Operations Research

OPERATIONS RESEARCH

CDR NK NATARAJAN

ORIGIN OF OR • DATES BACK TO 1800s

• EARLY CONTRIBUTORS INCLUDE FW TAYLOR, HENRY L GANTT, AK ERLANG ETC.

• DURING WW II, IT WAS WIDELY USED TO FIND OUT EFFECTIVE ALLOCATION OF LIMITED MILITARY RESOURCES TO VARIOUS OPERATIONS.

ORIGIN OF OR • POST WW II, INDUSTRIES STARTED ADOPTING THE TECHNIQUE FOR OPTIMIZATION PROBLEMS • IN INDIA OR CAME IN TO EXISTANCE IN 1949 AT REGIONAL RESEARCH LABORATORY IN HYERABAD.

CHARECTERISTICS OF OR

•SYSTEM ORIENTATION

• USES INTER DISCIPLINARY TEAMS

• APPLICATION OF SCIENTIFIC METHODS

• IMPROVEMENT IN THE QUALUTY OF DECISION

• USE OF COMPUTER

• QUANTITATIVE SOLUTIONS

SCOPE OF OR • Industry – product mix, blending, inventory control, demand forecasting, transportation, repair and maintenance, scheduling and sequencing, control of projects etc.• Defence – Allocation of resources to various operations, logistics, movement of personnel etc.•Planning – growth of per capita income, greatest impact of various projects, employment etc.

SCOPE OF OR • Agriculture – Optimum allocation of land for various crops, water, etc to maximize yield• Public Utilities – Hospitals to reduce waiting time, telephone services, public transportation, insurance sector etc. • Management –

– Allocation and Distribution– Production and facility Planning– Procurement– Marketing– Finance– Personnel– Research and Development

OBJECTIVES OF OR

• To provide a scientific basis to the managers of an organization for solving problems involving interaction of the components of the system, by employing a systems approach by a team of scientists drawn from different disciplines for finding a solution which is in the best interest of the organization as a whole.

PHASES OF OR

• Formulate the problem

• Construct a model to represent the system under study

• Deriving a solution from the model

• Testing the model and the solution derived from it

• Establishing controls over the solution

• Implementing the solution

CONSIDER THIS

• You have five week business commitment between Mumbai and Kochi. You fly out of Mumbai on Mondays and return on Wednesdays. A regular round trip ticket costs Rs 4000, but a 20% discount is granted if the dates of the ticket span a weekend. A one way ticket in either direction costs 75% of the regular price. How should you buy the tickets for the five week period.

– What are the decision alternatives?– Under what restrictions is the decision made?– What is an appropriate objective criteria for evaluating the alternatives?

CONSIDER THIS

• Decision Alternatives– 5 regular Mumbai- Kochi- Mumbai tickets for departure on Monday and return on Wednesday of the same week– One Mumbai – Kochi, Four Kochi – Mumbai _ Kochi that span weekends and one Kochi – Mumbai ticket– One Mumbai – Kochi – Mumbai to cover Monday of the first Week and Wednesday of the last week and four Kochi - Mumbai - Kochi to cover the remaining legs.

CONSIDER THIS

• Cost Alternatives– Alternative 1 - 5X4000 = 20000– Alternative 2 - .75 X 4000 + 4 (.80 X 4000) + .75 X 4000 = 18800 – Alternative 3 – 5 X (.8 X 4000) = 16000

• Alternative 3 is the best option as it minimizes the travel cost.

CONSIDER THIS

• Consider making a maximum area rectangle out of a piece of wire of length ‘L’ inches. What should be the width and height of the rectangle.

– Let ‘W’ be the width of the rectangle in inches and – ‘H” be the height of the rectangle in inches

• Based on these – Width + Height = Half the length of the wire– Width and Height can not be negative

• Algebraically– 2(W+H)=L– W ≥ 0; H ≥ 0

CONSIDER THIS

• What is the objective? – Maximization of the area of the rectangle. Let ‘Z’ be the area of the rectangle. Then the model becomes–Maximize Z =WH– Subject to

• 2(W+H) = L• W,H ≥ 0

APPROACH OF OR

• In OR we either maximize or minimize an objective function subject to the various constraints.

• A solution of the model is feasible only if it satisfied all the constraints.

• It is optimal if in addition to being feasible, it yields the best (Max or Min) value of the objective function

•

TYPES OF OR MODELS

• Mathematical models• Statistical models• Inventory Models• Allocation Models• Sequencing Models• Project Scheduling Models• Routing Models• Competitive Models• Queuing Models• Simulation Techniques• Decision Theory• Replacement Models• Reliability Theory• Markov Analysis• Advanced OR Models• Combined Methods

LINEAR PROGRAMMING

• Can be used only if – There is a well defined objective function (Profit, Cost or quantities produced) which is to be maximized or minimized and which can be expressed as a linear function of decision variables. – There must be constraints on the amount or extent of attainment of the objective and these constraints must be capable of being expressed as linear equations or inequalities in terms of variables.

LINEAR PROGRAMMING

• Can be used only if – There must be alternative courses of action. For example, a given product may be processed by two different machines and problem may be as to how much of the product to allocate to which machine. – Decision variables should be interrelated and non-negative. The non-negativity condition shows that the linear programming deals with real life situations– Resources must be in limited supply

FORMULATION OF LP PROBLEMProduction allocation problem •A firm produces three products. These products are processed on three different machines. The time required to manufacture one unit of each of the three products and the daily capacity of the three machines are given in the next slide. • It is required to determine the daily number of units to be manufactured for each product. The profit per unit of product 1,2 and 3 is Rs 4,3&6 respectively. It is assumed that all the amounts produced are consumed in the market. Formulate the LP model that will maximize the daily profit.

