6 and is to be crashed at extra cost of $20 per day. 10 9 6 9 5 10 7 4 Duration of project = 34 days and Total cost = $555 + $20 * 2 = $595.
1 0 0
5 29 29
4 25 25
2 10 10
3 19 19
6 36 36
1 0 0
5 29 29
4 25 25
2 10 10
3 19 19
6 34 34
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3rd Crashing: Now crash job 2 –3 for 3 days and is to be crashed at extra cost of $25. 10 6 6 9 5 10 7 4 Duration of project = 31 days and Total cost = $595 + $25 * 3 = $670. 4th Crashing: Now crash job 1 –2 for 2 days and is to be crashed at extra cost of $30. 10 6 6 9 5 8 7 4 Duration of project = 29 days and Total cost = $670 + $30 * 2 = $730.
1 0 0
5 26 26
4 22 22
2 10 10
3 16 16
6 31 31
1 0 0
5 24 24
4 20 20
2 8 8
3 14 14
6 29 29
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5th Crashing: Final crash job 3 –4 for 1 day and it is to be crashed at extra cost of $40 and two critical paths occurs. 10 6 5 9 5 8 7 4 Duration of project = 28 days and Total cost = $730+ $40 * 1 = $770. Optimum Duration of project = 28 days and Total Cost = $770.
1 0 0
5 23 23
4 19 19
2 8 8
3 14 14
6 28 28
OPERATIONS RESEARCH
UNIT 5: RANDOM VARIABLES AND THEIR PROBABILITY: HOURS: 20.
� RANDOM VARIABLES AND PROBABILITY FUNCTIONS (DISCRETE): Given a Sample space S = {1,2,3,4,5,6} we may therefore use the variable such as X to represent an outcome in the sample space such a variable is called Random variable. When the outcome in a sample space are represented by values in a random variable the assignment of probabilities to the outcome can be thought of as a function for which the domain is the sample space, we refer to this as the probability function written as Pr. We use the following notation with probability function Pr {X = a} which means the probability associated with the outcome a while Pr {X in E} means the probability associated with event E. Given S = {1,2,3,4,5,6}
a. Find the probability of S = 3. b. Find the probability of S = 5. c. Find the probability of X in E when E = 1,2.3.
Solution:
i. P (S = 3) = 1/6. ii. P (S = 5) = 1/6. iii. P (X in E) = ½.
� PROBABILITY DENSITY FUNCTIONS:
2 PROPERTIES OF DISCRETE RANDOM VARIABLE:
a. It is a discrete variable. b. It can only assume values x1, x2. …xn. c. The probabilities associated with these values are p1, p2. …pn. Where P(X = x1) = p1. P(X = x2) = p2. . . P(X = xn) = pn. Then X is a discrete random variable if p1 + p2. …pn = 1. This can be written as ∑ P(X = x) =1.
all x
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Question 1: The P.d.f. of a discrete random variable Y is given by P (Y=y) = cy2, for y = 0,1,2,3,4. Given that c is a constant, find the value of c. Solution: Y 0 1 2 3 4 P(Y =y) 0 c 4c 9c 16c = ∑ P(X = x) =1. all x
� 1 = c + 4c + 9c + 16c � 1 = 30c � c = 1/30.
Question 2: The Pdf. of a discrete random variable X is given by P (X=x) = a(¾)x , for x = 0,1,2,3... Find the value of the constant a. Solution: = ∑P(X = x) =1. all x
P(X = 0) = a(¾)0. P(X = 1) = a(¾)1. P(X = 2) = a(¾)2. P(X = 3) = a(¾)3 and so on.
So ∑P(X = x) =a + a(¾) + a(¾)2 + a(¾)3 + … all x = a( 1 + ¾ + (¾)2 + (¾)3 + …) = a ( 1/1- ¾) -> (sum of an infinite G.P with first term 1 and common ratio ¾) = a(4) 4a = 1 a = ¼
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� EXPECTED VALUE/ MEAN / AVERAGE: For a random variable X associated with a sample space {x1, x2. …xn} the concept of expected value is the generalization of the average of numbers {x1, x2. …xn}.
2 EXPECTED VALUE WITH SAME PROBABILITIES: The expected value of X with same probabilities is given by: E(x) = X1 + X2 + …Xn/n. Question 3: Given that an die is thrown 6 times and the recordings are as follows then calculate the expected mean or mean score Score x 1 2 3 4 5 6 P(X = x) 1/6
1/6 1/6
1/6 1/6
1/6 Solution: E(x) = X1 + X2 + …Xn/n. = 1 + 2 + 3 + 4 + 5 + 6/6 = 21/6 = 7/2 = 3.5
2 EXPECTED VALUE WITH DIFFERENT PROBABILITIES: The expected value of X with different probabilities is given by: E(x) = P1 * X1 + P2 * X2 + …Pn * Xn. Question 4: Given a random variable X which has a Pdf shown below. Calculate the expected mean.
X -2 -1 0 1 2 P(X = x) 0.3 0.1 0.15 0.4 0.05
Solution:
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn = -2 * 0.3 + -1 * 0.1 + 0 * 0.15 + 1 * 0.4 + 2 * 0.05 = -0.2
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Question 5: A venture capital firm is determined based on the past experience that for each $100 invested in a high technology startup company; a return of $400 is experienced 20% of time. A return of $100 is experienced 40% of the time and zero (0) total loss is experienced 40% of the time. What is the firm’s expected return based on this data?. Solution: S = {400, 100, 0} E(x) = P1 * X1 + P2 * X2 + …Pn * Xn = 400 * 0.2 + 100 * 0.4 + 0 * 0.4. = 80 + 40 + 0 = $120 Question 6: Given that an unbiased die was thrown 120 times and the recordings are as follows then calculate the expected mean or mean score. Score x 1 2 3 4 5 6 Frequency f 15 22 23 19 23 18 Total = 120
Solution: E(x) = ∑fx/∑f. = (15 + 44 + 69 + 76 + 115 +108)/120. = 3.558 Question 7: The random variable X has Pdf P(X=x) for x = 1,2,3.
X 1 2 3 P(X = x) 0.1 0.6 0.3
Calculate:
a. E(3). b. E(x). c. E(5x). d. E(5x + 3). e. 5E(x) + 3. f. E(x2). g. E(4x2 - 3). h. 4E(x2) – 3.
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Solution: X 1 2 3 5x 5 10 15 5x + 3 8 13 18 x2 1 4 9 4x2 – 3 1 13 33 P(X = x) 0.1 0.6 0.3
a. E(3) = ∑P(X = x). all x = ∑3P(X = x). all x = 3(0.1) + 3(0.6) + 3(0.3). = 3.
b. E(x) =∑xP(X = x). all x =1(0.1) +2(0.6) +3(0.3) = 2.2
c. E(5x) = ∑5xP(X = x).
all x = 5(0.1) + 10(0.6) +15(0.3) = 11 d. E(5x + 3) = ∑(5x + 3)P(X = x).
all x = 8(0.1) + 13(0.6) + 18(0.3) = 14
e. 5E(x) + 3 = 5(2.2) + 3 = 14 f. E(x2) = ∑ x2P(X = x).
all x = 1(0.1) + 4(0.6) + 9(0.3) = 5.2
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g. E(4x2 - 3) = ∑(4x2 – 3)P(X = x). all x = 1(0.3) + 13(0.6) + 33(0.3) = 17.8
h. 4E(x2) – 3 = 4(5.2) – 3 =20.8 – 3 = 17.8
� VARIANCE: The expected value of a random variable is a measure of central tendency i.e. what values are mostly likely to occur while Variance is a measure of how far apart the possible values are spread again weighted by their respective probabilities. The formula for variance is given by: Var (x) = E(x – µ)2 this can be reduced to Var (x) = E(x2) - µ2 Question 8: The random variable X has probability distribution shown below.
x 1 2 3 4 5 P(X =x) 0.1 0.3 0.2 0.3 0.1
Find: i. µ = E(x). ii. Var(x) using the formula E(x – µ)2 iii. E(x2) iv. Var(x) using the formula E(x2) - µ2
Solution:
i. E(x) =µ = ∑xP(X = x).
all x =1(0.1) + 2(0.3) + 3(0.2) + 4(0.3) + 5(0.1) = 3
ii. Var (x) = E(x – µ)2
=∑(x – 3)2P(X = x). all x
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X 1 2 3 4 5 (x – 3) -2 -1 0 1 2 (x – 3)2 4 1 0 1 4 P(X = x) 0.1 0.3 0.2 0.3 0.1 = 4(0.1) + 1(0.3) + 0(0.2) + 1(0.3) + 4(0.1) = 1.4 iii. E(x2) = ∑x2P(X = x).
all x = 1(0.1) + 4(0.3) + 9(0.2) + 16(0.3) + 25(0.1) = 10.4
iv. Var(x) = E(x2) - µ2 = 10.4 – 9 = 1.4
� STANDARD DEVIATION: Is the square root of its variance given by the following formula:
δ = √Var (x).
Question 9: From the question given above find the standard deviation for part (iv). δ = √Var (x). δ = √1.4 = 1.183215957 = 1.18 � CUMULATIVE DISTRIBUTION FUNCTION: When we had a frequency distribution, the corresponding Cumulative frequencies were obtained by summing all the frequencies up to a particular value. In the same way if X is a discrete random variable, the corresponding Cumulative Probabilities are obtained by summing all the probabilities up to a particular value. If X is a discrete random variable with Pdf P(X = x) for x = x1, x2. …xn then the Cumulative distribution function is given by: F(t) = P(X <= t)
= ∑t P(X = x). x = x1
The Cumulative Distribution is sometimes called Distribution function.
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Question 10: The probability distribution for the random variable X is given below then constructs the Cumulative distribution table. X 0 1 2 3 4 5 6 P(X =x) 0.03 0.04 0.06 0.12 0.4 0.15 0.2
Solution: F(t) = ∑t P(X = x).
x = x1 So F(0) = P(X <= 0) = 0.03 F(1) = P(X <= 1) = 0.03 + 0.04 = 0.07 F(2) = P(X <= 2) = 0.03 + 0.04 + 0.06 = 0.13 and so on. The Cumulative Distribution table will be as follows:
X 0 1 2 3 4 5 6 F(x) 0.03 0.07 0.13 0.25 0.65 0.8 1
Question 11: For a discrete random variable X the Cumulative distribution function F(x) is given below:
X 1 2 3 4 5 F(x) 0.2 0.32 0.67 0.9 1
Find:
a) P(x = 3). b) P(x > 2).
Solution: a. F(3) = P(x = 3) => P(x = 1) + P(x = 2) + P(x = 3) = 0.67 F(2) = P(x <= 2) => P(x = 1) + P(x = 2) = 0.32 Therefore P(x = 3) = 0.67 – 0.32 = 0.35 b. P(x > 2) = 1 – P(x <= 2) = 1 – F(2) = 1 – 0.32 = 0.68
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� PROBABILITY DISTRIBUTION (CONTINUOUS RANDOM VARIABLES): A random variable X that can be equal to any number in an interval, which can be either finite or infinite length, is called a Continuous Random Variable.
PROBABILITY DENSITY FUNCTIONS: There are 2 essential properties of Pdf:
� Because probabilities cannot be negative. The integral of a function must be non-negative for all choices of interval [a, b] i.e. f(x) >= 0 for all values in the sample space for the random variable X.
� Since the probability associated with the entire sample space is always 1. The integral of f(x) of the entire sample space = 1.
Question 1: A continuous random variable has Pdf f(x) where f(x) = kx, 0<= x <= 4.
i. Find the value of constant k. ii. Sketch y = f(x). iii. Find P(1 <= X <= 2½).
Solution:
i. ∫ f(x) ∂x = 1. all x ∫0
4 kx ∂x = 1. [kx2/2]0
4 = 1. 8k = 1 k = ⅛
ii. Sketch of y = f(x). ½ y = ⅛x 0 4 P(1<= x <= 2½) = ∫1
2½[⅛x]∂x. = [x2/16]1
2½ = 0.328
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Question 2: A continuous random variable has Pdf f(x) where Kx 0<= x <= 4. f(x)= k(4 – x) 2<= x <= 4 0 otherwise
a) Find the value of constant k. b) Sketch y = f(x).
Solution:
D =>∫ab P ∂x + ∫a
b Q ∂x = 1. =>∫0
2 kx ∂x + ∫24 k(4 – x) ∂x = 1.
=> [kx2/2]0
2 + [4xk - kx2/2]24 = 1.
=> [4k/2] – [0] + {[16k - 16k/2] – [8k - 4k/2]}
= 1.
� [2k] + {[8k] – [6k]} = 1. � 4k = 1 � k = ¼
c. Sketch y = f(x).
