logistics. example
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AACIMP 2009 Summer School lecture by Boris Goldengorin. "Logistics" course. 3rd - 4th hour. Part 2.TRANSCRIPT
Operations Research 1 – 06/07 – Chapter 3 1
Operations Research 1
Chapter 3Introduction to Operations Research
Hillier & LiebermannEighth EditionMcGrawHill
Operations Research 1 – 06/07 – Chapter 3 2
Contents• Linear Programming
– Example– Formulations– Graphical solution– The linear programming model– Additional examples
Operations Research 1 – 06/07 – Chapter 3 3
The Wyndor Glass Co. Example• Three plants for glass products (windows and doors):
– Aluminium frames and hardware are made in Plant 1.– Wood frames are made in Plant 2.– Plant 3 produces the glass and assembles the products.
• Capacity is released voor new products:– Two products are an option: product 1 and product 2.
• Product 1 requires capacity in Plants 1 and 3 only.• Product 2 requires capacity in Plants 2 and 3 only.
• Both products need capacity in plant 3. What is the most profitable mix to produce?
Operations Research 1 – 06/07 – Chapter 3 4
Formulation as a Linear Programming Problem
Plant
Production Time per batch, Hours
Production Time Available per Week,
Hours
Product 1 2
1 1 0 42 0 2 123 3 2 18
Profit per batch $ 3,000 $ 5,000
X1 = number of batches of product 1 produced per week
X2 = number of batches of product 2 produced per week
Z = total profit per week (in thousands of dollars) from producing these two products
X1 and X2 are the decision variables for the model. Using the bottom row of table 3.1 we obtain Z = 3X1 + 5X2
Table 3.1
Operations Research 1 – 06/07 – Chapter 3 5
Mathematical formulation of the LP problemMaximize Z = 3X1 + 5X2
Subject to restrictions X1 ≤ 4
2X2 ≤ 123X1 + 2X2 ≤ 18
AndX1 ≥ 0, X2 ≥ 0.
Operations Research 1 – 06/07 – Chapter 3 6
Graphical Solution figure 3.1
Figure 3.1
Shaded area shows the values of (X1,X2) allowed by:
X1 ≥ 0,
X2 ≥ 0,
X1 ≤ 4
X1 ≤ 4
X2 ≥ 0
X1 ≥ 0
Operations Research 1 – 06/07 – Chapter 3 7
Graphical Solution figure 3.2
Figure 3.2
Shaded area shows the set of permissible values of (X1,X2), called the feasible region.2X2 = 12
X1 =4
3X1 + 2X2 = 18
Operations Research 1 – 06/07 – Chapter 3 8
Graphical Solution figure 3.3
Z = 10 = 3X1 + 5X2
Z = 20 = 3X1 + 5 X2
Z = 36 = 3X1 + 5X2 (2,6)
Operations Research 1 – 06/07 – Chapter 3 9
Terminology for Linear Programming
Prototype Example General ProblemProduction capacities of plants3 plants
Production of products2 productsProduction rate of product j, xj
Profit Z
Resourcesm resources
Activitiesn activitiesLevel of activity j, Xj
Objective function (Overall measure of performance Z)
TABLE 3.2
Operations Research 1 – 06/07 – Chapter 3 10
Data needed for a LP Model
Resource
Resource Usage per Unit of Activity
Amount of Resource available
Activity 1 2 … n
12..
m
a11 a12 … a1n
a21 a22 … a2n
… … … … am1 am2 … amn
b1
b2
.
.bm
Contribution to Z per unit of activity
c1 c2 … cn
TABLE 3.3 Data needed for a linear programming model involving the allocation of resources to activities
Operations Research 1 – 06/07 – Chapter 3 11
Symbols used as notation of the various components commonly used for a LP Model
• Below certain Symbols are listed, along with their interpretation of the general problem.– Z = Value of overall measure of performance.– xj = level of activity j (for j = 1, 2, …, n).– cj = increase in Z that would result from each unit increase in level of
activity j.– bi = amount of resource i that is available for allocation to activities (for i
= 1, 2, …, m).– aij = amount of resource i consumed by each other unit of acivity j.
• The model poses the problem in terms of making decisions about the levels of the activitys, so x1, x2, …, xn are called the decision variables. In Table 3.3 the values of ci, bi, and aij (for i = 1, 2, …, m and j = 1, 2, … , n) are the input constants also referred to as the parameters of the model.
Operations Research 1 – 06/07 – Chapter 3 12
A standard form of the model• We can now formulate the mathematical model for this
general problem of allocation resources to activities. The model is to select the values of x1, x2, …, xn so as to
Maximize Z = c1x1 + c2x2 + … + cnxn, subject to restrictions
a11x1 + a12x2 + … + a1nxn ≤ b1
a21x1 + a22x2 + … + a2nxn ≤ b2
… … … … … … … … … …am1x1 + am2x2
+ … + amnxn ≤ bm,and
x1 ≥ 0, x2 ≥ 0 , …, xn ≥ 0
Operations Research 1 – 06/07 – Chapter 3 13
Summary of terminology in LP 1objective function = the function being maximized: c1x1 +
c2x2 + … + cnxn.constraints = restrictions for the objective function.functional constraints = the first m constraints with a
function of all the variables (ai1x1 + ...+ ainxn) on the left-hand side.
nonnegativity constraints = the xj ≥ 0 restrictionsFeasible solution = a solution for which all the constraints
are satisfied.Infeasible solution = a solution for which at least one
constraint is violated.Feasible region = the collection of all feasible solutions.