FORMULATION OF LP PROBLEM

Machine Time per unit (minutes) Machine Capacity (Minutes

/Day)

Product 1

Product 2

Product 3

M1

M2

M3

2

4

2

3

-

5

2

3

-

440

470

430


Step 1

Find the key decision to be made. The key decision is to decide the extent of product 1,2&3 to be produced as this can vary.

Step 2

Assume symbols for the extent of production. Let the extent of Product 1,2&3 be X1, X2 & X3.

Step 3

Express the feasible alternatives mathematically in terms of variables. Feasible alternatives are those which are physically, economically and financially possible. In this example, feasible alternatives are sets of values of x1, x2 & x3, where x1,x2 &x3 ≥ 0 since negative production has no meaning and is not feasible.


Step 4Mention the object quantitatively and express it as a linear function of variables. IN the present example, objective is to maximize the profit. i.e. Maximize Z = 4x1+3x2+6x3

Step 5Express the constraints as linear equations/inequalities in terms of variables. Here, constraints are o the machine capacities and can be mathematically expressed as

2x1 + 3x2 + 2x3 ≤ 440,4x1 + 0x2 + 3x3 ≤ 470,2x1 + 5x2 + 0x3 ≤ 430.

FORMULATION OF LP PROBLEMDiet Problem

• A person wants to decide the constituents of a diet which will fulfill daily requirements of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods. The yields per unit of these foods are given in a table in the next slide.

• Formulate a linear programming model for the problem.


Food Type

Yield per unit Cost per unit (Rs)Protein Fats Carbohydrates

1

2

3

4

3

4

8

6

2

2

7

5

6

4

7

4

45

40

85

65

Min Requirement

800 200 700


Let x1, x2, x3 and x4 denote the number of units of food of type 1,2,3 & 4 respectively. Objective is to minimize the cost i.e.

Minimize Z = 45x1+40x2+85x3+65x4

Constraints are on the fulfillment of the daily requirements of various constituents i.e.

Proteins - 3x1 + 4x2 + 8x3 + 6x4 ≥ 800Fats - 2x1 + 2x2 + 7x3 + 5x4 ≥ 200,Carbohydrates - 6x1 + 4x2 + 7x3 + 4x4 ≥ 700.

Where x1,x2,x3,x4 each ≥ 0

FORMULATION OF LP PROBLEMBlending Problem • A firm produces an alloy having the following specifications:

– Specific gravity ≤ 0.98– Chromium ≥ 8%– Melting Point ≥ 450ºC

• Raw materials A,B and C having the properties shown in the table can be used to make the alloy. Cost of various raw materials per ton are : Rs. 90 for A, Rs. 280 for B and Rs. 40 for C. Formulate the LP model to find the proportions in which A,B and C be used to obtain an alloy of desired properties while the costs of raw materials is minimum.


Property Properties of raw materials

A B C

Specific Gravity

Chromium

Melting Point

0.92

7%

440ºC

0.97

13%

490ºC

1.04

16%

480ºC


Let the percentage contents of raw materials A,B and C to be used for making the alloy be x1, x2 and x3 respectively. Objective is to minimize the cost i.e.

Minimize Z = 90x1+280x2+40x3

Constraints are imposed by the specifications required for the alloy. i.e.

0.92x1 + 0.97x2 + 1.04x3 ≤ 0.987x1 + 13x2 + 16x3 ≥ 8,440x1 + 490x2 + 480x3 ≥ 450. x1+x2+x3 = 100 as x1,x2 and x3 are the

percentage contents of materials A,B and C in making the alloy. Also x1,x2,x3 each ≥ 0

FORMULATION OF LP PROBLEMProduct Mix Problem • A truck company has Rs. 50 Lakhs to invest and is to choose among three types of trucks A,B & C. Truck A has 12 tonne payload and is expected to average 50 Km per hour. It costs Rs 80,000. Truck B has a 20 tonne payload, is expected to average 45 Km per hour and costs Rs. 1,00,000. Truck C is a modified form of B. It has sleeping space for the driver, which reduces its payload capacity to 17 tonnes, while raising the cost to Rs 1,20,000. • Truck A requires a crew of one man and if driven on three shifts per day, could run for an average of 20 hours a day. Truck B & C require a crew of two men each and if driven on three shifts a day could be run for an average of 19 hours and 22 hours respectively. The company has a fleet of 120 crewmen available to it. If the total number of trucks are not to exceed 40, how many trucks of each type should be purchased if the company wants to maximize its capacity in Tonne - KM per day? Formulate the problem as LP problem.


Let the number of trucks of type A,B and C to be purchased be x1, x2 and x3 respectively. Constraints are :On the number of crew:x1+2x2+2x3 ≤ 120On the number of trucks:x1 + x2 + x3 ≤ 40On the money to be invested:80,000x1 + 1,00,000x2 + 1,20,000x3 ≤ 50,00000Objective is to maximize the tonne Km/per day. Tonne Km per Day = Pay load in tonne X Km/Hour X hours / DayA = 20 x 50 x 20 = 12,000B = 20 x 45 x 19 = 17,100C = 17 x 45 x 22 = 16,830Objective function is Maximize Z = 12,000 x1 + 17,100 x2 + 16830 x3 Where x1,x2,x3 ≥ 0

GRAPHICAL METHOD A firm manufactures two products A & B on which the profits earned per unit are Rs. 3 and Rs. 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2. b requires 2 minute on M1 and M2. Machine M1 is available for not more than 7 hrs. 30 minutes. While machine M2 is available for 10 hrs. during any working day. Find the number of units of product A and B to be manufactured to get maximum profit.