X 0 1 2 3 4 Y 0 ¼ ½ ¾ 1
F(x) = kx.
X 2 3 4 Y ½ ¼ 0
F(x) = k(4 – x)
1 ¾ ½ ¼ 0 1 2 3 4
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� EXPECTED VALUE OR MEAN (CONTINUOUS RANDOM VARIABLE): For a continuous random variable X defined within finite interval [a, b] with continuous Pdf f(x) then the expected value or mean is given by: E(x) = ∫a
b x. f(x) ∂x
Question 3: A continuous random variable has Pdf f(x) where 6/7x 0<= x <= 1. f(x)= 6/7x(2 – x) 1<= x <= 2 0 otherwise
i. Find E(x). ii. Find E(x2).
Solution:
D =>∫ab P ∂x + ∫a
b Q ∂x = 1.
E(x) = ∫ab x. f(x) ∂x
E(x) = ∫0
1 6/7 x2 ∂x + ∫1
2 6/7 x2(2 – x)∂x
= 6/7[x
3/3]01 + 6/7[
2/3x3 – x4/4]1
2 = 6/7[⅓] + 6/7{
16/3 – 4 – (⅔ - ¼)} = 6/7[
5/4] = 15/14 E(x2) = ∫a
b x2. f(x) ∂x E(x2) = ∫0
1 6/7 x3 ∂x + ∫1
2 6/7 x3(2 – x)∂x
= 6/7[x
4/4]01 + 6/7[x
4/2 – x5/5]12
= 6/7[¼] + 6/7{8 - 32/5 – (½ - 1/5)} = 6/7[
31/20] = 93/70
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VARIANCE AND STANDARD DEVIATION:
The variance and Standard Deviation associated with a continuous random variable X on the sample space [a, b] is given by:
Var (x) = ∫ab [x - E(x)]2 . f(x) ∂x or Var(x) = E(x2) - µ2 ∂x
and Standard deviation = √Var (x). E(x) = µ.
Question 4: A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4. Find:
a) E(x). b) E(x2). c) Var (x). d) The standard deviation of x. e) Var(3x +2). Solution:
i. E(x) = ∫ab x. f(x) ∂x
E(x) = ∫0
4 ⅛x2 ∂x = ⅛[x3/3]0
4
= 8/3
ii. E(x2) = ∫a
b x2. f(x) ∂x
=∫04 ⅛x3 ∂x
= ⅛[x4/4]0
4 = ⅛(64)
= 8
iii. Var(x) = E(x2) - µ2 ∂x = E(x2) - E2(x) ∂x
= 8 – (8/3)
2 = 8/9
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iv. Standard Deviation => δ = √Var (x). = √8/9
= 2√2/3
v. Var(3x + 2) = 9 Var(x) this has been obtained form the concept Var (ax)= Var a2(x). = 9 (8/9) = 8
� MODE: The Mode is the value of X for which f(x) is greatest in the given range of X. It is usually to draw a sketch of y = f(x) and this will give an idea of the location of the Mode. For some Probability Density functions it is possible to determine the mode by finding the maximum point of the curve y = f(x) from the relationship f1(x) = 0. f1(x) = d/∂x * f(x). Question 5: A continuous random variable has Pdf f(x) where f(x) = 3/80(2 + x)(4 – x), 0<= x<= 4.
a) Sketch y = f(x). b) Find the mode. Solution:
X 0 1 2 3 4 Y 24/80
27/80 24/80
15/80 0 a) 27/80 Mode 24/80 f(x) = 3/80(2 + x)(4 – x) 15/80 0 1 2 3 4
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b) The mode => f(x) = 3/80(2 + x)(4 – x) = 3/80(8 + 2x – x2) f1(x) = (2+ 2x)
f1(x) = 0. 0 = 2+ 2x 2x = 2 x = 1
� MEDIAN: The median splits the area under the curve y = f(x) into 2 halves so if the value of the Median is m. Therefore the formula for the median is given by:
∫am f(x) ∂x = 0.5.
F(m) = 0.5
Question 6: A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4. Find:
a. The median m.
Solution: m => ∫a
m f(x) ∂x = 0.5. F(m) = 0.5 f(x) = ⅛x ∂x
0.5 = m2/16
m2 = 8
m = 2.83
� CUMULATIVE DISTRIBUTION FUNCTION: F(x) When considering a frequency distribution the corresponding cumulative frequencies were obtained by summing all the frequencies up to a particular value. In the same way if X is a continuous random variable with Pdf f(x) defined for a<=x<=b then the Cumulative Distribution Function is given by F(t):
F(t) = P(X <= t) = ∫at f(x) ∂x
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2 PROPERTIES OF CDF:
� F(b) = ∫ab f(x) ∂x = 1.
� If f(x) is valid for -∞ <= x <= ∞ then F(t) = ∫-∞t f(x) ∂x where the interval is taken
over all values of x <= t. � The Cumulative distribution function is sometimes known as just as the distribution
function. Question 6:
A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4. Find:
i. The Cumulative distribution function F(x). ii. Sketch y = F(x). iii. Find P(0.3 <=x<= 1.8).
Solution:
i. F(t) = ∫at f(x) ∂x
F(t) = ∫0t ⅛x ∂x
= ⅛[x2/2]0
t = t2/16 F (t) = t2/16 0<=t<=4
NB: (1) F(4) = 42/16 = 1
0 x <= 0. F(x) = x2/16 0<= x <= 4 1 x >= 4
ii. Sketch y = F(x).
X 0 1 2 3 4 Y 0 1/16
1/2 9/16 1
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1 F(x) = 1 9/16 1/2 F(x)= x2/16 1/16 0 1 2 3 4
iii. P(0.3 <= x <= 1.8) = F(1.8) – F(0.3) F(1.8) = (1.8)2/16 = 0.2025 F(0.3) = (0.3)2/16 = 0.005625 Therefore P(0.3 <= x <= 1.8) = F(1.8) – F(0.3) = 0.2025 – 0.005625 = 0.196875 = 0.197 Question 7:
A continuous random variable has Pdf f(x) where x/3 0<= x <= 2. f(x)= -2x/3 +2 2<= x <= 3
0 otherwise
a. Sketch y = f(x). b. Find the Cumulative distribution function F(x). c. Sketch y = F(x). d. Find P(1 <= X <= 2.5) e. Find the median m.
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Solution:
i. Sketch y = f(x). X 0 1 2 Y 0 ⅓ ⅔
X 2 3 Y ⅔ 0
⅔ y = x/3 ⅓ y =-2x/3 +2 0 1 2 3
ii. CDF = F(t) = ∫0
t x/3∂x
= [x2/6]0t
= t2/6 F (t) = x2/6 0<=x<=2
NB: F (2) = 22/6 = ⅔ F(t) = F(2) + (Area under the curve y = -2x/3 +2 between 2 and t) So
F(t) = F(2) + ∫2t (-2x/3 +2) ∂x
= F(2) + [-x2/3 + 2x]2
t = ⅔ + {-t2/3 +2t – ( -4/3 + 4)} = -t2/3 +2t – 2 2<= t <= 3 NB: F(2) = -9/3 + 6 – 2 = 1
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Therefore CDF = x2/6 0<= x <= 2. f(x)= - x2/3 +2x -2 2<= x <= 3.
1 x >= 3.
iii. Sketch of y = F(x).
y = 1 1 y = - x2/3 +2x -2 2/3 y = x2/6 1/3 0 1 2 3
iv. P(1 <= X <= 2.5) = F(2.5) – F(1) as 2.5 is in the range 2<= x <=3.
� F(2.5) = - x2/3 +2x –2 � F(2.5) = - (2.5)2/3 +2(2.5) –2 � = 11/12
F(1) = x2/6 as 1 is in the range 0 <= x <= 2.
� F(1) = x2/6
� F(1) = 12/6
� = 1/6 Therefore P(1 <= X <= 2.5) = F(2.5) – F(1) = 11/12 -
1/6 = 0.75
v. m => ∫am f(x) ∂x = 0.5 where m is the median.
F(2) = ⅔ so the median must lie in the range 0 <= x <= 2. F(m) = m2/6 m2/6 = 0.5
m2 = 3. m = 1.73
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� OBTAINING THE PDF FROM THE CDF: The Probability Density Function can be obtained from the Cumulative Distribution function as follows: Now F(t) = ∫a
t f(x) ∂x a<= t <= b. So f(x) = d/∂x * F(x). = F1(x). NB: The gradient of the F(x) curve gives the value of f(x).
Question 8: A continuous random variable has Pdf f(x) where 0 x <= 0. F(x)= x3/27 0<= x <= 3
1 x >= 3. Find the Pdf of X, f(x) and sketch y = f(x).
Solution: a. f(x) = d/∂x * F(x). = d/∂x(x3/27). = 3x2/27
= x2/9
Therefore the Pdf is equal to: x2/9 0<=x<=3 f(x) = 0 otherwise. b. Sketch of y = f(x).
1 y = x2/9 0 1 2 3
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Question 9: A continuous random variable X takes values in the interval 0 to 3. It is given that P(X > x) = a + bx3, 0 <= x <= 3.
i. Find the values of the constants a and b. ii. Find the Cumulative distribution function F(x). iii. Find the Probability density function f(x). iv. Show that E(x) = 2.25. v. Find the Standard deviation.
Solution:
a. P(X > x) = a + bx3, 0 <= x <= 3. So P(X > 0) = 1 and P(X > 3) = 0. i.e. a + b(0) = 1 and a + b(27) = 0 Therefore a = 1 and 1 + 27b = 0. B = -1/27. So P(X > x) = 1 - x3/27, 0 <= x <= 3.
b. Now P(X <= x) = x3/27 (CDF)
X3/27 0<=x<=3 F(x) =
1 x > 3.
c. f(x) = d/∂x * F(x). = d/∂x(x3/27). = 3x2/27
= x2/9
d. E(x) = ∫a
b x. f(x) ∂x =∫0
3 x. x2/9∂x =∫0
3 x3/27∂x = [x4/36]0
3
= 2.25
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e. Var(x) = ∫a
b [x - E(x)]2 . f(x) ∂x = ∫ab x2.f(x)∂x - E2(X)
=∫0
3 x4/9∂x – 2.252. =[x5/45]0
3 - 5.0625. = 0.3375 f. δ = √ Var (x). = √ 0.3375 = 0.581
� RELATIONSHIPS AMONG PROBABILITY DISTRIBUTIONS:
� JOINT PROBABILITY DISTRIBUTION: Question 1:
2(X + Y - 2XY)0<= X<=1, 0<= Y<=1 Given f(X, Y) = 0 Otherwise
i. Show that this is a PDF. ii. Find P(0 <= X <=½), (0 <= Y <=¼). iii. Find CDF.
Solution:
a) =∫ab∂X∫a
b∂Y =∫0
1∂X∫01∂Y [2(X + Y - 2XY)]
= 2∫01∂X [(XY + Y2/2 - XY2)]0
1
= 2∫01∂X [(X + ½ - X)]
= 2[(X2/2 + ½X - X2/2)]01
= 2[½ + ½ - ½] = 2 * ½ = 1 b) =∫a
b∂X∫ab∂Y
=∫0½∂X∫0
¼∂Y [2(X + Y - 2XY)] = 2∫0
½∂X[(XY + Y2/2 - XY2)]0¼
= 2∫0½∂X[(¼X + 1/32 -
1/16X)] = 2∫0
½∂X[(3/16X + 1/32)] = 2[(3/32X
2 + 1/32X)] 0½
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= 2[(3/128 + 1/64)]
= 2 * 5/128
= 5/64 c) F(u, v) =∫a
b∂u∫ab∂v
=∫0x∂U∫0
y∂V[2(U + V – 2UV)] = 2∫0
x∂U[(UV + V2/2 - UV2)]0y
= 2∫0x∂U[(UY + Y2/2 - UY2)]
= 2[(YU2/2 + UY2/2 - U2X2/2)]0X
= 2[(X2Y/2 + XY2/2 - X2X2/2)] = X2Y + XY2 - X2X2
NB: Given CDF to get PDF we differentiate the function. Given PDF to get CDF we integrate the function. Given a PDF to prove that it is a PDF integrate until the answer is 1. Given PDF the marginal distribution is found by integrating. Given CDF the marginal distribution is found by differentiating.
Question 1: Given the CDF = x2 + 3xyz + z. Find the corresponding PDF.
Solution: To get the corresponding PDF we differentiate with respect to x, y, z. f(x, y, z) =x2 + 3xyz + z = d/∂x(2x + 3yz). = d/∂y(3z). = d/∂z (3).
= 3
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UNIT 6 DECISION THEORY HOURS: 20
� DECISION THEORY:
There are four types of criteria that we will look at.
Expected Value (Realist) Compute the expected value under each action and then pick the action with the largest expected value. This is the only method of the four that incorporates the probabilities of the states of nature. The expected value criterion is also called the Bayesian principle.
Maximax (Optimist)
The maximax looks at the best that could happen under each action and then chooses the action with the largest value. They assume that they will get the most possible and then they take the action with the best best case scenario. The maximum of the maximums or the "best of the best". This is the lotto player; they see large payoffs and ignore the probabilities.
Maximin (Pessimist)
The maximin person looks at the worst that could happen under each action and then choose the action with the largest payoff. They assume that the worst that can happen will, and then they take the action with the best worst case scenario. The maximum of the minimums or the "best of the worst". This is the person who puts their money into a savings account because they could lose money at the stock market.
Minimax (Opportunist)
Minimax decision making is based on opportunistic loss. They are the kind that look back after the state of nature has occurred and say "Now that I know what happened, if I had only picked this other action instead of the one I actually did, I could have done better". So, to make their decision (before the event occurs), they create an opportunistic loss (or regret) table. Then they take the minimum of the maximum. That sounds backwards, but remember, this is a loss table. This similar to the maximin principle in theory; they want the best of the worst losses.
Example: A bicycle shop
Zed and Adrian and run a small bicycle shop called "Z to A Bicycles". They must order bicycles for the coming season. Orders for the bicycles must be placed in quantities of twenty (20). The cost per bicycle is $70 if they order 20, $67 if they order 40, $65 if they order 60, and $64 if they order 80. The bicycles will be sold for $100 each. Any bicycles left over at the end of the season can be sold (for certain) at $45 each. If Zed and Adrian run out of bicycles during the season, then they will suffer a loss of "goodwill" among their customers. They estimate this goodwill loss to be $5 per
OPERATIONS RESEARCH
customer who was unable to buy a bicycle. Zed and Adrian estimate that the demand for bicycles this season will be 10, 30, 50, or 70 bicycles with probabilities of 0.2, 0.4, 0.3, and 0.1 respectively.
Actions
There are four actions available to Zed and Adrian. They have to decide which of the actions is the best one under each criteria.
1. Buy 20 bicycles 2. Buy 40 bicycles 3. Buy 60 bicycles 4. Buy 80 bicycles
Zed and Adrian have control over which action they choose. That is the whole point of decision theory - deciding which action to take.
States of Nature
There are four possible states of nature. A state of nature is an outcome.
1. The demand is 10 bicycles 2. The demand is 30 bicycles 3. The demand is 50 bicycles 4. The demand is 70 bicycles
Zed and Adrian have no control over which state of nature will occur. They can only plan and make the best decision based on the appropriate decision criteria.
Payoff Table
After deciding on each action and state of nature, create a payoff table. The numbers in parentheses for each state of nature represent the probability of that state occurring.
Action
State of Nature Buy 20 Buy 40 Buy 60 Buy 80
Demand 10 (0.2)
50 -330 -650 -970
Demand 30 (0.4)
550 770 450 130
Demand 50 (0.3)
450 1270 1550 1230
Demand 70 (0.1)
350 1170 2050 2330
OPERATIONS RESEARCH
Ok, the question on your mind is probably "How the [expletive deleted] did you come up with those numbers?". Let's take a look at a couple of examples.
Demand is 50, buy 60: They bought 60 at $65 each for $3900. That is -$3900 since that is money they spent. Now, they sell 50 bicycles at $100 each for $5000. They had 10 bicycles left over at the end of the season, and they sold those at $45 each of $450. That makes $5000 + 450 - 3900 = $1550.
Demand is 70, buy 40: They bought 40 at $67 each for $2680. That is a negative $2680 since that is money they spent. Now, they sell 40 bicycles (that's all they had) at $100 each for $4000. The other 30 customers that wanted a bicycle, but couldn't get one, left mad and Zed and Adrian lost $5 in goodwill for each of them. That's 30 customers at -$5 each or -$150. That makes $4000 - 2680 - 150 = $1170.
Opportunistic Loss Table
The opportunistic loss (regret) table is calculated from the payoff table. It is only needed for the minimax criteria, but let's go ahead and calculate it now while we're thinking about it.
The maximum payoffs under each state of nature are shown in bold in the payoff table above. For example, the best that Zed and Adrian could do if the demand was 30 bicycles is to make $770.
Each element in the opportunistic loss table is found taking each state of nature, one at a time, and subtracting each payoff from the largest payoff for that state of nature. In the way we have the table written above, we would subtract each number in the row from the largest number in the row.
Action
State of Nature Buy 20 Buy 40 Buy 60 Buy 80
Demand 10 0 380 700 1020
Demand 30 220 0 320 640
Demand 50 1100 280 0 320
Demand 70 1980 1160 280 0
Remember that the numbers in this table are losses and so the smaller the number, the better.
Expected Value Criterion
Compute the expected value for each action.
For each action, do the following: Multiply the payoff by the probability of that payoff occurring. Then add those values together. Matrix multiplication works really well for this as it multiplied pairs of numbers together and adds them. If you place the probabilities into a 1x4 matrix and use the 4x4 matrix shown above, then you can multiply the matrices to get a 1x4 matrix with the expected value for each action.
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Here is an example of the "Buy 60" action if you wish to do it by hand. 0.2(-650) + 0.4(450) + 0.3(1550) + 0.1(2050) = 720
The expected values for buying 20, 40, 60, and 80 bicycles are $400, 740, 720, and 460 respectively. Since the best that you could expect to do is $740, you would buy 40 bicycles.
Maximax Criterion
The maximax criterion is much easier to do than the expected value. You simply look at the best you could do under each action (the largest number in each column). You then take the best (largest) of these.
The largest payoff if you buy 20, 40, 60, and 80 bicycles are $550, 1270, 2050, and 2330 respectively. Since the largest of those is $2330, you would buy 80 bicycles.
Maximin Criterion
The maximin criterion is as easy to do as the maximax. Except instead of taking the largest number under each action, you take the smallest payoff under each action (smallest number in each column). You then take the best (largest of these).
The smallest payoff if you buy 20, 40, 60, and 80 bicycles are $50, -330, -650, and -970 respectively. Since the largest of those is $50, you would buy 20 bicycles.
Minimax Criterion
Be sure to use the opportunistic loss (regret) table for the minimax criterion. You take the largest loss under each action (largest number in each column). You then take the smallest of these (it is loss, afterall).
The largest losses if you buy 20, 40, 60, and 80 bicycles are $1980, 1160, 700, and 1020 respectively. Since the smallest of those is $700, you would buy 60 bicycles.
Putting it all together.
Here is a table that summarizes each criteria and the best decision.
Action
Criterion Buy 20 Buy 40 Buy 60 Buy 80 Best Action
Expected Value 400 740 720 460 Buy 40
Maximax 550 1270 2050 2330 Buy 80
Maximin 50 -330 -650 -970 Buy 20
Minimax 1980 1160 700 1020 Buy 60
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Practice Problem
Since there aren't any problems of this kind in the text, work this problem out, and then you can check your answer with the instructor.
Finicky's Jewelers sells watches for $50 each. During the next month, they estimate that they will sell 15, 25, 35, or 45 watches with respective probabilities of 0.35, 0.25, 0.20, and ... (figure it out). They can only buy watches in lots of ten from their dealer. 10, 20, 30, 40, and 50 watches cost $40, 39, 37, 36, and 34 per watch respectively. Every month, Finicky's has a clearance sale and will get rid of any unsold watches for $24 (watches are only in style for a month and so they have to buy the latest model each month). Any customer that comes in during the month to buy a watch, but is unable to, costs Finicky's $6 in lost goodwill.
Find the best action under each of the four decision criteria.
� DECISION UNDER RISK:
A. THE EXPECTED VALUE CRITERION: Muchadura buys and sell tomatoes at $3 and $8 a case respectively. Because tomatoes are perishable, they must be bought a day and sold after that, they become valuable. A 90 days observation of a business reveals the following information in cases. Daily Sales Number Of Days Sold Probability Of Demand
10 cases 18 (18 / 90) = 0.2 11 cases 36 (36 / 90) = 0.4 12 cases 27 (27 / 90) = 0.3 13 cases 9 (9 / 90) = 0.1 Total number of days 90 = 1
The Probability of demand = Number of days sold / Total number of days. There are only 4 sales volumes but their sequence in UNKNOWN. The problem is how many cases must Muchadura stock for selling the following day. If he stocks more or less than the demand on any day he will suffer some losses. Example: If he stocks 13 cases on a particular day and the customer demand on that particular day is 10 cases, then he gets a profit of 3 cases. Let buying price = $3. Let Selling price = $8.
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NB: This method if getting Maximum profit = Selling Price – Buying Price per case = $5. Then 10 * 5 = $50 and perishable ones = 3 * 3 = $9 and then the Maximum profit = $50 - $9 = $41. For the 3 unsold cases he suffer a loss for each equivalent to the Buying price. All such losses are calculated in a conditional profit table. I.e. a table of profit on a given supply and demand levels as illustrated below. STOCK ACTION EXPECTED PROFIT Possible Probability Demand of Demand Profits 10 11 12 13 10 11 12 13 10 50 47 44 41 0.2 10 9.4 8.8 8.2 11 50 55 52 49 0.4 20 22 20.8 19.6 12 50 55 60 57 0.3 15 16.5 18 17.1 13 50 55 60 65 0.1 5 5.5 6 6.5 Total 1 50 53.4 53.6 51.4 Muchadura must stock 12 cases which give the highest Expected profit of $53.60 on a daily bases to get the maximum profit over a long time.
B. EXPECTED PROFIT WITH PERFECT INFORMATION: Suppose Muchadura had perfect information i.e. complete and accurate information about the future. Sales demand would still vary from 10 to 13 cases in their respective probabilities. But Muchadura would not know in advance how many cases would be needed each day. The table below shows the conditional profits value in such situation. STOCK ACTION EXPECTED PROFIT Possible Probability Demand of Demand Profits 10 11 12 13 10 11 12 13 10 50 0.2 10 11 55 0.4 22 12 60 0.3 18 13 65 0.1 6.5 Total 1 10 22 18 6.5 Maximum possible profit = 10 + 22 + 18 + 6.5. = $56.50
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C. MINIMISING EXPECTED LOSS:
There are two types of Losses:
� Overstocking � Under stocking.
OVERSTOCKING: Means you will loss the buying amount. UNDER STOCKING: It means you will loss the profit. STOCK ACTION EXPECTED LOSS Possible Probability Demand of Demand Loss 10 11 12 13 10 11 12 13 10 0 3 6 9 0.2 0 0.6 1.2 1.8 11 5 0 3 6 0.4 2 0 1.2 2.4 12 10 5 0 3 0.3 3 1.5 0 0.9 13 15 10 5 0 0.1 1.5 1 0.5 0 Total 1 6.5 3.1 2.9 5.1 The Expected Minimum Loss is $2.9. The values above the zero diagonal are due to overstocking and those below are due to under stocking.
D. THE EXPECTED VALUE OF PERFECT INFORMATION: The expected value of perfect information is calculated as: The expected Profit with perfect information – The expected value Criterion. = 56.50 – 53.60 = 2.9 In general the expected value of perfect information = Minimum Expected Loss.
OPERATIONS RESEARCH
E. ITEMS WITH A SALVAGE VALUE: Main items do not become completely worthless after their prime deaths. They will have reduced values (Salvage values). The Salvage values must be considered when computing conditional profits or losses. Consider the previous case when the cost price was $3 and selling price was $8. Any unsold cases will be disposed / salvage of at $4 the profit will be $4 - $3 = $1. STOCK ACTION EXPECTED PROFIT Possible Probability Demand of Demand Profits 10 11 12 13 10 11 12 13 10 50 51 52 53 0.2 10 10.2 10.4 10.6 11 50 55 56 57 0.4 20 22 22.4 22.8 12 50 55 60 61 0.3 15 16.5 18 18.3 13 50 55 60 65 0.1 5 5.5 6 6.8 Total 1 50 54.2 56.8 58.2 Muchadura must stock 13 cases, which give the maximum profit of $58.20.
� DECISION MAKING UNDER UNCERTAINITY: Under conditions of uncertainty, only payoffs are known and nothing is known about the likelihood of each state of nature. Different persons have suggested several decisions rules for making decisions under such situations:
A. MAXIMIN (PESSIMISTIC): It is based upon the consecutive approach to assume that the worst possible is going to happen. The decision maker considers each alternative and locates the minimum payoff for each and then selects that alternative which maximizes the minimum payoffs. STEPS INVOLVED:
i. Determine the minimum assured payoffs for each alternative.
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ii. Choose that alternative which corresponds to the Maximum of the above minimum payoff.
