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BHARATI VIDYAPEETH UNIVERSITY, PUNE Prof. Nilesh Mahajan M.Sc., M.C.A., M.Phil. COMPUTER ORIENTED DECISION MODELS

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C O N T E N T S

Chapter 1 Operation Research

Chapter 2 Decision Making

Chapter 3 Linear Programming Problem

Chapter 4 Solution of Linear Programming Problem

Chapter 5 Transportation Problem

Chapter 6 Assignment Problem & Job Sequence

Chapter 7 Network Analysis

Chapter 8 Simulation

CHAPTER - 1

OPERATIONS RESEARCH

Introduction to operations Research. Models in operation Research. Applications and Limitations of operations Research.

Introduction to Operation Research“Operations Research” is the application of scientific methods, technique and tools to problems involving the operations of systems so as to provide those in control of operations with optimum solution to the problems.

OR - Operations Research is a scientific approach to problems solving for execute management.

OR - Operations Research is applied decision theory. It uses any scientific, mathematical, or logical means to attempt to cope with the problems that confront the executive, when he tries to achieve a through-going rationality problems.

OR - Operation Research is the systematic application of the quantitative methods, techniques and tools to the analysis of problems involving the operation of systems.

OR – is a scientific method of providing executive departments with a quantitative basis for decisions under their control..

Application of Operation Research1. Finance and Accounting2. Marketing3. Purchasing, Procurement and Exploration4. Production Management5. Personnel Management.6. Technique and General Management7. Government.

MODELS IN OPERATION RESEARCH

1) Physical ModelsWhen we utilize all forms of drawings sketches, diagrams, groups

or charts to describe a situation or problem specific in nature. There are two types of physical models.a) Iconic Models:

An icon is the depiction of an object as its image or likeness. It is useful tool but the application in management problem areas is restrictive and narrow.

b) Analog Models:These models are similar to Iconic Models but not the exact replica of the actual system.

2) Symbolic Models:-These models are used to represent actual problems. There are

two types of symbolic models.a) Verbal Model:

When we represent the problem, and parameter inter-relationships written or spoken in words, these are called verbal models.e.g. A magazine

b) Mathematical Model:When we represent the problem and its parameter by a set of mathematical expression these are called Mathematical Models.

These are:(i) Deterministic Models:

It can be represented as:R = S x PWhere, R = Profit earned

S = Sales volumeP = Profit per unit product

Similarly, Total Cost = Fixed cost + sales variable cost

(ii) Probabilistic Models:When risks or uncertainties can be represented in mathematical relationship from the models are called ‘Probabilistic Models’.

(iii) Heuristic Models:These models use intuititive rules to solve a particular problem. It is solving based on past experience. These models need an amount of creativity and experience by the decision maker.

DECISION THEORY & DECISION TREESThe success and failure that an individual or an organization

experiences, depends to a large extent on the ability of making appropriate decision. Decision theory provides an analytical & systematic approach to the study of decision – making. Decision

models useful in helping decision – makers makes the best possible decisions.

Steps in Decision Theory:The decision making process involves the following steps:(1) Identify and define the problem.(2) Listing of all possible future events, called states of nature, which

can occur in the context of the decision problem.(3) Identification of all the courses of action which are available to

the decision maker.(4) Expressing the payoffs (Pij) resulting from each pair of course of

action & state of nature.(5) Apply an appropriate mathematical decision. Theory model to

select best course of action from the given list.

Decision Tree AnalysisDecision tree analysis involves construction of a diagram

showing all the possible courses of action, states of nature and the probabilities associated with the states of nature. The decision diagram looks very much like a drawing of a tree; therefore this is also called Decision Tree.

A decision tree consists of nodes, branches, probability estimates and payoffs.

Steps Involved in Drawing a Decision Tree

Step 1 : Identify all decisions to be made and the order in which they must be made.

Sep 2 : Identify the chance events that can occur after each decision.

Step 3 : Develop a tree diagram showing the sequence of decisions and chance events. The tree is constructed starting from left and moving towards right. The square box ‘’ denotes

a decision point at which the available strategies are considered. The circle ‘’ represents the chance Node or Event, the various state of nature or outcomes emanate

from this chance event.

Step 4 : Obtain a probability estimate of the chances of each outcomes occurrence.

Step 5 : Obtain estimates of the consequences of all possible outcomes and actions.

Step 6 : Calculate the expected value of all possible actions.

Step 7 : Select the action offering the most attractive expected value.

PROBLEM 1Pay-offs three acts X, Y, Z and states of nature L, M, N are given below:

States of Nature

Pay-offs (in Rs.)

X Y ZL -40 -50 220M 250 -120 -50N 400 650 350

The probabilities of the states of nature are 0.3, 0.4 and 0.3 respectively. Calculate the expected monetary values for different acts and select the best act.

Solution:Expected Pay Off

States of

nature A

Probability

B

ActsX Y Z

C D = B C

E F = B E

G H = G B

L 0.3 -40 -12 -50 -15 220

66

M 0.4 250 100 -120 -48 -50 -20N 0.3 400 120 650 195 35

0105

TOTAL

208 132 151

Since the expected payoff is maximum at Act X, therefore Act X should be adopted.

PROBLEM 2

Following table gives Pay – Offs for actions A1, A2 and A3 corresponding to states of nature S1 and S2 whose chances are 0.6 and 0.4 respectively.

States of Nature

ActionsA1 A2 A3

S1 16 20 18

S2 19 15 12

Required: Find decisions under:(i) Maximum criterion(ii) EMV Criterion

Solution:Pay off Table

States of Nature

ActionsA1 A2 A3

S1 16 20 18S2 19 15 12

Col. Min 16 15 12Maximum criteria – A1

Pay off TableStates

of Natur

eA

Prob.

B

ActionA1 A2 A3

C D = B C

E F = B E

G H = B G

S1 0.6 16 9.6 20 12 18 10.8S2 0.4 19 7.6 15 6 12 4.8

Total 17.2 18 15.6

CHAPTER - 3

LINEAR PROGRAMMING PROBLEM

IntroductionThe linear programming method is a technique of choosing the

best alternative form a set of feasible alternatives in situations in which the objective function as well as constraints can be expressed as linear mathematical function.

Components of Linear Programming Problem:-1. Decision Variable (activities)2. Objective Function3. Constraint

Decision Variable:We pursue certain activities usually denoted by x1, x2,….., xn

The value of these activities represents the extent to which each of these is performed.

All decision variables are continuous and non-negative i.e. x1 0, x2 0 …, xn 0.

Objective Function:The objective function of each LPP is expressed in terms of

decision variables to optimize the criterion of optimally such as profit, cost etc.The general form is optimizing (max or min.)

Z = C1x1 + C2x2 + ……….. + Cnxn

Constraints:There are always certain limitations on the use of resources e..g

labour, machine, money, raw material etc. that limit the degree to which an objective can be achieved.

The solution of LP model must satisfy these constrains.

FORMULATION OF LPP:There are three basic steps in formulation of linear programming

model.

Step 1 : Identify the decision variables to be determined. These should be brought into algebraic relation form for

utilization.

Step 2 : Clearly define all the limitations for a given situation. These limitations or constraints also need to be expressed in algebraic form either as linear equations or inequalities

in terms of the decision variables so identified in step.

Step 3 : Identify the objective to be optimized and it also should be expressed in terms of Linear function of decision

variables.

APPLICATIONS OF LPPSome of the application are given below:

Define1. Transportation costs.2. Optimum weaponry system.3. Optimum level of force development.

Finance1. Profit planning2. Investment policy for maximum between.3. Investment risk analysis.4. Auditing.

Marketing1. Traveling salesman cost.2. Plant locations.3. Media selection.

Personnel1. Determination of optimum organization level.2. Job evaluation and allocations.3. Salary criteria

Production1. Production mix and product proportionity2. Production planning.3. Assembly line balancing.4. Crude oil refinery5. Paper trimming.6. Agriculture7. Hospital & scheduling.

Advantages of Linear Programming1. It is a useful technique to obtain optimum use of productive

resources. It helps a decision maker to ensure effective use of scarce resources by their proper development.

2. Due to its structured form, linear programming technique improves the quality of decision making.

3. This technique can also cater for changing situations. The changed conditions can be used to read just the plan decided for execution.

4. This technique also indicates ideal capacity of machines or materials in a production process.

Limitations of Linear Programming1. L.P. treated all relationship among decision variables as

linear.2. While solving an LP model, there is no guarantee that we will

get integer valued solutions.3. LP model does not take into consideration the effect of time

and uncertainty.4. Parameters appearing in the model are assumed to be

constant but in real life situations, they are frequently neither known or constant.

LINEAR PROGRAMMING FORMULATION

Linear ProgrammingLinear programming is a powerful quantitative technique designs to solve allocation problem. The term ‘linear programming’ consists of the two words ‘Linear’ and ‘Programming’.

‘Linear’ is used to describe the relationship between decision variables which are directly proportional.

‘Programming’ means planning of activities in a manner that achieves some ‘optimal’ result with available resources.

Decision Variables The decision variables refer to the economic or physical

quantities which are competing with one another for sharing the given limited resources. The relationship among these variables must be linear programming. The numerical values of decision variables indicate the solution of the linear programming problem.

Objective Function

The objective function of a linear programming problem is a linear function of the decision variable expressing the objective of the decision maker.Constraints

The constraints indicate limitations on the resources which are to be allocated among various decision variables. These resources may be production capacity, manpower, time, space or machinery. These must be capable of being expressed as linear equation (i.e. =) on inequalities (i.e. is or type) in terms of decision variables. Thus, constraints of a linear programming problem are linear equalities or inequalities arising out of practical limitations.

Non-negativity RestrictionNon – negativity restriction indicates that all decision variables must take on values equal to or grater than zero.

Step Involved in the Formulation of LP Problem

Step 1 : Identify the Decision Variables of interest to the decision make and express them as x1, x2, x3.

Step 2 : Ascertain the Objective of the decision maker whether he wants to minimize or to maximize.

Step 3 : Ascertain the cost (in case of minimization problem) or the profit (in case of maximization problem) per unit of each of the decision variables

Step 4 : Ascertain the constraints representing the maximum availability or minimum commitment or equality and represent them as less than or equal to () type inequality or greater than or equal to () type inequality or ‘equal to’ (=) type equality respectively.

Step 5 : Put non-negativity restriction as under:xj 0; j = 1, 2, …. n (non-negativity restriction)

Step 6 : Now formulate the LP problem as under:Maximize (or Minimize) Z = c1x1 + c2x2 + ….cnxn

Subject to constraints:a11x1 + a12x2…..a1nxn b1 (Maximum availability)a21x1 + a22x2….a2nxn b2 (Minimum commitment)a31x1 + a32x2….a3nxn = b3 (Equality)

am1x1 + am2x2, …….amxn

x1; x2 …. xn 0 (Non – negativity restriction)where,xj = Decision variables i.e. quantity of jth variable of interest to

the decision maker.

cj = Constant representing per unit contribution (in case of Maximization Problem) or Cost (in case of Minimization Problem) of the jth decision variable.

aij = Constant representing exchange coefficients of the jth

decision variable in the ith constant.bi = Constant representing ith constraint requirement or

availability.

PROBLEM 1A garment manufacturer has a production line making two styles of shirts. Style I requires 200 grams of cotton thread, 300 grams of Dacron thread, and 300 grams of linen thread. Style II requires 200 grams of cotton thread, 200 grams of Dacron thread and 100 grams of linen thread. The manufacturer makes a net profit of Rs. 19.50 on Style 1, Rs. 15.90 on Style II. He has in hand an inventory of 24 kg of cotton thread, 26 kg of Dacron thread and 22 kg of linen thread. His immediate problem is to determine a production schedule, given the current inventory to make a maximum profit. Formulate the LPP model.

SolutionLet x1 = Number of Style I shirts,

x2 = Number of Style II shifts,Since the objective is to maximize the profit, the objective function is given by-Maximize Z = 19.50x1 + 15.90x2

Subject to constraints:200x1 + 200x2 24,000 [Maximum Qty. of Cotton thread available]300x1 + 200x2 26,000 [Maximum Qty. of Dacron thread available]300x1 + 100x2 22,000 [Maximum Qty. of Lines thread available]x1, x2 0 [Non-Negativity constraint]

PROBLEM 2A firm makes two types of furniture: chairs and tables. The contribution for each product as calculated by the accounting department is Rs. 20 per chair and Rs. 30 per table. Both products are processed on three machines M1, M2 and M3. The time required by each product and total time available per week on each machine is as follows:

Machine

Chair Table Available Hours

M1 3 3 36M2 5 2 50M3 2 6 60

How should the manufacturer schedule his production in order to maximize contribution?

Solution:Let x1 = Number of chairs to be produced

x2 = Number of tables to be producedSince the objective is to maximise the profit, the objective function is given by-

Maximise Z = 20x1 + 30x2 Subject to constraints:3x1 + 3x2 36 (Total time of machine M1)5x1 + 2x2 50 (Total time of machine M2)2x1 + 6x2 60 (Total time of machine M3)x1, x2 0 (Non-negativity constraint)

PROBLEM 3The ABC manufacturing company can make two products P1 and P2. Each of the products requires time on a cutting machine and a finishing machine. Relevant data are:

P1 P2

Cutting Hours (per unit)

2 1

Finishing Hours (per unit)

3 3

Profit (Rs. per unit) 6 4Maximum sales (unit per week)

200

The number of cutting hours available per week is 390 and the number of finishing hours available per week is 810. How much should be produced of each product in order to achieve maximum profit for the company?

Solution:Let x1 = Number of Product P1 to be produced

x2 = Number of Product P2 to be producedSince the objective is to maximize profit, the objective function is

given by – Maximize Z = 6x1 + 4x2

Subject to constraints:2x1 + x2 390 (Availability of cutting hours)3x1 + 3x1 810 (Availability of finishing hours)x2 200 (Maximum sales)x1, x2 0 (Non-negativity constraint)

PROBLEM 4An animal feed company must produce 200 kg of a mixture consisting of ingredients X1, and X2 daily. X1 cost Rs. 3 per kg and X2 Rs. 8 per

kg. Not more than 80 kg of X1 can be used, and at least 60 kg of X2 must be used. Find how much of each ingredient should be used if the company wants to minimise cost.

Solution:Let x1 = kg of ingredient X1 to be used

x1 = kg of ingredient X2 to be usedSince the objective is to minimise the cost, the objective function is given by –

Minimise Z = 3x1 + 8x2

Subject to constraints:x1 + x2 = 200 (Total mixture to be produced)x1 80 (Maximum use of x1)x2 60 (Minimum use of x2)x1 0 (Non-negativity constraint)

PROBLEM 5A company manufacturing television sets and radios has four major departments: chasis, cabinet, assembly and final testing. Monthly capacities are as follows:

Department

Television capacity

Chasis 1,500 or 4,500Cabinet 1,000 or 8,000

Assembly 2,000 or 4,000Testing 3,000 or 9,000

The contribution of television is Rs. 150 each and the contribution of radio is Rs. 250 each. Assuming that the company can sell any quantity of either product, determine the optimal combination of output.

Solution:Let x1 = Number of televisions to be produced

x2 = Number of radio to be producedSince the objective is to maximise the profit, the objective

function is given by – Maximise Z = 150x1 + 250x2

Subject to constraints:3x1 + x2 4,500 (Maximum capacity in chasis department)8x1 + x2 8,000 (Maximum capacity in cabinet department)2x1 + x2 4,000 (Maximum capacity in assembly department)3x1 + x2 9,000 (Maximum capacity in testing department)x1, x2 0 (Non – negativity)

PROBLEM 6

Two products A and B are to be manufactured. A single unit of product A requires 2.4 minutes of punch press time and 5 minutes of assembly time. The profit for product A is Rs. 0.60 per unit. A single unit of product B requires 3 minutes of punch press time and 2.5 minutes of welding time. The profit for product B is Rs. 0.70 per unit. The capacity of the punch press department available for these products is 1.200 minutes / week. The welding department has an idle capacity of 600 minutes / week and assembly department has 1,500 minutes week.

a) Formulate the problem as linear programming problem.b) Determine the quantities of products A and B so that total

profit is maximised.

Solution:Let x1 = Number of product A to be manufactured

x2 = Number of product B to be manufacturedSince the objective is to maximise the profit, the objective function is given by-

Maximise Z = 60x1 + 70x2

Subject to constraints:2.4x1 + 3x2 1200 (Availability of punch press time)5x1 1500 (Availability of assembly time)2.5x2 600 (Availability of welding time)

x1, x2 0 (Non-negativity constraint)

CHAPTER IVA. GRAPHICAL METHOD

________________________________________________________________________

Graphical Method is used for solving linear programming problems that involve only two variables.

FEASIBLE SOLUTIONAny non-negative solution which satisfies all the constraints is known as a feasible solution of the problem.

FEASIBLE REGIONThe collection of all feasible solutions is known as a feasible region.

UNBOUNDED SOLUTIONAn unbounded solution of a linear programming problem is a solution whose objective function is infinite. A linear programming problem is said to have unbounded solution if its solution can be made infinitely large without violating any of the constraints in the problem.

INFEASIBLE PROBLEMA linear programming problem is said to be infeasible if there is no solution that satisfies all the constraints. It represents a state of inconsistency in the set of constraints.

PRACTICAL STEPS INVOLVED IN SOLVING LPP BY GRAPHICAL METHOD

Step 1 Consider each inequality constraint as equation.Step 2 Take one variable (say x) in a given equation equal to zero

and find the value of other variable (say y) by solving that equation to get one coordinate [say (0,y)] for that equation.

Step 3 Take the second variable (say y) as zero in the said equation and find the value of first variable (say x) to get another coordinate [say (x,0)] for that equation.

Step 4 Plot both the coordinates so obtained [i.e., (0,y) and (x,0)] on the graph and join them by a straight line. This straight line shows that any point on that line satisfies the equality and any point below or above that line shows inequality. Shade the feasible region which may be either convex to the origin in case of less than type of inequality (<) or opposite to the origin in case of more than type of inequality (>).

Step 5 Repeat Steps 2 to 4 for other constraints.Step 6 Find the common shaded feasible region and mark the

coordinates of its corner points.Step 7 Put the coordinates o each of such vertex in the Objective

Function. Choose that vertex which achieves the most optimal solution (i.e., in the case of Maximisation, the vertex that gives the maximum value of ‘Z’ & in case of Minimisation the vertex that gives the minimum value of ‘Z’).

Optimal Solution:-Type of Problem Optimal Solution

(a) In case of maximisation problem

The vertex which gives the maximum value of Z is the optimal solution

(b) In case of minimisation problem.

The vertex which gives the minimum value of Z is the optimal solution

PROBLEM 1PARLOK Ltd. Has two products Heaven and Hell. To produce one unit of Heaven, 2 units of material X and 4 units of material Y are required. To produce on unit of Hell, 3 units of material X and 2 units of material Y are required. Only 16 units of material X and 16 units of material Y are available. Material X cost Rs. 2.50 per unit and Material Y cost Rs. 0.25 per unit respectively.

Required : Formulate an LP Model and solve it graphically

Solution:Formulation of LP Model

Calculation of Cost per unit of each productHeave

nHell

A. Material X (2.50 x 2)

5.00 (Rs. 2.50 x 3)

7.50

B. Material Y (0.25 x 4)

1.00 (Rs. 0.25 x 2)

0.50

6.00 8.00Let, x represent the no. of units of product ‘Heaven’ to produced

y represent the no. of units of product ‘Hell’ to be produced.Since the object is to minimise, the objective function is given by,Minimise Z = 6x + 8ySubject to constraints2x1 + 3y 16 [Maximum material X constraint] …….(i)

4x1 + 2y 16 [Maximum material Y constraint] …….(ii)

Step 1: Finding the vertex for each constraint by treating the constraint of inequality nature as equality,Constraint (i) in limiting form 2x1 + 3y = 16.

When x = 0,

When y = 0, = 8

Thus, the vertices are & (8, 0)

Constraint (ii) in limiting form 4x1 + 2y = 16

When x = 0, = 8

When y = 0, = 4

Thus, the vertices are (0, 8) & (4, 0)

Step 2: Plotting both the coordinates of the 1st constraint on the graph and joining them by straight line and shading the feasible region which is convex to origin in case of less than type of inequality. Similarly drawing straight line and shading feasible region for other constraint.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective function. Selecting those vertices which achieve the most optimal solution (i.e. in case of minimisation vertices which give the minimum value of Z).

Set No.

Co-ordinates of Vertices of Common shaded feasible region

Value of Z

1 F 0,0 6 x 0 + 8 x 0 = 0

2 A0, 6 x 0 + 8 x

3 E 2, 4 6 x 2 + 8 x 4 = 44.0

4 D 4, 0 6 x 4 + 8 x 0 = 24.0

Optimal Solution:-Thus, set No. 1 gives the minimum value of Z with x = 0 and y = 0.

PROBLEM 2Earth Ltd. has two products Sun and Moon. To produce one unit of Sun, 2 units of material X and 4 units of material Y are required. To produce one unit of Moon, 3 units of material X and 2 units of material Y are required. As the raw material X is in short supply so not more than 16 units of material X can be used. Atleast 16 units of material Y must be used in order to meet the committed sales of Sun and Moon. Cost per unit of material X and material Y are Rs. 2.50 and Rs. 0.25 respectively.You are required:

(i) to formulate mathematical model.(ii) To solve it for the minimum cost (Graphically)

Solution:Calculation of Cost per Unit of each Product

Particulars Sun(Rs.)

Moon(Rs.)

A. Material X (Rs. 2.50 x 2) 5.00 (Rs. 2.50 x 3) 7.50

B. Material Y (Rs. 0.25 x 4) 1.00 (Rs. 0.25 x 2) 0.50

C. Total Cost (A + B)

6.00 8.00

Part (i)Since the objective is to minimise the cost, the objective function is given by:Minimise Z = 6x + 8y,Subject to constraints:2x + 3y 16 (Maximum material X constraint) ………(i)4x + 26 16 (Minimum material Y constraint) ………(ii)x, y 0 (Non – Negativity constraint)Part (ii)Step 1: Finding the vertex of each constraint by treating the constraint of inequality nature as equality.Constraint (i) in limiting form 2x + 3y = 16

When x = 0,

When y = 0, = 0

Thus, the vertices are & (8, 0)

Constraint (ii) in limiting form 4x + 2y = 16

When x = 0, = 8

When y = 0, = 4

Thus, the vertices are (0, 8) & (4, 0)

Step 2: Plotting both the coordinates of the 1st constraint on the graph and joining them by straight line and shading the feasible region which is convex to origin in case of less than type of inequality. Similarly drawing straight line and shading feasible region.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective function. Selecting those vertices which achieve the most optimal solution (i.e. in case of minimisation vertices which gives the minimum value of Z).

Set No.

Coordinates of Vertex of common shaded feasible region

Value of Z

1 E (2, 4) Z1 = 6 x 2 + 8 x 4 = 44

2 B (8, 0) Z2 = 6 x 8 + 8 x 0 = 48

3 D (4, 0) Z3 = 6 x 4 + 8 x 0 = 24

Optimal Solution: Thus, set no. 3 gives the minimum value of 2 with x = 4 and y = 0.

PROBLEM 3Sky Ltd. Has two products Cloud and Wind. To produce one unit of Cloud, 2 units of material X and 4 units of material Y are required. To produce one unit of Wind, 3 units of material X and 2 units of material Y are required. As the raw material X is in short supply so not more than 16 units of material X can be used. Atleast 16 units of material Y must be used in order to meet the committed sales of Cloud and Wind.

Cost per unit of material X and material Y are Rs. 2.50 and Rs. 0.25 respectively. The selling price per unit of cloud and wind are Rs. 12 and Rs. 16 respectively.

You are required:(i) To formulate mathematical model.(ii) To solve it for maximum contribution (Graphically)

Solution:Calculation of Contribution per unit of each Product

Particular Cloud(Rs.)

Wind(Rs.)

A. Selling price per unit

12.00 16.00

B. Less: cost per unit

(i) Material X (Rs. 2.50 x 2)

5.00 (Rs. 2.50 x 3)

7.50

(ii) Material Y (Rs. 0.25 x 4)

1.00 (Rs. 0.25 x 2)

0.50

Total (i) + (ii) 6.00 8.00C. Contribution (A – B)

6.00 8.00

Part (i)Since the objective is to maximise the profit, the objective

function is given by,Maximise Z = 6x + 8ySubject to constraints:2x + 3y 16 (Maximum material X constraint) ………(i)4x + 2y 16 (Minimum material Y constraint) ………(ii)x, y 0 (Non – negativity constraint)

Part (ii)Step 1:

Finding the vertex for each constraint by treating the constraint of inequality nature as equality.Constraint (i) in limiting form 2x + 3y = 16.

When x = 0,

When y = 0,

Thus, the vertices are & (8, 0)

Constraint (ii) in limiting form 4x + 2y = 16

When x = 0, = 8.

When y = 0, = 4.

Thus, the vertices are (0, 8) & (4, 0)

Step 2: Plotting both the coordinates of the 1st constraint on the graph and joining them by straight line and shading the feasible region which is convex to origin in case of les type of inequality. Similarly drawing straight line and shading feasible region for other constraint.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective function. Selecting those vertices which achieve the most optimal solution (i.e. in case of maximisation vertices which give the maximum value of Z).

Set No.

Coordinates of Vertex of common shaded feasible region

Value of Z

1 E (2, 4) Z1 = 6 x 2 + 8 x 4 = 44

2 B (8, 0) Z2 = 6 x 8 + 8 x 0 = 48

3 D (4, 0) Z3 = 6 x 4 + 8 x 0 = 24

Optimal Solution: Thus, set no. 2 gives the max. value of 2 with x = 8 and y = 0.

PROBLEM 4Timber Ltd has two products Sofa and Chair. To produce one unit of Sofa, 2 units of material X and 4 units of material Y are required. To produce one unit of Chair, 3 units of material X and 2 units of material Y are required. As the raw material is in short supply not more than 16 units of each material can be used. The cost per unit of material X and material Y are Rs. 2.50 and Rs. 0.25 respectively. The cost per unit of material X and material Y are of Rs. 2.50 and Rs. 0.25 respectively. Atleast 2 units of sofa must be produced and sold.You are required:(i) to formulate mathematical model.(ii) To solve it for minimum cost (Graphically)

Solution:Calculation of Cost per Unit of each Product

Particular Sofa(Rs.)

Chair(Rs.)

A. Material (Rs. 2.50 x 2)

5.00 (Rs. 2.50 x 3)

7.50

B. Material Y (Rs. 0.25 x 4)

1.00 (Rs. 0.25 x 2)

0.50

C. Total (A + B) 6.00 8.00

Part (i)Since the objective is to minimise the cost, the objective function is given by,Minimise Z = 6x + 8ySubject to constraints:2x + 3y 16 (Maximum material X constraint) ………(i)4x + 2y 16 (Maximum material Y constraint) ………(ii)x 2 (Minimum sofas constraint) ………(iii)y 0 (Non – Negativity constraint)

Part (ii)Step 1: Finding the vertex for each constraint by treating the constraint of inequality nature as equality.

Constraint (i) in limiting form 2x + 3y = 16.

When x = 0,

When y = 0, x = 8

Thus, the vertices are & (8, 0)

Constraint (ii) in limiting form 4x + 2y = 16.Whenx = 0 y = 8Wheny = 0 x = 4

Thus, the vertices are (0, 8) & (4, 0)

Step 2: Plotting both the co-ordinates of the 1st constraint on the graph and joining them by straight line and shading the feasible region which is convex to origin in case of less than type of inequality. Similarly drawing straight line and shading feasible region for other constraint.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective functions. Choose those vertices which achieve the most optimal solution (i.e. in case of minimisation vertices which gives the minimum of Z).

Set No.

Coordinates of Vertex of common shaded feasible region

Value of Z

1 E (2, 0) Z1 = 6 x 2 + 8 x 0 = 12

2 D (4, 0) Z2 = 6 x 4 + 8 x 0 = 24

3 G (2, 4) Z3 = 6 x 2 + 8 x 4

= 44

Optimal Solution: Thus, set No. 1 gives the minimum value of Z with x = 2 and y = 0.

PROBLEM 5Johnson & Johnson has two products Deluxe Stayfree & Deluxe Carefree. To produce one unit of deluxe Stayfree, 2 units of material A, 4 units of material B & 2 units of material C are required. To produce one unit of Deluxe Carefree, 3 units of Material A & 2 units of Material B & 1 unit of Material ‘C’ are required. Not more than 16 units of material A can be used & at least 16 units of material B must be used & the use of material C in total should be equal to 16. The contribution per unit of deluxe Stayfree & Deluxe Carefree are Rs. 6 & Rs. 8 respectively.

You are required:(i) to formulate the mathematical model.(ii) to solve it for maximum contribution by Graphical Method.

Solution:

Products MaterialA B C

Deluxe Stayfree

2 4 2

Deluxe Carefree

3 2 1

Part (i)X = Number of units to be produced of Deluxe StayfreeY = Number of units to be produced of Deluxe CarefreeSince the objective is to maximise the profit, the objective function is given by,Maximise Z = 6x + 8ySubject to constraints2x + 3y 16 (Maximum material A constraint) ………. (i)4x + 2y 16 (Minimum material B constraint)

………. (ii)2x + y = 16 (Material C constraint)

………. (iii)x, y 0 (Non – Negativity constraint)

Part (ii)Step 1: Finding the vertex of each constraint by treating the constraint of inequality nature as equality.

Constraint (i) in limiting form 2x + 3y = 16.

When x = 0,

When y = 0, x = 8

Thus, the vertices are & (8, 0)

Constraint (ii) in limiting form 4x + 2y = 16.Whenx = 0 y = 8Wheny = 0 x = 4Thus, the vertices are (0, 8) & (4, 0)Constraint (iii): 2x + y = 16When x = 0, then y = 16When y = 0, then x = 8.Thus, the vertices are (0, 16) (8, 0)

Step 2: Plotting the both the coordinates of the 1st constraint on the graph and joining them by a straight line and shading the feasible region which is convex to origin in case of les than type of inequality. Similarly drawing straight line and shading feasible region for other constraint.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective function. Choosing the vertex which achieves the most optimal solution (i.e. in case of maximisation vertices which give the maximum value of Z).

Set No.

Coordinates of Vertex of common shaded feasible region

Value of Z

1 B (8, 0) 6 x 8 + 8 x 0 = 48

Optimal Solution: In this case, we obtain only one such vertex. Hence the value of Z will be maximum at x = 8 and y = 0.

PROBLEM 6Water Ltd. Has two products Drought and Flood. To produce one unit of Drought, 3 units of material A & 1 unit each of material B and material C respectively are required. To produce one unit of Flood, 1 unit each of material A & material B respectively & 2 units of material C are required. Not more than 40 units of material C can be used and atleast 27 units of material A must be used and the use of material B in total should be equal to 21. The selling price per unit of Drought & Flood are Rs. 16 & Rs. 8 respectively. The manufacturing cost per unit of Drought & flood are Rs. 8 & Rs. 4 respectively.You are required:

(i) to formulate the mathematical model, &(ii) to solve it for minimise cost.

Solution:Part (i)Since the objective is to minimise the cost, the objective function is given by,Minimise Z = 8x + 4ySubject to constraint:3x + y 27 (Minimum material A constraint)x + y = 21 (Material B constraint)x + 2y 40 (Maximum material C constraint)x, y 0 (Non – Negativity constraint)

Part (ii)Step 1: Finding the vertex for each constraint by treating the constraint of inequality nature as equality.Constraint (i) in limiting form 3x + y = 27When x = 0, y = 27When y = 0x = 9Thus, the vertices are (0, 27) & (9, 0)Constraint (ii) in limiting form x + y = 21

When x = 0, y = 21When y = 0, x = 21

Thus, the vertices are (0, 21) & (21, 0)

Constraint (iii) in the limiting form: x + 2y = 40.