Let x1 and x2 denote the number of units of products A and B to be produced per day. Objective FunctionMaximize Z = 3 x1+ 4 x2Subject to X1+x2 ≤ 4502X1 + x2 ≤ 600Where x1,x2 ≥ 0

GRAPHICAL SOLUTION

GRAPHICAL METHOD Find the maximum value of

Z = 2x1 + 3x2

Subject to

x1 + x2 ≤ 30

x2 ≥ 3

x2 ≤ 12

X1 – x2 ≥ 0

0 ≤ x1 ≤ 20

GRAPHICAL METHOD Find the maximum value of

Z = 2x1 + x2

Subject to

x1 + 2x2 ≤ 10

x1 + x2 ≤ 6

x1 - x2 ≤ 2

X1 – 2x2 ≤ 1

x1,x2 ≥ 0

GRAPHICAL SOLUTION

GRAPHICAL METHOD Find the minimum value of

Z = 5x1 - 2x2

Subject to

2x1 + 3x2 ≥ 1

x1,x2 ≥ 0

GRAPHICAL METHOD Find the minimum value of

Z = -x1 + 2x2

Subject to

-x1 + 3x2 ≤ 10

x1+x2 ≤ 6

x1-x2 ≤ 2

x1,x2 ≥ 0

GRAPHICAL SOLUTION

SIMPLEX METHOD A firm manufactures two products A & B on which the profits earned per unit are Rs. 3 and Rs. 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2. b requires minute on M1 and M2. Machine M1 is available for not more than 7 hrs. 30 minutes. While machine M2 is available for 10 hrs. during any working day. Find the number of units of product A and B to be manufactured to get maximum profit.

SIMPLEX METHOD Maximize Z = 3x1 + 4x2Subject to

x1 + x2 ≤ 450 2x1 + x2 ≤ 600x1,x2 ≥ 0

Step 1- Express the problem in standard formMaximize Z = 3x1 + 4x2 + 0s1 + 0s2Subject to

x1 + x2 +s1= 450 2x1 + x2 +s2 = 600x1,x2, s1, s2 ≥ 0

SIMPLEX METHODStep 2- Find initial basic feasible solution

– We start with a initial basic feasible solution which we get by assuming that the profit earned is zero. – Assume decision variable x1 andx2 as zero. – Substituting the values of x1 and x2 in the objective function we get s1= 450 and s2= 600.– this is called the initial basic feasible solution and s1 and s2 will form the basis.– This means that neither of the products have been produced and the time available fo both machines are unused.

SIMPLEX TABLE

Contribution/unit cj 3 4 0 0

Basis Body Matrix identity Matrix

F.R CB Basic variables x1 x2 s1 s2 b θ

0 s1 1 (1) 1 0 450 450

0 s2 2 1 0 1 600 600

Zj 0 0 0 0 0

Cj-Zj 3 4 0 0

K

SIMPLEX TABLE

Contribution/unit cj 3 4 0 0

Basis Body Matrix identity Matrix

F.R CB Basic variables x1 x2 s1 s2 b

4 X2 1 1 1 0 450

0 s2 1 0 -1 1 150

Zj 4 4 4 0 1800

Cj-Zj -1 0 -4 0

Optimum Solution

SIMPLEX METHOD Maximize Z = 2x1 + 5x2Subject to

x1 + 4x2 ≤ 24 3x1 + x2 ≤ 21x1 + x2 ≤ 9x1,x2 ≥ 0

Step 1- Express the problem in standard formMaximize Z = 2x1 + 5x2 + 0s1 + 0s2+0s3,Subject to

x1 + 4x2 +s1 = 24 3x1 + x2 +s2 = 21x1+x2+s3 = 9x1,x2, s1, s2, s3 ≥ 0

SIMPLEX TABLE

cj 2 5 0 0 0

F.R CB Basis x1 x2 s1 s2 s3 b θ

0 s1 1 (4) 1 0 0 24 6

1/4 0 s2 3 1 0 1 0 21 21

1/4 0 s3 1 1 0 0 1 9 9

Zj 0 0 0 0 0 0

Cj-Zj 2 5 0 0 0

KInitial Basic Feasible Solution

SIMPLEX METHOD Values of s2 Row Values of s3 Row

3 – ¼ x 1= 11/4 1 – ¼ x 1= ¾

1 - ¼ x 4 = 0 1 - ¼ x 4 = 0

0 – ¼ x 1 = -1/4 0 – ¼ x 1 = -1/4

1 – ¼ x 0 = 1 0 – ¼ x 0 = 0

0 – ¼ x 0 = 0 1 – ¼ x 0 = 1

21 – ¼ x 24 = 15 9 – ¼ x 24 = 3

SIMPLEX TABLE

cj 2 5 0 0 0

F.R CB Basis x1 x2 s1 s2 s3 b θ

1/3 5 x2 1/4 1 1/4 0 0 6 24

11/3 0 s2 11/4 0 -1/4 1 0 15 60/11

0 s3 3/4 0 -1/4 0 1 3 4

Zj 5/4 5 5/4 0 0 30

Cj-Zj 3/4 0 -5/4 0 0

KSecond Feasible Solution

SIMPLEX TABLE

cj 2 5 0 0 0

CB Basis x1 x2 s1 s2 s3 b

5 x2 0 1 1/3 0 -1/3 5

0 s2 0 0 2/3 1 -11/3 4

2 x1 1 0 -1/3 0 4/3 4

Zj 2 5 1 0 1 33

Cj-Zj 0 0 -1 0 -1

KThird Feasible Solution

Optimal Solution

SIMPLEX METHOD Maximize Z = 4x1 + 3x2 + 6x3Subject to

2x1 + 3x2 +2x3 ≤ 440 4x1 + 3x3 ≤ 4702x1 + 5x2 ≤ 430x1,x2, x3 ≥ 0

Step 1- Express the problem in standard formMaximize Z = 4x1 + 3x2 + 6x3 + 0s1 + 0s2 + 0s3,Subject to

2x1 + 3x2 + 2x3 + s1 = 440 4x1 + 3x3 + s2 = 4702x1 + 5x2 + s3 = 430x1,x2, x3, s1, s2, s3 ≥ 0

SIMPLEX TABLE

cj 4 3 6 0 0 0

FR CB Basis x1 x2 x3 s1 s2 s3 b θ

2/3 0 s1 2 3 2 1 0 0 440 220

0 s2 4 0 (3) 0 1 0 470 470/3

0 s3 2 5 0 0 0 1 430 ∞

Zj 0 0 0 0 0 0 0

Cj-Zj 4 3 6 0 0 0 K Third Feasible Solution

Optimal Solution

SIMPLEX TABLE

cj 4 3 6 0 0 0


0 s1 -2/3 (3) 0 1 -2/3 0 380/3 380/9

6 x3 4/3 0 1 0 1/3 0 470/3 ∞

5/3 0 s3 2 5 0 0 0 1 430 86

Zj 8 0 6 0 2 0 940

Cj-Zj -4 3 0 0 -2 0

KSecond Feasible Solution

SIMPLEX TABLE

cj 4 3 6 0 0 0

CB Basis x1 x2 x3 s1 s2 s3 b

3 x2 -2/9 1 0 1/3 -2/9 0 380/9

6 x3 4/3 0 1 0 1/3 0 470/3

0 s3 28/9 0 0 -5/3 10/9 1 1970/9

Zj 22/3 3 6 1 4/3 0 3200/3

Cj-Zj -10/3 0 0 -1 -4/3 0

Optimal solution

SIMPLEX METHODMinimize Z = x1 - 3x2 + 3x3Subject to

3x1 - x2 + 2x3 ≤ 7, 2x1 + 4x2 ≥ -12,-4x1 + 3x2 + 8 x3 ≤ 10x1,x2, x3 ≥ 0

Step 1- Express the problem in standard formMaximize Z = x1 - 3x2 + 3x3 + 0s1 + 0s2 + 0s3,Subject to

3x1 - x2 + 2x3 + s1 = 7 -2x1 - 4x2 + s2 = 12-4x1 + 3x2 + 8x3 + s3 = 10x1,x2, x3, s1, s2, s3 ≥ 0

SIMPLEX TABLE

cj 1 -3 3 0 0 0


1/3 0 s1 3 -1 2 1 0 0 7 -7

4/3 0 s2 -2 -4 0 0 1 0 12 -3

0 s3 -4 (3) 8 0 0 1 10 10/3

Zj 0 0 0 0 0 0

Cj-Zj 1 -3 3 0 0 0

K Initial Basic Feasible Solution

SIMPLEX TABLE

cj 1 -3 3 0 0 0


0 s1 (5/3) 0 14/3 1 0 1/3 31/3 31/3

22/5 0 s2 -22/3 0 32/3 0 1 4/3 76/3 -38/11

4/5 -3 x2 -4/3 1 8/3 0 0 1/3 10/3 -5/2

Zj 4 -3 -8 0 0 -1 -10

Cj-Zj -3 0 11 0 0 1

K Second Feasible Solution

SIMPLEX TABLE

Cj 1 -3 3 0 0 0

CB Basis x1 x2 x3 s1 s2 s3 b

1 x1 1 0 14/5 3/5 0 1/5 31/5

0 s2 0 0 156/5 22/5 1 14/5 354/5

-3 x2 0 1 32/5 4/5 0 3/5 58/5

Zj 1 -3 -82/5 -9/5 0 -8/5 -143/5

Cj-Zj 0 0 97/5 9/5 0 8/5

Optimal solution

BIG M METHOD• Minimize Z = 12x1 + 20x2 Subject to

6x1 + 8x2 ≥ 100, 7x1 + 12x2 ≥ 120,X1,x2, ≥ 0

Step 1- Express the problem in standard formMaximize Z = 12x1 + 20x2 + 0s1 + 0s2,Subject to

6x1 + 8x2 - s1 = 100 7x1 + 12x2 - s2 = 120x1,x2, s1, s2 ≥ 0

BIG M METHOD• Initial Basic Feasible Solution

Putting x1=x2 = 0, we get s1 = -100 and s2 = - 120 which is not feasible, as it does not fulfill the non negativity restriction.