B. MAXIMAX CRITERIA (OPTIMISTIC): Is based upon extreme optimistic, the decision maker selects that particular strategy which corresponds to the maximum of the maximum payoffs of each strategy. STEPS INVOLVED:
i. Determine the maximum possible payoff for each alternative. ii. Select that alternative which corresponds to the maximum of the above maximum
payoff.
C. HURWICZ: In order to overcome the disadvantage of extreme pessimistic (Maximin) and extreme Optimism (Maximax) criteria. Hurwicz introduced the concept of co-efficient of optimism or pessimism as α. Therefore the Hurwicz concept is given by: α Maximin θj {V (ai, θj)} + (1 - α) Maximax θj {V(ai, θj)} D. LAPLACE CRITERION: This criterion is based on what is known as the principal if insufficient reason. Since the probabilities associated with the occurrences of θ1, θ2….θn are unknown, we do not have enough information to conclude that these probabilities will be different. For if this is not the case, we should be able to determine these probabilities and the situation will no longer be a decision under uncertainty. Thus because of insufficient reason to believe otherwise; the states of θ1, θ2….θn are equally likely to occur. The Laplace principal assumes that of θ1, θ2….θn are likely to occur therefore the probability = 1/n where n = number of occurrences.
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E. MAXIMIN REGRET CRITERION (SALVAGE): This is a less conservative criterion. In this case a new loss matrix is created such that V(Ai, Qj) is replaced by R(Ai, Qj) which is defined by: Max {V (Ai, Qj) – V(Ai, Qj)} if V is profit R(Ai, Qj)= V(Ai, Qj) – Min {V(Ai, Qj)} if V is loss R(Ai, Qj) is the difference between the best choice in column Qj and the values of V(Ai, Qj) in the same column. R(Ai, Qj) is a regret matrix. Question 1: Given the following data: Customers Category Q1 Q2 Q3 Q4 Supplier A1 5 10 18 25 Level A2 8 7 8 23 A3 21 18 12 21 A4 30 22 9 25 Using the above data: Find
i. Maximin Criterion. ii. Maximax Criterion. iii. Hurwicz Criterion. iv. Laplace Criterion. v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion. Customers Category Q1 Q2 Q3 Q4 Maximin Supplier A1 5 10 18 25 5 Level A2 8 7 8 23 7 A3 21 18 12 21 12 A4 30 22 9 25 9
Maximin = 12 The optimal decision is to take supply level A3
OPERATIONS RESEARCH
b) Maximax Criterion.
Customers Category Q1 Q2 Q3 Q4 Maximax Supplier A1 5 10 18 25 25 Level A2 8 7 8 23 23 A3 21 18 12 21 21 A4 30 22 9 25 30
Maximax = 30 The optimal decision is to take supply level A4 c) Minimax Criterion.
Customers Category Q1 Q2 Q3 Q4 Minimax Supplier A1 5 10 18 25 25 Level A2 8 7 8 23 23 A3 21 18 12 21 21 A4 30 22 19 25 30
Minimax = 21 The optimal decision is to take supply level A3 d) Hurwicz Criterion.
Customers Category Q1 Q2 Q3 Q4 Supplier A1 5 10 18 25 Level A2 8 7 8 23 A3 21 18 12 21 A4 30 22 19 25
Hurwicz: = α Maximin θj {V (ai, θj)} + (1 - α) Maximax θj {V(ai, θj)}
Let assume that α = ½.
Maximin Maximax α Maximin + (1 - α) Maximax 5 25 15 7 23 15 12 21 16.5 9 30 19.5 The optimal decision is to take supply level A4
OPERATIONS RESEARCH
e) Laplace Criterion.
Customers Category Q1 Q2 Q3 Q4 Supplier A1 5 10 18 25 Level A2 8 7 8 23 A3 21 18 12 21 A4 30 22 19 25
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are: E(A1) = ¼(5 + 10 + 18 + 25) = 14.5 E(A2) = ¼(8 + 7 + 8 + 23) = 11.5 E(A3) = ¼(21 + 18 + 12 + 21) = 18 E(A1) = ¼(30 + 22 + 19 + 25) = 24
The optimal decision is to take supply level A2
f) Savage Minimax Regret Criterion. Customers Category Q1 Q2 Q3 Q4 Supplier A1 5 10 18 25 Level A2 8 7 8 23 A3 21 18 12 21 A4 30 22 19 25
Q1 Q2 Q3 Q4 Max{ R(Ai, Qj)} A1 0 3 10 4 10 A2 3 0 0 2 3 Minimax A3 16 11 4 0 16 A4 25 15 11 4 25 The optimal decision is to take supply level A2 which gives Minimax.
OPERATIONS RESEARCH
Question 2: A firm has to decide on its advertising campaign. It has a choice between Tv, Newspaper, Poster and Radio advertising. The return on the advertising medium is measured by the number of potential customers who have the opportunity to see each medium. This will depend on the type of the weather. The figures in the table are the numbers of potential customers in thousands. The firm has funds for using only to one medium. Poor Moderate Good Excellent Tv 200 190 170 130 Newspaper 180 160 150 130 Poster 110 140 140 190 Radio 210 190 160 110
Decide which medium the firm should use based on the following information: i. Maximin Criterion. ii. Maximax Criterion. iii. Hurwicz Criterion. iv. Laplace Criterion. v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion. Poor Mod Good Excellent Maximin Tv 200 190 170 130 130 Newspaper 180 160 150 130 130 Poster 110 140 140 190 110 Radio 210 190 160 110 110
Maximin = 130
The best medium to use for advertising is Tv or Newspaper since they have the highest number of (possible) potential customers of 130
\
OPERATIONS RESEARCH
b) Maximax Criterion. Poor Mod Good Excellent Maximax Tv 200 190 170 130 200 Newspaper 180 160 150 130 180 Poster 110 140 140 190 190 Radio 210 190 160 110 210
Maximax = 210
The best medium to use for advertising is the Radio since it has the highest number of (possible) potential customers of 210.
c) Minimax Criterion. Poor Mod Good Excellent Minimax Tv 200 190 170 130 200 Newspaper 180 160 150 130 180 Poster 110 140 140 190 190 Radio 210 190 160 110 210
Minimax = 180
The best medium to use for advertising is the Newspaper with 180 potential customers. d) Hurwicz Criterion. Poor Mod Good Excellent Tv 200 190 170 130 Newspaper 180 160 150 130 Poster 110 140 140 190 Radio 210 190 160 110
OPERATIONS RESEARCH
Hurwicz: = α Maximin θj {V (ai, θj)} + (1 - α) Maximax θj {V(ai, θj)}
Let assume that α = 0.7
Maximin Maximax α Maximin + (1 - α) Maximax 130 200 151 130 180 145 110 190 109.7 110 210 140 The best medium to use for advertising is the Tv since it has the highest number of (possible) potential customers of 151
g) Laplace Criterion.
Poor Mod Good Excellent Tv 200 190 170 130 Newspaper 180 160 150 130 Poster 110 140 140 190 Radio 210 190 160 110
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are: E(Tv) = ¼(200 + 190 + 170 + 130) = 172 500 E(Newspaper) = ¼(180 +160 + 150 + 130) = 155 000 E(Poster) = ¼(110 + 140 + 140 + 190) = 145 000 E(Radio) = ¼(210 + 190 + 160 + 110) = 167 500
The best medium to use for advertising is Tv since it has the highest number of (possible) potential customers who amount to 172 500
OPERATIONS RESEARCH
h) Savage Minimax Regret Criterion. Poor Mod Good Excellent Tv 200 190 170 130 Newspaper 180 160 150 130 Poster 110 140 140 190 Radio 210 190 160 110 Poor Mod Good Excellent Max {R(Ai, Qj)} Tv 90 50 30 20 90 Newspaper 70 20 10 20 70 Minimax Poster 0 0 0 80 80 Radio 100 50 20 0 100
The best medium to use for advertising is Newspaper, which minimize maximum view ship.
OPERATIONS RESEARCH
UNIT 7: THEORY OF GAMES: HOURS: 20 CHARACTERISTICS OF A COMPETITIVE GAME
OPERATIONS RESEARCH
ALTERNATIVELY
� GAME THEORY: It deals with Decision under Uncertainty which involves 2 or more intelligent opponents in which each opponent aim to optimize his / her own decision at the expense of the other opponent. Typical examples include launching an advertisement campaign. (2) Competing for products or planning war techniques for opposing enemy. In game theory an opponent is referred to as a Player. Each player has the number of choices, limited or unlimited called Strategies. The outcomes payoffs of a game are summarized as functions of different strategies for each player. A game with 2 players were a gain of one player is equal to the loss of another player is called a Two Person Zero Sum game. To illustrate two person zero sum game considers a coin-matching situation in which each of the players A and B select a Head H or Tail T. If the outcome match Head or Tail Player A wins $1 from Player B otherwise Player A loses $1 to Player B. In this game each player has 2 strategies Head or Tail which yields the following 2*2 game matrix expressed in terms of payoff to Player A. Player B H T Player A H T � OPTIMUM SOLUTION OF TWO PERSON ZERO SUM GAME: Is obtained by using minimax Maximin criterion according to which Player A (whose strategies represents rows) select a strategy (mixed or Pure) which maximize his minimum gains, the minimum being taken over all the strategies of Player B. In similar way Player B selects his strategy that minimize his maximum loses.
1 -1 -1 1
OPERATIONS RESEARCH
1 VALUE OF THE GAME: Is the maximum guaranteed gain to Player A or the minimum possible loss to Player B denoted by V. When Maximin possible value is equal to Minimax value of the corresponding pure strategies, the game is said to have the Optimum Strategies and has a Saddle point.
Player B
1 2 3 4 Maximin Player A 1 2 2 5 3 -4 Minimax 8 5 9 18 The game has a Saddle point which exists at row 2 column 2 and the value of the game = 5.
Player B 1 2 3
Maximin Player A 1 1 2 1 3 1 Minimax 6 3 6 NB: The optimum solution which shows that Maximin <> Minimax means it is a mixed strategy i.e. there is no agreement (poor strategy). Below and above show that there is no Saddle point which means Maximin <> Minimax.
Player B 1 2 3 4
Maximin Player A 1 -10 2 1 3 2 4 -1 Minimax 8 7 15 4
8 2 9 5 6 7 18 7 3 -4 10
5
1 3 6 2 1 3 6 2 1
5 -10 9 0 6 7 8 1 8 7 15 2 3 4 -1 4
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Consider the following game G.
Player B B1 B2 Maximin Player A A1 2 A2 -1
Minimax 2 6 Maximin = Minimax which is a condition strictly determinable game hence the game is strictly determinable whatever λ maybe. The value of the game = 2 with the best strategy for Player A from A1 and the best strategy for Player B from B1. Question 1: Find the range of values of P and Q, which will render the entry (2,2) a Saddle point of the game.
Player B B1 B2 B3
Maximin Player A A1 2 A2 7 A3 4 Minimax 10 7 8 Ignoring values of P and Q determine the Maximin and Minimax value of the payoff matrix. The Maximin =f and Minimax = f thus there exists a Saddle point at position (2,2). This impose a condition on P as P<= f and on Q as Q >= f. Hence the required range for P and Q is = 7<= Q, P<= 7. Question 2: For what values of λ is the game with the following payoff. Payoff is strictly determinable.
Player B B1 B2 B3
Maximin Player A A1 2 A2 -7 A3 -2 Minimax -1 6 2 This shows that the value of the game (V) lies between –1 and 2. That is –1 <= V <= 2. For a strictly determinable we have –1 <= λ <= 2.
2 6 -2 -1
2 4 5 10 7 q 4 p 8
λλλλ 6 2 -1 λλλλ -7 -2 4 λλλλ
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� GRAPHICAL SOLUTION OF 2 BY N & M BY 2 GAMES: The graphical solution of 2 by N and M by 2 is only applicable to games in which at least on of the Players have 2 strategies only. Consider the following 2 by N games.
Player B Y1 Y2 Y3
Player A X1 X2 The expected payoff of the corresponding to the pure strategies of B are given below: This shows that A’s average payoff varies linearly with X1. According to the Minimax criterion for mixed strategies games Player A should select the value V that minimize his maximum expected playoffs. This may be done by plotting a straight line as function of X1. Question 3: Consider the following 2 by 4 games.
Player B 1 2 3 4
Player A 1 2 Solution:
A11 A12 A1n A21 A22 A2n
B’s Pure Strategies A’s Expected Payoff
1 (A11 –A12) X1 + A21 2 (A21 – A22) X1 + A22 N (A2n –A2n) X1 + A2n
2 2 3 -1 4 3 2 6
B’s Pure Strategies A’s Expected Payoff
1 (2 – 4) X1 + 4 2 (2 – 3) X1 + 3 3 (3 – 2) X1 + 2
4 (-1 –7) X1 + 6
OPERATIONS RESEARCH
6 6 5 5 4 4 3 3 2 2 1 1 X1 0 1/2 0 X1 - 1 -1 -2 -2 This is a point of integration of any 2 of the lines or more i.e. lines 2, 3 and 4. A’s optimum strategy is X1 = ½ and X2 = 5/2 and the value of the game is obtained by substituting X1 in the equations of any of the lines passing the intersection of the point. The value of the game = 5/2. Question 4: Consider the following 2 by 3 game, find B’s pure strategies and A’s expected payoff using the graphically method.