When x = 0, then y = 20When y = 0, then x = 40

Thus, the vertices are (0, 20) & (40, 0)

Step 2: Plotting both the coordinates of the 1st constraint on the graph and joining them by straight line and shading the feasible region. Similarly drawing straight line and shading feasible region for other constraints.

Step 3: Reading the coordinates of the vertices of common shaded feasible region and putting the coordinates of each of such vertex in the objective function. Selecting those vertices which achieve the most optimal solution (i.e. in case of minimisation, vertices which gives the minimum value of Z).

Set No.

Coordinates of Vertex of common shaded feasible

region

Value of Z

1 G (3, 18) Z1 = 8 x 3 + 4 x 18 = 96

2 B (21, 0) Z2 = 8 x 21 + 4 x 0 = 168

Optimal Solution: Thus, Set No. 1 gives the minimum value of Z with x = 3 and y = 18.

LINEAR PROGRAMMING – SIMPLEX METHOD

Step 1 : Add Non-negative Slack Variables (say S1, S2 etc.) in each constraint to convert inequalities into equations. A

Slack Variable indicates under utilization of capacity of the constraint; hence, its contribution to Objective Function is assumed to be zero (or negative, if given). This Slack Variable is also called as Basic Variable. Other variables are called Non- basic Variables.

Step 2 : Assuming the value of each Non-basic Variable equal to ‘zero’ (i.e. assuming that no performance has taken place), calculate the value of each Basic Variable from the equations.

Step 3 : Find the value of ‘Z’ by putting the value of each basic and non- basic variable in the Objective Function.

Step 4 : Draw the Initial Simplex Table as follows:

Contribution per unit Ci C1 C2 0 0 Replacement

Ratio Qty. /

value of key

column

Ci Variables

Quantity

x1 x2 S1 S2

C1 S1 …. …. …. 1 0C2 S2 …. …. …. 0 1

Total Contribution Zi 0 0* 0 ….. ….Opportunity Cost (Ci – Zj) Ci C2 0 0

*Note: The value in the Zj Row for x1 variable is computed by the formula = C1x1 + C2x2. The values in the Zj Row in the column for other variables are computed by the same formula as stated.

Step 5 : Key Column – Mark the Column having maximum positive value in (Ci – Zj) row by sign representing the Opportunity Cost or Loss of not introducing one unit of the variable of that column. This Column is also known as Key Column. This column indicates the selection of incoming variable in the Next Simplex Table.

Step 6 : Replacement Ratio – Find out Replacement Ratio (also known as Key Ratio) by dividing value in the Quantity Column of each row by corresponding key Column value. Replacement Ratio represents how much quantum of variable can be produced based on that row taking the Key Column value.

Step 7 : Key Row – Mark the Row having minimum non-negative Replacement Ratio by sign. Minimum Replacement Ratio ensures that no basic variable will ever be negative. (This

can be verified by putting higher Replacement Ratio in all constraints). The Row is known as Key Row. This row indicates the selection of outgoing variable from the Current Simplex Table.

Step 8 : Key Element – Encircle the element at the intersection of Key Row and Key Column. This value is known as Key Element or Pivot Element

Step 9 : Replace the outgoing variable by the incoming variable with its contribution per unit, i.e., Ci in the first Column for

Contribution Per unit.Step 10 : Calculate the new values of the Key Row as under:

New Values of Key Row =

Step 11 : Calculate the new values of other rows (i.e. Non-Key Rows) as under:

A. Old ValuesB. New values of Key RowC. Key Column ElementD. Product of B & C

Step 12 : Draw another Simplex Table based on new values of each row.

Step 13 : Repeat Steps 4 to 12 till all values in (C i – Zj) row becomes zero or negative.

Step 14 : Find Optimal Solution from the Final Simplex Table, i.e., the value of Z and quantity of Non – basic Variables.

Value of Z = Ci.(Qty of xi) + Cj(Qty of xj)

UNBOUNDED SOLUTIONAn unbounded solution of a linear programming problem is a situation where objective function is infinite. A linear programming problem is said to have unbounded solution if its solution can be made infinitely large without violating any of its constraints in the problem.

INFEASIBLE PROBLEMA linear programming problem is said to be infeasible if there is no solution that satisfies all the constraints. It represents a state of inconsistency in the set of constraints. Under the Simplex Method, the problem is said to have no feasible solution if at least one of the artificial variable remains in the final table as basic variable with non-zero quantity.

PROBLEM 1Use Simplex Method to solve the following L.P. Problem.Max. Z = 6x1 + 8x2

Subject to: 30x1 + 20x2 3005x1 + 10x2 110

Solution:Step 1 : Formulation of LP problem after introducing slack variables:Max. Z = 6x1 + 8x2 + 0S1 + 0S2

Subject to: 30x1 + 20x2 + S1 = 300

5x1 + 10x2 + S2 = 100x1, x2, S1, S2 0

Step 2 : Preparing Initial Simplex Table:

Simplex Table 1

Ci Contribution per unit

6 8 0 0 Replacement Ratio Qty./Value of Key Column

Variables

Qty. x1 x2 x3 x4

0 S1 300 30 20 1 0 300/20 = 15

0 S2 110 5 10 0 1 110/10=11

Total Contribution (Zj) 0 0 0 0

Opportunity Loss Ci - Zj 6 8 0 0

Key Element Key Column Key Row

Key Row indicates that outgoing variable is S2 and Key Column indicates that incoming variable is x2.

Step 3 : Replacing the outgoing variable (S2) by the incoming variable (x2) together with its contribution per unit.Step 4 : Calculating the new values of Key Row as under:

Step 5 : Calculating the new values of other Rows (i.e. Non – Key Row) as under:Step 6 : Preparing Second Simplex Table.

Simplex Table II

Ci Contribution per unit

6 8 0 0 Replacement Ratio Qty./value of Key Column

Variables Qty x1 x2 s1 S2

0 S1 80 20 0 1 -2 80/20 = 4

8 x2 11 1 0 11 5/10 = 22

Total Contribution Zj 4 8 0

Opportunity Loss (Ci – Zj) 2 0 0-

Key Element Key Column Key Row

Step 7 : Replacing the outgoing variable (S1) by the incoming variable (x1) together with its contribution per unit.Step 8 : Calculating the new values of Key Row as under:

4 1 0

Step 9 : Calculating the new values of other Rows (i.e. Non – Key Rows) as under:

A Old Values 11

B New Values of Key Row

4 1 0-

C Key Column Element

D Product of B & C 2 0

E New Values (A – D) 9 0 1-

Step 10 : Preparing Simplex Table III:Ci Contribution per

unit6 8 0 0

Variables Qty. x1 x2 s1 s2

6 x1 4 1 0-

8 x2 9 0 1

Total Contribution Zj 6 8

Opportunity Loss Ci - Zj 0 0- -

Optimal Solution: Since all Ci - Zj are 0, this table gives the optimal solution, i.e. x1 = 4 and x2 = 4. The optimal value of Z = 6 x 4 + 8 x 9 = Rs. 96.

PROBLEM 2Use Simplex method to solve the following L.P. Problem:

Max. Z = 6x1 + 8x2

Subject to Constraints:2x1 + 3x2 164x1 + 2x2 16

Solution:

Step 1 : Formulation of LP Problem after introducing slack variables:

Max. Z = 6x1 + 8x2 + OS1 + OS2

Subject to constraints:2x1 + 3x2 + S1 = 164x1 + 2x2 + S2 = 16x1, x2, S1, S2 0

Step 2 : Preparing Initial Simplex Table:

Simplex Table

Contribution per unit Ci

6 8 0 0 Replacement Ratio Qty. / Value of Key Column

Ci

Variable

Qty. X1 x2 s1 s2

0 S1 16 2 3 1 0 = 5.33

0 S2 16 4 2 0 1 = 8

Total Contribution (Zj) 0 0 0 0

Opportunity Loss (Ci – Zj) 6 8 0 0

Key Column Key element Key Row

Hence, the outgoing variable is S1 and the incoming variable is x2Step 3 : Replacing the outgoing variable (S1) by incoming variable (x2) together with its contribution per unit.Step 4 : Calculating the new values of Key Row as under:

=

Step 5 : Calculating the new values of other Rows (i.e. Non – key Row) as under:

A Old Values 16 4 2 0 1B New Values of Key

Row1 0

C Key Column Element 2 2 2 2 2D Product of B & C 2 0

E New values (A – D) 0-

1

Step 6 : Preparing Second Simplex Table

Simplex Table II

Contribution per unit Ci 6 8 0 0 Replacement Ratio

Qty./Value of Key Column

Ci

Variabl

eQty. x1 x2 S1 S2

8 x2 1 0 = 8

0 S2 0 1 = 2

Total Contribution (Zj) 5.33 8 0

Opportunity Loss (Ci – Zj) -67 0 - 0

Key Column Key Element Key Row

Step 7 : Replacing The outgoing variable (S2) by incoming variable (x1) together with its contribution per unit.Step 8 : Calculating the new values of Key Row as under:

Step 9 : Calculation of the new values of non Key Rows as under:

A Old Values 1 0

B New Values of Key

Row

2 1 0

C Key Column Element

D Product of B & C 0

E New Values (A – D) 4 0 1

Step 10 : Preparing Simplex Table IIIContribution per unit Ci 6 8 0 0

Ci

Variabl

eQty. x1 x2 S1 S2

8 x2 4 0 1

6 x1 2 1 0

Total Contribution (Zj) 44

6 8 2.5 0.25

Opportunity Loss (Ci – Zj) 0 0 -2.5 -0.25

Optimal Solution: Since, the values of Ci - Zj are 0, the solution is optimal at x1 = 2 and x2 = 4. The optimal value of Z = 2 x 6 + 4 x 8 = Rs. 44

PROBLEM 3Use simplex method to solve the following LP ProblemMax Z = 4x1 + 5x2 + 8x3

Subject to: x1 + x2 + x3 1003x1 + 2x2 + 4x3 500x1, x2, x3 0

Solution:Step 1 : Formulation of LP problem after introducing slack variables:

Max. Z = 4x1 + 5x2 + 8x3 + 0S1.0S2

Subject to constraints:x1 + x2 + x3 + S1 = 1003x1 + 2x2 + 4x3 + S2 = 500x1, x2, x3, S1, S2 0

Step 2 : Preparing Initial Simplex Table:

Simplex Table 1

Contribution per unit Ci

4 5 8 0 0 Replacement Ratio Qty./Value of Key ColumnCi

Variabl

eQty. x1 x2 x3 S

1

S2

0 S1 16 1 1 1 1 0 = 100

0 S2 16 3 2 4 0 1 = 125

Total Contribution Zj 0 0 0 0 0

Opportunity Loss (Ci – Zj)

4 5 8 0 0

Key element

Key Colum

n

Key Row

Step 3 : Replacing the outgoing variable (S1) by the incoming variable (x3) together with the contribution per unit.Step 4 : Calculating the new values of key Row as under:

Step 5 : Calculating the new values of other rows (i.e. Non-Key Rows) as under:

A Old Values 500 3 2 4 0 1B New Values of Key

Row100 1 1 1 1 0

C Key Column Element 4 4 4 4 4 4

D Product of B & C 400 4 4 4 4 0E New values (A – D) 100 -1 -2 0 -4 1

Step 6 : Preparing Second Simplex Table

Simplex Table IIContribution per unit Ci 4 5 8 0 0

Ci

Variable Qty x1 x2 x3 S1 S2

8 x3 100 1 1 1 1 0

0 S2 100 -1 -2 0 -4 1

Total Contribution Zj 8 8 8 8 0Opportunity Loss (Ci - Zj) -4 -3 0 -8 0

Optimal Solution: Since all Ci – Zj are 0, this table gives the optimal solution, i.e. x3 = 100 and x1 = 0, x2 = 0. The optimal value of Z is = 100 x 8 = 800.

PROBLEM 4Use Simplex Method to solve the following LP Problem:

Maximise Z = 5x1 + 3x2

Subject to constraints:x1 + x2 62x1 + 3x2 12x1 3x2 3x1, x2, x3 0

Solution:Step 1 : Formulation of LP problem after introducing slack variables:

Max. 5x1 + 3x2 + 0S1 + 0S3 + 0S4

Subject to constraints:x1 + x2 + S1 = 62x1 + 3x2 + S2 = 12x1 + S3 = 3x2 + S4 = 3x1, x2, S1, S3, S4 0

Step 2 : Preparing Initial Simplex Table 1:

Simplex Table 1

Contribution per unit Ci 5 3 0 0 0 0 Replacement Ratio

Ci

Variable Quantity

x1 x2

S1 S2 S3 S4 Qty./Value of Key Column

0 S1 6 1 1 1 0 0 0 = 6

0 S2 12 2 3 0 1 0 0 = 6

0 S3 3 1 0 0 0 1 0 = 3

0 S4 3 0 1 0 0 0 1 =

Total Contribution (Zj)

0 0 0 0 0 0 0

Opportunity Loss(Ci – Zj) 5 3 0 0 0 0

Key Column Key Element Key RowHence the outgoing variable is S3 and the incoming variable is x1.

Step 3 : Replacing the outgoing variable (S3) by the incoming variable (x1) together with its contribution per unit.Step 4 : Calculating the new values of Key Row as under:

Step 5 : Calculation of the new values of other Non – Key Rows:Ist Non-Key Row

A Old Values 6 1 1 1 0 0 0B New Values of Key

Row3 1 0 0 0 1 0

C Key Column Element 1 1 1 1 1 1 1D Product of B & C 3 1 0 0 0 1 0

E New Values (A – D) 3 0 1 1 0 -1 0IInd Non Key Row

A Old Values 12

2 3 0 1 0 0

B New Values of Key Row

3 1 0 0 0 1 0

C Key Column Element 2 2 2 2 2 2 2D Product of B & C 6 2 0 0 0 2 0E New Values (A – D) 6 0 3 0 1 -2 0

IIIrd non Key Row:A Old Values 3 0 1 0 0 0 1B New Values of Key

Row3 1 0 0 0 1 0

C Key Column Element 0 0 0 0 0 0 0D Product of B & C 0 0 0 0 0 0 0E New Values (A – D) 3 0 1 0 0 0 1

Step 6 : Preparing Second Simplex Table II

Simplex Table II

Contribution per unit Ci 5 3 0 0 0 0 Replacement Ratio

Ci

Variable Quantity

x1 x2 S1 S2 S3 S4 Qty./Value of Key Column

0 S1 3 0 1 1 0 -1 0 = 3

0 S2 6 0 3 0 1 -2 0 = 2

5 x1 3 1 0 0 0 1 0 =

0 S4 3 0 1 0 0 0 1 = 3

Total Contribution (Zi)

15 5 0 0 0 5

Opportunity Loss (Ci – Zj) 0 3 0 0 -5 0

Key Column Key Element Key Row

Hence the outgoing variable is S2 and the incoming variable is x2.

Step 7 : Replacing the outgoing variable (S2) by incoming variable (x2) together with its contribution per unit.

Step 8 : Calculating the new values of Key Row as under:

Step 9 : Calculation of the new values of 1st Non Key Rows:

Ist Non Key RowA Old Values 3 0 1 1 0 -1 0B New Values of Key

Row2 0 1 0 0

C Key Column Element 1 1 1 1 1 1 1D Product of B & C 2 1 1 1 0

E New Values (A – D) 1 -1 0 0 0

IInd Non Key RowA Old Values 3 1 0 0 0 1 0B New Values of Key

Row2 0 1 0 0

C Key Column Element 0 0 0 0 0 0 0D Product of B & C 0 0 0 0 0 0 0E New Values (A – D) 3 1 0 0 0 1 0

IIIrd Non Key RowA Old Values 3 0 1 0 0 0 1B New Values of Key

Row2 0 1 0 0

C Key Column Element 1 1 1 1 1 1 1D Product of B & C 2 0 1 0 0

E New Values (A – D) 1 0 0 0 0

Step 10 : Preparing Simplex Table III.

