LINEAR PROGRAMMING - WordPress.com 02, 2015 · LINEAR PROGRAMMING 2.1. ... Formulate this problem as an LP model so as to maximize the total profit Formulation of LP model

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2.1. INTRODUCTION Decision making, in today's complex environment, has become a very difficult task. Two approaches, qualitative analysis and quantitative analysis, are widely used for decision making. Qualitative analysis refers personal experience, knowledge etc. while quantitative analysis, for the evaluation of every possible decision alternatives so as to provide some aid for sound decision making. Among prevalence of different quantitative techniques, one of the widely used techniques in different perspectives is linear programming. The word linear refers the existence of linear relationship among the variables in a model which is developed to solve the problem. The word programming refers planning the problem mathematically in order to get the particular strategy using the available limited and scarce resources. Hence, linear programming (LP) is one of the mathematical models which is designed to allocate some limited or scarce resources to optimize (either minimize or maximize) a single stated criterion. The LP technique was first invented by the Russian mathematician C.V. Kantorovich and Cater developed this technique by GB Dantzing while working with US Air force during World War II for solving military logistics problems. Nowadays, this technique is widely used in almost in all functional areas of management, agriculture, economics, education, hospitals, planning and controlling, etc. 1.2.1 COMPONENTS OF LP MODEL For the development of linear programming model, following three components are used. Decision Variables Decision variables are activities, usually denoted by X1, X2, ... Xn which are being evaluated as per the nature of objective function and availability of limited resources. These are always under the control of decision makers. Simply, decision variables are strategies. Decision variables also help to link up between objective function and the constraint equations. For example, the manager has to decide how many units of each product be produced so as to minimize the total operation costs while planning the production. Objective Function Objective function is a linear equation involving the decision variables which is either maximized or minimized (optimized) subject to given situation. It is usually represented as

Z = C1X1 + C2X2 + ... + CnXn

where, Z is the effect of decision variables X1, X2, ...., Xn. C1, C2, ....., Cn are the respective coefficients of variables X1, X2, ...., Xn. For example, if Z is defined as total revenue, it is maximized. If X is defined as total production cost, it is minimized.

Constraints: Constraint are certain limitations in using the resources like raw material, space, labour hour, machine hour, money, etc. These are always expressed as linear inequalities. In other words constraints are the restriction

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imposed on the available resources.For example, Mr. Nurbu buys x1 and x2 articles of type A and B @ Rs. 100 and Rs. 200 respectively. He has only Rs. 5000. Then, the required constraint equation :

100X1 + 200X2 < 5000 Here Rs. 5,000 is budgetary constraints. 2.1.2 CONSTRUCTION OF LP MODEL While constructing the LP model, we will use three components of LP model as follows. For this, we take an example of two variables. Example 1

A manufacturing company produces two types of products A and B. The production department can produce 50 units of A and 25 units of B per day. However, the management has to confront with the problem of optimizing the daily production of the products as only 60 man-hours are available daily to assemble the products. The following additional information is provided.

T y p e o f t h e p r o d u c t P r o f i t / u n i t ( R s . )

A s s e m b l y t i m e p e r p r o d u c t ( h r . )

A B

3 0 2 0

2 . 5 1 . 5

The company assures the client at least 25 products in total can be produced. Formulate this problem as an LP model so as to maximize the total profit

Formulation of LP model Defining decision variables:

Let X1 units of A and X2 units of B should be produced. Defining objective function:

As the profit per unit of product A and product B are Rs. 30 and Rs. 20 respectively, the total contribution from the X1 units of A and X2 units of B are Rs. 30X1 and Rs. 20X2. The objective function is to maximize the sum of the total contribution from the production of A and B i.e. Max Z = 30 X1 + 20 X2

Defining constraints: Production constraint:

As the production department can produce 50 units of A and 25 units of B per day, the constraints are

X1 < 50 and X2 < 25 That is the department can't exceed the production of A and B by 50 and 25. Assembly time constraint:

As the assembly time (hour) for the product A and B are respectively 2.5 and 1.5 and only 60-man hours are available for assembly, the constraint is

2.5 X1 + 1.5 X2 < 60 Assurance constraint:

As the company assures at least 25 products in total can be produced, the constraint is X1 + X2 > 25 Minimum production constraints

The minimum production for each product is zero. Hence, X1 > 0 Y1 > 0 This is also known as non-negative constraint. Hence, the LP model can be summarized as follows: Max. Z = 30 X1 + 20 X2

Subject to X1 < 50 X2 < 25 2.5 X1 + 1.5 X2 < 60 X1 + X2 > 25 and X1, X2 > 0

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2.2 ASSUMPTIONS OF LINEAR PROGRAMMING While formulating the linear programming model, following assumptions are considered. Certainty : All parameters of a LP model such as availability of resources, profit contribution, cost of an unit, pattern of consumption of resources should be well known constants. However, in real world problem, this assumption may not be completely applied. In the preceding formulation we have been given marginal profit, production capacity, assembly time per-product along with total assemble time in the assembling department and order commitment of the company.

Linearity : In a LP model, the objective and constraints are all linear. Moreover, the effect of each decision variable in objective and constraints should be proportional to the value.In the preceding formulation assembly time (hour) per unit of A is 2.5, then producing 4 units of that product takes 2.5 × 4 = 10 hours of the assembly time available per day in the assembly department. Additivity : The total effect of all the decision variables in the objective and constraints should be the sum of the effect of each decision variable. In the preceding formulation, the objective function is to maximize the sum of the total contribution from the production of A and B. That is,

Max. Z = 30 X1 + 20 X2

Divisibility : The solution set in linear programming can be any fractional value. If only integer values are needed, integer programming method may be applied. 2.3. APPLICATION AREAS OF LPP The following are the some prominent application area of linear programming: Agricultural applications These applications fall in to categories of farm economics and farm management. The former deals with agricultural economy of a nation or region, while the later is concerned with the problem of individual farm. Linear programming can be applied in agricultural planning e.g.. Allocation of limited resources, such as acreage, labor, water supply and working capital etc. in a way so as to maximize net revenue. Military applications Military applications include the problem of selecting an air weapon system against enemy so as to keep them pinned down and at the same time minimizing the amount of fuel used. The variation transportation problems that maximize the total tonnage of bombs dropped on a set of targets and LP could solve the many more. Production management In production management, LP has the wide use on determination of product mix, or in production planning so as to maximize the total revenue and minimize the cost respectively. Similarly, LP has application on areas like : – Assembly line balancing. – Blending problems. – Product formualtions etc.

Financial management In financial management, LP is used in portfolio selection and profit planning. In portfolio selection, LP deals with the selection of specific investment activity among several other activities. The objective is to find the allocation, which maximizes the total expected return or minimizes risk under certain limitation. In profit planning LP deals with the maximization of profit margin from investment in plant facilities and equipment, cash in hand in inventory. Marketing management In marketing LP is used for : – Media selection. – Traveling sales man problem. – Physical distribution. Personnel management In personnel management LP is used for: – Staffing problem – Determination of equitable salaries. – Job evaluation and selection.

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Other applications of linear programming lie in the area of administration, education, fleet utilization, contract awarding, hospital administration and capital budgeting etc. (Fleet means vehicles like ships, aircraft etc.) 2.4. ADVANTAGE OF LP Following are few prominent advantage of LP. 1. LP helps in attaining the optimum use of productive resources. It also indicates how a decision maker

can employ his productive factors effectively by selecting and distributing (allocating) these resources. 2. LP techniques improve the quality of decisions. The decision making approach of the user of this

technique become more objective and less subjective. 3. LP techniques provide possible and practical solutions, since there might be other constraints operating

outside the problem, which must be taken in to account. Just because we can produce so many units does not mean that they can be sold. Thus, necessary modification of its mathematical solution is required for the sake of convenience to the decision maker.

4. Highlighting the bottlenecks in the production process is the most significant advantage of this technique. For example, when bottle neck occur, some machine can not meet demand while other remains ideal for some of the time.

5. LP also helps in re-evaluation of a basic plan for changing conditions. If conditions changed when the plan is partly carried out.

2.5 GRAPHICAL SOLUTION OF LP PROBLEMS Graphical method of solving LP is basically designed for two variable models. In real world problem there could be even more than thousands of variables and constraints. For the solution of LP model consisting many variables and constraints we will use software. Graphical solution of LPP will give us core idea of solving the problem leading the development of advance solution technique. 2.5.1 PROCEDURES OF GRAPHICAL SOLUTION Formulation of LP model as per the provided information Determination of feasible solution space or feasible region:

The given inequalities are being replaced by equality sign resulting into straight lines. The straight lines are plotted on graph paper. For plotting the graph of the straight line of the equality equation, the properties of axis are used. In ordinate the value of abscissa is zero and in abscissa value of the ordinate is zero. The area of the given constraint is decided as per the respective inequality sign. The common portion of the graph is shaded which satisfies all given constraints, known as feasible solution space.

Determination of the optimal solution: Next is to identify the extreme points of the feasible region and determine the optimal solution among all

the extreme points. It is done because the optimum solution of LP will occur at only extreme point. 2.5.2 GRAPHICAL SOLUTION , EXAMPLE OF A MAXIMIZATION MODEL

Considering the example 1, Let the horizontal axis (abscissa) X1 and the vertical axis (ordinate) X2 represent the number of units of the products A and B that should be produced by a company to maximize the profit. As the minimum production for each product is zero, we have

X1 > 0, X2 > 0 (non-negative constraints) The non negative constraints restrict the feasible solution space area to the first quadrant of the graph. Considering only equality signs, we rewrite the constraint equation as follows:

X1 = 50 ... (i) X2 = 25 ... (ii) 2.5X1 + 1.5X2 = 60 ... (iii) X1 + X2 = 25 ... (iv)

From (i) X1 = 50 indicates that the straight line is parallel to vertical axis and taking (0, 0) as a reference point for X1 < 50, we get 0 < 50, this satisfied condition concludes that the feasible side of the constraint X1 < 50 includes the origin.


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From (ii) X2 = 25 indicates that the straight line is parallel to horizontal axis and taking (0, 0) as a reference point for X2 < 25, we get 0 < 25, this satisfied condition concludes that the feasible side of the constraint X2 < 25 includes the origin. From (iii) 2.5X1 + 1.5X2 = 60, when X1 = 0, X2 = 40 and when X2 = 0, X1 = 24. Hence, the straight line passes through (0, 40) and (24, 0). Taking (0, 0) as a reference point for 2.5X1 + 1.5X2 < 60, we get 0 < 60, this satisfied condition concludes that the feasible side of the constraint includes the origin. From (iv) X1 + X2 = 25 when X1 = 0, X2 = 25 and when X2 = 0, X1 = 25. Hence, the straight line passes through (0, 25) and (25, 0). Taking (0, 0) as a reference point for X1 + X2 > 25, we get 0 > 25, this unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin.

Graphical representation of given LP problem

The feasible region is ABCD as shown in the graph which is shaded. The extreme points are A(0, 25), B(9, 25) and C(25, 0) respectively. These extreme points could also be determined by solving the simultaneous equations if graph paper is not used or fraction value is observed.. To find the optimal solution from among all the extreme points, let’s evaluate each extreme point in objective function.

At A(0,2 5), z = 30 × 0 + 25 × 20 = 500 At B(9, 25), z = 30 × 9 + 25 × 20 = 770 At C(25, 0), z = 30 × 25 + 0 × 20 = 750

Hence, the company should produce the product mix of 9 units of product A and 25 units of product B to obtain maximum profit Rs. 770. 2.5.3 GRAPHICAL SOLUTION OF LP OF A MINIMIZATION MODEL Example 2

Dhital Breweries Ltd. has two bottling plants, one located at 'G' and the other at 'J'. Each plant produces three drinks: Whisky, Beer and Brandy, named A, B and C respectively. The number of bottles produced per day are as follows:

P l a n t L o c a t i o n D r i n k G J

W h i s k y 1 5 0 0 1 5 0 0 B e e r 3 0 0 0 1 0 0 0 B r a n d y 2 0 0 0 5 0 0 0

A market survey indicates that during the month of July, there will be a demand of 20,000 bottles of whisky, 40,000 bottles of beer and 44,000 bottles of brandy. The operating cost per day for plants at G and

10 20 30 40 50 60 O

X2

X1 70 80 90

10

20

30

40

50

60

70

80

90

100

X1 = 50

100

X2 = 25

C

BA

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J are 600 and 400 monetary units. For how many days should each plant be run in July so as to minimize the production cost, while still meeting the market demand?

F o r m u l a t i o n o f L P M o d e l Defining decision variables

Let X1 and X2 be the number of days that plants at G and J should be run in July. Defining objective function

As the operating cost per day for plants at G and J are 600 and 400 monetary units respectively, the total operating cost from the X1, working days in G and X2 working days in J are 600 X1 and 400 X2. The objective function is to minimize the sum of the total cost from the working days of plants at G and J. That is, Min. Z = 600X1 + 400X2

Defining constraints

Whisky constraint: As the number of bottles of whisky produced at plant G and J are 1500 each and there will be a demand of 20,000 bottles per day, the constraint is

1500X1 + 1500X2 > 20,000 Beer constraint: As the number of bottles of beer produced at plant G and J are 3000 and 1000 and there will be a demand of 40,000 bottles per day, the constraint is

3000X1 + 1000X2 > 40,000 Brandy constraint: As the number of bottles of Brandy produced at plant G and J are 2,000 and 5,000 and there will be a demand of 44,000 bottles per day, the constraint is

2000X1 + 5000X2 > 44,000 Minimum operating constraints. The minimum operating days of each plant is zero, hence, X1 > 0 and X2 > 0. This is also known as non-negative constraints.

Hence, the formulated LP model can be summarized as follows: Min. Z = 600 X1 + 400 X2

Subject to 1500X1 + 1500X2 > 20,000 3000X1 + 1000X2 > 40,000 2000X1 + 5000X2 > 44,000 and X1, X2 > 0

G r a p h i c a l S o l u t i o n Let the horizontal axis X1 and the vertical axis X2 represent the number of days that plant at G and J should be operated. As the minimum operating days of each plant is zero, we have

X1 > 0, X2 > 0 These two constraints restrict the feasible solution space area to the first quadrant. Considering only equality signs, we rewrite the constraint equation as follows:

1500X1 + 1500X2 = 20,000 ... (i) 3000X1 + 1000X2 = 40,000 ... (ii) 2000X1 + 5000X2 = 44,000 ... (iii)

From (i) 1500X1 + 1500X2 = 20,000 when X1 = 0, X2 = 403 and when X2 = 0, X1 =

403 . Hence, the straight line passes

through

0‚

403 and

40

3 ‚ 0 . Taking (0, 0) as a reference point for 1500X1 + 1500X2 > 20,000, we get 0 > 20,000, this

unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin.

From (ii) 3000X1 + 1000X2 = 40,000, when X1 = 0, X2 = 40 and when X2 = 0, X1 = 403 . Hence, the straight line passes

through (0, 40) and

40

3 ‚ 0 . Taking (0, 0) as a reference point for 3000X1 + 1000X2 > 40,000, we get 0 > 40,000, this

unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin.


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From (iii) 2000X1 + 5000X2 = 44,000, when X1 = 0, X2 = 445 and when X2 = 0, X1 = 22. Hence, the straight line passes

through

0‚

445 and (22, 0). Taking (0, 0) as a reference point for 2000X1 + 5000X2 > 44,000, we get 0 > 44,000, this

unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin. The feasible solution space ABC is shown in the graph. The extreme points are A(0, 40), B(12, 4) and C(22, 0) respectively. To determine the optimal solution from among all extreme points, let's evaluate each extreme point in objective function.

At A(0, 40), z = 600 × 0 + 400 × 40 = 16,000 At B(12, 4), z = 600 × 12 + 400 × 4 = 8,800 At C(22, 0), z = 600 × 22 + 400 × 0 = 13,200

Hence, the company should operate the plants at G and J respectively for 12 days and 4 day at the minimum production cost Rs. 8,800. 2.5.4 SPECIAL CASES IN LINEAR PROGRAMMING Alternative (or multiple) optimal solutions In LP problem we may have various extreme points (solution sets) resulting the same objective value. Each solution set is known as alternative optimal solution. The alternative optimal solution exists off the slope of the objective function is equal to the slope of a constraint which forms a boundary line of the feasible solution space. Example 3

Solve graphically Max Z = 5X1 + 3X2

Subject to 10X1 + 6X2 < 60 X1 + 2X2 < 18 and X1, X2 > 0

Solution:

Let horizontal axis and vertical axis be X1 and X2. Let's consider non-negative constraints X1 > 0, and X2 > 0 which restricts the feasible solution space to the first quadrant.

Considering only equality signs, we rewrite the constraint equations as follows: 10X1 + 6X2 = 60 ...(i)

10 20 30 40 O

X2

X1

10

20

30

40

C B

A

B

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X1 + 2X2 = 18 ...(ii) From (i) 10X1 + 6X2 = 60 when X1 = 0, X2 = 10 and when X2 = 0, X1 = 6. Hence the straight line passes through (0, 10) and (6, 0). Taking (0, 0) as a reference point for 10X1 + 6X2 < 60 we get 0 < 60, this satisfied condition concludes that the feasible side of the constraint includes, the origin. From (ii) X1 + 2X2 = 18, when X1 = 0, X2 = 9 and when X2 = 0, X1 = 18. Hence, the straight line passes through (0, 9) and (18, 0). Taking (0, 0) as a reference point for X1 + 2X2 < 18, we get 0 < 18, this satisfied condition concludes that the feasible side of the constraint includes the origin.

The feasible solution space OABC is shown in the graph.

The extreme points are O(0, 0), A(0, 9), B

6

7‚ 607 and C(6, 0).

To determine the optimal solution from among all extreme points, let's evaluate each extreme point in the objective function.

At O(0, 0), Z = 0 At A(0, 9), Z = 5 × 0 + 3 × 9 = 27

At B

6

7‚ 607 , Z = 5 ×

67 + 3 ×

607 = 30

At C(6, 0), Z = 5 × 6 + 3 × 0 = 30

As two extreme points B

6

7‚ 607 and C(6, 0) results the max value 30, the points are alternative optimal

solutions.

U n b o u n d e d S o l u t i o n The LP problem with maximization as objective function may not have a finite solution. The case is said to have unbounded solution Example 4

Solve graphically Max. Z = 5X1 + 7X2 Subject to 2X1 - 2X2 > 2 2X1 + 2X2 > 8 and X1, X2 > 0

Solution: Let horizontal axis and vertical axis be X1 and X2. Let's consider non-negative constraints X1 > 0, and X2 > 0 which restricts the feasible solution space to the first quadrant.

Considering only equality signs. We rewrite the constraint equation as follows: 2X1 - 2X2 = 2 ...(i) 2X1 + 2X2 = 8 ...(ii)

From (i) 2X1 - 2X2 = 2 when X1 = 0, X2 = - 1 and when X2 = 0, X1 = 1. The straight line passes through (0, -1) and (1, 0).

2 4 6 8 10 12 O 14 16 18

2

4

6

8

10

C

A

B A

X2

X1


20

-1 1 4

4

Taking (0, 0) as a reference point for 2X1 - 2X2 > 2, we get 0 > 2, this unsatisfied condition conclude that the feasible side of the constraint does not include the origin. From (ii) 2X1 + 2X2 = 8, when X1 = 0, X2 = 4 and when X2 = 0, X1 = 4. Hence, the straight line passes through (0, 4) and (4, 0). Taking (0, 0) as a reference point for 2X1 + 2X2 > 8, we get 0 > 8, this unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin.

The feasible solution space in unbounded. In which, the value of Z increases as the variables X1 and X2 increase. Thus, the problem has an unbounded solution. Infeasible solution All the linear programming problem can not be solved provided that there is no contradiction in the given constraints. However, if some or all the constraints of the LPP are contradictions to each other, the problem will not have common feasible region. Therefore, the problem is said to be infeasible.

Example 5 Solve graphically Max. Z = 12X1 - 4X2

Subject to X1 + 2X2 < 2 2X1 + 4X2 > 8 X1 = 6 ands X1, X2 > 0

Solution Let horizontal axis and vertical axis be X1 and X2. Let's consider non-negative constraints X1 > 0, and X2 > 0 which restricts the feasible solution space to the first quadrant.

Considering only equality signs, we rewrite the constraint equation as follows:

X1 + 2X2 = 2 ... (i) 2X1 + 4X2 = 8 ... (ii) X1 = 6 ... (iii)

From (i) X1 + 2X2 = 2, when X1 = 0, X2 = 1 and when X2 = 0, X1 = 2. Hence, the straight line passes through (0, 1) and (2, 0). Taking (0, 0) as a reference point for X1 + 2X2 < 2, we get 0 < 2, this satisfied condition concludes that the feasible side of the constraint includes the origin. From (ii) 2X1 + 4X2 = 8, when X1 = 0, X2 = 2 and when X2 = 0, X1 = 4. Hence, the straight line passes through (0, 2) and (4, 0). Taking (0, 0) as a reference point for 2X1 + 4X2 > 8, we get 0 > 8. This unsatisfied condition concludes that the feasible side of the constraint doesn't include the origin. From (iii) X1 = 6 indicates that the straight line is parallel to vertical axis.

X1O

B

A

X2

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The graph shows two shaded areas where we can't find any solution set (X1, X2) satisfying all the given constraints. Hence, the LPP has an infeasible solution.

2.6 SIMPLEX METHOD The graphical methods of linear programming are limited to problems which have only a few variables and constraints. This method has some limitations. By this graphical method only two or at the most three variables linear programming problem can be solved. As we know that more than three variables need more dimensions, which cannot be graphed. However, it should be noted that linear programming problem having two variables is easier to plot in two dimensions graph rather than three dimensions graph for three variables. As the number of variables and constraints of linear programming are increased, it will be quite difficult to handle the problems by the graphical method. The linear programming is relating to decision problem which needs to optimize. Most of the decision problems involve not two but many more variables. Hence, to wave limitation of graphical solution we need to look alternative technique to handle linear programming problem having many more variables. Various methods have been developed for solving optimization problem but simplex method is efficient means of solving complex linear programming involving large number of variables. Although this method is a powerful analytical mechanism but the basic idea is quite simple. The simplex method provides a systematic algorithm (a rule of iterative procedure) which consists of moving from one basic feasible solution to another (from one corner point to another corner point of the feasible area) in such a way that the value of the objective function at the succeeding corner is more than the preceding corner of feasible area in case of maximization and minimum in case of minimization. The iterative process of moving from corner to corner i.e., one basic feasible solution to another basic feasible solution is repeated untill the best solution is reached i.e., we get extreme point or optimal corner of feasible solution. Therefore, the essence of the simplex method is to start with some initial basic feasible solution and computes the value of objective function and then see whether the value of the objective function can be improved upon by moving to adjacent basic feasible solution. The peculiarity of the method is that it always moves towards the optimal solution. Finally, this method of solution provides an indicator which determines when an optimum solution has been reached. We begin our discussion of the problem of linear programming and the simplex method by considering the example given below.

2.6.1 SOLVING MAXIMIZATION PROBLEM BY SIMPLEX METHOD

We will be discussing the solving of the maximization linear programming problem with the help of example 6

Example 6 Time available on each of the two machines M1 and M2 of a firm is to be allocated to the production of two products P1 and P2. The machines viz. M1 and M2 have 80 are 60 hours of time available on them respectively. The two products P1

and P2, require for their production different amount of time on each of two machines as shown in the table below.

1 2 3 4 5 6 O

1

2

C

X2

X1

X1 = 6


22

Time required per unit of product

P r o d u c t O n m a c h i n e M 1 O n m a c h i n e M 2 P 1 P 2

2 h r s . 4 h r s .

3 h r s . 2 h r s .

T i m e a v a i l a b l e 8 0 h r s . 6 0 h r s . Each unit of product P1 is yielding a profit of Rs. 60 and each unit of product P2 is yielding a profit of Rs. 50. The firm wishes to determine the product mix so as to maximize profit.

Clearly, it is a problem of resource allocation which can be stated as follows. "The firm wishes to determine the number of units of product P1 and the number of units of product P2

which maximize the profit without violating the constraints arising out of limited machine time." Since, the data are entirely quantitative, they can be translated into a group of mathematical expressions. We employ the following decision variables, X1 = the number of units of product P1 that we shall produce X2 = the number of units of product P2 that we shall produce Z = the value of the total profit Now, Using the linearity assumption, above problem can be portrayed by an algebraic function as follows:

Objective function Max. Z = (Profit/unit × number of units of product P1) + (Profit/unit × number of unit of product P2

= 60 X1 + 50 X2 ... (1) Constraints:

The limited times available on-machine M1 and M2 possess two resource constraints. Each constraint generate one algebraic functions. For machine M1 linear constraint function can be formulated as follows: (2 hrs./unit of product P1) × (number of units of product P1)+ (4 hrs./unit of product P2) × (number of units of product P2) must not exceed 80 hours, or 2X1 + 4X2 < 80 ... (2) Similarly, For machine M2, linear constraint function ca be formulated as follows. (3 hrs./unit of product P1) × (number of units of product P1) + (2 hrs./unit of product P2) (number of units of product P2) must not exceed 60 hours, or 3X1 + 2X2 < 60 ... (3) Besides, above two linearity constraints inequalities pertaining to machine times 80 hours and 60 hrs. of M1 and M2 respectively, we need to add one more constraint known as non-negativity. That is, neither X1

nor X2 can be negative, o r all variables > 0 ... (4) Since, X1 and X2 are physical quantities, they ca not be negative. In precise notation the mathematical formulations of the above linear programming problem can be written as follows: Max. Z = 60X1 + 50X2

Subject to 2X1 + 4X2 < 80 3X1 + 2X2 < 60 ... (5) X1, X2 > 0

Step 1: Strict equality equation

Arrange the problem in a set of exact equation:

In order to solve the problem by iterative process under simplex method for arriving at optimal solution, for maximization problem, we first transform all the inequality constraints into strict exact equations by introducing extra variable (slack variable for inequality functions lesser than or equal to "<" type) for each of the constraints and modify the objective function adding slack variables with zero coefficients.

Slack variables: The economic meaning of the slack variable is the amount of the available resources unused. Therefore, the slack variable represents idle resources which serve to indicate how much of a particular resource is unused in any solution. It can be calculated by taking differences between available resource and requirement obviously the idle resource has no effect and contributing zero to the objective function. Therefore, zero coefficients is used while adding slack variable in objective function to eliminate its effect.

Note that slack variables must also be constrained to be non-negative. If they were allowed to become negative, the given original constraint (for instant, 2X1 + 4X2 < 80) would be violated. Further, slack variable can not be larger than constant to the right of the equal sign. Also note that when slack variable is zero, it should be understood that all of the available resource is used during production process. After introducing slack variables, our LP problem (5) becomes

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Max. Z = 60X1 + 50X2

Subject to 2X1 + 4X2 + S1 = 80 ... (6) 3X1 + 2X2 + S2 = 60 X1, X2, S1, S2 > 0 where, s1 and s2 are slack variables which correspond to appropriate unused resources of machine times M1 and M2 respectively.

In order to apply the simplex method in solving LP problem, it is necessary to include all the unknown variables in each equation. Therefore, it is necessary to include slack variables S1 and S2 in each equation. However, it is also necessary to eliminate effect of any variables on the equation to which it does not apply. For this purpose a zero coefficient is used to any variable on the equation which has no effect. In case of slack variables which have an effect on the equation, a coefficient of 1 is used. Considering all the points mentioned as above, the reorganized LP problem appears as follows:

Max. Z = 60X1 + 50X2 + 0S1 + 0S2, Row 0 s.t. 2X1 + 4X2 + 1S1 + 0S2 = 80, Row 1 (7) 3X1 +2X2 + 0S1 + 1S2 = 60, Row 2

This reorganized problem is also known as standard format of linear programming.

Since, simplex method is an iterative technique, the optimal solution of (7) repetitive scheme over and again until an optimal solution is reached. Clearly this method needs one feasible solution to the problem as a starting (as reference) or initial solution. Keeping this starting solution as basis, rules are used to guide computations to obtain improve feasible solutions and finally optimal solution. An optimal answer of all the variables is conveniently read from the table.

Step 2: Tabular Format: The initial solution and all the iterative of the linear programming problem under simplex method can be arranged in the tabular format, which is a major advantage of this method. Once the problem is properly organized, the coefficients of the equations can be presented in the simplex tableau. This tabular form makes relatively easier to handle the equations. To explain meaning and significance of the simplex tabular, equations R0 through R2 of (7) are presented in the following initial tableau. Initial simplex tableau Besides, the number of equations, labels and two extra rows are introduced in the table. Each and every portion of the tableau has significance meaning. Now, it's role and some other important concepts will be described as follows. Decision variables: Each and every decision variable is listed as headings. The decision variables are the variables of the given equations of LP problems. All the coefficients (numeral values of the equations) are put in the column headings accordingly.

60 50 0 0

x1 x2 s1 s2

Decision variables column

2 4 1 0

3 2 0 1

80 60

S1 S2

Row 1: Row 2:

0 0 0 0 0 Zj (Z1) (Z2) (Z3) (Z4)(Z0)

60 50 0 0 Index row Cj - Zj

Identity matrix

Constraints

Profit is lost if one unit is brought into

the solutionPotential profit

per unit

Basic quantity (Constant) = b

Units of activity in the solutions

Basic variables - XB

Basic activity

Cj column - profit/unit

Cj ← Cj Row

← Variable row


24

Constraints: All number of constraints are shown in rows of the simplex tableau. In our particular problem Row 1 and Row 2 represent the time available in machine M1 an machine M2 respectively. It is noted that the coefficients of all the constraint equations are shown in the body of the simplex tableau. The variable corresponding to any coefficient of the equation is identified by the column heading. The constants of each constraint equation are placed at the left of the table in the quantities column (b). Basic Variables The basic variable columns contains all the variables which form on identity matrix. The identity matrix is a square matrix whose diagonal elements are positive 1 and all off diagonal elements are zero and its order is equal to the number of constraints. Note that the basic variables columns contains the variables in the solutions. In other words the variables which have solutions are shown in the basic variable column of the simplex tableau. The respective constant values of the variables are obtained from the quantities column. The slack variables s1 and s2

form the identity matrix. Thus, slack variables s1 and s2 defined as basic variables and corresponding constant values 80 and 60 of quantities column are the solutions of s1 and s2 respectively. In addition, the variable which are not corresponding to identity matrix are called non-basic variables. Thus, X1 and X2 are non-basic variable which are not corresponding to identity matrix. The non-basic variables have zero values. Thus, two types of variables are used in the simplex tableau basic and non-basic. The variable corresponding to identity matrix are called basic variables and remaining ones are non-basic variables. Basic variables generally have, non-zero values while non-basic variables have zero values. Basic Solution A basic solution means the solution at the extreme or corner point. Since non basic variables take zero and basic variables take non-zero values, therefore, only those variables which lies on the basis of the solution are basic variables. The solution due to such basic variables constitute basic solution. Some time, the basic solution is also called the basis. The underlying concept of simplex method is that an orderly search of corner points is made in the beginning and proceeds towards the optimum solution step by step. This is done by bringing one of the non-basic variables into basis and one of the former basic variable into non-basic (set equal to zero). At each step the values of the basic variables are determined and using their effect the value of the objective function is calculated. Initial Basic Solution The simplex method starts form an initial solution by setting non-basic variables equal to zero and solving for basic variables and then proceeds step by step to subsequent solution in a systematic manner. In case of maximization at each step it ensured that (i) the profit is increase in each new solution. (ii) the solution will always satisfy the given constraints (always remain feasible). For the maximization problem one can get the initial solution at the origin i.e. X1 = 0 and X2 = 0. In other words, one can start making decision of no production initially meaning there by that keeping the capacity on each machine totally idle i.e. S1 = 80 and S2 = 60 which yielding zero profit. This solution is, some time, called zero solution. The simplex table 10.1 is itself the initial basic solution where the non basic variables X1 and X2 are equal to zero and quantities column identifies the values of the basic variables. In the simplex tableau, the column Cj indicates the profit per unit of each basic variable in the solution. It can be seen in the Table 10.1 that Rs. 0 profit is earned from each unit of slack variable S1 and S2. This is because, the slack variable mean unused resource and assists in earning only when it is used to make product. The Cj row indicates the profit per unit attainable from each of the variables. The values corresponding this Cj

row are coefficients of the objective function. It is used to calculate Zj - Cj row (index row) as explained below: In the simplex tableau, the Zj row refers to the values of the objective function. The quantity column shows the values of the objective function in the basic solution which will be used to calculate the amount of Zj from the basic variable as follows:

Units of S1 times profit per unit 80 hrs. × Rs. 0 = Rs. 0 Units of S2 times profit per unit 80 hrs. × Rs. 0 = Rs. 0 Total profit due to S1 and S2 = Rs. 0

Every elements of the Zj rows in the variable columns indicates the amount by which profit would be reduced if one unit of associated variable is brought into the basic (its meanings will be clear after next improve solution). For the initial solution, it can be seen that the introduction of non-basic variable will not reduce total profit. For

example, if we want to produce 1 unit of X1 (product P1), the entries. 23 under X1 column of the initial solution

of the simplex tableau 10.1 indicates that 2 hours of slack variable S1 and 3 hours of slack variable S2 must be given up. Since, unused time contributes zero value to the objective function Z, therefore, there will be no reduction in profit. The Zj value of the variable column X1 is calculated as follows:

Linear Programming

25

Number of hours of slack variable s1 given up = 2 times the profit of S1 × Rs. 0 = Rs. 0 Number of hours of slack variable s2 given up = 3 times the profit of S2 x Rs. 0 = Rs. 0 Total profit reduce (z0) = Rs. 0

Proceeding on the same line the values of Zj row for other variable columns are also calculated. Therefore, the entries of Zj row are calculated by multiplying the Cj of the row by the number in the row and jth column, and summing (adding). The detail computation of Zj's of table 2.1 are shown below.

Z0 = 80 × (0) + 60 (0) = 0 Z1 = 2 × 0 + 3 (0) = 0 Z2 = 4 × 0 + 2 (0) = 0 Z3 = 1 × 0 + 0 (0) = 0 Z4 = 0 × 0 + 1 (0) = 0

Here Z0 represents the profit of the current solution or no production. Finally, a index row Cj - Zj is added in the simplex tableau. The entries of this row is obtained by subtracting Zj

amount from Cj to find out Cj - Zj row values. The corresponding entries of Cj - Zj row indicate the profit that will result from adding one unit of a non-basic variable to the solution. For instance, it can be seen that Rs. 60 is added to objective function i.e. to the profit by producing one unit of X1 (product P1). By producing one unit of X1, Rs. 60 is gained and no other current profit is lost. That is, available resources can be used as per its requirement in order to produce one unit of X1 without reducing current production of any product. Hence, the net change is (Cj - Zj ) = Rs. 60. Proceeding on the same line, net profit per unit of each decision variables can be obtained. In the nut shell, little efforts are required for the simplex method. In this method again LP problem have to be reorganized property into standard format and put into the simplex tableau for the initial solution. The next step is searching for improved solution. Before, proceeding toward improve solution let us rewrite the initial simplex tableau in simple form as given in table 1 Simplex tableau 1

C j R s . 6 0 R s . 5 0 R s . 0 R s . 0 R a t i o B a s i c

v a r i ab l e s

Q u a n t i t i e s ( c o n s t a n t s )

X 1 X 2 S 1 S 2

R o w 1 : R s . 0 S 1 8 0 2 4 1 0 8 0 / 2 = 4 0 R o w 2 : R s . 0 S 2 6 0 3 2 0 1 6 0 / 3 = 1 5 m i n .

Z j R s . 0 R s . 0 R s . 0 R s . 0 R s . 0 C j - Z j R s . 6 0 R s . 5 0 R s . 0 R s . 0

Step 3: Computing improved solution: Simplex methods examine fewer corner points in solving the LP problem. The discussion now basically focused on the detail procedure for obtaining improved situations. (a) : Table 1 is used for detail calculation. The underlying concept of simplex method is searching for some new solution relatively better than the solution already found. For this purpose identify the variable (a product) which can be brought into the solution which makes the profit more. The Cj - Zj row of the simplex tableau identifies such variable. The rule is to select a decision variable which has the greatest potential profit when variable is brought into a solution. Obviously it would be the variable of the column which has maximum number with a positive sign in index row (Cj - Zj ). Here, X1 is the variable selected because it makes highest profit per unit Rs. 60. Therefore, the column X1 of the decision variable is named as key column (pivot column). It is noted that key column is always selected from one of the none basic variable. Thus, the role of key column guides the choosing the most beneficial product i.e., high profit product. (b) : After choosing key column next question arises how many units of X1 (product P1) allowed to produce. However, the number of units produce is decided by the constraints given in the problem. By examining initial solution it has been found that if 1 unit of X1 is added, 2 units of S1 must also be given up and 3 units of S2 must be given up. Note that a variable can not be withdrawn more than it's availability. For instance, if 50 units of X1 is produced it requires 50 × 2 = 100 units of S1, which is not possible as only 80 unit of S1

Key row

Pivot element (key element)

Key column


26

is available. Thus, to make proper decision about the maximum number of units of variable that can be produced, the rule of key row can be used. The smallest ratio which is obtained by dividing value of the quantity column by corresponding key column entries is called key row. Since, X1 column is the key column, the ratio (displacement quotient) are calculated as follows:

Row 1: 802 = 40

Row 2: 603 = 15 smallest

The row 2 has smaller ratio, therefore, it is called key row. The reason for selecting row 2 as key row is that any other row would cause the S2 variable value negative and consequently infeasible. Therefore, the key row selection rule is designed to maintain feasibility by selecting the most limiting constraint. In other words, selection of key row rule always makes most plausible solution of the selected non basic variable brought into the basis without violating all rest of the constraints. However, it should be further noted that the negative coefficient of the key column should not be used for computing the ratio. As it makes the solution of the selected variable negative this makes contradictory to all the variables positive. Similarly, the zero coefficient of key column is taken as positive that makes the ratio infinite and consequently never becomes key row because it is undefined. Table 2.2 is the initial tableau where the key column and key row are identified and indicated by column and row respectively. The coefficient which lies at the intersection of key column and key row is termed as key element or pivotal element. This element is the focal point of the exchange variables basic and non basic status. For instance, the key element is 3 which lies at intersection of key column of variable X1 and key row 2 of tableau 2.2. (c) : After determining key element next step is to bring the selected non basic variable X1 of key column into basic variable. For that it is necessary to convert key element into positive one and rest of the elements of key column into zero which forms unit vector. This brings X1 into basic solution which replace the old basic variable S2 corresponding to key row (Row 2). Therefore, X1 is some time, called entering variable (choice variable) and S2

out going or exiting variable. Once non basic variable X1 enter into the basic replacing old basic variable S2, the solution of LP gets improved. Improve solution can be developed by this simple rule and presented in next tableau. First, the coefficients key row is updated. To update the entries of key row, all the elements of old row are divided by key element and named it as replacing row.

Thus, for example, on dividing all the coefficients of old row by 2 we get : 60, 3, 2, 0, 1 . On dividing by by key element 3, we obtain replacing row as : 20, 1, 2/3, 0, 1/3 respectively. The updated new row is shown in next improve tableau 2. It should be noted that X1 will be now basic variable in the new tableau and S2 will be listed as non basic variable. (d) : Update the entire reaming basic variable rows of the old table for obtaining improve table with the help of replacing row. For instance to update all the coefficients of basic variable row 1, we proceed as follow.

Row 1: 80 2 4 1 0 Replacing row or New row 2: 20 1 2/3 0 1/3 Row 1: 80 2 4 1 0 2 × Replacing row 40 2 4/3 0 2/3 – – – – – New row 1: 40 0 8/3 1 -2/3

It is simply the simultaneous solving of two equations Row 1 and Row 2. This makes the computation little bit time consuming. However, the coefficients of the basic variable row 1 can be calculated by using following formula.

× 2

Linear Programming

27

Elements of old row –

Intersecting

elementof old rowwith key column

×

Corresponding

elementsof replacing row

= Elements of new row

For instance R1 → R1 - 2 Repeated row (Old row 1) – 2 × (Replacing row) = New row 1 80 – 2 × 20 = 40 2 – 2 × 1 = 0 4 – 2 × 2/3 = 8/3 1 – 2 × 0 = 1 0 – 2 × 1/3 = – 2/3 Above formula will be used for updating coefficient of the basic variable rows now and onward. This formula, once mastered, greatly simplify the updating are presented in the improve tableau 10.3. (e): Finally, the fifth sub step is updating the Zj and Cj - Zj row of the old tableau for obtaining new improve tableau. As mentioned earlier in initial tableau 2.2. Each Zj is calculated by summing the products of Cj column coefficients times the corresponding coefficient in jth column,

Z0 = 50 × 0 + 60 × 20 = 1200 Z1 = 0 × 0 + 60 × 1 = 60 Z2 = 0 × 8/3 + 60 × 2/3 = 40 Z3 = 0 × 1 + 60 × 0 = 0 Z4 = 0 × -2/3 + 60 × 1/3 = 20

These Zi's values are used to calculate index row Cj - Zj . The new values of each Cj - Zj are found by simply subtracting Cj from Zj values. The complete improve second tableau 2 along with new key column new key row and new key element as follows. Simplex tableau 2

C j R s . 6 0 R s . 5 0 R s . 0 R s . 0 R a t i o

B a s i c V a r i a b l e s

Q u a n t i t i e s ( c o n s t a n t s )

X 1 X 2 S 1 S 2

R o w 1 : 0 S 1 4 0 0 8 / 3 1 - 2 / 3 4 08 / 3 = 1 5 m i n .

R o w 2 : 6 0 X 1 2 0 1 2 / 3 0 1 / 3 2 02 / 3 = 3 0

Z j 1 2 0 0 6 0 4 0 0 2 0

C j - Z j 0 1 0 0 - 2 0

Step 4: Search for future improve solution.

The fourth step of simplex method is searching for another future improve solution. This step determine whether profit can be improve further or not. Since, there is one more high value with positive sign (+ 10) in the index row Cj - Zj second improve tableau 2.3 is not optimal. By inserting non basic variable X2 in the basis and replacing S1, the total profit Z can be further improved. It suggests next third table can be developed for obtaining improved solution. For this step 3 is repeated until no further improvement can be made. That is, step 3 is repeated until all the values of index row Cj - Zj are less than or equal to zero . Since second tableau 2.3 is not optimal proceeding on the same line as earlier, third improve tableau is developed as follows. Key column: X2 column of variable column

Key row, Row 1: corresponding to minimum ratio 15. Key element: 8/3

Dividing key row Row 1 by 8/3 or multiplying key row Row 1 by 3/8, we get replacing row Row 1: 15 0 1 3/8 – 1/4

Updating Row 2:


28

Old row 2 — 2/3 × Replacing row = New row 2 20 — 2/3 × 15 = 10 1 — 2/3 × 0 = 1 2/3 — 2/3 × 1 = 0 0 — 2/3 × 3/8 = - 1/4 1/3 — 2/3 × – ¼ = 1/2

The procedure of computing Zj is same. Each value of Zj is sum of products of Cj column coefficients times the corresponding coefficient in jth column. The index row Zj - Cj is the difference of calculated Zj and Cj. All updated rows thus, obtained, are summarized in the following completed third tableau. Simplex tableau 3

C j R s . 6 0 R s . 5 0 R s . 0 R s . 0 B a s i c

a c t i v i t y Q u a n t i t i e s ( c o n s t a n t s )

X 1 X 2 S 1 S 2

R o w 1 : 5 0 X 2 1 5 0 1 3 / 8 - 1 / 4 R o w 2 : 6 0 X 1 1 0 1 0 - ¼ ½

Z j 1 3 5 0 0 6 0 5 0 1 5 / 4 3 5 / 2 C j - Z j 0 0 - 1 5 / 4 - 3 5 / 2

Since all the coefficients of index row (Cj - Zj ) < 0, this indicates an optimum solution has been arrived at. Therefore, the optimum solution is Maximum profit Z = Rs. 13500

When, X1 = 50 X2 = 60 S1 = 0 S2 = 0

2.6.2 BIG M METHOD

So far we have discussed the simplex method for solving linear programming problem of the constraint sign lesser–than–or–equal–to type. However, the constraints sign of constraints some time may be greater–than–or–equal to type as well as initially equality type. Then the solution the problem will be some how initially differ from the lesser–than–or–equal–to type LP problem. Greater–than–equal–to type constraint sign (>) When a constraint sign is of the type greater–than–or–equal–to (>) two steps are necessary to convert it into the standard format in the beginning. First subtract surplus variable to convert the constraint greater–than–or–equal–to to exact equality equation. Second add artificial (dummy variable) to obtain the initial basic feasible solution.For explanation of the two steps needed in case of greater–than–or–equal–to type we can take a simple example. Example 7

Max. Z = X1 + 2X2

Subject to X1 + X2 < 10 (1) 7X1 + X2 < 40 X1 + X2 > 12 X1, X2 > 0 The problem can be solved by simplex method. The problem is reformulated initially as follows. Insert the slack variables S1 and S2 respectively in first two constraint equations of (1) in the usual manner. X1 + X2 + 1S1 = 10 (2) 7X1 + X2 + 1S2 = 40 and the third constraint requires to add surplus variable S3 in the right hand side of the sign (>). The surplus variable represents the shortage of the capacity of the given constraint. Then the equation is written in equality form as X1 + X2 = 12 + 1S3 (3) However, for simplex tableau all the unknown variables must be expressed to the left side of the equal sign. Therefore, third constraint (3) becomes X1 + X2 - 1S3 = 12 (4)

Linear Programming

29

Clearly incase of the > sign, to convert it into equality surplus variable is subtracted. Since, surplus variable means shortage of the capacity, it can be calculated by taking difference between required resource and available resource (surplus = required - available). But the simplex method requires some feasible starting tableau and then interacting we move towards optimal solution is reached. It both X1 and X2 are set equal to zero in the above constraint the surplus variable is S3 = -12 + 1(0) + (0) = –12, which makes the starting basic solution infeasible. As it is also assumed that all the variables should be non-negative. Artificial Variable: Due to subtracting surplus variable in the constraint of the type greater–than–or–equal–to (>) sign, the necessary basic feasible tableau of the simplex method will not be available. To short out this problem, the additional artificial variable is introduced (A1). The artificial variable (A1) is introduced. The artificial variable (A1) is just assume to be imaginary substitute product of X1 or X2 which may be used totally interchangeably. By adding artificial variable (A1) the equation (13) can be written as

X1 + X2 - 1S3 + 1A1 = 12

Now A1 can be used as basic variable in the initial solution satisfying the constraint equation which yield the solution A1 = 12 - 1(0) - 1(0) = 12. Any solution with A1 included must be imaginary. Therefore, the first solution which includes A1 is purely imaginary. No real product or resource exist for A1. As such, the initial solution is only considered to make the solution mathematically feasible (but not feasible in reality).

On the other hand A1 may not have any effect in the solution as long as its value is zero. Therefore, the value of A1 must be made zero in the final solution. It can be done by converting A1 non basic variable in the final optimal tableau.

When A1 = 0, the constraint equation is X1 + X2 - S1 = 12, which makes no effect in the problem's solution. The 'Big M' method ensures the artificial variable A1 never appears in the optimal solution if it exist.. It is also noted that the substitute product always increases the cost of the production very high.

In case of maximization problem we assign relatively large negative coefficient to the artificial variable in the objective function so that artificial variable is not optimally chosen. For example the objective function with high negative coefficient of A1 looks the equation

Max. Z = X1 + 2X2 + 0S1 + 0S2 - MA1

Where, M is relatively high negative coefficient which can be replaced by numerical value M = 100 for solving LP problem. The logic of use of relatively high negative coefficient in objective function is as follows.

The coefficient of A1 in the objective function is the assumption of relatively high cost price per unit. Let this high cost price per unit be M(= Rs. 100). Then, MA1 = 100A1 is the production cost of A1 imaginary quantity. Therefore, the cost of A1 make heavy decrease in the profit function. To make heavy decrease in the profit high negative coefficient of artificial variable A1 is used in the objective function. After introducing slack surplus and artificial variables the problem can be written completely as follows:

Max. Z = X1 + 2X2 + 0S1 + 0S2 + 0S3 – 100A1

Subject to X1 + X2 + S1 = 10 7X1 + 3X2 + S2 = 40 X1 + 2X2 - S3 + A1 = 12

For the simplex tableau, all the variables should be included in each equations. As usual zero coefficient is used to include the variables which are not applicable and coefficient 1 is used to which the variables are applicable. Then, the standard format of LP problem is written as

Max. Z = X1 + 2X2 + 0S1 + 0S2 + 0S3 - 100 A1

Subject to Row 1: X1 + X2 + 1S1 + 0S2 + 0S3 + 0A1 = 10 Row 2: 7X1 + 3X2 + 0S1 + 0S2 + 0S3 + 0A1 = 40 Row 3: X1 + 2X2 + 0S1 + 0S2 + 1S3 + 1A1 = 12

Following table represents the initial tableau with the key column and key row identifiable.


30

Initial simplex tableau - I

C j R s . 1 R s . 2 R s . 0 R s . 0 R s . 0 R s . -1 0 0

R a t i o

Basic Variables

Quantity (Constants)

X 1 X 2 S 1 S 2 S 3 A 1

R o w 1 0 S 1 1 0 1 1 1 0 0 0 1 01 = 1 0

R o w 2 0 S 2 4 0 7 3 0 1 0 0 4 03 = 1 3

13

R o w 3 - 1 0 0 A 1 1 2 1 2 0 0 – 1 1 1 22 = 6 m i n .

Z j - 1 2 0 0 - 1 0 0 - 2 0 0 0 0 + 1 0 0 - 1 0 0 C j - Z j 1 0 1 2 0 2 0 0 - 1 0 0 0

All the coefficients of key row are updated as usual: Key R3 ÷ 2. The elements of replacing row (R3) are:

6 ½ 1 0 0 - ½ ½ Other basic variable rows (R1) and (R2) are updated in usual way Old (R1) - (- 1) × Rep. (R3) = New R1 Old R2 - (-3) × Rep.(R3) = New R2

10 - 1 × 6 = 4 40 - 3 × 6 = 22 1 - 1 × ½ = ½ 7 - 3 × ½ = 11/2 1 - 1 × 11 = 0 3 - 3 × 1 = 0 1 - 1 × 0 = 1 0 - 3 × 0 = 0 0 - 1 × 0 = 0 1 – 3 × 0 = 1 0 - 1 × - ½ = ½ 0 - 3 × - ½ = 3/2 0 - 1 × ½ = - ½ 0 - 3 × ½ = -3/2 Improve Simplex Tableau II

C j R s . 1 R s . 2 R s . 0 R s . 0 R s . 0 R s . -1 0 0

R a t i o

Basic Variables

Quantity (Constants)

X 1 X 2 S 1 S 2 S 3 A 1

R o w 1 0 S 1 4 ½ 0 1 0 ½ - ½ 41 / 2 =

8 m i n . R o w 2 0 S 2 2 2 1 1 / 2 0 0 1 3 / 2 - 3 / 2 2 2

3 / 2 = 1 4

23

R o w 3 2 X 2 6 ½ 1 0 0 - ½ ½ –

Z j 1 2 1 2 0 0 – 1 1 C j - Z j 0 0 0 0 1 - 1 0 1

Replacing row R1 → Key row R3 × 21

Elements of replacing row R1

8 1 0 2 0 1 - 1 Old (R2) - (- 3/2) × Rep. (R1) = New R2 Old R3 - (-1/2)× Rep.(R1) = New R3

22 - 3/2 × 8 = 10 6 + ½ × 8 = 10 11/2 - 3/2 × 1 = 4 ½ + ½ × 1 = 1 0 - 3/2 × 0 = 0 1 + ½ × 0 = 1 0 - 3/2 × 2 = - 3 0 + ½ × 2 = 1 1 - 3/2 × 0 = 1 0 + ½ × 0 = 0 3/2 - 3/2 × 1 = 0 - ½ + ½ × 1 = 0

Linear Programming

31

Improve Simplex Tableau III C j 1 2 0 0 0 - 1 0 0


Q u a n t i t y ( C o n s t a n t )

X 1 X 2 S 1 S 2 S 3 A 1

R o w 1 0 S 3 8 1 0 2 0 1 - 1 R o w 2 0 S 2 1 0 4 0 - 3 1 0 0

R o w 3 2 X 2 1 0 1 1 1 0 0 0 Z j 2 0 2 2 2 0 0 0 C j - Z j - 1 0 - 2 0 0 - 1 0 0

Decision: Since all the coefficient of index row Cj - Zj < 0, hence optimal solution has been reached. Max. Z = 20, X1 = 0, X2 = 0, X3 = 10 S1 = 0, S2 = 10, S3 = 8

2.6.3 MINIMIZATION PROGRAMMING So far we have discussed in detail the maximization problem in earlier section. It has been observed that there was no need to search for an initial basic feasible solution. This is mainly because of the fact that the point of origin (0, 0) normally lies inside the feasible area and a readymade one was available in the point of origin. Since, the origin point normally lies in the feasible region of maximization problem; it can usually serve as an initial basic feasible solution. The minimization linear programming problem is slightly more cumbersome than the solution of maximization problem. In case of minimization problem, the feasible region often does not include the point of origin. So we do not have a ready-made basic feasible region and thus, we can not have the convenience of starting from the point of origin. Due to this fact the solution of minimization problem with the help of simplex method makes complex. In order to handle this problem of having an initial basic feasible solution, we do the method of pivoting taking a solution with a set of artificial variables. To explain the detail method of simplex method for solving minimization linear programming problem let us consider following example. Example 8 A Oil and Mining company is under contract to deliver at least 24 million gallons of crude oil. Company owns two mines located in two different fields A and B respectively. Each mine has different crude to sledge (salt, water and silt) characteristics. For each barrel of material pumped from field A, 3 gallons of crude oil can be extracted. From each barrel of material pumped from field B, 4 gallons of crude can be obtained. Company engineer states that the pumping station only has 48 units of capacity available during one month. The capacity is based on many factors, such as the specific gravity of materials being pumped, distance, etc. Engineering reports indicate that each million barrels of substance pumped from field A requires 2 units of capacity. Each million barrel from field B requires 6 units. Finally, it costs a net of Rs. 5 to extract a barrel of material from A and Rs. 3 for barrel from field B. How many barrels of materials to extract from each field so as to minimize the costs satisfying the given conditions? Solution: Since, the objective is to minimize the costs, the problem is formulated mathematically as follows: Let X1 and X2 be the barrels of crude oil needed to extract from fields A and B respectively which minimize the costs. Minimize: Total cost = Rs. 5 + Rs. 3 Subject to: 3X1 + 4X2 > 24 2X1 + 6X2 < 48 ... (i) X1, X2 > 0 Since first inequality constraint is in the form of greater than equal (>), it is converted into exact equation using non-negativity surplus variable S1. Second constraint is converted into equality equation by using slack variable S2. So we rewrite the problem as Min. Z = 5X1 + 3X2 + OS1 + OS2

Subject to Row 1: 3X1 + 4X2 - S1 = 24 Row 2: 2X1 + 6X2 + S2 = 48 ... (ii) X1, X2, S1, S2 > 0 The problem can be rewrite as


32

Min. Z = 5X + 3X2 + OS1 to S2

Subject to Row 1: -3X1 - 4X2 + S1 = - 24 Row 2: 2X1 + 6X2 + S2 = 48 ... (iii) X1, X2, S1, S2 > 0 On expressing above minimization problem in simplex tableau, we get the basic feasible in terms of negative value S1. But the solution is not acceptable as S1 should be non-negative.

Simplex Tableau I

C j 5 3 0 0 B a s i c

V a r i a b le s

Q u a n t i t i es

( c o n s t a n ts )

X 1 X 2 S 1 S 2

R o w 1 : 0 S 1 - 2 4 - 3 - 4 1 0 R o w 2 : 0 S 2 4 8 2 6 0 1

Z j 0 0 0 0 0 0 C j - Z j 5 3 0 0

The initial basic feasible solution give Z = 0, S1 = - 24 and S2 = 48. To resolve this negative value of S1 (surplus variable) we add a non-negative artificial or dummy variable A1 to constraint whose inequality constraint is greater than equal (>). We also assign relatively large coefficient to the artificial variable A1 in the objective function M (say; M = Rs. 10 in our example) so that the cost of the production will be very high initially and not of optimally chosen. No one will prefer the artificial variable optimally as their price M = Rs. 10 compare to the prices of choice variables X1 and X2. Now introducing artificial variable A1 the problem can be rewritten as Min. Z = 5X1 + 3X2 + 0S1+ 0S2 + 10A1

Subject to 3X1 + 4X2 - 1S1 + OS2 + A1 = 24 2X1 + 6X2 + 0S1 + 1S2 + 0A1 = 48 ... (iv) X1, X2, S1, S2, A1 > 0 Now, presenting the problem into simplex tabular form, we get

Simplex Tableau (II) C j 5 3 0 0 1 0


Q u a n t i t y ( C o n s t a n t s )

X 1 X 2 S 1 S 2 A 1

R a t i o

R o w 1 1 0 A 1 2 4 3 4 - 1 0 1 2 4 / 4 = 6 ( m i n . ) R o w 2 0 S 2 4 8 2 6 0 1 0 4 8 / 6 = 8

Z j 2 4 0 3 0 4 0 - 1 0 0 1 0 C j - Z j - 2 5 - 3 7 1 0 0 0

Since, the objective is to minimize the cost, the key column is identified with some modified criteria. In this case, we look for key column having largest negative value in Cj - Zj index row. The column in Cj - Zj which contains the highest value with negative sign is identified as key column. The identification of key row and key element are exactly same as in case of maximization problem. The row with minimum ratio i.e., minimum displacement quotient is key row. The element lying at the intersecting point of key column and row is key element. In simplex tableau II, X2 is key column, R1 is key row and encircled element 4 is key element as in the index row (Cj - Zj ) has highest negative 37 which lies at X2 column and R1 has smallest ratio. We use the following mathematical operation to up date key row elements to obtain replacing row. Replacing row R1 → R1 ÷ 4 then updated replacing row elements are R1: 6 ¾ 1 - ¼ 0 ¼ Now, we update the elements of other row as follows.

Linear Programming

33

Elements ofold row 2 —

Intersecting

elements ofold row 2 withkey column

×

Corresponding

elements ofreplacing of row 1

= New R0

48 — 6 × 6 = 12 2 — 6 × 3/4 = - 5/2 6 — 6 × 1 = 0 0 — 6 × -1/4 = 3/2 1 — 6 × 60 = 1 0 — 6 × 1/4 = 3/2 The resulting values of updated rows are furnished in simplex tableau III below. Simplex Tableau III

C j 5 3 0 0 1 0 B a s i c

V a r i a b l e s Q u a n t i t y

( C o n s t a n t s ) X 1 X 2 S 1 S 2 A 1

R o w 1 3 X 2 6 ¾ 1 - ¼ 0 1 / 4 R o w 2 0 S 2 1 2 - 5 / 2 0 3 / 2 1 3 / 2

Z j 1 8 9 / 4 3 - ¾ 0 ¾ C j - Z j 1 1 / 4 0 ¾ 0 3 7 / 4

Since, the index row (Cj - Zj ) does not contain any negative elements in the above simplex tableau III there is no further scope of reducing the cost. Therefore, we arrive at the optimal solution. We have, Min. Z = Rs. 18, X2 = 6, S2 = 12, X1 = 0 and S1 = 0

2.6.4 SPECIAL CASES IN LINEAR PROGRAMMING IN SIMPLEX METHOD

Alternative (or Multiple) Optimal Solution: We will know the existence of alternative optimal solution if we find zero as the value of any non-basic decision variable in the index row (Cj - Zj ). For having alternative optimal solution we consider this variable as entering variable. Example 9 :

Max Z = 5X1 + 3X2

Subject to 10X1 + 6X2 < 60 X1 + 2X2 < 18 and X1, X2 > 0

Solution: The given LPP is converted into standard form of LPP by adding slack variables to given constraints, we get Max. Z = 5X1 + 3X + OS1 + OS2

Subject to 10X1 + 6X2 + S1 + OS2 = 60 X1 + 2X2 + OS1 + S2 = 18 and X1, X2, S1, S2 > 0 To solve the problem we furnish the above equations in the following standard simplex table.

Simplex Tableau 1

C j 5 3 0 0 R a t i o B a s i c

V a r i a bl e

Q u a n t i t i e s ( c o n s t a n t )

X 1 X 2 S 1 S 2

R o w 1 : 0 S 1 6 0 1 0 6 1 0 6 0 / 1 0 = 6 R o w 2 : 0 S 2 1 8 1 2 0 1 1 8

Z j 0 0 0 0 0 C j - Z j 5 3 0 0


34

Here, X1 is entering variable and S1 is leaving variable All the coefficients of key row are updated as usual: R1 ÷ 10 The elements of replacing row (R1) are:

6010 = 6,

1010 = 1,

610 =

35,

110,

010 = 0

Updating R2 Old R2 – 1 × Rep. (R1) = New (R2) 18 – 1 × 6 = 12 1 – 1 × 1 = 0 2 – 1 × 3/5 = 7/5 0 – 1 × 1/10 = –1/10 1 – 1 × 0 = 1 Simplex table 2

C j 5 3 0 0 B a s i c

V a r i ab l e s

Q u a n t i t i e s ( C o n s t a n t s )

X 1 X 2 S 1 S 2

R a t i o

R o w 1 : 5 X 1 6 1 3 / 5 1 / 1 0 0 1 0 R o w 2 : 0 S 2 1 2 0 7 / 5 - 1 / 1 0 1 6 0 / 7

Z j 3 0 5 3 ½ 0 C j - Z j 0 0 - ½ 0

Since all (Cj - Zj ) is < 0, the optimal solution has been obtained i.e. X1 = 6, X2 = 0, Max. Zj = 30. Note: Since the value of X2 (non-basic decision variable) in (Cj - Zj ) is zero, we might take it as entering variable to generate alternative optimal solution which is shown below.All the coefficients of key row are updated as usual: R2 ÷ 7/5.The elements of replacing row (R2) are: 0/7 0 1 -1/14 5/7

Updating R1

Old R1 – 1 × Rep. (R2) = New (R1) 6 – 3/5 × 60/7 = 6/7 1 – 3/5 × 0 = 1 3/5 – 3/5 × 1/14 = 1/7 1/10 – 3/5 × -1/14 = 1/7 0 – 3/5 × 5/7 = -3/7 Simplex table 3

C j 5 3 0 0 B a s i c

v a r i a b le

Q u a n t i t y ( c o n s t a n t )

X 1 X 2 S 1 S 2

R o w 1 : 5 X 1 6 / 7 1 0 1 / 7 - 3 / 7 R o w 2 : 3 X 2 6 0 / 7 0 1 - 1 / 1 4 5 / 7

Z j 3 0 5 3 ½ 0 C j - Z j 0 0 - ½ 0

Since all (Cj - Zj ) is < 0, the optimal solution has been obtained.

i.e., X1 = 67, X2 =

607 , Max. Zj = 30

Hence we have found two solution sets (6, 0) and 6/7, 60/7 for (X1, X2) decision variables which yield the same maximum profit Rs. 30.

Unbounded Solution We will know the existence of unbounded solution if we find that optimum solution is not achieved but all values of ratio column are negative and infinity. For example, Let's see the same example which was done previously in graphical solution. Example 10:

Linear Programming

35

Max. Z = 5X1 + 7X2

Subject to 2X1 - 2X2 > 2 2X1 + 2X2 > 8 and X1, X2 > 0

Solution: The given LPP is converted into standard form of LPP by introducing surplus and artificial variables, we set Max. 2 = 5X1 + 7X2 + OS1 - MA1 + OS2 - MA2

Subject to 2X1 - 2X2 - S1 + A1 + OS2 + OA2 = 2 2X1 + 2X2 + OS1 + OA1 - S2 + A2 = 8 and X1, X2, S1, A1, S2, A2 > 0 To solve the problem we furnish the above equations in the following standard simplex table.

Simplex table 1

C j 5 7 0 - M 0 - M B a s i c

V a r i a b l e sQ u a n t i t i e s( c o n s t a n t )

X 1 X 2 S 1 A 1 S 2 A 2

R a t i o

R o w 1 : - M

A 1 2 2 - 2 - 1 1 0 0 2 / 2 = 1

R o w 2 : - M

A 2 8 2 2 0 0 - 1 1 8 / 2 = 4

Z j - 1 0 M - 4 M 0 M - M M - M C j - Z j 4 M + 5 7 - M 0 - M 0

Here, X1 is entering variable and A1 is leaving variable.

All the coefficients of key row are updated as usual: R1 ÷ 2, then the elements of replacing row (R1) : 1 1 - 1 -1/2 1/2 0 0

Simplex table 2

C j 5 7 0 - M 0 - M B a s i c

V a r i a b l e sQ u a n t i t i e s

( C o n s t a n t s )X 1 X 2 S 1 A 1 S 2 A 2

R a t i o

R o w 1 : 5 X 1 1 1 - 1 - 1 / 2 1 / 2 0 0 - 1 R o w 2 : -

M A 2 6 0 4 1 - 1 - 1 1 3 / 2

Z j - 6 M + 5 5 - 4 M - 5 - M - 2 . 5

M + 5 / 2 M - M

C j - Z j j 0 4 M + 1 2 M + 2 . 5 - 2 M - 5 / 2

- M 0

X1 is entering variable, A1 is leaving variable. All the coefficients of key row are updated as usual: R2 ÷ 4 The elements of replacing row (R2) are:

64 =

32,

04 = 0,

44 = 1,

14,

- 14 ,

- 14 ,

14

Updating R1

Old R1 – (-1) × Rep. (R2) = New (R1) 1 – (-1) × 3/2 = 5/2 1 – (-1) × 0 = 1 -1 – (-1) × 1 = 0 -1/2 – (-1) × 1/4 = -1/4 1/2 – (-1) × -1/4 = 1/4 0 – (-1) × -1/4 = -1/4


36

0 – (-1) × 1/4 = 1/4

Simplex table 3

C j 5 7 ` 0 - M 0 - M B a s i c


( C o n s t a n t s )X 1 X 2 S 1 A 1 S 2 A 2

R a t i o

R o w 1 : 5 X 1 5 / 2 1 0 - 1 / 4 1 / 4 - 1 / 4 ¼ - 1 0 R o w 2 : 7 X 2 3 / 2 0 1 ¼ - 1 / 4 - 1 / 4 ¼ - 6

Z j 2 3 5 7 ½ - 1 / 2 - 3 3 C j - Z j 0 0 - ½ - M + ½ + 3 - M - 3

Here, all values of ratio column are negative but optimum solution is not achieved, the Linear programming problem is unbounded.

Infeasible Solution LPP is said to have infeasible solution if artificial variable appeared as the basic variable with some positive value at the optimal stage of the solution. To fixed this idea let's take the previous example which was done in graphical solution.

Example 11 : Max. Z = 12X1 - 4X2

Subject to X1 + 2X2 < 2 2X1 + 4X2 > 8 X1 = 6 and X1, X2 > 0

Solution: The given LPP is converted into standard form of LPP by introducing slack, surplus and artificial variables to given constraints, we get Max. Z = 12X1 - 4X2 + OS1 + OS2 - MA2 - MA3

Subject to X1 + 2X2 + S1 + OS2 + OA1 + OA2 = 2 2X1 + 4X2 + OS1 - S2 + A1 + OA2 = 8 X1 + 0X2 + OS1 + OS2 + OA1 + A2 = 6 and X1, X2, S1, S2, A1, A2 > 0 To solve the problem we furnish the above equations in the following standard simplex table. Simplex table 1

C j 1 2 - 4 0 0 - M - M B a s i c V a r i a b l e s Q u a n t i t i e s

( C o n s t a n t s )

X 1 X 2 S 1 S 2 A 1 A 2

R a t i o

R o w 1 : 0

S 1 2 1 2 1 0 0 0 1

R o w 2 : -M

A 1 8 2 4 0 - 1 1 0 2

R o w 3 : -M

A 2 6 1 0 0 0 0 1 ∞

Z j - 1 4 M - 3 M - 4 M 0 M - M - M C j - Z j 3 M + 1 2 4 M +

4 0 - M 0 0

Here, X2 is entering variable and S1 is leaving variable. All the coefficients of key row are updated as usual: R1 ÷ 2 The elements of replacing row (R1) are:

22 = 1,

12,

22 = 1,

12,

02 = 0,

02 = 0,

02 = 0

Updating R2 and R3

Linear Programming

37

Old (R2) - (4) × Rep. (R1) = New R2 Old R3 - (0) × Rep.(R1) = New R3

8 – 4 × 1 = 4 6 – 0 × 1 = 6 2 – 4 × 1/2 = 0 1 – 0 × ½ = 1 4 – 4 × 1 = 0 0 – 0 × 1 = 0 0 – 4 × 1/2 = -2 0 – 0 × 1/2 = 0 -1 – 4 × 0 = -1 0 – 0 × 0 = 0 1 – 4 × 0 = 1 0 – 0 × 0 = 0 0 – 4 × 0 = 0 1 – 0 × 0 = 0 Simplex table 2

C j 1 2 - 4 0 0 - M - M B a s i c


( C o n s t a n t s )X 1 X 2 S 1 S 2 A 1 A 2

R a t i o

R o w 1 : - 4

X 2 1 ½ 1 1 / 2 1 / 2 0 0 2

R o w 2 : - M

A 1 4 0 0 - 2 - 1 1 0 ∞

R o w 3 : - M

A 2 6 1 0 0 0 0 1 6

Z j - 1 0 M - 4 - M - 2 - 4 2 M - 2 M - M - M C j - Z j M + 1 4 0 - 2 M + 2 - M 0 0

Here, X1 is entering variable and X2 is leaving variable. All the coefficients of key row are updated as usual: R1 ÷ 1/2,.The elements of replacing row (R1) are:

1

1/2 = 2, 1/21/2 = 1,

11/2 = 2,

1/21/2 = 1, 0, 0, 0

Updating R2 and R3


4 – 0 × 2 = 4 6 – 1 × 2 = 4 0 – 0 × 1 = 0 1 – 1 × 1 = 0 0 – 0 × 2 = 0 0 – 1 × 2 = 0 -2 – 0 × 1 = -2 0 – 1 × 1 = -2 -1 – 0 × 0 = -1 0 – 1 × 0 = -1 1 – 0 × 0 = 1 0 – 1 × 0 = 1 0 – 0 × 0 = 0 1 – 1 × 0 = 0 Simplex table 3

C j 1 2 - 4 0 0 - M - M B a s i c


( C o n s t a n t s )

X 1 X 2 S 1 S 2 A 1 A 2

R o w 1 : 1 2 X 1 2 1 2 1 0 0 0 R o w 2 : -

M A 1 4 0 0 - 2 - 1 1 0

R o w 3 : - M

A 2 4 0 0 - 2 - 1 1 0

Z j - 6 M + 2 M 1 2 M 2 4 + 2 M 1 2 + 3 M M - M - M Z j - C j 1 2 - 1 2 M - 2 M - 2 8 - 1 2 - 3 M 0 0 0


38

Here, The solution seems to have in optimal stage as all the values in index row are all negative. But both artificial variables A2 and A3 appeared as basic variables at positive level, i.e. A2 = 4 and A3 = 4 and M appears in Zj . Hence, the solution is infeasible.

Degeneracy In a linear programming problem degeneracy occurs when one or more of the basic variables assume a value of zero. Degeneracy is identified when two or more rows will have equal minimum positive values in the ratio column of simplex tableau. To avoid degeneracy one of the widely used methods is to consider the row at the top for a key row. However, if a corresponding variable to the row at the top has already been non-basic variable, we might select next row for a key row. Remarks: In introducing additional variables, we need to pay attention to the types of constraints, their coefficients in objective function and need to know their presence in initial basic solution. The table below shows the treatment of additional variables for various inequality/equality signs.

Coefficients of additional variables in the objective

function

Basic variables Types of constraint

Additional variables

Max. Min. (i) Less tan or equal to (<)

A slack variable is added 0 0 Yes

(ii) Greater than or equal to (>)

A surplus a variable is subtracted and at the same time artificial variable is added

0 for surplus – M for artificial

0 for surplus + M for artificial

No for surplus Yes for artificial

(iii) Equal to (=) Only artificial variable is added – M + M Yes 2.7 EXAMPLE WORKED OUT Example 12 The Greencastle Company manufactures three products: A, B and C. Their contribution to profit is Rs. 3, Rs. 5 and Rs. 2 per unit respectively. The sales department currently has firm orders for 150 units of A and 100 units of C for the week (those orders must be filled). All three products must go through a bottleneck machine (a bottleneck machine controls the output of a facility), which has only 80 hours of processing time available each week. The time requirements are 10, 10 and 5 minutes per unit for products A, B and C respectively. Formulate the problem and use the simplex method to find the optimal product mix that maximizes profit. Solution: Let X1 units of A, X2 units of B and X3 units of C be manufactured. As the profit per unit of A, B and C are Rs. 3, Rs. 5 and Rs. 2 respectively, the total contribution from X1 units of A, X2 units of B and X3 units of C are Rs. 3X1, Rs. 5X2 and Rs. 2X3. The objective function is to maximize the sum of the total contribution from the production of A, B and C. That is, Max. Z = 3X1 + 5X2 + 2X3 Defining constraints: Orders constraint: As the rules department currently has firm order for 150 units of A and 100 units of C for the week, the constraints are : X1 = 150 , X3 = 100 Processing time constraint As the processing time (minute) for the product A, B and C are 10, 1 and 5 and only 80 hr. = 80 × 60 = 4800 minutes processing time available each week, the constraint is 10x1 + 10x2 + 5x3 < 4800 Minimum production constraints The minimum production for each product is zero. Hence, X1, X2, X3 > 0.

Linear Programming

39

This is also known as non-negative constraint. Hence, the LP model can be summarized as follows: Max. Z = 3X1 + 5X2 + 2X3

Subject to X1 = 150 X3 = 100 10X1 + 10X2 + 5X3 < 4800 and X1, X2, X3 > 0 Simplex method The LPP is converted to standard form of LPP by introducing slack and artificial variables to given constraints, we get

Max. Z = 3X1 + 5X2 + 2X3 - MA1 - MA2 + OS3

Subject to X1 + 0X2 + 0X3 + A1 + OA2 + OS3 = 150 0X1 + 0X2 + X3 + OA1 + A2 + 0S3 = 100 10X1 + 10X2 + 5X3 + 0A1 + 0A2 + S3 = 4800 and X1,X2, X3, A1, A2, S3 > 0 To solve the problem we furnish the above equations in the following standard simplex table.

Simplex table 1

C j 3 5 2 - M - M 0 B a s i c

V a r i ab l e s


X 1 X 2 X 3 A 1 A 2 S 3

R a t i o

R o w 1 : - M A 1 1 5 0 1 0 0 1 0 0 1 5 0 R o w 2 : - M A 2 1 0 0 0 0 1 0 1 0 ∞ R o w 3 : 0 S 3 4 8 0 0 1 0 1 0 5 0 0 1 4 8 0

Z j - 2 5 0 M - M 0 - M - M - M 0 C j - Z j M + 3

5 M + 2 0 0 0

Here, X1 is entering variable and A1 is leaving variable. All the coefficients of key row are updated as usual: R1 ÷ 1 The elements of replacing row (R1) are: 150, 1, 0, 0, 1, 0, 0

Updating R2 and R3


100 – 0 × 150 = 100 4800 – 10 × 150 = 3300 0 – 0 × 1 = 0 10 – 10 × 1 = 0 0 – 0 × 0 = 0 10 – 10 × 0 = 10 1 – 0 × 0 = 1 5 – 10 × 0 = 5 0 – 0 × 1 = 0 0 – 10 × 1 = 10 1 – 0 × 0 = 1 0 – 10 × 0 = 0 0 – 0 × 0 = 0 1 – 10 × 0 = 1


40

Simplex table 2

C j 3 5 2 - M - M 0 B a s i c

V a r i a b le s


X 1 X 2 X 3 A 1 A 2 S 3

R a t i o

R o w 1 : 3 X 1 1 5 0 1 0 0 1 0 0 ∞ R o w 2 : - M A 2 1 0 0 0 0 1 0 1 0 1 0 0 R o w 3 : 0 S 3 3 3 0 0 0 1 0 - 1 0 0 0 1 6 6 0

Z j 4 5 0 - 1 0 0 M 3 0 - M 3 - M 0 C j - Z j 0 5 M + 2 - 3 - M 0 0 Here, X3 is entering variable and A2 is leaving variable. All the coefficients of key row are updated as usual: R2 ÷ 1 The elements of replacing row (R2) are: 100, 0, 0, 1, 0, 1, 0

Updating R1 and R3

Old (R1) - (0) × Rep. (R2) = New R1 Old R3 - 5 × Rep.(R2) = New R3

150 – 0 × 100 = 150 3300 – 5 × 100 = 2800 1 – 0 × 0 = 1 0 – 5 × 0 = 0 0 – 0 × 0 = 0 10 – 5 × 0 = 10 0 – 0 × 1 = 0 5 – 5 × 1 = 0 1 – 0 × 0 = 1 -10 – 5 × 0 = -10 0 – 0 × 1 = 0 0 – 5 × 1 = -5 0 – 0 × 0 = 0 1 – 5 × 0 = 1 Simplex table 3

C j 3 5 2 - M - M 0 B a s i c


( C o n s t a n t s )X 1 X 2 X 3 A 1 A 2 S 3

R a t i o

R o w 1 : 3 X 1 1 5 0 1 0 0 1 0 0 ∞ R o w 2 : 2 X 3 1 0 0 0 0 1 0 1 0 ∞ R o w 3 : 0 S 3 2 8 0 0 0 1 0 0 - 1 0 - 5 1 2 8 0

Z j 6 5 0 3 0 2 3 2 0 C j - Z j 0 5 0 - 3 - M - 2 - M 0

Here, X2 is entering variable and S3 is leaving variable. All the coefficients of key row are updated as usual: R3 ÷ 10 The elements of replacing row (R3) are: 280, 0, 1, 0, -1, -1/2, 1/10

Updating R1 and R2

Old (R1) - (0) × Rep. (R3) = New R1 Old R2 - 0 × Rep.(R3) = New R2

150 – 0 × 280 = 150 2800 – 0 × 280 = 2800 1 – 0 × 0 = 1 0 – 0 × 0 = 0 0 – 0 × 1 = 0 10 – 0 × 1 = 10 0 – 0 × 0 = 0 0 – 0 × 0 = 0 1 – 0 × -1 = 1 -10 – 0 × -1 = -10 0 – 0 × -1/2 = 0 -5 – 0 × -1/2 = -5 0 – 0 × 1/10 = 0 1 – 0 × 1/10 = 1

Linear Programming

41

Simplex table 4

C j 3 5 2 - M - M B a s i c


( C o n s t a n t s )X 1 X 2 X 3 A 1 A 2

0 S 3

R o w 1 : 3 X 1 1 5 0 1 0 0 1 0 0 R o w 2 : 2 X 3 1 0 0 0 0 1 0 1 0 R o w 3 : 5 X 2 2 8 0 0 1 0 - 1 - ½ 1 / 1 0

Z j 2 0 5 0 3 5 2 - 2 - ½ 2 C j - Z j 0 0 0 - M + 2 - M + ½ - 2

Since all the values in index row are negative, the optimal solution has been obtained. That is x1 = 150, X3 = 100 and X2 = 280 and Max. Z = Rs. 2050

Example 13 Solve by simplex method P = 10a + 12b S.t. 2a + b ≤ 60 3a + 4b ≤ 120 A, b ≥ 0

a. Converting the inequalities to equalities by adding slack variables. P = 10a + 12b S.t. 2a + b + S1 = 60 3a + 4b + S2 = 120 a, b ≥ 0 b. Arrange the constraints equation for simplex solution. P = 10a + 12b + 0S1 + 0S2 S.t. 2a + b + S1 + 0S2 = 60 3a + 4b + 0S1 + S2 = 120 a,b ≥ 0 c. Presenting equation in simplex table. Simplex table 1

C j 1 0 1 2 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 S 1 S 1

R a t i o

R o w 1 : 0 S 1 6 0 2 1 1 0 6 0 R o w 2 : 0 S 2 1 2 0 3 4 0 1 3 0

Z j 0 0 0 0 0 C j - Z j 1 0 1 2 0 0

Simplex Table 2

C j 1 0 1 2 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 S 1 S 1

R a t i o

R o w 1 : 0 S 1 3 0 5 / 4 0 1 - 1 / 4 2 4 R o w 2 : 1 2 b 3 0 3 / 4 1 0 1 / 4 4 0

Z j 3 6 0 9 1 2 0 3 C j - Z j 1 0 0 - 3


42

Simplex Table 3

C j 1 0 1 2 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 S 1 S 1

R a t i o

R o w 1 : 1 0 a 2 4 1 0 4 / 5 - 1 / 5 R o w 2 : 1 2 b 1 2 0 1 - 3 / 5 2 / 5

Z j 3 8 4 1 0 1 2 6 / 5 1 4 / 5 C j - Z j 0 0 - 6 / 5 - 1 4 / 5

Since all the values in index row are negative or , the optimal solution has been obtained.

That is a = 24, b = 12 and z = 384

Example 14: Maximize: P = X1 – X2 + 3X3 s.t. X1 + X2 + X3 ≤ 10 2X1 – X3 ≤ 2 2 2X1 – 2X2 + 3X3 ≤ 3 X1, X2, X3 ≥ 0

Solution : (a) introducing the slack variable S1, S2, S3 in the objective functions and constraints, we get the following

equation. P = X1 – X2 + 3X3 + 0S1 + 0S2 + 0S3 s.t. X1 + X2 + X3 + S1 = 10 2X1 – X3 + S2 =2 2X1 – 2X2 + 3X3 + S3 = 6 X1, X2, X3 ≥ 0 (b) Arranging the constraints equation for simplex solution.

P = X1 - X2 + 3X3 + 0S1 + 0S2 + 0S3 s.t. X1 + X2 + X3 + S1 + 0S2 + 0S3 =10 2X1 + 0X2 – X3 + 0S1 + S2 + 0S3 =2 2X1 – 2X2 + 3X3 + 0S1 + 0S2 + S3 = 6 X1, X2, X3 ≥ 0

(c) Presenting the equation of b in simplex table. Simplex table 1

C j 1 - 1 3 0 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 X 3 S 1 S 2 S 3

R a t i o

R o w 1 : 0 S 1 1 0 1 1 1 1 0 0 1 0 R o w 2 : 0 S 2 2 2 0 - 1 0 1 0 - 2 R o w 3 : 0 S 3 6 2 - 2 3 0 0 1 2

Z j 0 0 0 0 0 0 0 C j - Z j 1 - 1 3 0 0 0

Simplex table 2

C j 1 - 1 3 0 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 X 3 S 1 S 2 S 3

R a t i o

R o w 1 : 0 S 1 8 1 / 3 5 / 3 0 1 0 - 1 / 3 2 4 / 5 R o w 2 : 0 S 2 4 8 / 3 - 2 / 3 0 0 1 1 / 3 - 1 2 / 2 R o w 3 : 3 X 2 2 2 / 3 - 2 / 3 1 0 0 1 / 3 - 6 / 2

Z j 6 2 - 2 3 0 0 1 C j - Z j - 1 1 0 0 0 - 1

Linear Programming

43

Simplex table 3 C j 1 - 1 3 0 0 0


Q u a n t i t i e s( C o n s t a n t

s )

X 1 X 2 X 3 S 1 S 2 S 3

R a t i o

R o w 1 : - 1 X 2 2 4 / 5 1 / 5 1 0 3 / 5 0 - 1 / 5 R o w 2 : 0 S 2 3 6 / 5 1 4 / 5 0 0 2 / 5 1 1 / 5 R o w 3 : 3 X 2 2 6 / 5 4 / 5 0 1 2 / 5 0 1 / 5

Z j 5 4 / 5 1 1 / 5 - 1 3 3 / 5 0 4 / 5 C j - Z j - 6 / 5 0 0 - 3 / 5 0 - 4 / 5

Since all the values in index row are negative or zero , the optimal solution has been obtained. That is X1 =0, X2 =24/5, X3 =26/5 and z = 54/5 Example: 15 Maximize, Z = 6X1 + 5X2 s.t. X1 + X2 ≥ 10 3X1 + 2X2 ≤ 35 X1, X2 ≥ 0 Here the first constraints have ≥ sign. Thus in these conditions to remove ≥ sign, surplus variable S1 is

subtracted and artificial variable A1 is added to LHS of constraints. By doing these we get, X1 + X2 + A1 – S1 = 10 The artificial variable A1 is added to meet the conditions of non-negativity of variables. The coefficient A1 is

objective function is M where, M is very large number. This is done so that M is not appeared in final solution.