Player B 1 2 3
Player A 1 2 Solution:
B’s Pure Strategies A’s Expected Payoff
1 - 2X1 + 4 2 -X1 + 3 3 X1 + 2
4 - 7X1 + 6
1 3 11 8 5 2
B’s Pure Strategies A’s Expected Payoff
1 - 7X1 + 8 2 - 2X1 + 5 3 9X1 + 2
OPERATIONS RESEARCH
11 11
10 10 8 9 8 8 7 7
6 6 5 5 4 H 4 3 3 2 2 1 1 X1 0 0 X1 - 1 Maximin -1 -2 -2 Since Player A wishes to Maximize his Minimum expected payoff we consider the highest point of intersection on the lower elevation of A’s expected payoff equations. This point it represents the maximum value of the game for A. Lines 2 and 3 passes through it defines the relative moves B2 and B3 it along B needs to play. The solution of the original 2 * 3 game reduces to a 2 by 2 payoff matrix.
Player B B1 B2 Player A A1 X1+ X2 = 1 A2 Y1 + Y2 = 1
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X1 = 3 – 11 / (3 + 2) – (5 + 11) = 8 / 11 X2 = 1 - 8 / 11 = 3 / 11 The value of the game V = (2 * 3) – (5 * 11) / (3 + 2) – (5 + 11) = 49 / 11
3 11 5 2
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Question 4: Consider the following 2 by 4 game, find B’s pure strategies and A’s expected payoff using the graphically method.
Player B
1 2 3 4 Player A 1 2 Solution: 11 11
10 10 9 9
8 8 7 7
6 6 5 5 4 4 3 3 2 2 1 1 X1 0 0 X1 -1 H -1 -2 -2 -3 Maximin -3 -4 -4 -5 -5 -6 -6
1 3 -3 7 2 5 4 -6
B’s Pure Strategies A’s Expected Payoff
1 - X1 + 2 2 - 2X1 + 5 3 - 7X1 + 4
4 13X1- 6
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The solution of the original 2 * 4 game reduces to a 2 by 2 payoff matrix.
Player B B1 B2 Player A A1 X1+ X2 = 1 A2 Y1 + Y2 = 1
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X1 = -3 – 7 / (-3 + -6) – (4 + 7) = 11 / 20 X2 = 1 - 11 / 20 = 9 / 20 Y1 = -6 – 7 / (-3 + -6) – (4 + 7) = 13 / 20 Y2 = 1 - 13 / 20 = 7 / 20 The value of the game V = (-2 * -6) – (7 * 4) / (3 + 2) – (5 + 11) = 40 / 20 = 2
� MINIMAX PROBLEMS: Obtain the optimum strategy for both persons and the value of the game for zero person sum game whose payoff matrix is as follows: Player B
1 2
Player A 1 2 3 4 5 6
-3 7 4 -6
1 -3 3 5
-1 6 4 1 2 2 -5 0
OPERATIONS RESEARCH
Solution: 11 11
10 10 9 Minimax 9
8 8 7 H 7
6 6 5 5 4 4 3 3 2 2 1 1 Y1 0 0 Y1 - 1 -1 -2 -2 -3 -3 -4 -4 -5 -5 -6 -6 -7 -7 The solution of the original 6 * 2 game reduces to a 2 by 2 payoff matrix.
Player B B1 B2 Player A A1 X1+ X2 = 1 A2 Y1 + Y2 = 1
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X2 = 3 – 5 / (3 + 1) – (4 + 5) = 2 / 5
A’s Pure Strategies B’s Expected Payoff
1 - 4Y1 - 3 2 - 2Y1 + 5 3 - 6Y1 + 6
4 5Y1+ 1 5 2Y1+ 2 6 -5Y1+ 0
3 5 4 1
OPERATIONS RESEARCH
X1 = 1 - 2 / 5 = 3 / 5 Y1 = 1 – 5 / (3 + 1) – (4 + 5) = 4 / 5 Y2 = 1 - 4 / 5 = 1 / 5 The value of the game V = (1 * 3) – (4 * 5) / (3 + 1) – (4 + 5) = 17 / 5
� GENERAL SOLUTION OF M * N RECTANGULAR GAMES: In a rectangular game with M * N payoff matrix if there does exists any saddle point and is not possible to reduce the size of the game to 2 by 2 payoff matrix. The following methods are generally used to solve the game:
� Linear programming method. � Iterative Method.
Question 5: Solve the following game; consider the game problem in which B’s linear programming problem.
Player B 1 2 3
Maximin Player A 1 2 2 2 3 1 Minimax 8 8 8
Player B Y1 Y2 Y3
Player A X1 X2 X3 B’s linear programming problem is as follows: Maximize Z = Y1 + Y2 + Y3 Subject to: 8Y1 + 4Y2 + 2Y3 <= 1 {1} 2Y1 + 8Y2 + 4Y3 <= 1 {2} Y1 + 2Y2 + 8Y3 < = 1 {3} Y1, Y2, Y3 >= 0
8 4 2 2 8 4 1 2 8
8 4 2 2 8 4 2 2 8
OPERATIONS RESEARCH
TABLEAU 1: Y1 Y2 Y3 S1 S2 S3 Solution S1 4 2 1 0 0 1 S2 2 8 4 0 1 0 1 S3 1 2 8 0 0 1 1 Z 1 1 1 0 0 0 0
PC
TABLEAU 2: Y1 Y2 Y3 S1 S2 S3 Solution Y1 0.5 0.25 0.125 0 0 0.125 S2 0 7 3.5 -0.25 1 0 0.75 S3 0 1.5 7.75 -0.125 0 1 0.875 Z 0 0.5 0.75 -0.125 0 0 -0.125 PC TABLEAU 3: Y1 Y2 Y3 S1 S2 S3 Solution Y1 1 0 0 0.107 -0.072 0 0.072 Y2 0 0.5 0.036 0.143 0 0.107 S3 0 0 7 -0.179 -0.215 1 0.715
Z 0 0 0.5 -0.143 -0.072 0 -0.1785
8
1
1
OPERATIONS RESEARCH
TABLEAU 3: Y1 Y2 Y3 S1 S2 S3 Solution Y1 1 0 0 0.107 -0.072 0 0.072 Y2 0 1 0 0.049 0.159 -0.072 0.057 Y3 0 0 -0.026 -0.031 0.143 0.102
Z 0 0 0 -0.156 -0.087 -0.072 -0.2295 Conclusion: Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, Y1= 0.072, Y2= 0.057 and Y1= 0.102 producing Z = $0.2295. Solving a Matrix Game Using the Simplex Method
A game may be solved using the simplex method as follows.
Before you start, check for saddle points. If you find one, you have already solved the game; the optimal strategies are the pure strategies passing through a saddle point. Otherwise, continue with the following steps.
1. Reduce the payoff matrix by dominance. 2. Add a fixed number k to each of the entries so that they all become non-negative. 3. Set up and solve the associated linear programming problem using the simplex method. 4. Obtain the optimal strategies and the expected value as follows:
Column Strategy
a. Express the solution to the linear programming problem as a column vector. b. Normalize by dividing each entry of the solution vector by the sum of the entries, or
by the value of the objective variable p. c. Insert zeros in positions corresponding to the columns deleted during reduction.
1
OPERATIONS RESEARCH
Row Strategy
d. Write down the row vector whose entries are the numbers appearing in the bottom row of the final tableau underneath the slack variables.
e. Normalize by dividing each entry of the solution vector by the sum of the entries. f. Insert zeros in positions corresponding to the rows deleted during reduction.
Value of the Game
e = 1/p - k.
OPERATIONS RESEARCH
UNIT 8: INVENTORY MODELLING: HOURS: 20
� INVENTORY SYSTEM MODEL: Inventory deals with manufacturing sufficient stock of goods (parts and raw materials) to ensure a smooth operation of a business activity.
1 GENERALIZED INVENTORY MODEL: The ultimate objective of inventory model is to answer the following two questions:
⇒ How much to order. ⇒ When to order.
The answer to the 1st question is expressed in terms of economic order quantity (EOQ), which is the optimum amount that should be ordered every time an order is placed and may vary with time depending on the situation under consideration. The 2nd part depends on the type of the inventory type. If the inventory system requires periodic review i.e. equal time intervals (e.g. every week or month). The time for acquiring a new order usually coincides with the beginning of each time interval. If a system is of continuous review type a reorder point is usually specified by an inventory level at which a new order must be placed. We can express the solution of the general inventory problems as follows: PERIODIC REVIEW CASE:
Receive a new order of the amount specified by the order quantity at equal intervals of order time, which can be weekly or monthly.
CONTINUOUS REVIEW CASE: When the inventory level reaches the re-order point places an order whose size is equal to the quantity. The order quantity and re-order point are normally determined by minimizing the total inventory cost, the point which can be expressed as a function of its principle components in the following manner. Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.
1 PURCHASING COST:
OPERATIONS RESEARCH
Purchasing cost is the price of the quantity of goods to be supplied. The purchasing cost becomes an important factor when the commodity unit prices become dependant on the size of the order. This situation is normally expressed in terms of quantity discount or price break where the unit price of an item decrease with the increase of order quantity.
1 SETUP COST / ORDERING COST: Represents a fixed charge incurred when an
order is placed. Thus to satisfy the demand for a given time period, ordering of smaller quantities will result in a higher set up cost during the period than if the demand is satisfied by placing large orders.
1 HOLDING COST / CARRYING COST: Represents the cost of carrying inventory
or represents the cost of maintaining inventory e.g. warehouse rental, security, handling, depreciation, stock maintenance, insurance and loses of interest on capital.
1 SHORTAGE COST: Is a penalty incurred when we run out of stock of the needed
commodity. It generally includes costs due to the loss of customer’s goodwill as well as potential loss in income. We need to minimize the total cost involved in stocking operations by determining the optimum quantity for the replenishment order Q.
� INVENTORY NOTATION:
2 Tc – Total inventory cost per unit time. 2 D – Total demand per unit time.
2 Ci – Cost of production or purchase per unit expressed as $ per unit.
2 Q – Quantity per order.
2 Q* - Optimum quantity = EOQ.
2 H – Holding cost or Carrying cost per unit of inventory per time.
2 T – Time between orders or length of inventory cycle Q/D.
2 K – Cost of processing an order or Ordering cost / Setup cost.
2 L – Leading time.
2 Cs – Shortage cost.
Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.
Tc = CiD + QH/2 + DK/Q + Cs.
OPERATIONS RESEARCH
We need to minimize Tc depending on the quantity Q hence by differentiation we get δ(Tc)/δQ = δ(QH/2)/ δQ + δ(DK/Q)/δQ + δ(CiD)/δQ + δ(Cs)/δQ 0 = H/2 – DKQ2 0 = H/2 – DK/Q
2
Multiply both sides by 2Q2 we get 0 = HQ2 – 2DK 0 = HQ2 – 2DK HQ2 = 2DK Q2 = 2DK/H
Q* = √2DK/H (This is the formula for EOQ)
Economic Order Quantity is a deterministic model referred to as Wilson Economic Lot Size.
� EOQ ASSUMPTIONS:
Tc = CiD + QH/2 + DK/Q + Cs. The above model will hold for the following assumptions:
1) Demand should be constant. 2) Replenishment should be instantaneous. 3) Shortages are not allowed. 4) Fixed quantity order per inventory order level should be maintained.
Question 1: The demand for an item is 18000 units per month and the holding cost per unit is $14.40 per year and the cost of ordering is $400. No shortages are allowed and the replenishment rate is instantaneous.
a) Determine the optimum order quantity. b) The total cost per year of the inventory if the cost of one unit is $1. c) The number of orders per year. d) The time between orders.
Solution: Since all the information asked for are years. Convert all information in months to years. D = 18000 * 12 = 216000 per year. H = $14.40 per year. K = $400. EOQ => Q* = √2DK/H
� √2* 21600 * 400/14.40
� √172800000/14.40
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� √12000000
� 3464.101615
� 3464
Tc = CiD + QH/2 + DK/Q. = (1 * 216000) + (3464 * 14.40)/2 + (216000 * 400)/3464
= 216000 + 24940.80 + 24942.26
= 265883.06
The number of orders per year = D/Q
=216000/3464 = 62.36 orders
The time between orders = Q/D
= 3464/216000 = 0.016 = 0.02 Question 2: The demand for a particular item is 18000 units per year. Holding cost per unit is $1.20 per year and the cost of procurement is $400. No shortages are allowed and the replenishment rate is instantaneous.
a. Determine the optimum order quantity. b. The number of orders per year. c. The time between orders. d. The total cost per year of the inventory if the cost of one unit is $1.