Contribution per unit Ci 5 3 0 0 0 0

Ci

Variable Quantity

x1 x2 S1 S2 S3 S4

0 S1 1 -1 0 0 0

3 x2 2 0 1 0 0

5 x1 3 1 0 0 0 1 0

0 S4 1 0 0 0 1

Total Contribution (Zj)

21 5 3 0 1 3 0

Opportunity Loss (Ci

– Zj)0 0 0 -1 -3 0

Optimal Solution: Since all Ci - Zj are 0, the solution is optimum with x1 = 3 and x2 = 2. Thus, value of Z = 3 x 5 + 2 x 3 = 21.

PROBLEM 5Use Simplex Method to solve the following LP Problem:

Maximise Z = 2x1 + 3x2

Subject to constraints:- x1 + 2x2 4x1 + x2 6x1 + 3x2 9x1, x2 0

Solution:Step 1 : Formulation of LP problem after introducing slack variables:

Max. 2x1 + 3x2 + 0S1 + 0S2 + 0S3

Subject to Constraints-x1 + 2x2 + S1 = 4x1 + x2 + S2 = 6x1 + 3x2 + S3 = 9

Step 2 : Preparing Initial Simplex Table I

Simplex Table I

Contribution per unit Ci 2 3 0 0 0 Replacement Ratio

Ci

Variable Quantit

yx1

x2 S1 S2

S3 Qty./Value of Key Column

0 S1 4 -1

2 1 0 0 = 2

0 S2 6 1 1 0 1 0 = 6

0 S3 9 1 3 0 0 1 = 3

Total Contribution (Zj)

0 0 0 0 0 0

Opportunity Loss (Ci – Zj) 2 3 0 0 0

Key Column Key Element Key Row

Hence the outgoing variable is S1 and the incoming variable is x2.Step 3 : Replacing the outgoing variable (S1) by incoming variable (x2) together with its contribution per unit.Step 4 : Calculating the new values of key Row as under:

Step 5 : Calculation of the new values of other Non – Key Row:Ist Non Key Row

A Old Values 6 1 1 0 1 0

B New Values of Key

Row

2 1 0 0

C Key Column Element 1 1 1 1 1 1

D Product of B & C 2 1 0 0

E New Values (A – D) 4 0 1 0

IInd Non Key RowA Old Values 9 1 3 0 0 1

B New Values of Key

Row

2 1 0 0

C Key Column Element 3 3 3 3 3 3

D Product of B & C 6 3 0 0

E New Values (A – D) 3 0 0 1

Step 6 : Preparing Second Simplex Table II.

Simplex Table II

Contribution per unit Ci 2 3 0 0 0 Replacement Ratio

Ci

Variable Quantit

yx1 x2 S1 S2 S3 Qty./Value of

Key Column

3 x1 2 1 0 0 = -4

0 S2 4 0 1 0 =

0 S3 3 3 0 0 1 =

Total Contribution (Zj)

6 3 0 0

Opportunity Loss (Ci – Zj) 0 0 0

Key Column Key Element Key Row

Hence, the outgoing variable is S3 and the incoming variable is x1.

Step 7 : Replacing the outgoing variable (S3) by incoming variable (x1) together with its contribution per unit.Step 8 : Calculating the new values of Key Row as under:

Step 9 : Calculation of the new values of Non – Key Rows:Ist Non Key Row

A Old Values 2 1 0 0

B New Values of Key Row

1 0 0

C Key Column Element

D Product of B & C 0 0

E New Values (A – D) 0 1 0

IInd Non Key RowA Old Values 4 0 1 0

B New Values of Key Row

1 0 0

C Key Column Element

D Product of B & C 0 0

E New Values (A – D) 0 0 1

Step 10 : Preparing Simplex Table III

Simple Table III

Contribution per unit Ci 2 3 0 0 0

Ci

Variable Quantit

yx1 x2 S1 S2 S3

3 x2 0 1 0 0

0 S2 0 0 1 0

2 x1 1 0 0 1

Total Contribution (Zj)

10.20 2 3 0

Opportunity Loss (Ci – Zj) 0 0 0

Optimal Solution: Since all Ci - Zj are 0, the solution is optimum

at x1 = and x2 = . Thus value of Z =

CHAPTER - 6ASSIGNMENT PROBLEMS

Assignment Problem is a special type of linear programming problem where the objective is to minimise the cost or time of completing a number of jobs by a number of persons.

The assignment problem in the general form can be stated as follows:“Given n facilities, n jobs and the effectiveness of each facility for each job, the problem is to assign each facility to one and only one job in such a way that the measure of effectiveness is optimised (Maximised or Minimised)”.

Hungarian Method:-Although an assignment problem can be formulated as a linear programming problem, it is solved by a special method known as Hungarian Method because of its special structure.

Balanced Assignment Problem:-Balanced Assignment Problem is an assignment problem where the number of facilities is equal to the number of jobs.

Unbalanced Assignment Problem:-Unbalanced Assignment problem is an assignment problem where the number of facilities is not equal to the number of jobs. To make unbalanced assignment problem, a balanced one, a dummy facility(s) or a dummy job(s) (as thee case may be) is introduced with zero cost of time.

Practical Steps Involved in Solving Minimisation Step 1 : See whether the number of Rows are equal to Number of Column. If yes, problem is balanced one; if not, then add a Dummy Row or Column to make the problem a balanced one; if not, then add a Dummy Row or Column to make the problem a balanced one by allotting zero value or specific value (if any given) to each cell of the Dummy Row or Column, as the case may be.Step 2 : Row Subtraction: Subtract the minimum element of each row from all elements of that row.Step 3 : Column Subtraction: Subtract the minimum element of each column from all elements of that column.

Note: If there is zero in each column, there is no need for column subtraction.

Step 4 : Draw minimum number of Horizontal and/or Vertical Lines to cover all zeros.

To draw minimum number of lines the following procedure may be followed:

1. Select a row containing exactly one uncovered zero and draw a vertical line through the column containing this zero and repeat the process till no such row is left.

2. Select a column containing exactly one uncovered zero and draw a horizontal line through the row containing the zero and repeat the process till no such column is left.

Step 5 : If the total lines covering all zeros are equal to the size of the matrix of the Table, we have got the optimal solution: if not, subtract the minimum uncovered element from all uncovered elements and add this element to all elements at the intersection point of the lines covering zeros.Step 6 : Repeat Steps 4 and 5 till minimum number of lines covering all zeros is equal to the size of the matrix of the Table.Step 7 : Assignment: Select a row containing exactly one unmarked zero and surround it by ‘ ‘ and draw a vertical line through the column containing this zero. Repeat this process till no such row is left; then select a column containing exactly one unmarked zero and surround it by ‘ ‘ and draw a horizontal line through the row

containing this zero and repeat this process till no such column is left.Step 8 : Add up the value attributable to the allocation which shall be the minimum value.Step 9 : Alternate Solution: If there are more than one unmarked zero in any row or column, select the other one (i.e., other than the one selected in Step 7) and pass two lines horizontally and vertically. Add up the value attributable to the allocation which shall be the

minimum value.

MINIMIZATION PROBLEMS

PROBLEM 1Carefree Corporation has four plants each of which can manufacture any one of the four products. Product costs differ from one plant to another as follow:

Plant Product1 2 3 4

A 33 40 43 32

B 45 28 31 23C 42 29 36 29D 27 42 44 38

You are required:(a) to obtain which product each plant should produce to minimise

cost,(b) to build a Linear Programming Model.

Solution:Part (a)Step 1 : Row Deduction: Subtracting the minimum element of each row from all the elements of that row.

1 8 11 022 5 8 013 0 7 00 15 17 11

Step 2 : Column Deduction: Subtracting the minimum element of each column of above matrix from all the elements of that column and then drawing the minimum number of lines (whether horizontal/vertical) to cover all the zeros:

1 8 4 022 5 1 013 0 0 00 15 10 11

Since number of lines drawn = 3 and order of matrix = 4, we will have to take the step to increase the number of zeros.

Step 3 : Subtracting the minimum uncovered element (1 in this case) from all the uncovered elements and adding to the elements at intersection points, and then drawing the minimum numbers of lines to cover all zeros.

0 7 3 021 4 0 013 0 0 10 15 10 12

Since the number of lines drawn (4) = order of matrix (4), the above matrix will provide the optimal solution.

Step 4 : Assignment: Selecting a row containing exactly one unmarked zero and surrounding it by ‘’ and draw a vertical line thorough the column containing this zero. Repeating this process till no such row is left; then

selecting a column containing exactly one unmarked zero and surrounding it by ‘’ and draw a horizontal

line through the row containing this zero and repeating this process till no such column is left.

0 7 3 0

21 4 0 0

13 0 0 11

0 15 10 12

Step 5 : Computing the minimumThus, the optimal assignment pattern is as follows:

Plant

Product

Cost

A 4 Rs. 32B 3 Rs. 31C 2 Rs. 29D 1 Rs. 27

Rs. 119

Part (b)Formulation of the LP model for the given assignment problem:

33 x11 40 x12 43 x13 32 x14

45 x21 28 x22 31 x23 23 x24

42 x31 29 x32 36 x33 29 x34

27 x41 42 x42 44 x43 38 x44

Minimise Z = 33x11 + 40x12 + 43x13 + 32x14 + 45x21 + 31x23 + 23x24 + 42x31 +

29x32 + 36x33 + 27x41 + 42x42 + 44x43 + 38x44

Subject to x11 + x12 + x13 + x24 = 1x21 + x22 +x23 + x24 = 1x31 + x32 + x33 + x34 = 1

x41 + x42 + x43 + x44 = 1x11 + x21 + x31 + x41 = 1x12 + x22 + x32 + x42 = 1x13 + x23 + x33 + x43 = 1x14 + x24 + x34 + x44 = 1& xij 0

PROBLEM 2The XYZ company has five jobs A, B, C, D, E to be done and five men L, M, N, O, P to do these jobs. The number of hours each man would take to accomplish each job is given by the following table:

L M N O PA 4 6 11 16 9B 5 8 16 19 9C 9 13 21 21 13D 6 6 9 11 7E 11 11 16 26 11

Required: Find the optimal schedule with time for the above assignment problem.

Solution:Step 1 : Row Subtraction: Subtracting the minimum element of each row from all elements of that row.

0 2 7 12 50 3 11 14 40 4 12 12 40 0 3 5 10 0 5 15 0

Step 2 : Column Subtraction: Subtracting the minimum element of each column from all elements of that column and then drawing the minimum number of lines to cover all zeros.

0 2 4 7 50 3 8 9 40 4 9 7 40 0 0 0 10 0 2 10 0

Since number of lines = 3 and order of matrix = 5. we will have to take step to increase the number of zeros.

Step 3 : Subtracting the minimum uncovered element (2 in the case) from all the uncovered elements and adding to the elements at the intersection poinl of two lines and then drawing the minimum

number of lines to cover all zeros.0 0 2 5 3

0 1 6 7 20 2 7 5 22 0 0 0 12 0 2 10 0

Since number o!' lines = 4 and order of matrix = 5. we will have to take step to increase the number of zeros.Step 4 : Subtracting the minimum uncovered element (1 in the case) from all the uncovered elements and adding to the elements at the intersection point of two lines and then drawing the minimum number of lines to cover all zeros.

1 0 2 5 30 0 5 6 10 1 6 4 13 0 0 0 13 0 2 10 0

Since number of lines = 4 and order of matrix - 5, we will have to take step to increase the number of zeros, we will have to take step to increase the number of zeros.

Step 5 : Subtracting the minimum uncovered element (I in the case) from all the uncovered elements and adding to the elements at the

intersection point of two lines and then drawing the minimum number of lines to cover all zeros.

1 0 1 4 20 0 4 5 00 1 5 3 04 1 0 0 14 1 2 10 0

Since number of lines = 4 and order of matrix = 5, we will have to take step to increase the number of zeros.

Step 6 : Subtracting the minimum uncovered element (1 in this case) from all the uncovered elements and adding to the elements at the intersection points of two lines and then drawing the minimum

number of lines to cover all zeros.10 0 0 3 20 0 3 4 00 1 4 2 05 2 0 0 24 1 1 9 0

Since number of lines drawn (5) = order of matrix (5), the above matrix will provide optimal solution.

Step 7 : Assignment: Selecting a row containing exactly one unmarked zero and surrounding it by '0' and draw a vertical line thorough the column containing this zero. Repeating this process till no such row is left; then selecting a column containing exactly one

unmarked zero and surrounding it by ' D' and draw a horizontal line through the row containing this zero and repeating the process till no such column is left.

1 0 0 3 2

0 0 3 4 0

0 1 4 2 0

5 2 0 0 2

4 1 1 9 0

Step 8 : Computing the minimum timeJob Men TimeA N 11B M 8C L 9D O 11E P 11

50

PROBLEM 3The assignment problem represented by the following effective matrix.

a b c d e fA 9 22 58 11 19 27B 43 78 72 50 63 48C 41 28 91 37 45 3D 74 42 27 49 39 32E 36 11 57 22 25 18F 3 56 53 31 17 28

Solution:Step 1 : Row Subtraction: Subtracting the smallest element of each row from all the elements of that row.

a b c d e fA 0 13 49 2 10 18

B 0 35 29 7 20 5C 13 0 63 9 17 5D 47 15 0 22 12 5E 25 0 46 11 14 7F 0 53 50 28 14 25

Step 2 : Column Subtraction: Subtracting the smallest element of each column from all the elements of that column and then drawing the minimum number of lines to cover all zeros.

0 13 49 2 10 180 35 29 7 20 513 0 63 9 17 547 15 0 22 12 525 0 46 11 14 70 53 50 28 14 25

Since number of lines drawn (5) order of matrix (6), we will have to take the step to increase the no. of zeros.

Step 3 : Subtracting the minimum uncovered element (2 in this case) from all uncovered elements and adding it to the elements at the interaction points of two liens and then drawing the minimum no. of lines to cover all zeros.

2 15 51 0 0 150 35 29 3 8 013 0 63 5 5 047 15 0 18 0 025 0 40 7 2 26 53 50 24 2 20

Since number of lines drawn (5) order of matrix (6), we will have to take the step to increase the no. of zeros.

Step 4 : Subtracting the minimum uncovered element (2 in this case) from all uncovered elements and adding it to the elements at the intersection points of two lines and then drawing the minimum no. of lines to cover all zeros.

4 7 51 0 0 170 35 27 1 6 013 0 61 3 3 049 17 0 18 0 225 0 40 5 0 20 53 48 22 0 20

Since the number of lines drawn (6) = order of matrix (6), the above matrix will provide the optimal solution.

Step 5 : Assignment: Selecting a row containing exactly one unmarked zero and surrounding it by ‘’ and draw a vertical line

thorough the column containing this zero. Repeating this process till no such row is left; then selecting a column containing exactly one unmarked zero and surrounding it by ‘’ and draw a horizontal line through the row containing this zero and repeating the process till no such column is left.

a b c d e f

A 4 17 51 0 0 17

B 0 35 27 1 6 0

C 13 0 61 3 3 0

D 49 17 0 18 0 2

E 25 0 44 5 0 2

F 0 53 48 22 0 20

Step 6 : Computing the minimumColum

nRow Cost

A d 11B f 48C b 28D c 27E d 25F a 3

142

Note: Alternative solution also exists.

See whether Total Requirements are equal to Total Availability; if yes, go to Step 2; if not, introduce a Dummy Origin / Destination, as the case may be, to make the problem a balanced one, taking Transportation Cost per unit as zero for each Cell of Dummy Origin / Destination or as otherwise indicated

Find Initial Feasible Solution by following either the Least Cost Method (or LCM) or North-West Corner Method (or NWCM) or Vogel’s Approximation Method (or VAM), (Refer to Page 5.5, 5.11 and 5.15)

After obtaining the Initial Feasible Solution Table, see whether Total Number of Allocations are equal to “m + n – 1”, ; if yes, go to Step 4; if not, introduce an infinitely small quantity ‘e’ to the Least Cost Independent Cell, (i.e., for which no Loop can be formed).Note: Introduce as many number of ‘e’ as the total number of allocated cells falls below “m + n – 1”.