• Introduce artificial variables in the constraints

• 6x1 + 8x2 - s1 + A1 = 100

7x1 + 12 x2 - s2 + A2 = 120

x1,x2, s1, s2,A1, A2 ≥ 0

BIG M METHOD• Since they are artificial variables A1 and A2 should not arrear in the final solution. Hence they are assigned a large unit penalty ( a large + value M) in he objective function which can be written as Minimize Z = 12x1+20x2+0s1+0s2+MA1+MA2• Now there are 6 variables and two constraints and hence we have to zeroise four variables to get the initial basic feasible solution. Assume x1=x2=s1=s2=0•A1 = 100, A2 = 120, Z = 220 M

SIMPLEX TABLE

Cj 12 20 0 0 M M

FR CB Basis x1 x2 s1 s2 A1 A2 b θ

2/3 M A1 6 8 -1 0 1 0 100 25/2

M A2 7 (12) 0 -1 0 1 120 10

Zj 13M 20M -M -M M M 220M

Cj-Zj 12-13M 20-20M M M 0 0


SIMPLEX TABLE

Cj 12 20 0 0 M

FR CB Basis x1 x2 s1 s2 A1 b θ

M A1 4/3 0 -1 2/3 1 20 15

7/16 20 X2 7/12 1 0 -1/12 0 10 120/7

Zj 35/4+4/3M 20 -M -5/3+2/3M M 200+20M

Cj-Zj 1/3-4/3M 0 M 5/3 – 2/3M 0


SIMPLEX TABLE

Cj 12 20 0 0

CB Basis x1 x2 s1 s2 b

12 X1 1 0 -3/4 1/2 15

20 X2 0 1 7/16 -3/4 5/4

Zj 12 20 -1/4 -9 205

Cj-Zj 0 0 ¼ 9

Optimum Solution

BIG M METHOD• Maximize Z = 3x1 - x2 Subject to

2x1 + x2 ≤ 2, x1 + 3x2 ≥ 3,x2 ≤ 4X1,x2, ≥ 0

Step 1- Express the problem in standard formMaximize Z = 3x1 - x2 + 0s1 + 0s2 + 0s3 – MA1,Subject to

2x1 + x2 + s1 = 2 x1 + 3x2 - s2 + A1 = 3x2 + s3 = 4x1,x2, s1, s2, s3, A1 ≥ 0

SIMPLEX TABLE

Cj 3 -1 0 0 0 -M

FR CB Basis x1 x2 s1 s2 s3 A1 b θ

0 s1 2 1 1 0 0 0 2 2

-M A1 1 (3) 0 -1 0 1 3 1

0 s3 0 1 0 0 1 0 4 4

Zj -M -3M 0 M 0 -M -3M

Cj-Zj 3+M -1+3M 0 -M 0 0


SIMPLEX TABLE

Cj 3 -1 0 0 0

FR CB Basis x1 x2 s1 s2 s3 b θ

0 s1 (5/3) 0 1 1/3 0 1 3/5

-1 x2 1/3 1 0 -1/3 0 1 3

0 s3 -1/3 0 0 1/3 1 3 -9

Zj -1/3 -1 0 1/3 0 -1

Cj-Zj 10/3 0 0 -1/3 0

K second Feasible Solution

SIMPLEX TABLE

Cj 3 -1 0 0 0

FR CB Basis x1 x2 s1 s2 s3 b

3 x1 1 0 3/5 1/5 0 3/5

-1 x2 0 1 -1/5 -2/5 0 4/5

0 s3 0 0 1/5 2/5 1 16/5

Zj 3 -1 2 1 0 1

Cj-Zj 0 0 -2 -1 0

Optimal Solution

BIG M METHOD• Maximize Z = x1 + 2x2 + 3x3 – x4 Subject to

x1 + 2x2 + 3x3 = 15, 2x1 + x2 + 5x3 = 20,x1 + 2x2 + x3 + x4 = 10x1,x2, x3, x4 ≥ 0

Step 1- Express the problem in standard formMaximize Z = x1 + 2x2 + 3x3 – x4 - MA1 – MA2 – MA3,Subject to

x1 + 2x2 + 3x3 +0x4 +A1 + 0A2 + 0A3 = 15 2x1 + x2 + 5 X3 + 0A1 +A2 + 0A3 = 20X1 + 2x2 + X3 + X4 + 0A1 + 0A2 + A3 = 10x1,x2, X3, X4, A1, A2, A3 ≥ 0

SIMPLEX TABLE

Cj 1 2 3 -1 -M -M -M

CB Basis x1 x2 X3 X4 A1 A2 A3 b θ

-M A1 1 2 3 0 1 0 0 15 5

-M A2 2 1 (5) 0 0 1 0 20 4

-M A3 1 2 1 1 0 0 1 10 10

Zj -4M -5M -9M -M -M -M -M -45M

Cj-Zj 1+4M 2+5M 3+9M -1+M 0 0 0


SIMPLEX TABLE

Cj 1 2 3 -1 -M -M

FR CB Basis x1 x2 X3 X4 A1 A3 b θ

-M A1 -1/5 (7/5) 0 0 1 0 3 15/7

3 X3 2/5 1/5 1 0 0 0 4 20

-M A3 3/5 9/5 0 1 0 1 6 10/3

Zj 6-2M/5 3-16M/5 3 -M -M -M 12-9M

Cj-Zj -1+2M/5 7+16M/5 0 -1+M 0 0

K Second Feasible Solution

SIMPLEX TABLE

Cj 1 2 3 -1 -M

FR CB Basis x1 x2 X3 X4 A3 b θ

2 x2 -1/7 1 0 0 0 15/7 ∞

3 X3 3/7 0 1 0 0 25/7 ∞

-M A3 6/7 0 0 (1) 1 15/7 15/7

Zj 7-6M/7 2 3 -M -M 105-15M/7

Cj-Zj 6M/7 0 0 -1+M 0

K

Third Feasible Solution

SIMPLEX TABLE

Cj 1 2 3 -1

FR CB Basis x1 x2 X3 X4 b θ

2 x2 -1/7 1 0 0 15/7 -15

3 X3 3/7 0 1 0 25/7 25/3

-1 x4 (6/7) 0 0 1 15/7 5/2

Zj 1/7 2 3 -1 90/7

Cj-Zj 6/7 0 0 0

K

Fourth Feasible Solution

TANSPORTATION MODEL• A dairy has three plants located throughout the state. Daily milk production at each plant is as follows:

• Plant 1 - 6 million litres

• Plant 2 – 1 million litres

• Plant 3 – 10 million litres

• Each day the firm must fulfill the needs of its four distribution centres. Milk requirement at each centre is as ollows:

• Distribution centre 1 – 7 million litres




• Cost of shipping one million litres of milk from each plant to each distribution centre is given in the following table in hundreds of rupees.

• Formulate the mathematical model for the problem.

• Determine the optimal transportation policy.

SIMPLEX TABLE

Cj 1 2 3 -1

FR CB Basis x1 x2 X3 X4 b

2 x2 0 1 0 1/6 5/2

3 X3 0 0 1 -1/2 5/2

1 x1 1 0 0 7/6 5/2

Zj 1 2 3 0 15

Cj-Zj 0 0 0 -1

Optimal Solution

TANSPORTATION MODEL• Deals with transportation of a product available at several sources a number of different destinations in such a way that the total transportation cost is minimum.

• The origin of this model dates back to 1941 when FL Hitchcock presented a study on the subject.

• Assumptions in the transportation model:-

• Total quantities of an item item available at different sources is equal to total requirement at different destinations.

• Items can be transported conveniently from all sources to destinations.

• The unit transportation cost of the item from all sources to destinations is certainly and precisely known.

• The transportation cost on a given route is directly proportional to the number of units shipped on that route.

• The objective is to minimize the total transportation cost for the organization as a whole and not for individual supply and distribution centres.

TANSPORTATION MODEL

2 3 11 7

1 0 6 1

5 8 15 9

Distribution Center

1

2

3

Plant

1

2

3

4

TANSPORTATION MODEL

X11 X12 X13 X14

X21 X22 X23 X24

X31 X32 X33 X34

Distribution Center

1

2

3

Plant

1

2

3

4

TANSPORTATION MODEL• Objective is to minimize the cost of transportation

i.e. Minimize Z = 2X11 + 3X12 + 11X13 + 7 X14 + X21 + 0X22 + 6X23 + X24 + 5X31 + 8X32 + 15X33 + 9X34

Subject to

X11 + X12 + X13 + X14 = 6 (For plant 1)

X21 + X22 + X23 + X24 = 1 (For plant 2)

X31 + X32 + X33 + X34 = 10 (For plant 3)

X11 + X21 + X31 = 7, (For DC 1)

X12 + X22 + X32 = 5, (For DC 2)

X14 + X24 + X34 = 2 (For DC 3)

TANSPORTATION MODEL

2 3 11 7

1 0 6 1

5 8 15 9

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

6

5

3

2

1

10

17

NORTH WEST CORNER METHOD

2

(6)

3 11 7

1

(1)

0

(0)

6 1

5 8

(5)

15

(3)

9

(2)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

6

5

3

2

1

10

17

ROW MINIMA METHOD

2

(6)

3 11 7

1 0

(1)

6 1

5

(1)

8

(4)

15

(3)

9

(2)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

6

5

3

2

1

10

17

COLUMN MINIMA METHOD

2

(6)

3 11 7

1

(1)

0 6 1

5 8

(5)

15

(3)

9

(2)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

6

5

3

2

1

10

17

VAM METHOD

2 3 11 7

1 0 6 1

(1)

5 8 15 9

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(1)

6 (1)

5

(3)

3

(5)

2

(6)

1 (1)

10 (3)

17

VAM METHOD

2 3

(5)

11 7

1 0 6 1

(1)

5 8 15 9

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (1)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

VAM METHOD

2

(1)

3

(5)

11 7

1 0 6 1

(1)

5 8 15 9

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (5)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

VAM METHOD

2

(1)

3

(5)

11 7

1 0 6 1

(1)

5

(6)

8 15

(3)

9

(1)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (5)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

VAM METHOD

2

(1)

3

(5)

11 7

1 0 6 1

(1)

5

(6)

8 15

(3)

9

(1)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (5)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

TRANSPORTATION MODELImportant Note:

In case of tie among the highest penalties, select the row or column having minimum cost. In case of tie in the minimum cost also, select the cell which can have maximum allocation. If there is tie among maximum allocation cells also, select the cell arbitrarily or allocation. This will give the best initial basic feasible solution requiring less number of iterations subsequently.

Optimality Test

Optimality test can be performed only on that feasible solution in which

(a)Number of allocations is m + n – 1, where m is the number of rows and n is the number of columns.

(b)These (m + n - 1) allocations should be in independent positions.

TRANSPORTATION MODELOptimality Test

The test procedure for optimality involves examination of each vacant cell to find whether or not making an allocation in it reduces the total transportation cost.

Two Methods:

Stepping Stone method:

Modified Distribution Method (MODI)

STEPPING STONE METHOD

2

(1)

3

(5)

11 7

1 0 6 1

(1)

5

(6)

8 15

(3)

9

(1)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (5)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

TRANSPORTATION MODELOptimality Test

Cell Evaluations

Cell (1,3) = 11-15+5-2 = -1

Cell (1,4) = 7-9+5-2 = +1

Cell (2,1) = 1-1+9-5 = 4

Cell (2,2) = 0-1+9-5+2-3 = 2

Cell (2,3) = 6-1+9-15 = -1

Cell (3,2) = 8-5+2-3 =2

MODI METHOD

2

(1)

3

(5)

11 7

1 0 6 1

(1)

5

(6)

8 15

(3)

9

(1)

Distribution Center

1

2

3

Plant

1

2

3

4

Supply

Requirement7

(3)

6 (5)

5

(5)

3

(4)

2

(2)

1 (1)

10 (3)

17

ui

vj

MODI METHODU1+V1=2

U1+V2=3

U2+V4=1

U3+V1=5

U3+V3=15

U3+V4=9

Let v1 = 0 (Arbitrary) – Then

U1=2, v2=1, u3=5, v3=10, v4=4, u2=-3

MODI METHOD

2 3

1

5 15 9

Distribution Center

2

-3

5

Plant

0

1

10

4

ui

vj

MODI METHOD

12 6

-3 -2 7

6

Distribution Center

2

-3

5

Plant

0

1

10

4

ui

vj

Compute ui+vj for each empty cell

MODI METHOD

11-12=-1 7-6=1

1+3=4 0+2=2 6-7=-1

8-6=2

Distribution Center

2

-3

5

Plant

0

1

10

4

ui

vj

Compute cij-(ui+vj) for each empty cell

MODI METHODOptimality Test

A –Ve value in an unoccupied cell indicates that a better solution can be obtained by allocating units to this cell.

A +ve value in an unoccupied cell indicates that a poorer solution will result by allocating units to the cell.

A zero value in an unoccupied cell indicates that another solution of the total value can be obtained by allocating units to this cell.

Iterate for optimality by allocating to most –ve cell. Recheck for optimality. If still notoptimal allocate to most –ve cell and go on.

EXAMPLEIs

50 20

55

30 35 25

An optimal solution for the transportation problem:

6 1 9 3

11 5 2 8

10 12 4 7

70

55

90

85 35 50 45 If not iterate it for optimal solution

EXAMPLESolve the following transportation problem

68 35 4 74 15

57 88 91 3 8

91 60 75 45 60

52 53 24 7 82

51 18 82 13 7

D1 D2 D3 D4 D5

O1

O2

03

O4

O5

Available

18

17

19

13

15

Required 16 18 20 14 14 82

(P-264)

EXAMPLEA product is produced by four factories A,B,C,D. The unit production

costs in them are Rs. 2, 3, 1 and 5 respectively. Their production capacities are 50,70,30,50 units respectively. These factories supply the product to four stores, demands of which are 25,35,105,20 units respectively. Unit transportation cost in rupees from each factory to each store is given in the table below:

2 4 6 11

10 8 7 5

13 3 9 12

4 6 8 3

Determine the extent of deliveries from each of the factories to each of the stores so that the total production and transportation cost is minimum. (p-267)

ASSIGNMENT PROBLEMASSIGNMENT MODEL CAN BE REGARDED AS A SPECIAL CASE OF

TRANSPORTATION PROBLEM.

25 15 22

31 20 19

35 24 17

Determine the assignment that will optimize the total processing cost. (325)

X Y Z

A

B

C

JOBS

ASSIGNMENT PROBLEMMake sure the matrix is balanced. If not add a dummy row or column to

balance it.

Subtract each row element from the lowest value in that row to arrive at the machine opportunity cost.

25 – 15 = 10

0 7

12 1 0

18 7 0

X Y Z

A

B

C

JOBS

ASSIGNMENT PROBLEMSubtracting each column element from the lowest value in that column

will give job opportunity cost.

The total opportunity cost (combined job and machine opportunity cost) is given by subtraction the lowest column value in table 2 from each of the column values.

10 – 10 = 0

0 7

2 1 0

8 7 0

X Y Z

A

B

C

JOBS

ASSIGNMENT PROBLEM

Checking for optimality. Hungarian Method

Draw vertical and horizontal lines to cover all zeros in the matrix. If the number of lines are equal to n then the assignment is optimal. If no subtract the smallest value of the uncovered cells from each uncovered cell and add it to all intersecting values. Again draw vertical and horizontal lines to cover all zeros.

Repeat till the number of lines are equal to n.

EXAMPLEA machine tool company decides to make four subassemblies through

four contractors. Each contractor is to receive only one sub assembly. The cost of each sub assembly is determined by the bids submitted by each contractor and is shown in table below in hundreds of rupees.

15 13 14 17

11 12 15 13

13 12 10 11

15 17 14 16

(1) Formulate the mathematical model for the problem

(2) Show that the assignment model is special case of the transportation model

(3) Assign the different subassemblies to contractors so as to minimize the total cost (329)

EXAMPLEA machine tool company decides to make four subassemblies through

four contractors. Each contractor is to receive only one sub assembly. The cost of each sub assembly is determined by the bids submitted by each contractor and is shown in table below in hundreds of rupees.

15 13 14 17

11 12 15 13

13 12 10 11

15 17 14 16

(1) Formulate the mathematical model for the problem

(2) Show that the assignment model is special case of the transportation model

(3) Assign the different subassemblies to contractors so as to minimize the total cost (329)

EXAMPLEFour different jobs can be done on four different machines. The set up

and take down time costs are assumed to prohibitively high for change over. The matrix below gives the cost in rupees of producing job i on machine j.

5 7 11 6

8 5 9 6

4 7 10 7

10 4 8 3

(1) How should the jobs be assigned to the various machines so that the total cost is minimum. Also formulate the mathematical model for the problem. (332)

machines

EXAMPLEFour different jobs can be done on four different machines. The set up

and take down time costs are assumed to prohibitively high for change over. The matrix below gives the cost in rupees of producing job i on machine j.

5 7 11 6

8 5 9 6

4 7 10 7

10 4 8 3

(1) How should the jobs be assigned to the various machines so that the total cost is minimum. Also formulate the mathematical model for the problem. (332)

machines

EXAMPLESolve the following assignment problem.

11 17 8 16 20

9 7 12 6 15

13 16 15 12 16

21 24 17 28 26

14 10 12 11 13

machines

EXAMPLEFive wagons are available at stations 1,2,3,4&5. These are required at

five stations I, II , III , IV & V. The mileages between various between various stations are given below. How should the wagons be transferred so that the total mileage covered is minimum.

10 5 9 18 11

13 9 6 12 14

3 2 4 4 5

18 9 12 17 15

11 6 14 19 10

Stations

EXAMPLEUnbalanced Matrix

A company has one surplus truck in each of the cities A,B,C,D and E and one deficit truck in each of the cities 1,2,3,4,5&6. The distance between the cities in kilometers is shown in the matrix below. Find the assignment of trucks from cities in surplus to cities in deficit so that the total distance covered by the vehicles is minimum. (341)

12 10 15 22 18 8

10 18 25 15 16 12

11 10 3 8 5 9

6 14 10 13 13 12

8 12 11 7 13 10

1,2,3,4,5&6

EXAMPLEMaximization problem

A company has a team of four salesmen and there are four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts, the company estimates that the profit per day in rupees for each salesman in each district is as below. Find the assignment of salesmen to various districts which will yield maximum profit.

16 10 14 11

14 11 15 15

15 15 13 12

136 12 14 15

District

DECISION THEORYDecision taking environments

(1)Certainty – Various out comes are known with certainty (CVP, LP, Assignment Tpt Model etc)

(2)Uncertainty – where information required to assign probability is not available, though various possible outcomes are known (

(3)Risk – Information for assigning probability is available

(4)Conflict – Where two opponents are involved.

DECISION THEORYDecision making under conditions of uncertainty

MaxiMax Criterion

Alternatives States of Nature (Product Demand)

Maximum of RowHigh Moder

ateLow Nil

Expand

Construct

Subcontract

50000

70000

30000

25000

30000

15000

-25000

-40000

-1000

-45000

-80000

-10000

50000

70000

30000


MaxiMin Criterion


Minimum of RowHigh Moder

ateLow Nil

Expand

Construct

Subcontract

50000

70000

30000

25000

30000

15000

-25000

-40000

-1000

-45000

-80000

-10000

-45000

-80000

-10000


MiniMax Regret Criterion


Maximum of RowHigh Moder

ateLow Nil

Expand

Construct

Subcontract

20000

0

40000

5000

0

15000

24000

39000

0

35000

70000

0

35000

70000

40000


Hurwicz Criterion (Criterion of realism or Weighted Average Criterion)


Maximum of Row

Minimum of Row

P= α. Max +

(1 – α) min

α = .8High Moder

ateLow Nil

Expand

Construct

Subcontract

50000

70000

30000

25000

30000

15000

-25000

-40000

-1000

-45000

-80000

-10000

50000

70000

30000

-45000

-80000

-10000

31000

40000

22000

α = Appropriate degree of optimism


Laplace Criterion (Criterion of rationality or Bayes’ Criterion OR Equal Probability)


Expected Pay Offs Rs

= 1/n (P1+P2+P3+…….Pn)

High Moderate

Low Nil

Expand

Construct

Subcontract

50000

70000

30000

25000

30000

15000

-25000

-40000

-1000

-45000

-80000

-10000

1000/4 (50 + 25 - 25 – 45) = -1250

1000/4 (70+30-40-80) = -5000

1000/4 (30 + 15 - 1 -10) = 8500

EXAMPLEThe following matrix gives the payoff of different strategies

(alternatives) S1,S2,S3 against conditions (events) N1, N2, N3 & N4.

N1 N2 N3 N4

S1

S2

S3

4000

20000

20000

-100

5000

15000

6000

400

-2000

18000

0

1000

Indicate the decision taken under the following approaches

(a) Pessimistic (b) Optimistic (c) Regret (d) Equal Probability

EXAMPLEThe following matrix gives the payoff of different strategies

(alternatives) S1,S2,S3 against conditions (events) N1, N2, N3 & N4.

N1 N2 N3 N4

S1

S2

S3

4000

20000

20000

-100

5000

15000

6000

400

-2000

18000

0

1000

Indicate the decision taken under the following approaches

(a) Pessimistic (b) Optimistic (c) Regret (d) Equal Probability

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Operations Research