Solution: D = 18000 per year. H = $1.20 per year. K = $400. EOQ => Q* = √2DK/H
� √2* 18000 * 400/1.20
� √14400000/1.20
� √12000000
� 3464.101615
� 3464
OPERATIONS RESEARCH
The number of orders per year = D/Q
=18000/3464 = 5.196 orders
The time between orders = Q/D or 365/5.196 = 70.24 days = 3464/18000 = 0.1924 = 0.19 days Tc = CiD + QH/2 + DK/Q. = (1 * 18000) + (3464 * 1.20)/2 + (18000 * 400)/3461
= 18000 + 206.40 + 2078.52
= 22156.92
Question 3: Suppose a company has soft drinks products. It has a constant annual demand rate of 3600 cases. A case of soft drinks cost the Harare drinks company $3. Ordering cost is $20 per order and holding cost is charged at 25% of the cost per unit. There are 250 working days per year and the lead-time is 5 days. Identify the following aspects of inventory policy:
i. Economic order quantity. ii. Re-order point. iii. Cycle time (Length) = Q/D. iv. The total annual cost.
Solution: D = 3600. H = 25% of $3. K = $20 per order. Ci = $3 per unit.
i. EOQ => Q* = √2DK/H
� √2* 3600 * 20/0.75
� √144000/0.75
� √192000
� 438.178
� 438
ii. Re-order point = Daily Demand * Lead-time. = 3600/250 * 5 = 72 crates
OPERATIONS RESEARCH
iii. Cycle time = Q/Daily demand.
Daily Demand is given by = 3600/250 = 14.40 Cycle Time = 438/14.40 = 30.42 days
iv. Tc = CiD + QH/2 + DK/Q. = (3 * 3600) + (438 * 0.75)/2 + (20 * 3600)/438 = 10800 + 164.25 + 164.38356 = 11128.633 = 11128.63
� RE-ORDER POINT: Is a point in which a new order should be placed?
1 GENERAL RULES TO DETERMINE THE RE-ORDER POINT:
i. Determine the number of days the order should be placed (Cycle length) = Q/D. ii. Determine the number of days which is the minimum level to re-order new entities and is
given by Lead-time – Cycle Length (Q/D) where Lead-time is the time between the placement of an order and its receipt.
iii. Determine the level of inventory (stock) to re-order new items and is given by: Number of days which gives minimum level to order new items * Demand.
Question 4: The daily demand for a commodity is approximately 100 units. Every time an order is placed a fixed cost of $100 is incurred and daily holding cost per unit inventory $0.02. If the lead-time is 12 days. Determine:
i. Economic order quantity. ii. Re-order point. iii. Cycle time. iv. Minimum inventory level. v. Level of inventory stock.
OPERATIONS RESEARCH
Solution: D = 100. H = $0.02. K = $100.
i. EOQ => Q* = √2DK/H
� √2* 100 * 100/0.02
� √20000/0.02
� √1000000
� 1000
ii. Re-order point = Demand * Lead-time = 100 * 12 = 1200
iii. Cycle time = Q/D
= 1000/100 = 10
vi. Minimum inventory level = Lead-time – Cycle Length. = 12 – 10 = 2 days
vii. Level of inventory stock = Minimum inventory level * Demand.
= 2 * 100 = 200 orders. � EOQ WITH GRADUAL REPLENISHMENT / ECONOMIC PRODUCTION LOT
SIZE: Suppose a demand of an item in a company is D units per unit time and the company can produce them at a rate of R units per unit time where R>D. The cost of set-up is $K and the holding cost per unit time is $H. We wish to determine the Optimum manufacturing quantity and the total cost per year assuming the cost of one unit is $Ci. It is assumed that shortages are not allowed. EOQ => Q* = √2DK/H(1 –
D/R)
OPERATIONS RESEARCH
Question 5: A company uses 100000 units per year that cost $3 each. The carrying costs are 1% per month and the ordering cost are $2.50 per year. What is the economic order quantity made the items themselves on a machine with a potential capacity of 600000 units per year. Solution: D = 100000. K = $2.50. H = 1% = 0.01 * 12 * 3 = 0.36
EOQ => Q* = √2DK/H(1 –
D/R)
= √2* 100000 * 2.50/0.36(1 –
100000/600000)
= √500000/0.36(1 –
1/6)
=√1666666.667 = 1290.994449 =1291.
Question 6: A company manufactures an item, which is also used in the company. The demand of this item is 18000 units per year and the production rate is 3000 per month. The cost of set up is $500. The holding cost of 1 unit per month is $0.15. Determine the optimum manufacturing quantity and the total cost per year assuming the cost of 1 unit is $2 and shortages are not allowed. Solution: D = 18000 per year. H = $0.15 * 12 = $1.8 per year. K = 500 Ci = $2 R = 3000 * 12 = 36000 per year.
EOQ => Q* = √2DK/H(1 – D
/R)
= √2* 18000 * 500/1.8 *(1 –
18000/36000)
= √18000000/0.36(1 – 0.5)
=√18000000/0.18
=3162.27766 = 3162
OPERATIONS RESEARCH
Tc = CiD + QH/2 + DK/Q. = (2 * 18000) + (3162 * 1.8)/2 + (18000 * 500)/3162 = 36000 + 2845.8 + 2846.29981 = 41692.09981 = 41692
� INVENTORY CONTROL (TYPES OF CONTROL SYSTEMS): There are 2 two basic inventory control systems:
� Reorder Level � Periodic Review system.
1 PERIODIC REVIEW SYSTEM: This is the reviewing of all stocks at fixed intervals periodically. Therefore absolute stocks can be eliminated. Variable quantities can be ordered.
1 THE REORDER LEVEL SYSTEM: The reorder level system results in fixed quantities being ordered at variable intervals dependent upon demand. The reorder level system usually has three (3) control levels namely:
� The reorder level. � The minimum level. � The maximum level.
NB:
� The Reorder Level is calculated so that if the worst anticipated position occurs, Stock would be replenished in time.
� The Minimum Level is calculated so that management will be
warned when demand is above average. There may be no danger but the situation needs watching.
� The Maximum Level is calculated so that management will be
warned when demand is the minimum anticipated and consequently the stock level is likely to rise.
OPERATIONS RESEARCH
Question 7: The following is an illustration of a company with a simple manual reorder level system. Normal Usage = 110 per day. Minimum Usage = 50 per day. Maximum Usage = 140 per day. Lead time = 25 – 30 days. EOQ = 5000 Using the above data calculate the various control levels i.e.
� Reorder level. � Minimum Usage. � Maximum Usage. [20 marks].
Solution: a) Reorder Level = Maximum Usage * Maximum Lead time. = 140 * 30 = 4200 units b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage) = 4200 – (27.5 * 110) = 4200 – 3025 = 1175 units c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage) = 4200 + 5000 – (25 * 50) = 7950
Question 8: The following is an illustration of a company with a simple manual reorder level system. Normal Usage = 560 per day. Minimum Usage = 250 per day. Maximum Usage = 750 per day. Lead time = 15 – 20 days. EOQ = 10000 Using the above data calculate the various control levels i.e.
� Reorder level. � Minimum Usage. � Maximum Usage. [20 marks].
Solution:
a) Reorder Level = Maximum Usage * Maximum Lead time. = 750 * 20 = 15000 units
OPERATIONS RESEARCH
b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage) = 15000 – (17.5 * 560) = 15000 – 9800 = 5200 units c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage) = 15000 + 10000 – (15 * 250) = 21250
OPERATIONS RESEARCH
UNIT 9: REGRESSION ANALYSIS AND CORRELATION: HOURS: 20
� CORRELATION: Correlation is a measure of strength of a linear relationship between 2 sets of numbers. There are 2 measures of correlation namely:
� The Pearson Product Moment Correlation Coefficient represented by r. � The Spearman’s Rank Correlation Coefficient represented by R.
A coefficient of +1 indicates a perfect Positive relationship whilst a coefficient of – 1 shows perfect negative correlation. Hence the correlation coefficient will have values ranging between – 1 and + 1 inclusive. A coefficient of zero implies no correlation.
SCATTER DIAGRAM: It is a way of representing a set of data by a scatter of plots or dots. One variable is plotted on the X-axis and another on the Y-axis. Normally variable X is the one controlled over (independent variable) and Y variable is the variable that you are interested in (dependent variable).
POSITIVE CORRELATION: NEGATIVE CORRELATION: * * * * * * * Delivery * * * * Time * * * * * * *
* * * * * Number of deliveries Temperature
NO CORRELATION: NON-LINEAR CORRELATION: * * * * * * * * * * * * ** Salary * * * * * * ** * * * * * * ** * * * * * * * ** * * * * * * * ** Age * Quantity produced
The first diagram is a positive because number of deliveries increase so apparently does the delivery time. The second is a negative correlation because as air temperature increases the heating cost falls.
OPERATIONS RESEARCH
For the third correlation shows that there is no correlation existing between salary and age of an employee. For non-linear correlation it suggests that quantity produced and efficiency are correlated but not linearly. Given two sets of data represented by the variables X and Y the Product Moment Correlation Coefficient is given by: r = ∑ (X - r) (Y – Ý) √ ∑ (X - r)2 * (Y – Ý)2 OR
n ΣXY - ΣX * ΣY r =
2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]
Question 1: Calculate r for the data given below: X 15 24 25 30 35 40 45 65 70 75 Y 60 45 50 35 42 46 28 20 22 15 Solution: X Y XY X2 Y2 15 60 900 225 3600 24 45 1080 576 2025 25 50 1250 625 2500 30 35 1050 900 1225 35 42 1470 1225 1764 40 46 1840 1600 2116 45 28 1260 2025 784 65 20 1300 4225 400 70 22 1540 4900 484 75 15 1125 5625 225 424 363 12835 21926 15123
OPERATIONS RESEARCH
n ΣXY - ΣX • ΣY r =
2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]
10 * 12835 - 424 * 363 r =
10* 21926 - 179776 * 10 * 15123 - 131769
128350 - 153912 r =
39484 * 19461
- 25562 r =
27719.9959 r = - 0.922150238 r = - 0.92 Question 2: Calculate r for the data given below: X 1 2 3 6 5 6 Y 5 10.5 15.5 25 16 22.5 Solution: X Y XY X2 Y2 1 5 5 1 25 2 10.5 21 4 110.25 3 15.5 46.5 9 240.25 4 25 100 16 625 5 16 80 25 256 6 22.5 135 36 506.25 21 94.5 387.5 91 1762.75
OPERATIONS RESEARCH
n ΣXY - ΣX * ΣY r =
2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]
6 * 387.5 - 21 * 94.5 r =
6* 91 - 441 * 6 * 1762.75 – 8930.25
2325 – 1984.5 r =
546 - 441 * 10576 – 8930.25
340.5 r =
415.7598467 r = 0.819 (B) THE SPEARMAN’S RANK CORRELATION COEFFICIENT: When two sets of data are given positions or ranked, we use Spearman’s Rank Correlation Coefficient represented by R. It is given by: R = 1 - 6 Σd2 n(n2 -1) Where d = difference between pairs of ranked values. n = number of pairs of ranking. 1 = is an independent number. Question 2: A group of eight students were tested in Maths and English and the ranking in the two subjects are given below: Student A B C D E F G H Maths 2 7 6 1 4 3 5 8 English 3 6 4 2 5 1 8 7
OPERATIONS RESEARCH
Solution: Students Maths English D D2 A 2 3 - 1 1 B 7 6 1 1 C 6 4 2 4 D 1 2 -1 1 E 4 5 -1 1 F 3 1 2 4 G 5 8 -3 9 H 8 7 1 1 TOTAL = 22 R = 1 - 6 Σd2 n(n2 -1) R = 1 – 6 * 22 8(82 -1) R = 1 - 132 8* 63
R = 1 – 0.2619 = 0.738
� REGRESSION: This is a statistical technique that can be used for short to medium term forecasting. It seeks to establish the line of “best fit” for some observed data. These are 2 ways of obtaining the trend of a set of data:
2 The graphical method. 2 The method of least squares.
OPERATIONS RESEARCH
� THE GRAPHICAL METHOD: In this method a scatter diagram is plotted and a straight line is produced through the middle of the points representing the trend line. This line can then be used (when extended) to predict a future value. y * Trend line or line of best fit. * * * * * * * * * * x
Least Squares Regression Lines
This is a method of finding a regression line without estimating where the line should go by eye.