Optimality Test: Carry out the Optimality Test on the Initial Solution Table to find out the optimal solution. (Refer to page 5.21).

Calculate the Total Minimum Cost = (Xij x Cij),Where,X = Units Allocated to a Cell;

C = Shipping Cost per Unit of a Cell;i = Row Number;j = Column Number

CHAPTER - 6

TRANSPORTATION PROBLEMS_________________________________________________________

Practical Steps Involved in Solving Transportation Problems of Minimisation Type:The practical steps involved in solving transportation problems of minimsation type are given below:

Step 1

Step 2

Step 3

Step 4

Step 5

Mathematical Model for Transportation ProblemProblem 1 [Cost Matrix or Minimisation Problem]

Build the mathematical model for the following transportation problem:

Cost Matrix SupplyW1 W2 W3 W4

F1 1 2 4 4 6F2 4 3 2 0 8F3 0 2 2 1 10

Demand 4 5 8 6

Wj Warehouse, Fi Factory and cell entries are unit / costs

Solution:Step 1 : Introducing dummy warehouse with zero cost per unit as the toal demand is not equal to total supply in order to make the problem balanced one.

Let x0 represent the quantity transported from Fi to Wj

Formulation of the mathematical model for the transportation problems.

W1 W2 W3 W4 W5

Supply

F1 1 x11 2 x12 4 x13 4 x14 0 x15 6x23

F2 4 x21 3 x22 2 x23 0 x24 0 x25 8

F3 0 x31 2 x32 2 x33 1 x34 0 x35 10

Demand 4 5 8 6 1

Minimise Z = 1x11 + 2x12 + 4x13 + 4x14 +0x15 + 4x21 + 3x22 + 2x23 + 0x24 + 0x25

0x31 + 2x32 + 2x33 + 1x34 + 0x35

Subject to x11 + x12 + x13 + x14 + x15 = 6x21 + x22 + x23 + x24 + x25 = 8x31 + x32 + x33 + x34 + x35 = 10

x11 + x21 + x31 = 4 x12 + x22 + x32 = 5 x13 + x23 + x33 = 8 x14 + x24 + x34 = 6 x15 + x25 + x35 = 1

+ xij 0

PROBLEM 2 [Profit Matrix or Maximisation Problem]A multi-plant company has three manufacturing plants, A, B, and C, and two markets X and Y.

Production Cost of A, B and C is Rs. 1500, 1600 and 1700 per piece respectively Selling price in X and Y are Rs. 4400 and Rs. 4700 respectively. Demand in X and Y 3500 and 3600 pieces respectively.

Production capacity at A, B and C is 2000, 3000 and 4000 pieces respectively.

Transportation costs are as follows:From /

ToX Y

A 1000 1500B 2000 3000C 1500 2500

Build a mathematical model.

Solution:Step 1 : Introducing dummy market with zero profit as the total demand is not equal to total supply in order to make the problem balanced one.Calculating the profit as follows:Profit = Selling Price – Cost of Production – Transportation Cost

Let x0 represent the quantity transported from factory i to market j.

X Y Z Supply

A x11 x12 x13 20001900 1700 0

B x21 x22 x23 3000

From 800 100 0C x31 x33 x33 400

01200 500 0

Demand 3500 3600 1900

This mathematical model is formulated below from the above matrix.Maximise Z = 1900x11 + 1700x12 + 0x12 + 800x21 + 100x32 + 0x23 + 1200x31 + 500x32 + 0x33

Subject to x11 + x12 + x13 = 2,000x21 + x22 + x23 = 3,000x31 + x32 + x33 = 4,000

x11 + x21 + x31 = 3,500x12 + x22 + x32 = 3,600x13 + x23 + x33 = 1,900

xij 0

Make maximum possible Allocation to the Least Cost Cell depending upon the demand / supply for the Column / Row containing that Cell. In case of Tie in the Least Cost Cells, make allocation to the Cell by which maximum demand or capacity is exhausted

Make allocation to the Second Lowest Cost Cell depending upon the remaining demand / supply for the Row / Column containing that Cell

Repeat the above Steps till all Rim Requirements are exhausted i.e., entire demand and supply is exhausted

METHOD I : LEAST COST METHOD (OR LCM)The practical steps involved in the Least Cost Method are given below:

Step 1

Step 2

Step 3

PROBLEM 3 (Case of Cost Matrix)Find the initial basic feasible solution by Least Cost Method.

Wi W2 W3 W4 SuppliesF1 48 60 56 58 140F2 45 55 53 60 260F3 50 65 60 62 360F4 52 64 55 61 220

Demand 200 320 250 210Wj Warehouse, Fi Factory and cell entries are unit costs.

Solution:Initial Feasible Solution by Least Cost Method (LCM)

W1 W2 W3

W4 Supply

48 60 56 58140

F1 140

45200

55 5360

60F2 260

50 65320

60 6240

F3 360

52 64 55190

6130

F4 220Demand 200 320 250 210

PROBLEM 4 [In Case of Contribution Matrix]Find the initial feasible solution by Least Cost Method.

W1 W2 W3 W4 SuppliesF1 48 60 56 58 140F2 45 55 53 60 260F3 50 65 60 62 360F4 52 64 55 61 220

Demand 200 320 250 210

Note: Cell entries are the unit contributions.

Solution:Step 1: Deriving Loss Matrix by deducing each element from the maximum element (i.e. 65) in order to use minimisation technique and finding initial feasible solution by LCM.

W1 W2 W3 W4 Supply

17 5 9140

7F1 140

20200

10 1260

5F2 260

15 0320

5 340

F3 360

13 1 1050

4170

F4 220

Demand 200 320 250 210

Make maximum possible allocation to the Upper – Left Corner Cell (also known as North – West Corner Cell) in the First Row depending upon the availability of supply for that Row and demand requirement for the Column containing that Cell.Note: Unit transportation cost is completely ignored.

Move to the Next Cell of the First Row depending upon remaining supply for that Row and the demand requirement for the next Column. Go on till the Row total is exhausted.

Move to the next Row and make allocation to the Cell below the Cell of the preceding Row in which the last allocation was made and follow Steps 1 and 2.

Follow Steps 1 to 3 till all Rim requirements are exhausted, i.e., the entire demand and supply is exhausted.

METHOD II : North – West Corner Method (or NWCM)The practical steps involved in the North – West Corner Method are given below:

Step 1

Step 2

Step 3

Step 4

PROBLEM 5 [Case of Cost Matrix]Find the initial feasible solution by North – West Corner Method.

W1 W2 W3 W4 SuppliesF1 48 60 56 58 140F2 45 55 53 60 260F3 50 65 60 62 360F4 52 64 55 61 220

Demand 200 320 250 210Wj Warehouse, Fi Factory & cell entries are unit costs.

Solution:

Initial Feasible Solution by North West Corner Method (NWCM)W1 W2 W3 W4 Supplie

sF1

48140

60 56 58 140F2

4560

55200

53 60 260F3

50 65120

60240

62 360F4

52 64 5510

61210

220Deman

d200 320 250 210

PROBLEM 6 [Introduction of Dummy Destination in Case of Cost Matrix]Find the initial feasible solution by North – West Corner Method.

W1 W2 W3 SuppliesF1 48 60 56 140F2 45 55 53 260F3 50 65 60 360F4 52 64 55 220

Demand 200 320 250Note: Cell entries are the unit transportation costs.

Solution:Introducing a Dummy warehouse with zero cost per unit as the total demand is not equal to total supply in order to make the problem balanced one.

Initial Feasible Solution by North West Corner Method (NWCM)

W1 W2 W3 W4 Supplies F1

48140

60 56 0140

F245

6055

20053 0

260

F350 65

12060

2400

360

F452 64 55

100

210 220

Demand

200 320 250 210

Smallest element of each Row, representing the Opportunity Cost of not making the allocation to the Smallest Element Cell, and write the difference on the right-hand side of the concerned Row. In case of tie between two smallest elements, the difference should be taken as zero.

Smallest element of each column, representing the Opportunity Cost of not making the allocation to the Smallest Element Cell, and write the difference below the concerned Column. In case of tie between two smallest elements, the difference should be taken as zero.

Indicating the allocation to be made to the row/column having largest difference. Allocate maximum possible quantity to the Least Cost Cell of the Selected row/column depending upon the quantity available. In case of tie between the Differences, select the row or column having least cost cell. However, in case of tie even in case of Least Cost, make allocation to that Cell by which maximum requirements are exhausted.

Shade the Row / Column whose availability or requirement is exhausted so that it shall not be considered for any further allocation.

Repeat Step 3 and 4 till entire demand and supply is exhausted

Draw the Initial Feasible Solution Table obtained after the above

METHOD III : Vogel’s Approximation Method (or VAM)

The practical steps involved in Vogel’s Approximation Method (or VAM) are given below:

Step 1 Row Difference: Find the difference between Smallest and Second

Step 2 Column Difference: Find the difference between Smallest and Second

Step 3 Make the Largest Difference amongst all Differences by an arrow

Step 4

Step 5

Step 6

PROBLEM 7 [Case of Cost Matrix]Find the initial feasible solution by Vogel’s Approximation (VAM) Method.

W1 W2 W3 W4 SuppliesF1 48 60 56 58 140F2 45 55 53 60 260F3 50 65 60 62 360F4 52 64 55 61 220

Demand 200 320 250 210Wj Warehouse, Fi Factory & cell entries are unit costs.

Solution:Initial Feasible Solution by Vogel’s Approximation Method

(VAM)

W1 W2 W3 W4 Supply D1 D2 D3 D4 D5F1

48 60

6056

3058

50 140 8 2 2 2 2

F2

45 55

26053 60

260 8 2 2 - -

F3

50200

65

60 62160 360 10 2 2 2 2

F4

52 64

55220

61220 3 6 - - -

Demand 200 320 250 210D1 3 5 2 2D2 - 5 2 2D3 - 5 3 2D4 - 5 4 4D5 - - 4 4

PROBLEM 8 [Introduction of Dummy Destination in Case of Cost Matrix]Find the initial feasible solution by Vogel’s Approximation Method.

W1 W2 W3 SuppliesF1 48 60 56 140F2 45 55 53 260F3 50 65 60 360F4 52 64 55 220

Demand 200 320 250

Solution:Initial Feasible Solution by Vogel’s Approximation Method (VAM)

W1 W2 W3 W4 Supply D1 D2 D3 D4 D5F1

48 60

6056

800 140 48 8 4 4 4

F2

45 55

26053 0 260 45 8 2 2 -

F3

50200

65

60 0 360 50 10 5 5 5

F4

52 64

55 0210

220 52 3 9 - -

Demand 200 320 250 210D1 3 5 2 0D2 3 5 2 -D3 - 5 2 -D4 - 5 3 -D5 - 5 4 -

APPLICATION OF OPTIMALITY TESTThe practical steps involved in Optimality Test are given below:

Step 1

Step 2

Step 3

Compute “Ui” and “Vj” for all Rows and Columns respectively on the basis of Allocated Cells such that Cij = Ui + Vj after taking any Ui or Vj = 0,Where Cij = Shipping Cost per unit of Occupied Cell;

i = Row Numberj = Column Number;Ui = Shipping Cost per unit of Supplying

Station; andVj = Shipping Cost per unit of Receiving

Station.Note: While taking any Ui or Vj = 0, that row or column which is having maximum allocated cells, should preferably be selected

Compute Opportunity Cost, or say OC, for Unallocated Cells where, OC = Cij – (Uij + Vij)

If ‘OC” of each cell is either positive or zero, Initial Feasible Solution is the Optimal Solution. However, if ‘OC’ for any Cell is negative, Initial Feasible Solution is not optimal. In that case, Find Closed Loop (explained on page 5.24) for the Cell having negative ‘OC’ and transfer entire quantity from the Allocated Cell having minimum quantity, that is covered by that Loop amongst all Allocated Cells covered by that Loop, to the Unallocated Cell having negative “OC” (the procedure to transfer the quantity has been explained on page 5.24).Note: The above procedure will be followed even in case ‘OC’ of other Unallocated Cells is positive to get Alternate Solution.

Step 4

Step 5

PROBLEM 9 [When the number of allocation = “M + N – 1”]Test the following initial solution for optimality

W1 W2 W3 W4 Supplies

F148 60

6056

3058

50140

F245 55

26053 60 260

F350

20065 60 62

160360

F452 64 55

22061 220

Demand

200 320 250 210

Note: Cell entries are transportation costs per unit.

Solution:Step 1 : Calculation of Ui and Vj on the basis of costs of allocated cells.

W1 W2 W3 W4 Ui

F1 60 56 58 U0 = 0F2 55 U1 = -5F3 50 62 U2 = 4F4 55 U3 = -1Vj V0 = 46 Vi = 60 V2 = 56 V3 = 58

Step 2 : Writing Cij Matrix of Costs for unallocated cells

W1 W2 W3 W4F1 48F2 48 53 60F3 65 60F4 52 64 61

See whether total number of allocated cells after Step 3 is equal to “m + n – 1”; if yes, go to step 5 if not introduce an infinitely small quantity ‘e’ to the Least Cost Independent Cell, i.e., for which no Loop can be formed.Note: Introduce as many number of ‘e’ as the total number of Allocated Cells falls below “m + n – 1”

Repeat Steps 1 to 4 till ‘OC’ of all Unallocated Cells is positive

Put a ‘Tick’ mark in the most negative Opportunity Cost Cell. In case of tie, any one may be selected arbitrarily. Preferably select that one with which corners points is minimum

Draw at least four lines in the form of a rectangle covering at least four Cells of which One Cell will be the ticked Cell and the rest will be the Allocated Cells. This is called ‘Loop’. In other words all the corners (except the starting corner which lies in most negative unallocated cell) of the Loop will lie in the allocated cells.

Step 3 : Ui + Vj Matrix for unallocated CellsW1 W2 W3 W4

F1 46F2 41 51 53F3 64 50F4 45 59 57

Step 4 : ij Matrix where ij = Cij – (Uij + Vij)W1 W2 W3 W4

F1 2F2 4 2 7F3 1 0F4 7 5 4

Step 5 : Since all ij are positive, the above solution is optimal. The optimal solution is given below:

Factory Warehouse Qty. Cost per Unit

Total Cost

F1 W2 60 60 3,600F2 W2 260 55 14,300F3 W1 200 50 10,000F4 W3 220 55 12,100

40,000

LOOPING AND REALLOCATION MATRIXThe practical steps involved in the Looping are given below:

Step 1

Step 2

Put ‘+’ sign on ticked cell and ‘-‘ sign on the next allocated corner cell covered by the loop and ‘+’ sign on the next to next allocated corner cell covered by the loop and so on.In other words, ‘+’ or ‘-‘ sign should be put in each allocated corner cell in alternative order.

Transfer the entire quantity from the Least Allocated Cell (i.e. cell having the least quantity) which has ‘-‘ to the Cells containing ‘+’ and deduct the transferred quantity from the other cell containing ‘-‘.Note: A Loop can even contain more than Four Cells; however, Number will always remain ‘even’. In that case, one Cell will be the Unallocated Cell having most negative Opportunity Cost and the others will be the Allocated Cells.

Step 3

Step 4

PROBLEM 10Prepare a relocation matrix from the following information

1 2 3 4 5 1 2 3 4 5A 8 A 34 32 87 23B 3 4 B 7 69 71C 6 3 C 14 1 11D 2 1 D -2 30 70E 5 E E 2 63 11

Solution:Step 1 : Ticking the most negative cell (B, 4) of cell evaluation matrix from where looping is to be started and putting quantity therein and drawing lines in such a way that turning point of the line is in the allocated cells.

1 2 3 4 5

A8

B3+ 4-

C6- 3+

D 2 1-

E5- 6+

Step 2 : Calculating new quantity of each corner of the loop after subtracting from and adding to 1 quantity (being the least one in negative corner cells) to each of the loop.Corner (D, 1) = + 1 = 1Corner (C, 1) = 6 – 1 = 5Corner (C, 2) = 3 + 1 = 4Corner (E, 2) = 5 – 1 = 4Corner (E, 5) = e + 1 = 1Corner (B, 5) = 4 – 1 = 3Corner (B, 4) = 3 + 1 = 4

Step 3 : Now Rellocation Matrix becomes as under:

1 2 3 4 5

A 8

B 4 3

C 5 4

D 1 2

E 4 1

MINIMISATION PROBLEMS (NOT INVOLVING LOOPING

PROBLEM 11A company must ship from 3 factories to 7 warehouses. The transportation cost per unit from each factory to each warehouse, the requirements of each warehouse, and the capacity of each factory are:

Warehouses Factories Warehouses

requirements

1 2 3

A 6 11 8 100B 7 3 5 200C 5 4 3 450D 4 5 6 400E 8 4 5 200F 6 3 8 350G 5 2 4 300

Factory Capacity

600 400 1,000

Find the minimum cost schedule.

Solution:Step 1 : Finding Initial Feasible Solution by following the Vogel’s Approximation Method (or VAM)

6100

11 8100 2/2/2/2/2/2/2

7 3 5200 200/0 2/2/2/2/2

5 4 3450 450/0 1/1/2/2

4400

5 6400/0 1/1/2/2/2/2

8 4 5200 200/0 1/1/3

6 3350 8

350/0 3

5100

250

4150 300/250/1

002/2/1/1/1/1/1

600/200/0 400/50/0 1000/800/350/150/0

Diff.

1/1/1/1/1/1/1 1 /1 1/1/1/1/1/2/4

Step 2 : Optimality Test: Since the total number of allocations is equal to “M + N – 1” allocations, the initial solution is straight away tested for optimality.

Let us calculate “Ui’s” and “Vj’s” for all rows and columns respectively on the basis of allocated cells such that Cij = Ui + Vj assuming V3 = 0.

1 2 3 Ui Cij Matrix for Unallocated cells

A 6 5 11 8B 5 5 7 3C 3 3 5 4D 4 3 5 6E 5 5 8 4F 3 5 6 8G 5 2 4 4Vi 1 -2 0

Ui + Vj Ui ij = Cij – (Ui + Vj) Matrix3 5 5 8 3

6 3 5 1 04 1 3 1 3

1 3 3 4 36 3 5 2 16 5 5 0 3

4Vj 1 -2 0

Step 3 : Since all ij are positive, the above solution is optimal. The optimal solution is given below:Warehous

eFactory Qty. Transportati

on Cost per unit

Total Transportati

on CostA 1 100 6 600B 3 200 5 1,000C 3 450 3 1,350D 1 400 4 1,600E 3 200 5 1,000F 2 350 3 1,050G 1 100 5 500G 2 50 2 100G 3 150 4 600

Total 7,800

Note: Since some of ij are zero, the above solution is not unique. Hence an alternative solution exists. The alternative solution can be found by taking any cell with zero ij as the basic cell.

PROBLEM 12 [Use of High Cost Where there is Restriction on Transfer]Solve the following transportation problem:

1 2 3 4 5 6 Stock Available

F 1 7 5 7 7 5 3 60AC 2 9 11 6 11 - 5 20TO 3 11 10 6 2 2 8 90RY 4 9 10 9 6 9 12 50

Demand

60 20 40 20 40 40

Note: It is not possible to transport any quantity from factory 2 to godown 5.Required: State whether the solution derived by you is unique.

Solution:Step 1 : Finding Initial Feasible Solution by following the Vogel’s Approximisation method (or VAM) Factor

y

1

Godowns Availability

Diff.1 2 3 4 5 6

7 520 7 7 5 3

40 60/40/0 2/4/0

2 910 1

16

10 1

1 5 20/10/0 1/3

3 11

10

630 2

20 2

40 8 90/70/30/

00/4/2/5

4 950 1

09 6 9 1

250/0 3/0

60 20 40 20 40 40Demand

50 10

0 0 0 0 0 0Diff. 2 5 0/1 4 3 2 Step 2 : Since the total number of allocations is less than “M + N – 1” allocations, let us introduce an infinitely small quantity ‘e’ to the least cost independent cell to make the total number of allocations equal to “M + N – 1” allocations.

Let us calculate “Ui’s” and “Vj’s” for all rows and columns respectively on the basis of allocated cells such that Cij = Ui + Vj assuming U1 = 0,

Let us calculate ij (Opportunity Cost), for unallocated cells where ij = Cij – (Uij + Vij)

Ui & Vj Matrix Cij

Ui

Matrix for Unallocated Cells

53 -2

7 7 7 5

9 6 5e

011

11

6 2 2 011

10

8

9 010

9 6 9 12

Vj

9 7 6 2 2 5

Ui + Vj Matrix for Unallocated

cells

Ui ij = Cij – (Ui + Vj) Matrix

7 4 0 0 -2 0 3 7 5

7 2 0 4 9

9 0 2 3 3

7 6 2 2 5 0 3 3 4 7 7

Vj

9

7 6 2 2 5

Step 3 : Since all ij are positive, the above solution is optimal. The optimal solution is given below:

Factory to Godown Unit Cost Value (Rs.)

1 2 20 5 1001 6 40 3 1202 1 10 9 902 3 10 6 603 3 30 6 1803 4 20 2 403 5 40 2 804 1 50 9 450

Total cost Rs.

1120

Note: Since one of ij is zero, the above solution is not unique. Hence an alternative solution exists. The alternative solution can be found by taking any cell with zero ij as the basic cell.

Students may find that the alternate solution is as given below:

Factory to Godown Unit Cost Value (Rs.)

1 1 10 7 701 2 20 5 1001 6 30 3 902 3 10 6 602 6 10 5 503 3 30 6 1803 5 40 2 803 4 20 2 404 1 50 9 450

Total Cost Rs. 1120

CHAPTER - 7NETWORK ANALYSIS

_________________________________________________________

MEANING OF CPMA network is a graphical representation of a project, depicting

the flow as well as the sequence of well-defined activities and events. A network path consists of a set of activities that connects the networking beginning event to the network terminal event. The longest path through the network is called the critical path and its length determines the minimum duration in which the said project can be completed.

TUNEFULNESS OF CPMCPM plays an important role in project planning and control.1. Network indicates the specific activities required to complete a project.2. Network indicates the interdependence and sequence of specific activities.3. It indicates the start and finish time of each activity of the project.4. It indicates the critical path.5. It indicates the duration of critical path.6. It indicates those activities for which extra effort would not be beneficial in order to shorten the project duration.7. It indicates those activities for which extra effort would be beneficial in order to shorten the project duration.8. It enables the project manager to deploy resources from non-critical activities to critical activities without delaying the overall project duration.

9. It enables the project manager to assign responsibilities for each specific activity.10. It enables the project manager to allocate resources for each specific activity11. It can be used as a controlling device to monitor activities of the project by comparing the actual progress against planned progress.12. It can be used to determine possible alternative solutions.13. It can be used to determine various cost variances for initiating corrective action by comparing the actual costs against budgeted costs of the project.14. It enables the project manager to determine a revised plan and schedule of the project by recomputing the project's critical path and the slack of non- critical activities using the actual duration of activities already finished and the revised estimated duration of activities not yet finished.

ASSUMPTIONS OF CPM/PERT

1. A project can be sub-divided into a set of predicable, independent activities.

2. The precedence relationships of project activities can be completely represented by a non-cyclical network graph in which each activity connects directly into its immediate successors.

3. Activity times may be estimated either as single point estimates or as three point estimates (i.e. optimistic, pessimistic or most likely) and are independent of each other.

4. Activity duration is assured to follow the beta distribution, the standard deviation of the distribution is assumed to be 1/6 th of its range.

Mean (te) is assumed to be =

Variances in the length of a project is assumed to be = sum of variances of activities on the critical path.5. The duration of an activity is linearly related to the cost of

resources applied to the activity.

An activity or task or job is any portion of a project which consumes time or resources and has a definable beginning and ending.

Predecessor Activities: The activities which immediately come before another activity without any intervening activities are predecessor activities.

Successor Activities: The activities which follow another activity without any intervening activities are called successor activities to that activity.

EVENT (OR NODE OR CONNECTOR)

The starting and finishing point of an activity or a group of activities are called events.Tail Event : The starting point of an activity is called a tail event because it is connected to the tail of an activity. In a network, tail event is represented by symbol “i".Note : Head events should always have a number higher than that of the trail events.Merged Event: Merge event is an event which represents the joint completion of more than one activity.

Burst Event : Burst event is an event which represents the joint starting of more than one activity.

WORKING METHODOLOGY OF CRITICAL PATH ANALYSISThe working methodology of Critical Path Analysis (CPA) which includes both CPM and PERT, consists of following five steps:1. Analyze and breakdown the project in terms of specific activities

and/or events.2. Determine the interdependence and sequence of specific activities

and prepare a network.3. Assign estimate of time, cost or both to all the activities of the

network.4. Identify the longest or critical path through the network.5. Monitor, evaluate and control the progress of the project

replanning, rescheduling and reassignment of resources.

CONVENTIONS FOLLOWED IN DRAWING NETWORKThe conventions followed in drawing a network are as follows:1. Draw the arrow directing from left to right that is, time and

progress of the project flows from left to right.2. As far as possible avoid drawing arrows that cross each other.

Where ever crossing of arrows is unavoidable, bridging may be doe.

3. An activity is always bounded by two events, called the start event and the end event. No event should hang loosely on the network.

4. Each node should be numbered without ambiguity. The numbers are assigned to events in such a way that the number assigned to the ending event of an activity is greater than the number assigned to the beginning event of that activity.

Problem 1Draw the network for the following activities and find critical path and total duration of project

Activity Duration (Days)

1-2 201-3 252-3 102-4 123-4 54-5 10

Solution:

Various Paths

Duration of paths

1-2-4-5 20+12+10 = 421-2-3-4-5 20+10+5+10 =

451-3-4-5 25+5+10 = 40

Hence the critical path is 1-2-3-4-5 with project duration of 45 days.

PROBLEM 2Draw a network from the following activities and find critical path and duration of project.

Activity Duration (Days)

1-2 51-3 82-4 62-5 42-6 43-7 53-8 3

4-9 15-9 36-10 57-10 48-11 99-12 210-12 411-13 112-13 1

Solution:

Hence the critical path is 1-3-7-10-12-13 with duration of 28 days

PROBLEM 3Draw a network from the following activities and find a critical Path and duration of Project.

Activity

Duration (Days)

Various Paths

Duration of Paths

1-2-4-9-12-13 (5+6+1+2+7) = 21

1-2-5-9-12-13 (5+4+3+2+7) = 21

1-2-6-10-12-13

(5+4+5+4+7) = 25

1-3-7-10-12-13

(8+5+4+4+7) = 28

1-3-8-11-13 (8+3+9+1) = 21

1 – 2 132 – 3 83 – 4 103 – 5 114 – 6 95 – 6 105 – 7 66 – 8 87 – 8 78 – 9 149 – 10 18

Solution:

Various Paths

Duration of Paths

1-2-3-5-7-8-9-10

(13+8+11+6+7+14+18) = 77

1-2-3-5-6-8-9-10

(13+8+11+10+8+14+18) = 82

1-2-3-4-6-8-9-10

(13+8+10+9+8+14+18) = 80

Hence the critical path is 1-2-3-4-5-6-8-9-10 with total duration of 82 days.

PROBLEM 4Draw a network from the following activities and find a critical path and total project duration

Activity Duration (Days)

1 – 2 101 – 3 41 -4 62 – 3 52 – 5 122 – 6 93 – 7 124 – 5 155 – 6 66 – 7 57 – 8 47 – 8 7

Solution:

Various Paths

Duration of Paths

1–2–3–7–8 (10+5+12+7) = 34

1-2-6-7-8 (10+9+5+7) = 311-2-6-8 (10+9+4) = 23

1-2-5-6-7-8 (10+12+6+5+7) = 40

1-2-5-6-8 (10+12+6+4) = 32

1-3-7-8 (4+12+7) = 231-4-5-6-7-8 (6+15+6+5+7) =

39

1-4-5-6-8 (6+15+6+4) = 31

Hence the critical path is 1-2-5-6-7-8 with duration of 40 days.

PROBLEM 5Draw a network from the following activity and find critical path and total duration of project.

Activity Duration (Days)

1-2 32-3 52-4 83-5 43-6 24-6 94-7 35-8 126-8 107-8 6

Solution:

Various Paths

Duration of Paths

1-2-3-5-8 (3+5+4+12) = 24

1-2-3-6-8 (3+5+2+10) = 20

1-2-4-6-8 (3+8+9+10) = 30

1-2-4-7-8 (3+8+3+6) = 20

Hence the critical path is 1 – 2 – 4 – 6 – 8 with duration of 30 days.

PROBLEM 6Draw a network from the following activity and find a critical path and total project duration

Activity Duration (Days)

1-2 31-4 21-5 22-3 43-8 24-6 74-7 45-6 46-9 67-8 58-9 3

Solution:

Various Paths

Duration of Paths

1-2-3-8-9 3+4+2+3 = 121-4-7-8-9 2+4+5+3 = 141-4-6-9 2+7+6 = 151-5-6-9 2+4+6 = 12

Hence the critical path is 1-4-6-9 with duration of 15 weeks.

PROBLEM 7Draw a network from the following activity and find a critical path and total duration of project

Activity Duration (Days)

1-2 21-3 41-4 32-5 13-5 64-6 55-6 7

Solution:

Various Paths

Duration of Paths

1-2-5-6 2+1+7 = 101-3-5-6 4+6+7 = 171-4-6 3+5 = 18

Hence the critical Path is 1-3-5-6 with duration of 17 days.

PROBLEM 8Draw a network from the following activity and find a critical path and total duration of project.

Activity Duration (Days)

1-2 22-3 22-4 33-5 44-5 34-6 55-7 56-7 77-8 27-9 6

8-10 29-10 5

Various Paths

Duration of Paths

1-2-4-6-7-9-10 2+3+5+7+ +5 =

1-2-4-5-7-9-10 2+3+3+5+6 +5 =

24

1-2-4-5-7-8-10

2+3+3+5+2+2 = 17

1-2-3-5-7-9-10 2+2+4+5+6 +5 =

24

1-2-3-5-7-8-10

2+2+4+5+2+2 = 17

Hence the critical path is 1-2-4-6-7-9-10 with duration of 28 days.

PROBLEM 9Draw a network from the following activity and find a critical path and total duration of project

Activity

Duration (Days)

1-2 7.81-3 201-4 332-5 182-6 203-6 94-7 9.85-7 86-7 4

Solution:

Various Paths

Duration of Paths

1-3-6-7 20+9+4 = 331-2-6-7 7.80+20+4 =

31.801-2-5-7 7.80+18+8 =

33.801-4-7 33+9.80 =

42.80

Hence the critical path is 1-4-7 with total duration of 42.80 days.

PROBLEM 10Draw a network from the following activities and find a critical path and total duration of project.

Activity

Duration (Days)

1-2 71-6 62-3 142-4 53-5 114-5 75-8 46-7 117-8 18

Solution:

Various Paths

Duration of Paths

1-2-3-5-8 7+14+11+4 = 36

1-2-4-5-8 7+5+7+4 = 231-6-7-8 6+11+18 = 35

Hence the critical path is 1-2-3-5-8 with duration of 36 days.

PROBLEM 11Draw a network from the following activities and find a critical path and total project duration.

Activity

Duration (Weeks)

1-2 91-3 10.32-4 82-5 23-5 104-5 14-6 75-6 3

Solution:

Various Paths

Duration of Path

1-2-4-5-6 [9+8+1+3] = 21 weeks

1-2-4-6 [9+8+7] =24 weeks1-2-5-6 [9+2+3] = 14 weeks1-3-5-6 [10-3+10+3] = 23.3

weeks

Hence, the critical path is 1-2-4-6 with total duration of 24 weeks.

PROBLEM 12Draw a network from the following activities and find a critical path and total duration of project.

Activity

Duration (Days)

1-2 41-3 71-4 102-3 32-4 82-5 112-6 183-5 103-6 16

4-5 95-6 65-7 116-7 8

Solution:

Various Paths

Duration of Path

1-3-6-7 [7+16+18] = 311-3-5-6-7 [7+10+6+8] =

311-3-5-7 [7+10+11] = 28

1-2-3-6-7 [4+3+16+8] = 31

1-2-3-5-6-7 [4+3+10+6+8] = 31

1-2-3-5-7 [4+3+10+11] = 28

1-2-4-5-6-7 [4+8+9+6+8] = 35

1-2-4-5-7 [4+8+9+11] = 32

1-4-5-6-7 [10+9+6+8] = 33

1-4-5-7 [10+9+11] = 30

Hence the critical path is 1-2-4-5-6-7 with total duration of 35 days.PROBLEM 13Draw the network from the following activity and find critical path and total duration of project.

Activity

Duration (Days)

1-2 21-3 4

1-4 52-5 13-5 64-6 55-6 76-7 2

Solution:

Various Paths

Duration of Paths

1-2-5-6-7 [2+1+7+2] = 121-3-5-6-7 [4+6+7+2] = 191-4-6-7 [5+5+2] = 12

PROBLEM 14Draw a network from the following activity and find the critical path and total duration of project

Activity

Duration (Days)

1 61 5

2 43 43 44 65 6

Solution:

Various paths

Duration of path

1-2-4-6 [6+4+6] = 161-3-4-6 [5+4+6] = 151-3-5-6 [5+4+6] = 15

Hence the critical path is 1-2-4-6 wit duration of 16 days.

PROBLEM 15Draw a network from the following activity and find a critical path and total project duration.

Activity

Duration (Days)

1-2 21-3 21-4 12-5 43-6 83-7 54-6 35-8 16-9 57-8 48-9 3

Solution:

Various Paths

Duration of Path

1-2-5-8-9 2+4+1+3 = 101-3-7-8-9 2+5+4+3 = 141-3-6-9 2+8+5+151-4-6-9 1+3+5 = 9

Hence the critical path is 1-3-6-9 with total project duration of 15 days

PROBLEM 16Draw a network from the following activity and find critical path and total project duration.

Activity

Duration(Days)

1-2 3

2-3 22-4 42-5 63-6 34-7 65-8 45-9 26-8 27-8 48-10 48-10 2

Solution:

Various Paths

Duration of Paths

1-2-4-7-8-10 3+4+6+4+4 = 21

1-2-3-6-8-10 3+2+3+2+4 = 14

1-2-5-8-10 3+6+4+4 = 171-2-5-9-10 3+6+2+2 = 13

Hence the critical path is 1-2-4-7-8-10 with total project duration of 21 days.

PROBLEM 17Draw the network from the following activity and find critical path and total duration of project.

Activity

Duration(Days)

1-2 21-3 21-4 12-5 43-6 53-7 84-7 35-8 16-8 47-9 58-9 3

Solution:

Various Paths

Duration of Paths

1-3-7-9 2+8+5 = 151-3-6-8-9 2+5+4+3 = 141-2-5-8-9 2+4+1+3 = 101-4-7-9 1+3+5 = 9

Hence the critical path is 1-3-7-9 with project duration of 15 days.

PROBLEM 18Draw the network from the following activity and find critical path and total project duration:

Activity

Duration (Days)

1-2 41-3 72-4 32-6 33-4 23-5 24-5 25-6 3

Solution:Various Paths

Duration of Paths

1-2-6 4+3 = 71-2-4-5-6 4+3+2+3 = 121-3-4-5-6 7+2+2+3 = 141-3-5-6 7+2+3 = 12

Hence the critical path is 1-3-4-5-6 with total project duration of 14 days

PROBLEM 19Draw the network from the following activity and find critical path and total project duration

Activity

Duration (Days)

1-2 21-4 91-7 12-3 43-6 84-5 54-8 35-6 16-9 57-8 48-9 3

Solution:

Various Paths

Duration of Paths

1-2-3-6-9 2+4+8+5 = 191-4-5-6-9 9+5+1+5 =201-4-8-9 9+3+3 = 151-7-8-9 1+4+3 = 8

Hence the critical path is 1-4-5-6-9 with project duration of 20 days.

PROBLEM 20Draw the network from the following activity and find critical path and total project.

Activity

Duration (Days)

1-2 31-3 41-5 142-3 102-6 53-4 44-7 65-6 16-7 1

Solution:

Various Paths

Duration of Paths

1-2-6-7 3+5+1 = 91-2-3-4-7 3+10+4+6 = 231-5-6-7 14+1+1 = 16

Hence the critical path is 1-2-3-4-7 with project duration of 23 days.

PROBLEM 21Draw the network from the following activity and find critical path and total duration of project.

Activity

Duration (Days)

1-2 41-3 21-4 52-3 72-5 73-5 24-5 5

Solution:

Various Paths

Duration of Paths

1-2-5 4+7 = 111-2-3-5 4+7+2 = 131-4-5 5+5 = 10

Hence the critical path is 1-2-3-5 with project duration of 13 days.

JOB SEQUENCING PROBLEM________________________________________________________________________

Definition:-Suppose there are n jobs (1, 2, 3…., n), each of which has to be

processed one at a time at each of m machines A, B, C, …. The order of processing each job through machines is given (for example, job 1 is processed through machines A, C, B – in this order). The time that each job must require on each machine is known. The problem is to find a sequence among (n!)m number of all possible sequences (or combinations) (or order) for processing the jobs so that the total elapsed time for all the jobs will be minimum.

Mathematically, letAi = time for job I on machine A,Bi = time for job I on machine B, etc.T = time from start of first job to completion of the last job.Then, the problem is to determine for each machine a sequence of jobs i1, i2, i3, ….. in, where (i1, i2, i3, …..in) is the permutation of the integers which will minimise T.

TERMINOLOGY AND NOTATIONS

The following terminology and notations will be used in this chapter.(1) Number of Machines:

It means the service facilities through which a job must pass before it is completed.

For example, a book to be published has to be processed composing, printing, binding, etc. In this example, the book constitutes the job and the different processes constitute the number of machines.

(2) Processing Order:It refers to the order in which various machines are required for

completing the job.

PRINCIPAL ASSUMPTIONS:(1) No Machine can process more than one operation at a time.(2) Each operation, once started, must be performed till completion.(3) A job is an entity, i.e. even tough the job represents a lot of individual parts, no lot may be processed by more than one machine at a time.(4) Each operation must be completed before any other operation,

which it must precede, can begin.(5) Time intervals for processing are independent of the order in which

operations are performed.

(6) There is only one of each type of machine.(7) A job is processed as soon as possible subject to ordering

requirements.(8) All jobs are known and are ready to start processing before the

period under consideration begins.(9) The time required to transfer jobs between machines is negligible.

SOLUTION OF SEQUENCING PROBLEMA present solution of following cases is available:

1. n jobs and two machines A and B, all jobs processed in the order AB.

2. n jobs and three machines A, B and C, all jobs processed in the order ABC, other limitations are given in Section 24.6.

3. Two jobs and m machines. Each job is to be processed through the machines in a prescribed order (which is not necessarily the same for both the jobs).

4. Problems with n jobs and m-machines.

(I) Illustrative ExampleThe Johnson’s iterative procedure for determining the optimal

sequence for an n-job 2 – machine sequencing problem can be outlined as follows:

Step 1 : Examine the Ai’s and Bi’s for i = 1, 2, … n and find out min[Ai, Bi]Step 2 : (i) If this minimum be Ak for some i = k, do (process) the kth job first of all.

(ii) If this minimum be Br for some I = r, do (process) the rth job last of all.Step 3 : (i) If there is a tie for minima Ak = Br, process the kth job first of all and rth job in the last.

(ii) If the tie for the minimum occurs among the Ai’s, select the job corresponding to the minimum of Bi’s and process it first of all.

(iii) If the tie for minimum occurs among the Bi’s, select the job corresponding to the minimum of Ai’s and process it in the last. Go to next step.Step 4 : Cross-out the jobs already assigned and repeat steps 1 to 3 arranging the jobs next to fist or next to last, until the jobs have been assigned.

(II) Illustrative ExampleExample 1 There are five jobs, each of which must go through the two machines A and B in the order AB. Processing times are given (Table) below:

TableJob 1 2 3 4 5

Time for A

5 1 9 3 10

Time for B

2 6 7 8 4

Determine a sequence for five jobs that will minimise the elapsed time T.Calculate the total idle time for the machines in this period.

Solution: Apply steps I and II of solution procedure. It is seen that the smallest processing time is one hour for job 2 on the machine A. So list the job 2 at first place as shown below

2

Now, the reduced list or processing times becomes:Job A B

1 5 2

3 9 7

4 3 8

5 10 4

Again, the smallest processing time in the reduced list is 2 for job 1 on the machine B. So place job 1 last.

2 1

Continuing in the like manner, the next reduced list is obtained,Job A B

3 9 7

4 3 8

5 10 4

Leading to sequence and the list,2 4 1

Job A B

3 9 7

5 10 4

Gives rise to sequence,

2 4 5 1Finally, the optimal sequence is obtained,

2 4 3 5 1

Further, it is also possible to calculate the minimum elapsed time corresponding to the optimal sequencing, using the individual processing time given in the statement of the problem. The details are given in Table.

TableJob

Sequence

Machine A Machine B

Time in Time

out

Time in Time

out

2 0 1 1 7

4 1 4 7 15

3 4 13 15 22

5 13 23 23 27

1 23 28 28 30

Thus, the minimum time, i.e. the time for starting of job 2 to completion of the last job 1, is 30 hrs. only. During this time, the machine A remains idle for 2 hrs. (from 28 to 30 hrs) and the machine B remains idle for 3 hrs only (from 0-1, 22-23, and 27-28 hrs.)

The total elapsed time can also be calculated by using Gantt chart as follows:

From the figure it can be seen that the total elapsed time is 30 hrs, and the idle time of the machine B is 3 hrs.

In this problem, it is observed that job may be held in inventory before going to the machine. For example, 4th job will be free on machine A after 4th hour and will start on machine B after 7th hr. Therefore, it will be kept in inventory for 3 hrs. Here it is assumed that the storage space is available and the cost of holding the inventory for each job is either same or negligible. For short duration process problems, it is negligible. Second general assumption is that the order of completion of jobs has no significance, i.e. no job claims the priority.

(III) Illustrative Example:Example 2 There are five jobs, each of which must go through machines A, B and C in the order ABC, Processing times are given in Table.

TableJob i Processing Times

Ai Bi Ci1 8 5 42 10 6 93 6 2 84 7 3 65 11 4 5

Determine a sequence for five jobs that will minimise the elapsed time T.

[Banasthali (M.Sc.) 93; I.A.S. (Main)92; Meerut 83; Rohlikhand 81]

Solution: Here min Ai = 6, max Bi = 6 min Ci = 4Since one of two conditions is satisfied by min Ai = max Bi, so the

procedure adopted in Example 1 can be followed.The equivalent problem, involving five jobs and two fictitious

machine G and H, becomes

Table Job i Processing Times

Gi(=Ai + Bi) Hi(=Bi+Ci)1 13 92 16 153 8 104 10 95 15 9

This new problem can be solved by the procedure described earlier. Because of ties, possible optimal sequences are:

(i) 3 2 1 4 5 (ii) 3 2 4 1 5

(iii) 3 2 4 5 1 (iv) 3 2 5 4 1

(iv) 3 2 1 5 4 (vi) 3 2 5 1 4

It is possible to calculate the minimum elapsed time for first sequence as shown in Table

TableJob Machine A Machine B Machine C

Time in Time out

Time in Time out

Time in Time out

3 0 6 6 8 8 162 6 16 16 22 22 311 16 24 24 29 31 354 24 31 31 34 35 415 31 42 42 46 46 51Thus, any of the sequences from (i) to (vi) may be used to order

the jobs through machines A, B and C, and they all will give a minimum elapsed time of 51 hrs. Idle time for machine A is 9 hrs, for B 31 hrs, for C 19 hrs.

Remarks:1. If conditions min Ai max Bi and/or min Ci max Bi do not hold,

there is no general procedure yet available.2. The idle times for individual machines may be different for

alternative optimal sequences. Although, the min. elapsed time will obviously be the same for all alternative optimal sequences.

CHAPTER - 8

SIMULATION________________________________________________________________________

Definitions A simulation of a system or an organism is the operation of a model

or simulator which is a representation of the system or organism. The model is amenable to manipulation which would be impossible, too expensive or unpractical to perform on the entity it portrays. The operation of the model can be studied & for it, properties concerning the behaviour of the actual system can be inferred.

Simulation is the process of designing a model of a real system and conducting experiments with this model for the purpose of understanding the behaviour for the operation of the system.

‘X simulated Y’ is true if and only if:1) X and Y are formal systems2) Y is taken to be the real system.3) X is taken to be an approximation to the real system.4) The rules of validity in X are non-error-free otherwise X will

become the real system. Simulation is the use of a system model that has the designed

characteristics of reality in order to produce the essence of actual operation.

Steps of Simulation ProcessStep 1 : Identify the problem.Step 2 : (a) Identify the decision variable

(b) Decide the performance objective and the decision rulesStep 3 : Construct a numeral modelStep 4 : Validate the modelStep 5 : Design the experimentsStep 6 : Run the simulation modelStep 7 : Examine the results

Advantages of Simulation1. This approach is suitable to analyze large and complex real life

problems which cannot be solved by usual quantitative methods.

2. It is useful for sensitivity analysis of complex systems. It allows the decision – maker to study the interactive system variables and the effect of changes in these variables on the system performance in order to determine the desired one.

3. Simulation can be used to ‘experiment’ on a model of a real situation without incurring the costs of operating on the system.

4. Simulation can be used as a pre-service test to tryout new policies and decision rules for operating a system before running the risk of experimentation in the real system.

Disadvantages of Simulation Sometimes simulation models are expensive and take a long time to

develop.

Identify the Problem

Identify decision variables performance &

decision rules

Construct Simulation Model

Validate the Model

Design experiments

Run or conduct simulation

Is simulation process completed

Examine the results and select the best course of

action

Modify the model by changing the input data, i.e. values of

decision variables.

It is the trial and error approach that produces different solutions in repeated runs. This means it does not generate optimal solutions to problems.

Each application of simulation is ad-hoc to a great extent. The simulation model does not produce answers by itself. The user

has to provide all the constraints for the solutions which he wants to examine.

MONTE CARLO SIMULATION

The principle behind the Monte Carlo simulation technique is representative of the given system under analysis by a system described by some known probability distribution and then drawing random samples form probability distribution by means of random numbers:The Monte Carlo simulation technique consists of following steps:1. Setting up a probability distribution for variables to be analysed.2. Building a cumulative probability distribution for each random

variable.3. Generate Random numbers; Assign an appropriate set of random

numbers to represent value or range (interval) of values for each random variable.

4. Conduct the simulation experiment by means of the random sampling.

5. Repeat step 4 until the required number of simulation runs has been generated.

6. Design and Implement a course of action and to maintain control.

Role of Computers in Simulation:-The role of computers in simulation is vital. They are used to

generate random numbers, simulate the given problem with varying values of variables in few minutes and help the decision – maker to prepare reports which enable him to make decisions quickly as well as draw valid conclusions.

Computer languages available to help the simulation process can be divided into two categories:

(1) General Purpose Programming LanguagesFORTRAN, BASIC, PL/I, Pascal etc.