(a) Introducing slack / surplus / artificial variable in constraints and objective function. Z = 6X1 + 5X2 – MA1 + 0S1 + 0S2 s.t. X1 + X2 + A1 – S1 = 10 3X1 + 2X2 + S2 = 35 X1, X2 ≥ 0 (b) Arranging the constraints equations for simplex solution. Z = 6X1 + 5X2 – MA1 + 0S1 + 0S2 s.t. X1 + X2 + A1 – S1 + 0S2 = 10 3X1 + 2X2 + 0A1 + 0S1 + S2 = 35 X1, X2 ≥ 0 (c ) Presenting the equation in simplex table. Simplex table 1

C j 1 - 1 - M 0 0 B a s i c

V a r i a b l e s

Q u a n t i t i e s( C o n s t a n t

s )

X 1 X 2 A 1 S 1 S 2

R a t i o

R o w 1 : - M A 1 1 0 1 1 1 - 1 0 1 0 R o w 2 : 0 S 1 3 5 3 2 0 0 1 3 5 / 3

Z j - 1 0 M - M - M - M M 0 C j - Z j 6 +

M 5 + M 0 - M 0


44

Simplex table 2

C j 1 - 1 - M 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 A 1 S 1 S 2

R a t i o

R o w 1 : 6 X 1 1 0 1 1 1 - 1 0 - 1 0 R o w 2 : 0 S 1 5 0 - 1 - 3 3 1 5 / 3

Z j 6 0 6 6 6 - 6 0 C j - Z j 0 - 1 - M - 6 6 0

Simplex table 3

C j 1 - 1 - M 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 A 1 S 1 S 2

R a t i o

R o w 1 : 6 X 1 3 5 / 3 1 2 / 3 0 1 1 / 3 3 5 / 2 R o w 2 : 0 S 1 5 / 3 0 - 1 / 3 - 1 0 1 / 3 -

Z j 7 0 6 4 0 0 2 C j - Z j 0 1 - M 0 - 2

Simplex table 4

C j 1 - 1 - M 0 0 B a s i c


( C o n s t a n t s )X 1 X 2 A 1 S 1 S 2

R a t i o

R o w 1 : 5 X 1 3 5 / 2 3 / 2 1 0 0 ½ R o w 2 : 0 S 1 7 . 5 1 / 2 0 - 1 1 ½

Z j 8 7 . 5 7 . 5 5 0 0 2 / 5 C j - Z j - 1 . 5 0 - M 0 - 2 / 5

Since all the values in index row are negative or zero , the optimal solution has been obtained. That is X1 =0, X2 =17.5, and z = 87.5 (Note: This chapter contains only fundamentals of LPP, the more detail can be obtained from Sthapit , A.B., Yadav, R.P., Tamang, G., Dhital, S., and Adhikari, P. A Text Book of Operations Management. Asmita Books Publisher and Distributors, Katmandu, Nepal. 2005. or Sthapit , A.B., Yadav, R.P., Tamang, G., Dhital, S., and Adhikari, P. Production and Operations Management. Asmita Books Publisher and Distributors, Katmandu, Nepal. 2006.) 2.8 PROBLEMS 1. Solve graphically the following LP

a. Max. Z = 10X1 + 15X2

Subject to X2 + 3X2 < 50 250X1 + 150X2 < 5000 and X1, X2 > 0

Ans: X1 = 12.5, X2 = 12.5, Max Z = Rs. 312.50

b. Min. Z = 2X1 + 3X2

Subject to 10X2 + 15X2 > 500 3X1 + 4X2 < 48 and X1, X2 > 0

Ans: No feasible region exists

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c. Min.Z = 2X1 + 3X2

Subject to 10X1 + 15X2 > 500 3X1 + 4X2 > 48 and X1, X2 > 0

Ans: X1 = 0, X2 = 100/3, Min. Z = Rs. 100 d. Max Z = 2X1 + 6X2

Subject to 6X1 + 18X2 < 216 2X1 + 5X2 < 100 and X1, X2 > 0

Ans: Multiple optimal solutions exist e. Max. Z = 6X1 + 7X2

Subject to X1 - X2 < 12 X1 + 3X2 < 18 and X1, X2 > 0

Ans: X1 = 13.5, X2 = 1.5, Max Z = Rs. 91.50

f. Max. Z = 5X1 + 7X2

Subject to 2X1 - 2X2 > 2 2X1 + 2X2 > 8 and X1, X2 > 0

Ans: Unbounded solution 2. Solve following linear programming problems using simplex method.

a. Max. Z = X1 + 2X2 + 3X3

Subject to X1 + 1.5X2 + 2X3 < 50 10X1 + 12X2 + 18X3 < 375 X2 + X3 < 35 and X1, X2, X3 > 0

Ans: X1 = 0, X2 = 20, X3 = 7.5, Max Z = 62.50, Multiple optimal solutions exist

b. Min. Z = 3X1 + 2X2 + 3X3

Subject to 2X1 + X2 + X3 < 25 3X1 + 4X2 + 2X3 > 15 and X1, X2, X3 > 0

Ans: X1 = 0, X2 = 3.75, X3 = 0, Min. Z = Rs. 7.50

c. Max. Z = 10X1 + 15X2 + 20X3

Subject to X1 + X2 + X3 < 40 2X1 + 4X2 + 6X3 > 180 X1 + 2X2 + 3X3 = 80 and X1, X2, X3 > 0

Ans: No feasible solution exist d. Max. Z = X1 + 2X2 + 3X3

Subject to - X1 + 2X2 + 2X3 = 10 - X1 + 2X2 + 3X3 = 15 and X1, X2, X3 > 0


46

Ans: Unbounded solution

3. Solve the following LPP by simplex methods (i) Max. Z = X1 + 2X2 Subject to 2X1 + 3X2 < 8 4X1 + 5X2 < 10 and X1, X2 > 0 (ii) Min Z = X1 + 2X2

Subject to X1 + X2 > 10 3X1 + 4X2 > 50 and X1, X2 > 0 (iii) Min. Z = 3X1 + 2X2 + 3X3

Subject to 2X1 + X2 + X3 < 25 3X1 + 4X2 + 2X3 > 15 and X1, X2, X3 > 0 (iv) Max. Z = 10X1 + 15X2 + 20X3

Subject to X1 + X2 + X3 < 40 2X1 + 4X2 + 6X3 > 180 X1 + 2X2 + 3X3 = 80 and X1, X2, X3 > 0

4. A manufacturer makes two products P1 and P2 using two machines M1 and M2. Product P1 requires 5 hours on machine M1 and no time on machine M2, product P2 requires 1 hour on machine M1 and 3 hour on machine M2. There are 16 hours of time per day available on machine M1 and 30 hours on M2. Profit margin from P1 and P2 is Rs. 2 and Rs. 10 per unit respectively. What should be the daily production mix to optimize profit. Ans: X1 = 1.2, X2 = 10, Max. Z = Rs. 102.40

5. A manufacturer produces two items X1 and X2. X1 needs 2 hours on machine A and 2 hours on machine B.

X2 needs 3 hours on machine A and 1 hour on machine B. If machine A can run for a maximum of 1 hours per day and B for 8 hours per day and profits from X1 and X2 are Rs. 4 and Rs. 5 per item respectively, find by simplex method, how many items per day be produced to have maximum profit? Give the interpretation for the values of 'indicators' corresponding to slack variables in the final iteration.

Ans: X1 = 3, X2 = 2 and Max. Z = Rs. 22

6. Two materials A and B are required to construct tables and book cases. For one table 12 units of A and 16 units of B are needed while for a book case 16 units of A and 8 units of B are required. The profit on a book case is Rs. 25 and Rs. 20 on a table. 100 units of material A and 80 units of B are available. How many book cases and tables be produced to have maximum profit. Formulate this as a linear programming problem and solve by simplex method.

Ans: X1 = 3, X2 = 4 and Max. Z = Rs. 160

7. A firm has two grades of cashew nuts: Grade 1-750 kg. and Grade II-1,200 kg. These are to be mixed in two types of packets of 1 kg. each-economy and special. The economy pack consists of Grade I and Grade II in the proportion of 1:3, while the special pack combines the two in equal proportion. The contributions of the economy and the special packs are Rs. 5 and Rs. 8 per pack respectively. Formulate this as a linear programming problem to maximize contribution and solve it by simplex method.

Ans: X1 = 900, X2 = 1050 and Max. Z = Rs. 12,900

8. A marketing manager wishes to allocate his annual advertising budget of Rs. 20,000 in two media magazines A and B. The unit costs of a message in media A is Rs. 1,000 and that of B is Rs. 1,500. Media A is a monthly magazine and not more than one insertion is desired in one issue. At least 5 messages should

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appear in media B. The expected effective audience for unit messages on the media A is 40,000 and for media B is 50,000.

Develop a LPP and solve it for maximizing the total effective audience. Ans: Max. Z = 4000X + 5000Y, s.t. 1000X + 1500Y < 20,000 X < 12, Y > 5 X, Y > 0 X = 12, Y = 16/3

9. The Greencastle company manufactures three products: A, B and C. Their contribution to profit is Rs. 3, Rs. 5 and Rs. 2 per unit respectively. The sales department currently has firm orders for 150 units of A and 100 units of C for the week (those orders must be filled). All three products must go through a bottleneck machine (a bottleneck machine controls the output of a facility), which has only 80 hours of processing time available each week. The time requirements are 10, 10 and 5 minutes per unit for products A, B and C respectively.

Formulate the problem and use the simplex method to find the optimal product mix that maximizes profit.

10. A manufacturer has two products, both of which are made in two steps by machines A and B. The process times for the two products on the two machines are as follows:

P r o d u c t M a c h i n e A ( h r . ) M a c h i n e B ( h r . ) 1 2

4 5

5 2

For the coming period, machine A has 100 hours available and machine B has 80 hours available. The contribution for product 1 is 410 per unit, and for product 2 it is $5 per unit. Using the methods of the simplex algorithm, formulate and solve the problem for maximum contribution.

11. A manufacturing company manufactures two different products. The demand for both the products is

strong enough so that the firm can sell as many units of either product or of both, as it can produce and at such a price as to realize a per unit profit contribution of Rs. 16 on product A and Rs. 10 on product B. Unfortunately, the production capacity of the company's plant is several limited. This limitation items from the fact that the manufacture of the products involves the utilization of three scarce resource: raw material, labor and machine time. Each unit of product A requires four units of raw materials, three units of labor and two units of machine time. Each units of product B requires two units of raw materials, three units of labors and five units of machine time. The firm has a daily supply of 24 units of raw materials, 21 units of labors and 30 units of machine time.

Formulate a linear programming model and determine how much of each product should be

manufactured to maximize total profit contribution by using simplex method. [TU 2046 MBA] 12. A dealer deals in only two items cycles and scooters. He has Rs. 50,000 to invest and space to store at most

60 pieces. One scooter costs him Rs. 2,500 and a cycle costs him Rs. 500. He can sell a scooter at a profit of Rs. 500 and a cycle at a profit of Rs.150. Assuming that he can sell all the items that he buys how should he invest his money in order that he may maximize his profit?

Ans: X1 = 10, x2 = 50 profit = Rs. 12,500

13. A diet for a sick person must contain at least 4000 units of vitamins 50 units of minerals and 1400 of calories. Two foods A and B are available at a cost of Rs. 4 and Rs. 3 per unit respectively. If one unit of A contains 200 units of vitamins 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamins, 2 units of minerals and 40 calories, find by simplex method what combination of foods be used to have least cost.

Ans: X1 = 5, X2 = 30, Cost = Rs. 110

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LINEAR PROGRAMMING - WordPress.com 02, 2015 · LINEAR PROGRAMMING 2.1. ... Formulate this problem as an LP model so as to maximize the total profit Formulation of LP model