If the equation of the regression line is y = ax + b, we need to find what a and b are. We find these by solving the "normal equations".
Normal Equations
The "normal equations" for the line of regression of y on x are:
Σy = aΣx + nb and
Σxy = aΣx2 + bΣx
The values of a and b are found by solving these equations simultaneously.
For the line of regression of x on y, the "normal equations" are the same but with x and y swapped.
OPERATIONS RESEARCH
� METHOD OF LEAST SQUARES: This method gives an equation of the trend line or regression line. The equation is given by:
Y = A + Bx
Where B = n ΣXY - ΣX • ΣY or n ΣX2 - (ΣX)2 = ΣXY -n X Y n X2 - nX2 and A = Y - BX x is an independent variable. n is the number of items in the list. Question 3: Consider the following data: X 1 2 3 4 5 6 Y 6 4 3 5 4 2
i. Find the equation of the regression line. ii. Find Y when X = 4 using the equation.
Solution:
X Y XY X2 1 6 6 1 2 4 8 4 3 3 9 9 4 5 20 16 4 4 20 25 6 2 21 36 21 24 75 91
B = n ΣXY - ΣX • ΣY
n ΣX2 - (ΣX)2
OPERATIONS RESEARCH
B = 6(75) – 21 * 24 6 (91) – (21)2 = 450 – 504 546 – 441 = -18/35 X = ∑X/n
21/6 = 3.5 Y = ∑Y/n
24/6 = 4 A = Y - BX = 4 – (7/2 * - 54/105) = 4 + 9/5 =29/5
Y = A + Bx
= 29/5 – 18/35X (ii) When X = 4
Y = A + Bx = 29/5 – 18/35* 4 = 29/5 – 72/35
= 131/35 or 326/35
OPERATIONS RESEARCH
Question 4: For data below calculate the equation of the regression Y upon X. X 98 78 74 80 80 83 95 100 97 75 Y 14 9 10 11 10 11 12 13 11 9 Solution:
X Y XY X2 98 14 1372 9604 78 9 720 6084 74 10 740 5476 80 11 880 6400 80 10 800 6400 83 11 913 6889 95 12 1140 9025 100 13 1300 10000 97 11 1067 9409 75 9 675 5625 860 110 9589 74912
B = n ΣXY - ΣX • ΣY
n ΣX2 - (ΣX)2 B = 10(9589) – 860 * 110 10 (74912) – (860)2 = 95890 – 94600 749120 – 739600 = 0.1355 X = ∑X/n
860/10 = 86 Y = ∑Y/n
110/10 = 11
OPERATIONS RESEARCH
A = Y - BX = 11 – (0.1355 * 86) = 11 – 11.653 = - 0.6533 Y = A + Bx = - 0.6533 + 0.1355X
OPERATIONS RESEARCH
UNIT 10 INTRODUCTION TO QUEUEING THEORY HOURS: 20
Introduction to Queuing Theory Queueing theory is a set of mathematical tools for the analysis of probabilistic systems of customers and servers. In everyday life, it is seen that a number of people arrive at a cinema ticket window. If the people arrive “too frequently” they will have to wait for getting their tickets or sometimes do without it. Under such circumstances, the only alternative is to form a queue, called the waiting line, in order to maintain a proper discipline. Here the arriving people are called the customers and the person issuing the tickets is called a server. Specifically, the theory helps the manager to seek answer to the questions like the following when actual/expected arrival and service rates are known and a set of assumption is made.
• What is the fraction of time the system providing service is expected to be busy (or, idle)? • What are the chances of a certain number of customers present in the system at a random
point in time? • On the average, how many customers are expected to be present in the system? • What is the average length of queues formed from time to time? • How much time a customer is expected to spend in the system – in waiting for getting the
service? • How much time a customer typically spends in the queue? • What is the expected total cost of providing service for various alternative service levels and,
accordingly, what is the optimal level of service to provide?
Definitions:
• Arrival Rate refers to the average number of customers who require service within a specific period of time.
• A Capacitated Queue is limited as to the number of customers who are allowed to wait in line.
• Customers can be people, work-in-process inventory, raw materials, incoming digital messages, or any other entities that can be modeled as lining up to wait for some process to take place.
• A Queue is a set of customers waiting for service. • Queue Discipline refers to the priority system by which the next customer to receive
service is selected from a set of waiting customers. One common queue discipline is first-in-first-out, or FIFO.
OPERATIONS RESEARCH
• A Server can be a human worker, a machine, or any other entity that can be modeled as executing some process for waiting customers.
• Service Rate (or Service Capacity) refers to the overall average number of customers a system can handle in a given time period.
• Stochastic Processes are systems of events in which the times between events are random variables. In queueing models, the patterns of customer arrivals and service are modeled as stochastic processes based on probability distributions.
• Utilization refers to the proportion of time that a server (or system of servers) is busy handling customers.
Queueing Notation: In the literature, queueing models are described by a series of symbols and slashes, such as A/B/X/Y/Z, where A indicates the arrival pattern, B indicates the service pattern, X indicates the number of parallel servers, Y indicates the queue’s capacity, and Z indicates the queue discipline. We will be concerned primarily with the M/M/1 queue, in which the letter M indicates that times between arrivals and times between services both can be modeled as being exponentially distributed. The number 1 indicates that there is one server; we will also study some M/M/n queues, where n is some number greater than 1. Symbols:
Performance Measure Random Variable
Expected Value
Number of Customers in the System
N L
Number of Customers in the Queue
Nq Lq
Number of Customers in Service Ns Ls Time Spent in the System T W Time Spent in the Queue Tq Wq Time Spent in Service Ts Ws
System Parameters Number of Servers S Arrival Rate (number per unit of time) λ (Greek letter
lambda) Service Rate (number per unit of time) µ (Greek letter mu) Utilization Factor ρ (Greek letter rho)
OPERATIONS RESEARCH
Formulas: The utilization factor, or ρ (rho), also the probability that a server will be busy at any point in time:
(i) Idle time, or the proportion of time servers are not busy, or the probability that a server will be idle at any given time: (ii) The average time a customer spends in the system:
λµ −=+= 1
sq WWW (iii)
The average number of customers in the queue:
( ) ρρ
λµµλ
−=
−=
1
22
qL (iv)
The average number of customers in the system: sq LLL += (v)
The average time spent waiting in the queue:
( ) ( )ρµ
ρλµµ
λ−
=−
=1
qW (vi)
Those who would commune more directly with the Arab mathematician al Jebr (whose name is the source for our word algebra) will note that there is another formula for N:
Therefore:
(vii) The single most important formula in queueing theory is called Little’s Law: (viii) Little’s Law can be restated in several ways; for example, here is a formula for W:
(ix) There are also some formulas that are based on the special characteristics of the exponential distribution. These generally are beyond the scope of this course, but one useful example is offered here. In an M/M/n system, the probability of a customer spending longer than some specific time period t in the system is given by the following formula. e is the base of the natural logarithms (roughly 2.718):
(x)
ρλµ
=S
1− ρ
( )N N Nq s+ =
−+ =
−+ =
+ −
−
=−
=−
=ρ
ρ ρ
λµ
λµλµ
λµ
µλµ
λµ
λµ λ
µλµ
λµ
λµ
ρρ
2
11
1
1 11
L =−ρ
ρ1
L W= λ
WL
=λ
{ } ( )P T t e t≥ = −λ µ
OPERATIONS RESEARCH
MEASURES OF THE MODEL
i. To find expected (average) number of units in the system, Ls i.e. the average number of customers in the
system both waiting and in service.
SL , where 11 1
λ µ ρ ρ λ µλ µ ρ
= = = <− −
ii. To find expected (average) queue length, Lq i.e. average number of customers waiting in the queue.
2
q SL L1
λ ρµ ρ
= − =−
iii. To find mean (or expected) waiting time in the queue (excluding service time), Wq i.e. the average time spent by a customer in the queue before the commencement of his service.
qW( ) (1 )
λ ρµ µ λ µ ρ
= =− −
iv. To find expected waiting time in the system (including service time), Ws i.e. the total time spent by a customer in the system.
S
1W
µ λ=
−
v. To find expected waiting time of a customer who has to wait, (W | W > 0).
( ) 1W | W 0
(1 )µ ρ> =
−
vi. To find expected length of non-empty queue, (L | L > 0) or Lq’.
( ) 1L | L 0
1 ρ> =
−
vii. To find out the variance of queue length.
2
Var.{n}(1 )
ρρ
=−
Inter-Relationship between Ls, Lq, Ws, Wq.
Ls = λWs, Lq = λWq, Wq = Ws – 1/µ. Lq = Ls – λ/µ.
OPERATIONS RESEARCH
LIMITATIONS FOR APPLICATION OF QUEUEING MODEL:
1. The single channel queueing model can be fitted in situations where the following conditions are satisfied.
2. The service time has exponential distribution. The average service rate is denoted by µ. 3. Arrivals are from infinite population. 4. The queue discipline is FCFS i.e. the customers are served on first come, first served basis. 5. There is only a single service station. 6. The mean arrival rate is less than the mean service rate, i.e. λ < µ. 7. The waiting space available for customers in the queue is infinite.
� QUEUEING MODELS:
• It involves a customer’s standpoint where a waiting line is created. Customers arrive at the facility and they join a waiting line (or Queue). The server chooses a customer from the waiting line to begin service. Upon the completion of a service, the process of choosing a new (waiting) customer is repeated.
• It is assumed that no time is lost between the completion of a service and the admission of a new customer into the facility.
• The major actors in queuing situation are Customer and the Server. The interaction
between the Customer and the Serve are of interest only in as far as it relates to the period of time the customer needs to complete a service. From the standpoint of customer arrivals we will be interested in the time intervals that separate successive arrivals.
• The customer arrivals and service times are summarized in terms of probability
distributions normally referred to as: arrivals and service time distribution. • These distributions may represent situations where customers arrive and are served
individually {banks or supermarkets}. • Customers may arrive and or be served in groups are referred to as Bulk queues. • In queue model the manner of choosing customer from the waiting line to start service is
referred to as Service discipline. • The Service discipline occurs as FCFS rule (first come first service), LCFS rule (last come
first served) and SIRO (service in random order). • Customers arriving at a facility may be put in Priority queues such that those with higher
priority will receive preference to start service first.
• The facility may include more than one server thus allowing as many customers as the number of servers to be serviced simultaneously this is referred as Parallel service.
• Facilities which comprise a number of series stations through which the customer may pass before service is completed (e.g. processing of a product on a sequence of machines) is referred to as Queues in series or Tandem queues.
OPERATIONS RESEARCH
• A combination of Parallel and Series is called Network queues. • Calling source is the source from which calls for service (arrivals of customer) are
generated. • Queue size the number of limited customers that may be allowed at a facility may be
because of space limitation (e.g. car spaces in a drive-in bank). � PURE BIRTH MODEL / PROCESS: Pure birth model is process situation where customer arrive and never leave. E.g. issuing of birth certificates for newborn babies. These certificates are normally kept as permanent records in a Central office administered by the state health department. The birth of new babies and issuing of birth certificates is a completely random process that can be described ay a Poisson distribution. Assuming that λ is the rate at which birth certificates are issued, the pure birth process of having n arrivals (birth certificates) during the time period t is given as: Pn(t) = (λt)n e-λt n!
Where n = 0, 1, 2, 3… and λ is the rate of arrival per unit time with the expected number of arrivals during t being equal to λt. Question 1: Suppose that the birth in a state are spaced over time according to a an exponential distribution with one birth occurring every 7 minutes on the average. Solution: Since the average inter arrival (inter birth) time = 7 minutes the birth rate in the state is computed as: λ = 24{hours} * 60{minutes} = 205.7 births per day. 7 The number of birth in the state per year =: λt = 205.7 * 365 = 75.080 births per year The probability of no births in any one-day is computed as:
Pn(t) = (λt)n e-λt n! P0(1) = (205.7 * 1)0 e-205.7 * 1 ≈ 0. 0!
OPERATIONS RESEARCH
Suppose that we are interest in the probability of issuing 45 birth certificates by the end of a period of 3 hours given that 35 certificates were issued in the first 2 hours. We observe that since birth occur according to a Poisson process the required probability reduces to having 45 –35 = 10 births in one (3 – 2) hours. Given λ = 60/7 = 8.57 births per hour. P10(1) = (8.57 * 1)10 e-8.57 * 1 10! P10(1) = (2137018192) * (0.000189712) 3628800
≈ 0.11172 Question 1: Suppose that the clerk who enters the information from the birth certificate into a computer normally wait at least 5 certificates to be accumulated. What it the probability that the clerk will be waiting a new batch every hour. Solution:
λ = / = births per hour. P5(1) = (* 1)5 e- * 1 5!
≈ 0.92868
� PURE DEATH MODEL / PROCESS: Pure death model is process situation where customers are withdrawals. Consider the situation of stocking units of an item at start of the week to meet customer’s demand during the week. If we assume that customer demand occurs at the rate µ units per week and that the demand process is completely random, the associated probability of having n items remaining in stock after time t is given by the following Truncated Poisson distribution.
Pn(t) = (µt)N - n e-µt where n = 1, 2, 3 … N. (N – n)!
OPERATIONS RESEARCH
Pn(t) = 1 – N∑
n-1 * Pn(t) Question 2: At the beginning of each week, 15 units of an inventory item are stocked for use during the week. Withdrawals from stock occur only during the first 6 days (business is closed on Sundays) and follow a Poisson distribution with 3 units per day. When the stock level reaches 5 units, a new order of 15 units is placed for delivery at the beginning of next week. Because of the nature of the item all units left at the end of the week are discarded.
Solution:
We can recognize that the consumption rate is µ = 3 units per day. Suppose that we are interested in computing the probability of having 5 units (reorder level) on day t then we compute it as:
Pn(t) = (µt)N - n e-µt (N – n)!
P5(t) = (3t)15 - 5 e-3t (15 – 5)!
NB: P5(t) represents the probability of reordering on day t. This probability peaks at t = 3 and then declines as we advance through the week. If we are interested in the probability of reordering by day t we must compute the cumulative probability of having 5 units or less on day t. i.e. Pn <= 5(t) = P0(t) + P1(t) + …..+ P5(t).
� STEADY STATE MEASURES OF PERFORMANCE: The steady state measures of performance is used to analyze the operation of the queuing situation for the purpose of making recommendations about the design of the system. Among these measures of performance are:
� The expected number of customers waiting. � The expected waiting time per customer. � The expected utilization of the service facility.
The system comprises both the queue and the service facility. Let:
OPERATIONS RESEARCH
Ls = expected number of customers in system. Lq = expected number of customers in Queue. Ws = expected waiting time in system. Wq = expected waiting time in Queue. Suppose that we are considering a Service Facility with c parallel servers then from the definition of Pn we get: Ls = ∞∑n=0 * npn Lq = ∞∑n=c + 1 * (n – c)pn
NB: A strong relationship exists between Ls and Ws (also Lq and Wq) so that either measure is automatically determined from the other. Let λeff be the effective average arrival rate (independent of the number in the system n) then: Ls = λeff * Ws Lq = λeff * Wq The value of λeff is determined from the state dependent λn and the probabilities Pn is given as: λeff = ∞∑n=0 * λnpn NB: A direct relationship exists between Ws and Wq, which is equal to: Expected waiting time in system = Expected waiting time in queue + expected service time. Ws = Wq + 1/µ. Given that µ is the Service rate per busy server, the Expected Service time = 1/µ. Ws = Wq + 1/µ. Multiplying both sides by λeff we obtain: Ls = Lq + λeff/µ = expected number of customers in system. The expected utilization of a service is defined as a function of the average number of busy servers. Since the difference between Ls and Lq must equal the Expected number of busy servers we obtain: Expected number of busy servers = C = Ls – Lq = λeff/µ.
OPERATIONS RESEARCH
The Percent utilization of a service facility with c parallel servers is given by: Percent utilization = C/C * 100 NB: In summary Given Pn we can compute the system’s measure of performance in the following order: Pn Ls = ∞∑n=npn Ws = Ls/λeff Wq = Ws - 1/µ Lq = λeff * Wq C = Ls – Lq. Question 3: Job orders arriving at a production Facility is divided into three groups. Group1 will take the highest priority for processing; Group3 will be processed only if there are no waiting orders from group 1 and 2. It is assumed that a job once admitted to the facility must be completed before any new job is taken in. Orders from groups 1,2 and 3 occur according to Poisson distribution with mean 4, 3 and 2 per day respectively. The service times for the three groups are constant with rates 10, 9 and 10 per day respectively. Find the following:
I. The expected waiting time in the system for each of the three queues. II. The expected waiting time in the system for any customer.
III. The expected number of waiting jobs in each of the three groups. IV. The expected number waiting in the system.
Solution:
OPERATIONS RESEARCH
QUEUEING THEORY QUESTIONS Q1. Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between one arrival and the next. The length of a phone call assumed to be distributed exponentially with mean 3 minutes. Then, (a) What is the probability that a person arriving at the booth will have to wait?
(b) What is the average length of the queues that form from time to time?
(c) What is the average length of the queue?
(d) The telephone department will install a second booth when convinced that an arrival
would expect to have to wait at least 3 minutes for the phone. By how much must the
flow of arrivals be increased in order to justify a second booth?
(e) Find the probability that an arrival finds that 4 persons are waiting for their turn.
(f) Find the average number of persons waiting and making telephone calls.
Ans Here, 1 1
= and = 10 3
λ µ
(a) P(W > 0) = 1 −P0 = 1 3 3
= = = 0.310 1 10
λµ
×
(b) The average length of non empty queues or expected number of customers in the non empty queue.
(L | L > 0) = Lq, =
1
3 1.43 1 Person1 1
3 10
µµ λ
= = ≈− −
(c) The average length of queues (empty + non-empty) or expected number of customers in the queue.
Lq =
2
2
193
0 Customers11 7013
ρρ
= = ≈
− −
q
'q
''
'
' '
(d) W( )
1 Since W 3, , (say) for second booth
3
3 , giving 0.16.1 1
3 3
As, 0.16 0.16 60 9.6 10customers / hour
λµ µ λ
µ λ λ
λ λλ
λ λ
=−
= = =
∴ = = −
= ⇒ = × = ≈
Thus, as soon as the arrival rate increases to 10 customers/hour, another booth may be installed.
OPERATIONS RESEARCH
(e) With 4 people waiting for their turn in the queue implies a total of 5 people in the system. Thus, the probability that an arrival finds 4 persons in the queue is,
( ) n
5
P n 5 (1 )
0.3 (1 0.3)
0.0017
ρ ρ= = −
= −=
(f) The average number of people waiting and making calls is given by the expected length of the system. Thus,
sL1
ρρ
=−
0.3
1 0.3=
− 0.43=
Q2. Customers arrive at a sales counter manned by a single person according to a Poisson process with a mean rate of 20 per hour. The time required to serve a customer has an exponential distribution with a mean of 100 seconds. Find the average waiting time of a customer. Ans Here we are given:
60 60 = 20 per hour and = 36 per hour
100λ µ × =
The average waiting time of a customer in the queue is given by:
q
20 5 5 3600W hours or , i.e.,125seconds.
( ) 36(36 20) 36 4 36 4
λµ µ λ
×= = =− − × ×
The average waiting time of a customer in the system is given by:
s
1 1 1W or hour i.e., 225seconds.
(36 20) 16µ λ= =
− −
Q3. Customers arrive at a one-window drive according to a Poisson distribution with mean of 10 minutes and service time per customer is exponential with mean of 6 minutes. The space in front of the window can accommodate only three vehicles including the serviced one. Other vehicles have to wait outside this space. Calculate: (i) Probability that an arriving customer can drive directly to the space in front of the
window.
(ii) Probability that an arriving customer will have to wait outside the directed space.
(iii) How long an arriving customer is expected to wait before getting the service?
Ans From the given information, we find that: Mean arrival rate, λ = 6 customers per hour And mean service rate, µ = 10 customers per hour
(i) Probability that an arriving customer can drive directly to the space in front of the window is given by:
OPERATIONS RESEARCH
2
0 1 2
2
2
P P P 1 1 1
1 1
6 6 6 981 1 or 0.784
10 10 10 1225
λ λ λ λ λµ µ µ µ µ
λ λ λµ µ µ
+ + = − + − + −
= − + +
= − + + =
(ii) Probability that an arriving customer will have to wait outside the directed space is given by:
0 1 21 (P P P ) 1 0.784 0.216 or 21.6%− + + = − =
(iii) Expected waiting time of a customer being getting the service is given by:
q
6 3W hour or 9 min utes.
( ) 10(10 6) 20
λµ µ λ
= = =− −
Q4. A company distributes its products by trucks loaded at its only loading station. Both, company’s trucks and contractor’s trucks, are used for this purpose. It was found out that on an average every 5 minutes, one truck arrived and the average loading time was three minutes. 50% of the trucks belong to the contractor. Find out: (i) The probability that a truck has to wait,
(ii) The waiting time of truck that waits, and
(iii) The expected waiting time of contractor’s trucks per day, assuming a 24-hours shift.
Ans Here we are given:
Average arrival rate of trucks, 60
= 5
λ = 12 trucks/hr.
Average service rate of trucks, 60
= 3
µ = 20 trucks/hr.
(i) The probability that a truck has to wait is given by:
12
0.620
λρµ
= = =
(ii) The waiting time of a truck that waits is given by:
s
1 1 1W = hour i.e., 7.5 min utes.
(20 12) 8µ λ= =
− −
(iii) The expected waiting time of contractor’s trucks per day, assuming a 24-hours shift is given by: (No. of trucks per day) x (Contractor’s percentage) x (Expected waiting time of a truck)
5012 24
100 ( )
1 12 54288 or 10.8 hrs.
2 20 8 5
λµ µ λ
= × × ×−
= × × =×
Q5. Customers arrive at the First Class Ticket counter of a Theatre at a rate of 12 per hour. There is one clerk serving the customers at a rate of 30 per hour.
OPERATIONS RESEARCH
(i) What is the probability that there is no customer in the counter (i.e. the system is
idle)?
(ii) What is the probability that there are more than 2 customers in the counter?
(iii) What is the probability that there is no customer waiting to be served?
(iv) What is the probability that a customer is being served and nobody is waiting?
Ans From the given information, we find that: Mean arrival rate, λ = 12 customers per hour And mean service rate, µ = 30 customers per hour
12
= 0.430
λρµ
∴ = =
(i) P (system is idle) = P0 = 1 1 0.4 0.6ρ− = − = . (ii) P(n>2) = 1–P(n≤2)
[ ]
( ) ( )
( ) ( )
2
0 1 2
2
2
2
n 1
1 P P P 1 1 1 1
1 1 1
1 1 1
1 1 0.4 1 0.4 0.4
1 0.6 1.56
1 0.936
0.064
Also,P( n)
P( 2)
λ λ λ λ λµ µ µ µ µ
λ λ λµ µ µ
ρ ρ ρ
ρ +
= − + + = − − + − + −
= − − + +
= − − + +
= − − + +
= − ×= −=
> =∴ > 2 1 30.4 0.4 0.064+= = =
(iii) P(no customer waiting to be served) = P0 + P1 = 0.6 + 0.24 = 0.84
(iv) P (a customer is being served and none is waiting) = P1 = 1λ λµ µ −
= (1 )ρ ρ− = 0.4 x
0.6 = 0.24.
Q6. A TV repairman finds that the time spent on his job has an exponential distribution with mean 30 minutes. If he repairs sets in the order in which they come and if the arrival of sets is approximately Poisson with an average rate of 10 per 8-hour day, what is his expected idle time each day? How many jobs are ahead of the set just brought in? Ans From the given information, we find that: Mean arrival rate, λ = 10 sets per day. And mean service rate, µ = 16 sets per day.
10 5 =
16 8
λρµ
∴ = =
OPERATIONS RESEARCH
P (repairman is idle) = P0 =5 3
1 18 8
ρ− = − = .
∴Expected idle time per day = 3
88
× = 3 hours.
To determine the expected number of sets ahead of the set just brought in, we shall find the expected number of sets in the system. Thus,
s
55 28L 1 sets
51 3 318
ρρ
= = = =− −
.
Q7. Assume that at a bank teller window the customers arrive in their cars at the average rate of 20 per hour according to a Poisson distribution. Assume also that the bank teller spends an average of 2 minutes per customer to complete a service, and the service time is exponentially distributed. Customers, who arrive from an infinite population, are served on a FCFS basis, and there is no limit to possible queue length. (i) What is the expected waiting time in the system per customer?
(ii) What is the mean number of customers waiting in the system?
(iii) What is the probability of zero customers in the system?
(iv) What value is the utilization factor?
Ans From the given information, we find that: Mean arrival rate, λ = 20 customers per hour And mean service rate, µ = 30 customers per hour
20 2 =
30 3
λρµ
∴ = =
(i) Expected waiting time in the system per customer is given by:
s
1 1 1W = hour i.e., 6 min utes.
(30 20) 10µ λ= =
− −
(ii) Mean number of customers waiting in the system is given by: 2
2
q
243
L21 313
ρρ
= = =
− −
(iii) Probability of zero customers in the system is given by P0 =2 1
1 13 3
ρ− = − = .
(iv) Utilization factor, 2
3ρ = .