Operations Research 1 – 06/07 – Chapter 3 14
Summary of terminology in LP 2Optimal solution = a feasible solution that has the most
favorable value of the objective function.Most favorable value = the largest value if the objective
function is to be maximized or the smallest value if the objective function is to be minimized.
It is possible for a problem to have no feasible solutions, multiple optimal solutions. If there are no optimal solutions we speak of an unbounded Z or an unbounded objective
Operations Research 1 – 06/07 – Chapter 3 15
Figure 3.4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
x1 ≤ 0
X1 ≤ 4
x2 ≥ 0
3x1 + 5x2 ≥ 50
Maximize Z = 3x1 + 5x2,
Subject to X1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
3x1 + 5x2 ≥ 50
And x1 ≥ 0, x2 ≥ 0
Figure 3.4
The Wyndor Glass Co. Problem would have no solutions if the constraint 3x1 + 5x2 ≥ 50 were added to the problem.
Operations Research 1 – 06/07 – Chapter 3 16
Figure 3.5
Every point on this darker line segment is optimal, each with Z = 18
Maximize Z = 3x1 + 2x2,Subject to X1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18And x1 ≥ 0, x2 ≥ 0
Figure 3.5
The Wyndor Glass Co. Problem would have multiple solutions if the objective function were changed to Z = 3x1 + 2x2
Feasible region
Operations Research 1 – 06/07 – Chapter 3 17
Figure 3.6
Feasible region
(4,10), Z = 62
(4,8), Z = 52
(4,6), Z = 42
(4,4), Z = 32
(4,2), Z = 22
Maximize
Z = 3x1 + 5x2
Subject to x1 ≤ 4
And x1 ≥ 0, x2 ≥ 0
Figure 3.6
The Wyndor Glass Co. Problem would have no optimal solutions if the only functional constraint were x1 ≤ 4, because x2 then could be increased indefinitely in the feasible region without ever reaching the maximum value of Z = 3x1 + 5x2
Operations Research 1 – 06/07 – Chapter 3 18
Figure 3.7
(0,6) (2,6)
(4,3)
(0,0)
(4,0)
Figure 3.7
The five dots are the five CPF solutions for the Wyndor Glass Co. Problem.
Operations Research 1 – 06/07 – Chapter 3 19
Assumptions of LPProportionality assumption: The contribution of each activity to the value of the objective function Z is proportional to the level of the activity xj, as represented by the cjxj term in the objective function. Similarly, the contribution of each activity to the left-hand side of each functional constraint is proportional to the level of activity xj, as represented by the aijxj term in the constraint.
Consequently, this assumption rules out any exponent other than 1 for any variable in any term of any function (whether the objective function or the function on the left-hand side of a functional constraint) in a linear programming model.
Operations Research 1 – 06/07 – Chapter 3 20
Graphs of Table 3.4
0
2
5
8
11
0
3
7
12
18
0
3
56 6
0
2
4
6
8
10
12
14
16
18
20
0 1 2 3 4
ProportionalitySatisfiedCase 1
Case 2
Case 3
In case 1 through 3 proportionality is violated.
x1
x2
Operations Research 1 – 06/07 – Chapter 3 21
Additivity assumption & Table 3.5Additivity Assumption: Every function in a linear programming model (whether the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities.
(x1,x2)
Value of Z
Additivity SatisfiedAdditivity Violated
Case 1 Case 2
(1,0)(0,1)
35
3 35 5
(1,1) 8 9 7
• TABLE 3.5 Examples of satisfying or violating additivity for the objective function
Operations Research 1 – 06/07 – Chapter 3 22
Table 3.6
(x1,x2)
Amount of Resource Used
Additivity SatisfiedAdditivity Violated
Case 3 Case 4
(2,0)(0,3)
66
6 66 6
(2,3) 12 15 10.8
• TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint
Operations Research 1 – 06/07 – Chapter 3 23
Divisibility assumptionDivisibility Assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values. Since each decision variable presents the level of some activity, it is being assumed that the activities can be run at fractional levels.
Operations Research 1 – 06/07 – Chapter 3 24
Certainty assumptionCertainty assumption: The value assigned to each parameter of a linear programming model is assumed to be a known constant.
Operations Research 1 – 06/07 – Chapter 3 25
Regional Planning: The southern confederation of kibbutzim
Kibbutz Usable Land (Acres) Water Allocation (Acre Feet)123
400600300
600800375
Crop Maximum Quota (Acres)
Water Consumption (Acre Feet/ Acre)
Net Return ($ / Acre)
Sugar beetsCottonSorghum
600500325
321
1,000750250
• TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim
• TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim
Operations Research 1 – 06/07 – Chapter 3 26
Regional Planning (continued): Table 3.10
Crop
Allocation (Acres)Kibbutz
1 2 3
Sugar beetsCottonSorghum
X1
X4
X7
X2
X5
X8
X3
X6
X9
• TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem