Lpp Theory

LINEAR PROGRAMMING PROBLEMS PROF. VINAY PANDIT

LINEAR PROGRAMMING - I :

Formulation and Graphic Solution

DEFINITION AND APPLICATIONS:

The mathematical definition of linear programming (L.P.) can be stated as — “It is the

analysis of problems in which a linear function of a number of variables is to be

maximized (minimized), when those variables are subject to a number of restraints in the

form of linear inequalities”. Linear programming models thus belong to a class of

mathematical programming models concerned with efficient allocation of resources to

known activities with the objective of meeting a desired goal. Organizations can have

many goals. Hence, a wide variety of problems can be efficiently solved using L.P.

technique. Here are a few examples:

(1) A product mix problem: Decide the combination of various product quantities to

maximise profit or to minimise production cost.

(2) Allocation of bank funds: To achieve highest possible returns. This should be

achieved within liquidity limits set by RBI and maintaining flexibility to meet the

customers demand for loans.

(3) Manufacturing problem: To manufacture goods (say furniture) so as to give

maximum profits bearing in mind the time constrain and the market demand for the

goods.

(4) Advertising application: To achieve the best possible exposure to the client’s

product at the lowest possible advertising cost.

(5) Portfolio selection: Select specific investments among available alternatives so as to

maximise return or to minimise risk.

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(6) Staffing problem: Develop a work schedule that allows say, a large restaurant or a

hospital or a police station to meet their man power needs at all hours with minimum

number of employees.

(7) Trim loss problem: Find the combination of components to be produced from

standard sheets in order to keep trim loss to a minimum.

An extension of application of linear programming technique is found in transportation

and assignment problems.

TERMINOLOGY OF LINEAR PROGRAMMING:

A typical linear programme has the following components:

(1) An objective function.

(2) Constraints or restrictions.

(3) Non-negativity restriction.

And the following terms are commonly used to describe a typical L.P.P.

Decision variables: Decision variables are the unknowns whose values are to be

determined from the solution of the problem. E.g. decision variables in the furniture

manufacturing problem are say the tables and chairs whose values or actual units of

production are to be found from the solution of the problem.

These variables should be inter-related in terms of consumption of resources. For

example, both tables and chairs require carpenter’s time and also wood and other

resources and any change in the quantity produced of table affects the production

level of chairs. Secondly the relationship among the variables should be linear.

Objective function: A firms objectives are expressed as a function of decision

variables. It represents the mathematical equation of the goals of the firm in terms of

unknown values of the decision variables. Thus if the objective is to maximise net

profits in a furniture manufacturing problem, then profits are expressed as function of

(dependent on) the net per unit profits of table and chair and the number of units

produced (of tables and chairs).

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Constraints: A constraint represents the limitations imposed on the values of

decision variables in the solution. These limitations exist due to limited availability of

resources as well as the requirements of these resources in the production of each unit

of the decision variable. For example manufacturer of a table requires certain amount

of time in a certain department and the department works only for a given period, (say

8 hours in a day for 5 days in a week). The constraints may represent some other type

of limitations also. As in the production of a commodity the market demand can put

an upper limit on the value of the decision variable in the optimal solution.

Thus, the constraints define the limits within which a solution to the problem must be

found. These constraints must be capable of expression in mathematical form of an

equality or inequality.

Linear relationships: Linear programming deals with problems in which the

objective function and the constraints can be expressed as linear functions. Hence,

when the problem is solved graphically, in a two variable case the constraints the

objective function, gives a straight line on a two dimensional graph.