(2) Special Purpose Simulation Languagesi) GPSS (General Purpose System Simulation)ii) SIMSCRIPTiii) DYNAMO

PROBLEM 1 [Frequency Distribution of Probability Estimates]The investment required for introducing a new product is Rs. 10,000. The probability estimates for variable cost selling price and annual sales volume are given below:

Selling price Variable cost Annual volumeValue (Rs.)

Probability

Value(Rs.)

Probability

Value(Rs.)

Probability

3 0.20 1 0.20 4,000 0.204 0.60 2 0.50 6,000 0.505 0.20 3 0.30 8,000 0.30

Required: Simulate the situation by generating random number to find out selling price, variable cost and the annual sales volume. The random number schemes for the three factors are given below:

Selling PriceValue Cumulative

probabilityRandom number

3 0.2 00-194 0.8 20-795 1.0 80-99

Variable CostValue Cumulative

ProbabilityRandom number

1 0.2 00-192 0.7 20-693 1.0 70-99

Annual Sales VolumeValue Cumulative

probabilityRandom number

4,000 0.2 00-196,000 0.7 20-698,000 1.0 70-99

Also prepare the frequency distribution of the probability EstimatesUse the following sequence of Sixty random numbers (First 3 random numbers for the first trial etc.)97, 95, 12, 02, 92, 75, 80, 67, 14, 66, 24, 72, 86, 76, 20, 55, 64, 82, 50, 02, 74, 29, 53, 08, 58, 16, 01, 51, 16, 69, 14, 55, 36, 86, 54, 35, 24, 23, 52, 39, 36, 99, 47, 41, 41, 60, 71, 41, 65, 88, 48, 44, 74, 11, 93, 10, 95, 20, 46, 36.

Solution:

Net profit = (Selling price – Variable cost) Annual sales – Rs. 10,000 (being fixed cost)

Table : Simulation Work SheetObservati

on No.Selling Price Variable Cost Annual

volumeNet

Profit(Rs.)Rando

m No.

Value

(Rs.)

Random

No.

Value

(Rs.)

Random

No.

Value

(Rs.)1 97 5 95 3 12 4,000 -2,0002 02 3 92 3 75 8,000 -10,0003 80 5 67 2 14 4,000 2,0004 66 4 24 2 72 8,000 6,0005 86 5 76 3 20 6,000 2,0006 55 4 64 2 82 8,000 6,0007 50 4 2 1 74 8,000 14,0008 29 4 53 2 08 4,000 -2,0009 58 4 16 1 01 4,000 2,00010 51 4 16 1 69 6,000 8,00011 14 3 55 2 36 6,000 -4,00012 86 5 54 2 35 6,000 8,00013 24 4 23 2 52 6,000 2,00014 39 4 36 2 99 8,000 6,00015 47 4 41 2 41 6,000 2,00016 60 4 71 3 41 6,000 -4,00017 65 4 88 3 48 6,000 -4,00018 44 4 74 3 11 4,000 -6,00019 93 5 10 1 95 8,000 22,00020 20 4 46 2 36 6,000 2,000

Total Profit = Rs. 50,000.Average Profit = Rs. 50,000/20 = Rs. 2,500The results of 20 simulations are noted in the following table. We find that losses have occurred in some of the trials and the average profit = Rs. 2,500. The associated risk can be easily found out by taking the profitability figures and computing the associated probabilities.

Table Showing the Frequency Distribution of the Probability Estimates

Profit Frequency

Cumulative Frequency

Probability

22,000 1 1 1/2014,000 1 2 22/208,000 2 4 4/206,000 3 7 7/202,000 6 13 13/20

-2,000 2 15 15/20-4,000 3 18 18/20-6,000 1 19 19/20

-10,000

1 20 20/20

PROBLEM 2 [Simulating the Yield, Price and Revenue]The director of finance, for a farm cooperative is concerned about the yields per acre she can expect from this year’s corn crop. The probability distribution of the yields for the current weather conditions is given below:

Yield in kg. per acre

Probability

120 0.18140 0.26160 0.44180 0.12

She would like to see a simulation of the yields she might expect over the next 10 years for weather conditions similar to those she is now experiencing.

(i) Simulate the average yield she might expect per acre using the following random numbers : 20,72,34,54,30,22,48,74,76,02

She is also interested in the effect of market-price fluctuations on the cooperative’s farm revenue. She makes this estimate of per-kg prices for corn.

Price per kg. (Rs.)

Probability

2.00 0.052.10 0.152.20 0.302.30 0.252.40 0.152.50 0.10

(ii) Simulate the price she might expect to observe over the next 10 years using the following random numbers:82, 95, 18, 96, 20, 84, 56, 11, 52, 03

Solution:Table I – Random Number Coding (Yield)

Yields in kg.

per acre

Probability

Cumulative

probability

Random numbers assigned

120 0.18 0.18 00-17140 0.26 0.44 18-43160 0.44 0.88 44-87180 0.12 1.00 88-99

Table II – Random Number Coding (Price)

Price per kg.

Probability

Cumulative

Probability

Random Number Assigned

2.00 0.05 0.05 00-042.10 0.15 0.20 05-192.20 0.30 0.50 20-492.30 0.25 0.75 50-742.40 0.15 0.90 75-892.50 0.10 1.00 90-99

Table III – Simulation Work SheetA

YearB

Random No.

C(Simulated Yield)

DRandom Number

EPric

e

F = C x ERevenue/

acre1 20 140 82 2.40 3362 72 160 95 2.50 4003 34 140 18 2.10 2944 54 160 96 2.50 4005 30 140 20 2.20 3086 22 140 84 2.40 3367 48 160 56 2.30 3688 74 160 11 2.10 3369 76 160 52 2.30 36810 02 120 03 2.00 240

Total

3,386

Average revenue per acre is = Rs. 338.60

PROBLEM 3 [Simulating the Defected Items]The output of a production line is checked by an inspector for

one or more of three different types of defects, called defects A, B and C. If defect A occurs, the item is scrapped. If defect B or C occurs, the item must be reworked. The time required to rework a B defect is 15

minutes and the time required to rework a C defect is 30 minutes. The probabilities of an A, B and C defects are 0.15, 0.20 and 0.10 respectively. For ten items coming off the assembly line, determine the number of items without any defects, the number scrapped and the total minutes of rework time. Use the following random numbers:RN for defect A48 55 91 40 93 01 83 63 47 52RN for defect B47 36 57 04 79 18 10 13 57 09

RN for defect C82 95 18 96 20 8 56 11 52 03

Solution:Table I – Random Numbers Coding

Defect A Defect B Defect CWheth

er defect Exists?

Cum. Prob.

Random

numbers

assigned

Whether

defect Exists?

Cum. Prob

.

Random

numbers

assigned

Whether

defect Exists?

Cum.

Prob.

Random

numbers

assigned

Yes 0.15 00-14 Yes 0.20 00-19 Yes 0.10

00-09

No 1.00 15-99 No 1.00 20-99 No 1.00

10-99

Table II – Simulation WorksheetItem

No.

RN for

defect A

Whether

defect exist or not

RN for

defect B

Whether

defect exist or not

RN for

defect C exist

s

Whether

defect exist or not

Which defect exists

Rework time

(in minut

es

Remarks

1 48 No 47 No 82 No - - -2 55 No 36 No 95 No - - -3 91 No 57 No 18 No - - -4 40 No 04 No 96 No B 15 -5 93 No 79 No 20 No - - -6 01 No 18 No 8 No A, B, C - Scrap7 83 No 10 No 56 No B 15 -8 63 No 13 No 11 No B 15 -9 47 No 57 No 52 No - - -10 52 No 09 No 03 No B, C 15 +

30 = 45

-

During the simulated period,No. of items without any defect = 5

No. of items scrapped = 1No. of items required rework = 4Total rework time required = 90 minutes

PROBLEM 4 [Penalty Cost for Excess & Short Output]A company manufacturer 30 items per day. The sale of these items depends upon demand which has the following distribution.

Sales (Units)

Probability

27 0.1028 0.1529 0.2030 0.3531 0.1532 0.05

The production cost and selling price of each unit are Rs. 50 respectively. Any unsold product is to be disposed off at a loss of Rs. 15 per unit. There is a penalty of Rs. 5 per unit if the demand is not met.

Required : Using the following random numbers estimate total profit/loss for the company for the next 10 days: 10, 99, 65, 99, 95, 01, 79, 11, 16, 20.

If the company decides to produce 29 items per day, what is the advantage or disadvantage to the company?

Solution:Table 1 – Random Number Coding

Sales (units)

Probability

Cumulative probability

Random numbers assigned

27 0.10 0.10 00-0928 0.15 0.25 10-2429 0.20 0.45 25-4430 0.35 0.80 45-7931 0.15 0.95 80-9432 0.05 1.00 95-99

Profit + per unit = Selling Price per unit – Cost per unit= Rs. 50 – Rs. 40 = Rs. 10 per unit.

Table II – Simulation Work Sheet(i)Day

(ii)Random number

(iii)Estimate

d Sale

(iv)Profit / Loss per day

when production = 30

(v)Profit / Loss

per day when production =

items per day

29 items per day

1 10 28 28 x Rs. 10 – 2 x Rs. 15 = 250

28 x Rs. 10 – 1 x Rs. 15 =

2652 99 32 30 x Rs. 10

– 2 x Rs. 5 = 29029 x Rs. 10

– 3 x Rs. 5 = 275

3 65 30 30 x Rs. 10 = 300 29 x Rs. 10 – 1 x Rs.5 =

2854 99 32 30 x Rs. 10

– 2 x Rs. 5 = 29029 x Rs. 10

– 3 x Rs. 5 = 275

5 95 32 30 x Rs. 10 – 2 x Rs. 5 = 290

29 x Rs. 10 – 3 x Rs. 5 =

2756 01 27 27 x Rs. 10

– 3 x Rs. 15 = 22527 x Rs. 10

– 2 x Rs. 15 = 240

7 79 30 30 x Rs. 10 = 300 29 x Rs. 10 – 1 x Rs. 5 =

2858 11 28 28 x Rs. 10

– 2 x Rs. 15 = 25028 x Rs. 10

– 1 x Rs. 15 = 265

9 16 28 28 x Rs. 10 – 2 x Rs. 15 = 250

28 x Rs. 10 – 1 x Rs.15 =

26510 20 28 28 x Rs. 10

– 2 x Rs. 15 = 25028 x Rs. 10

– 1 x Rs. 15 = 265

Total Profit Rs. 2,695/- Rs. 2,695/-It is evident from this table that there is no additional profit or loss

if the production is reduced to 29 items per day since the total profit remains unchanged i.e. Rs. 2,695

Q U E S T I O N B A N K________________________________________________________________________Q.1 Define operation Research & explain briefly.

Q.2 Solve the following LPP graphicallyi) Maximise Z = 40x1 + 100x2

Subject to 12x1 + 6x2 30004x1 + 10x2 20002x1 + 3x2 900x1, x2 0

ii) Minimise Z = 4x1 + 5x2 + 2x3

Subject to 2x1 + x2 + 4x3 1503x1 + 4x2 + x3 100x1, x2, x3 0

iii) Minimise Z = 3x1 + 5x2

Subject to 3x1 + 4x2 122x1 – x2 -22x1 + 3x2 12x1 4, x2 2x1, x2 0

iv) Minimise Z = 4x + 3ySubject to 4x + 12 y 18

16x + 4y 248x + 6y 16

Q.3 Solve the following LPP by Simplex Method:i) Max Z = 3x1 + 5x2 + 4x3

Subject to constraints2x1 + 3x2 82x1 + 5x3 103x1 + 2x2 + 4x3 15x1, x2, x3 0

ii) Max Z = 6x1 + 3x2

Subject to 2x1 + x2 83x1 + 3x2 18

x3 3and x1, x2 0

iii) Minimise Z = x1 – 2x2 – 3x3

Subject to the constraints-2x1 + 3x2 + 3x3 = 22x1 +3x2 + 4x3 = 1x1, x2, x3 0

iv) Minimise Z = x1 + x2

Subject to constraints2x1 + 4x2 4

x1 + 7x2 7x1, x2 0

Q.4 Solve the following transportation problem to find its optimal solution:i)

W1 W2 W3 W4 Supply

P1 190 300 500 100 70P2 700 300 400 600 90P3 400 100 400 200 180

Demand

50 80 70 140 340

ii) Find the optimal solution for the following cost matrix

D1 D2 D3 D4 Suppl

y

O1 20 22 17 04 120

O2 24 37 09 07 70

O3 32 37 20 15 50

Deman

d

60 40 30 110 240

iii) A company has four factories F1, F2, F3, F4 manufacturing the same product. Production and raw material costs differ form factory to factor and are given in the following table in first two rows. The transportation cost from the factories to the sales deposits S1, S2, S3 are also given. The last two columns in the table give the sales price and the total requirement at each depot. The production capacity of each factory is given in the last row.

F1 F2 F3 F4 Sales price per

unit

Requirement

Production Cost/unit 15 18 14 13Raw material Cost/unit

10 9 12 9

Transportation Cost per unit

S1 3 9 5 5 34 80S2 1 7 4 5 32 120S3 5 8 3 6 31 150

Production Capacity 10 150 50 100

Determine the most profitable production and distribution schedule and the corresponding profit.

Q.5 Solve the following assignment problem for minimisation.

1 2 3 4 5A 8 8 8 1

112

B 4 5 6 3 4C 1

211

10

9 8

D 18

21

18

17

15

E 10

11

10

8 12

Q.6 Solve the following problem of assignment 4 computer programmers to 4 application programmes, where the estimated computer time in minutes required by each of them to develop the programmed is given.

ProgrammersProgrammers

1 2 3 4A 12

0100

80 90

B 80 90 110

70

C 110

140

120

100

D 90 90 80 90

Q.7 Consider the following profit table along with the given probabilities of each state

Strategies

N1 N2 N3Probability of

Sales0.3 0.3 0.1

S1 20 18 -9S2 25 15 10S3 40 -10 12

Q.8 Using various criteria for decision making find the optimal strategy for the marketing manager of an automobile company. The conditional pay-offs in crores of rupees for the two models of a car for the various likely sales figures are as follows.

Sales UnitsModel

1 Lac

2 Lacs

3 Lacs

X 30 10 10Y 55 20 3

Q.9 The management of ABC company is considering the question of marketing a new product. The fixed cost required in the project is Rs. 4,000. Three factors are uncertain viz. the selling price, the variable cost and the annual sales volume.

The product has a life of only one year. The management has the data on these three factors as under:

Selling Price (Rs.)

Probability

Variable Cost (Rs.)

Probability

Sales Volume

(Rs.)

Probability

3 0.2 1 0.3 2000 0.34 0.5 2 0.6 3000 0.35 0.3 3 0.1 5000 0.4

Consider the following sequence of thirty random numbers:81, 32, 60, 04, 46, 31, 67, 25, 25, 10, 40, 02, 39, 68, 08, 59, 66, 90, 12, 64, 79, 31, 86, 68, 82, 89, 25, 11, 98, 16

Using the sequence, simulate the average profit for the above project on the basis of 10 trials.

Q. What is simulation model. Give example.Q. Write a note on Monte Carlo Simulation.Q. Explain PERT & CPM. Also describe advantages and disadvantages.Q. Write a note on decision making under:

i) risk ii) uncertainty.Q. a) Five different m/c’s can do any of the fine required jobs

with different profit resulting from each assignment of jobs to m/c’s Machine.

Job

A B C D E

1 30

37

40

38

40

2 40

24

27

21

36

3 40

32

33

30

35

4 2 3 4 3 3

5 8 0 6 65 2

962

91

34

39

b) Two jobs J1 and J2 have the following processing times on two m/cs M1 & M2

Jobs Processing Times (Minutes)

M1 M2

J1 15 20J2 22 12

Work out the most time effective schedule

R E F E R E NC E B O O K S________________________________________________________________________

Operations Research: TAHA

Operations Research: Anand Saxena

Operations Research: Hira Gupta

Operations Research: S. D. Sharma

Operations Research: J. K. Sharma

Operations Research: P. C. Tulsi: Vishal Pandey