Equations and inequalities: Equations are represented by = (equality) sign. They are

specific statements. But many business problems cannot be neatly expressed in

equations (called strict equality). Instead of precise statements, we may have only

minimum or maximum requirements or availability. For example we may state that

available labour time is 40 hours per week, hence, labour time used in production

should be less than or equal to 40 hours per week. We thus need inequalities. Less

than or equal to relationship is written as () and () sign indicates greater than or equal

to relationship. Most of the constraints in a LPP are expressed as inequalities. They

indicate the upper or the lower limits of resource use or production level. They do not

express exact levels. Thus, they allow for many possibilities of the optimal values of

the decision variable i.e. more than one combination of the decision variables may

give the same optimal value of the objective function.

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Non-negativity restrictions: Linear programming technique is used to obtain

solution to real world problems. The solution to the problem implies finding values of

the decision variables. These must be non-negative. As one cannot think of

manufacture of -4 tables or -6 chairs i.e. negative production. Hence, decision

variable should assume either zero or positive values. If we denote two decision

variables as X1 and X2 then the non negativity restriction is expressed as X1 > = 0;

X2 >= 0.

FORMULATION OF LINEAR PROGRAMMING PROBLEM (LPP):

Formulation of a LPP involves constructing a mathematical model from the given data.

This can be done only if the following requirements are met:

(a) There should be a clearly identifiable objective and it should be measurable in

quantitative terms. E.g. In a manufacturing problem the objective can be maximisation of

profit or minimisation of cost.

(b) The resources to be allocated in the problem should be identifiable and quantitatively

measurable. E.g. The use of labour time, or raw material in the manufacturing process

should be clearly stated.

(c) The relationships representing the objective function and the constraints equations

must be linear.

(d) There should be a series of feasible alternative courses of action available to the

decision maker. These are determined by the resource constraints.

When all the above mentioned conditions are satisfied the problem can be expressed as

L.P. problem. Then solve it for an optimal solution.

Let us first illustrate the formulation of LPP.

We have seen that a typical L.P.P. has three components:

(1) Objective function

(2) Constraints

(3) Non-negativity restrictions

Formulation of LPP therefore implies translating the given descriptive problem into these

three sets of linear relationships between the decision variables.

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METHODS OF SOLUTION:

Solving an LP problem involves:

(i) Selection of appropriate method of solution and

(ii) Then obtain a solution to the problem with the help of selected method

(iii) Test whether this solution is optimal.

The problem can be solved by using:

(1) Graphical method: This method can be used if there are only t• decision variables in

the LPP.

(2) Simplex method: This method is useful in solving LP problems with two or more

than two decision variables.

We explain, in this chapter, the graphical method of solving LPP.

The Graphical method of solution: This method can be used in case where LPP has

only two decision variables. But there is no restriction on the number of constraints. The

method uses the familiar graphical presentation with two axes. The method becomes

unwieldy when there are three variables since we then need a three dimensional graph.

The method cannot be used if the number of decision variables is more than three. In

such a case we have to use a non graphical method to obtain a solution.

The graphical method of solution to L.P. problem uses all the equations in a given

problem, namely the equation expressing objective unction the constraints imposed in

achieving the objective. These constraints can be of (i) greater than (ii) less than or (iii)

strict equality type.

There is also a non-negativity restriction on the values of the decision variables. It

implies that the solution of the problem lies in the first quadrant of the graph. All these

relations are linear.

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SPECIAL CASES IN LPP

1. Infeasibility

2. Unboundedness

3. Redundancy

3. Alternate optima (Alternate optimum solution)

INFEASIBILITY

It is a case where there is no solution, which satisfies all the constraints at the same

time. This may occur if the problem is not correctly formulated. Graphically,

infeasibility is a case where there is no region, which satisfies all constraints

simultaneously.

UNBOUNDEDNESS

A LPP can fail to have an optimum solution if the objective can be made infinitely

large without violating any of the constraints. If we come across unboundedness in

solving real problems, then the problem is not correctly formulated. Since, no real

situation permits any management to have infinite production of goods and infinite

profits, unbounded solution results if in a maximization problem all constraints are of

greater than or equal to type. In such a situation there is no upper limit on feasible

region. Similarly, an unbounded solution occurs in a minimization problem if all

constraints are of less than or equal to type.

REDUNDANCY

A constraint, which does not affect the feasible region, is called a redundant

constraint. Such a constraint is not necessary for the solution of the problem. It can

therefore be omitted while formulating the problem. This will save the computation

time. In many LP problems, redundant constraints are not recognized as being

redundant until the problem is solved. However, when computers are used to solve

LPP, redundant constraints do not cause any difficulty.

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ALTERNATIVE OPTIMA

[The slope of a line ax + by + c = 0 is defined as –b ]

The solution to a LPP shall always be unique if the slope of the objective function

line is different from the slope of all of the constraint lines. Incase, the slope of

objective function line is same as the slope of one of its constraint line, then multiple

optimum solution might exist.

ISO-PROFIT/ISO-COST LINES:

An iso-profit line can be obtained by putting the objective function equal to some

numerical value and plotting that equation on the graph in the same way as the

constraints are plotted. Each point on an iso-profit line yields the same profit. The

lines parallel to the one obtained by moving away from the origin correspond to

higher and higher profit levels. The line whose one point touches the extreme corner

point is considered and the profit corresponding thereto is the highest profit

attainable. And the variable values at that point are optimal values of the decision

variables.

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LINEAR PROGRAMMING – II :

Simplex Method

THE SIMPLEX METHOD OF SOLUTION:

The simplex method uses a simplex algorithm; which is an iterative, procedure for

finding, in a systematic manner the optimal solution to a linear programming problem.

The procedure is based on the observation that if a feasible solution to a linear

programming exists; it is located at a corner point of the feasible region determined by

the constraints of the problem. The simplex method, selects the optimal solution from

among the set of feasible solution to the problem. The algorithm is very efficient as it

considers only those feasible solutions, which are provided by the corner points. Thus, we

need to consider a minimum number of feasible solutions to obtain an optimal one. The

method is quite simple and the first step requires the determination of basic feasible

solution. Then, with the help of a limited number of steps the optimum solution can be

determined.

Terminology of Simplex Method:

Algorithm: A formalised systematic procedure for solving problem.

Simplex Tableau: A table used to keep track of the calculations made b iteration of the

simplex procedure and to provide basis for tableau revision.

Basis: The set of basic variables which are not restricted to zero in the basic solution

and are listed in solution column.

The basic variables: The variables with non-zero positive values make up the basis are

called basic variables and the remaining variables are called non-basic variables.

Iteration: A sequence of steps taken in moving from one basic. The solution to another

basic feasible solution.

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Cj row: A row in the simplex tableau which contains the co-efficients variables in the

objective function.

Zj row: A row in the simplex tableau whose elements represent the decrease (increase)

of the value of the objective function if one unit of the jth variable is brought into the

solution.

Cj - Zj or j row: A row whose elements represent net per unit contribution of the

jth variable in the objective function, if the variable is brought into the new basic

solution. Positive value of j therefore indicates gain and negative value indicates

loss in the total value Z obtained of the objective function.

Key or pivot column: The column with the largest positive j and it indicates which

variable will enter the next solution in a maximization case.

Key or pivot row: The row with the smallest positive value of the, replacement ratio 0

of the constraint rows. The replacement ratio is obtained by dividing elements in the

solution column by the corresponding elements in the key column. The key row

indicates the variable that will leave the basis to make room for new entering variable.

Key (pivot) element: The element at the intersection of key row and key column.

In addition to these terms in a simplex tableau we have the follow terms which are

necessary to make a linear programming problem fit to be solved by simplex method.

Slack variable: A variable used to convert a less than or equal constraint () into

equality constraint. It is added to the left hand side of the constraint.

Surplus variable: A variable used to convert a greater than or equal to (?) constraint

into equality constraint. It is subtracted from the left hand side of the constraint.

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Artificial variable: It is a variable added to greater than or type () constraint. This is

in addition to surplus variables used.

These variables are used to obtain an initial feasible solution in simplex method. These

variables are reduced to zero at optimality.

SOME TECHNICAL ISSUES:

In the earlier chapter, we considered some special problems encountered in solving LPP

using graphical method. Here we discuss, how the presence of these problems - namely,

infeasibility, unboundedness, multiple solutions, degeneracy, is indicated in a simplex

tableau.

Infeasibility: A solution is called feasible if it satisfies all the constraints and the non-

negativity conditions. Sometimes it is possible that the constraints may be

inconsistent so that there is no feasible solution to the problem. Such a situation is

called infeasibility. In a graphical solution, the infeasibility is evident when there is

no feasible region in which all the constraints can be satisfied simultaneously.

However, problem involving more than two variables cannot be easily graphed and it

may not be immediately known that the problem is infeasible, when the model is

constructed.

The simplex method provides information as to where the infeasibility lies. If the

simplex algorithm terminates with one or more artificial variables at a positive value,

then there is no feasible solution to the original problem.

Unboundedness: It occurs when there are no constraints on the solution. So that one

or more of the decision variables can be increased indefinitely without violating any

of the restrictions. Graphically the objective function line can be moved in the desired

direction over the feasible region, without any limits.

How do we recognize unboundedness in a simplex method? We know the

replacement ratio determines the leaving variable in a simplex tableau. Now if there

are no non-negative ratios (i.e. ratios are negative) or they are equal to i.e. of

the type say 60/0, then we have unbounded solution.

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Alternative Optima: (Multiple optimum solution)

A solution to a linear programming problem may or may not be unique. This is

indicated in a graphical solution by the slope of the line of the objective function

which may coincide with the slope of one of the constraints.

In case of simplex method, Whenever a non basic variable (i.e. a variable which is

not in the solution ) has a zero value in the j (i.e. cj j – zj j) row of an optimal

tableau then bringing that variable into the solution will produce a solution

which is also optimal.( Alternative Optimal solution )

Degeneracy: It occurs when one or more of the basis variables assume zero value. In

conditions of degeneracy, the solution would contain a smaller number of non – zero

variables than the number of constraints i.e. if there are 3 constraints the number of

non-zero variables in the solution is less than 3.

Some obvious examples of degeneracy occur if:

a) One or more basic variable have a zero value in the optimal solution.

b) There is a tie in the replacement ratios for determining the leaving variable. The

next tableau gives the degenerate solution.

c) When algorithm pivots in a degenerate row, the objective function value in the

next tableau does not change i.e. there is the problem of cycling. - The system

moves along the same route and the cycle would be repeated forever. There are

sophisticated rules to handle the problem of cycling; however, they are outside the

scope of this book.

It is also observed that real life problems rarely cycle.

DUALITY IN LINEAR PROGRAMMING:

Corresponding to every linear programming problem, there is another linear

programming problem. The given problem is called the primal and the other its dual.

Although the idea of duality is essentially mathematical, it has important interpretations.

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This can help managers in answering questions about alternative courses of action and

their effect on values of the objective function.

When the primal problem is of the maximisation type the dual is of the minimisation type

and vice versa. It is an interesting feature of the simplex method that we can use it to

solve either the original problem — the primal — or the dual. Whichever problem we

start out to solve, it will also give us the solution to the other problem. Consider the

following general linear programming problem. The primal

Maximise Z = c1x1 + c2x2

Subject to

a11x1 + a12x2 <or = b1

a21x1 + b22x2 <or = b2

x1,x2 > or = 0

We can write this problem in a short form as

Maximise Z = cx

Subject to ax < or = b

x> or = 0

The dual corresponding to this problem is

Minimise Z* = b’y

Subject to a’y > or = c’

y> or = 0

where

b’ =transpose of matrix b

a’ = transpose of matrix a

c’ = column matrix of coefficients of objective function.

Y’ = matrix of dual variables.

Or

Minimise Z* = b1y1 +b2y2

Subject to

a11y1 +a21y2 < or = c1

a12y1 +a22y2 < or = c2

y1,y2> or = 0

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The dual problem is constructed from the primal as follows:

(1) Each constraint in primal problem will have a corresponding variable (dual variable

y) in dual problem.

(2) The elements of the right hand side of the constraints in the primal are equal to the

respective co efficient of the variables in objective function in the dual [i.e. y1 will have a

coefficient b1 in the objective function etc.].

(3)If the primal problem is that of maximisation the dual problem is of minimisation.

(4) The maximisation problem has (< or =) type constraints and the minimisation

problem has (> or =) type constraints. If the constraints in the primal are mixed type, they

are converted into constraints of the same type before formulating the dual.

(5) The coefficient matrix of the dual is the transpose of the coefficient matrix of the

primal.

(6) The variables in both the problems are non-negative.

The primal has 5 constraints and 3 variable hence dual will have 5 variables and 3

constraints. The Dual is

Minimise Z* = 4y1 + l2y2- 2y3 + 8y4 – 8y5

Subject to:

y1 +2y2- y3+3y4 -3y5 > or = 8

4y2-y3+2y4-2y5 > or = 10

3y1 --y3--y4.+y5 > or = 5

y1, y2, y3, y4, y5> or = 0

Dual and the optimum simplex tableau:

In the formulation of dual from a primal, we notice that dual and primal are structurally

related. This helps us in obtaining information about the dual variables from the solution

to the problem from the simplex algorithm.

If we look at the optimal table of the simplex algorithm of the primal, the ∆j values

corresponding to the slack variables give the optimum value of the dual variables. The

optimal values of the structural variables in either problem are always equal to the values

of the shadow prices of the corresponding constraints in the optimal solution of the other

(dual) problem.

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The value of the objective function in both the cases remains the same if a feasible

solution exists for both the primal and the dual. It is therefore not necessary to solve both

the primal and its dual. Another relation between the primal and the dual is, if one

problem has an unbounded solution its dual has no feasible solution.

ECONOMIC INTERPRETATION OF DUAL:

We noted that the primal and the dual are related mathematically, we can now show that

they are also related in economic sense. Consider the economic interpretation of the dual

— first for a maximisation problem and then for a minimisation problem.

Example 3.6: The maximisation problem: Consider the following

problem.

The optimal solution to this problem dives production of 18 units of

Xi and 8 units of x2 per week. It yields the maximum prof of a Rs. 1000,

Maximise Z = 40x1 + 35x2

Subject to

2x1 + 3X2 < or = 60 Raw materials constraint per week.

4x1 + 3X2 < or = 96 Capacity constraint per week.

x1,x2 > or = 0

The optimal solution to this problem gives production of 18 units of x1 and 8 units of x2

per week. It yields the maximum profit of a Rs. 1000.

Now, to rent the facilities of the firm for one week, the firm has 60 kg of raw material

and 96 capacity hours. If we let yl represent the rent per kg of raw material and y2 the rent

per capacity hour, the firm would receive a total rent equal to 6Oy1 + 96y2. We shall

compute the minimum value of the rent so that the firm will know what minimum offer

shall be economically acceptable to it. The lower limit can be set up after keeping in

mind that the alternative to renting must be at least as favourable as using the capacity

itself. The rent of the resources should be at least equal to the earnings from producing

products x1 and x2. We know that production of one unit of x1 requires 2 kg of raw

material and 4 capacity hours. Thus, the total rent for these amounts of resources should

be greater than, or equal to, the profit obtainable from one unit of the product, i.e. Rs. 40.

Hence

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2y1+4y2 > or = 40

similarly, the resources consumed in producing one unit of product x2 are 3 kg of raw

material and 3 capacity hours. The total rent of these resources should equal to atleast Rs.

35, the unit profit of product x2. i.e.,

3y1 + 3Y2 > or = 35

Besides, the rent cannot be negative. Therefore, y1 and y2 should both be non negative. In

complete form, the problem can be expressed as:

Minimise: Z* = 60y1 + 96y2

Subject to:

2y1 + 4y2 > or = 40

3y1 + 3y2 > or = 35

y1, y2 > or = 0

This problem is absolutely the same as the dual to the given problem. These rates y1 and

y2 are obtainable from the solution of the dual as y1 = 10/3 and y2 = 25/3. Also, these

values can be obtained from ∆j row of simplex tableau showing optimal solution to the

primal problem. As we have seen, values of the objective function of the primal and the

dual are identical. Naturally, the minimum total rent acceptable to the firm is equal to the

maximum profit that it can earn by producing the output itself using the given resources.

The individual rents of y1 and y2, are called the shadow prices or imputed prices. They

indicate the worth of the resources. These prices, of the two resources, materials and

capacity hours, are imputed from the profit obtained front utilizing their services, and are

not derived from the from the original cost of these resources.

Now we know that each unit of product x1 contributes Rs. 40 to the profit. The imputed

price of material and capacity is respectively, Rs. 10/3 per kg and 25/3 per hour, we can

find the total imputed cost of the resources used in making a unit of the product as

2kg at Rs. 10/3 per kg

i.e., 2 X 10/3 = 20/3 Rs.

4 hours at Rs. 25/3 per kg. = 4 X 25/3 = 100/3 Rs.

The total cost is 120/3 = 40 Rs.

Thus, the total imputed cost of producing one unit of product x1 equals the per unit profit

obtainable from it.

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Similarly, from each unit of product x2, the total imputed cost of resources employed

would be:

3kg at Rs.10/3 per kg = 3 x lO/3 = 10 Rs.

3 hours at Rs. 25/3 per hour 3 x 25/3 = 25 Rs.

The total cost is 10 + 25 = 35 Rs.

This obviously equals the profit per unit of the product. This proves that the valuation of

the resources is such that their total value equals the total profit obtained at the optimum

level of production.

The shadow prices are also called the marginal value products or marginal profitability

of the resources. Thus, if there were a market for renting resources, the firm would be

willing to take some materials if the price of the material were less than Rs. 10/3 per kg,

and capacity hours, if the price is less than Rs. 25/3 per hour.

If we denote marginal profitability of resources as MPR and the marginal profitability of

capacity as MPc, respectively, the shadow prices of the two resources, we can write the

dual as follows:

Minimise H = 50 MPR + 96 MPc

Subject to:

2MPR + 4MPc > or = 40

3MPR + 3MPc > or = 35

MPR, MPc > or = 0

Now let us consider the economic significance of the surplus variables S1 and S2 in the

dual. The numerical values of these variables can be obtained from ∆j row in the optimal

solution of the primal. The value of S1 in the optimal solution represents the opportunity

cost of the product x1 while the value of S2 represents the opportunity cost of product x2.

Production of an additional unit of x1 will give the firm a profit of Rs. 40 and, at the same

time, the firm would use up resources worth 2 x 10/3 + 4 x 25/3 = Rs. 40. Thus, the net

effect of producing one unit of product would be 40 - 40 = 0. Similarly, for product x2 the

opportunity cost equals zero.

Further if x1, x2 is a feasible solution to the primal and y1, y2 is the feasible solution to its

dual then c1x1 + c2x2 < or = b1y1 + b2y2 i.e. [profit obtained is less than or equal to the

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rents to be paid. This would induce the producer to rent the resources rather than produce

the goods himself.

The concept of dual and shadow prices help us in determining the upper and lower

bounds for changes in requirement vectors and coefficients in the objective function.

Such that the feasibility of the LPP is not disturbed.

SENSITIVITY ANALYSIS: REFER CLASS NOTES

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Lpp Theory