LINEAR PROGRAMMING

LINEAR PROGRAMMING

FORMULATION AND GRAPHIC METHOD

DEFINITIONMathematical technique for

optimizing the use of constrained resources, where the relationship between variables is linear.

Linear – relationship between two or more variables which is directly and precisly proportional.

Programming – refers to use of certain mathematical techniques to get the best possible solution to a given problem involving limited resources.

LP

Linear Programming

Basic Requirements Decision variables and their relationships should

be linear. Objective function should be well defined Presence of Constraints Alternative course of action Non negative restrictions Linearity: Both Objective and Constraints must

be expressed in Linear equations/inequalities.

Terminology Decision variables – variables directly under the

control of decision maker. Objective function – An expression showing

relationship between decision variables and organization's objectives.

Constraints – Limit or restriction on the availability of resources.

Non – negativity constraint – The decision variables must not assume negative values which is impossible situation.

Feasible region – The region which is common to all the constraints of given LPP.

Basic Solution – A solution of basic

variables, where the values of non-basic variables are zero.

Feasible Solution – A solution satisfying all constraints and non-negativity restriction of a LPP.

Basic Feasible Solution – A basic solution, where the values of basic variables are non-negative.

Optimal Solution – A basic feasible solution which also optimizes the objective function.

Assumptions Proportionality:

Proportionality/Linearity exists in Objective and Constraints

Additivity:         Total resources used must equal sum of resources

used in individual activity Divisibility:       

Solution need not be whole numbers and may take up fractional value.

Certainty:        Co-efficients are known (Deterministic)

Finiteness:      Finite number of alternative solutions

Applications of LPP Industry

Product Mix Production

Smoothing Blending Production

Scheduling Trim Loss

Miscellaneous Diet Problems Farm Planning Airline Routing Facility Location

Management Portfolio Selection Financial Mix Media Selection Manpower Planning Transportation Job Allocation Travelling Salesman

Problem Formulation

Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

LP Model FormulationMax/min z = c1x1 + c2x2 + ... + cnxn

subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1

a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2

: am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm

xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients

Guidelines for Model Formulation

Understand the problem thoroughly. Describe the objective. Describe each constraint. Define the decision variables. Write the objective in terms of the decision

variables. Write the constraints in terms of the

decision variables.

Linear Programming Problem A firm producing 2 products P1 and P2.

Each unit of P1 requires 1 kg material A, 2 kg of material B and 1 unit of labor. Each unit of P2 requires 3 kg of material B and 1 unit of labor. Each week firm has availability of 6 kg of material A, 19 kg of material B and 8 units of labor. One unit of P1 sold earn profit of Rs. 5 and P2 earns profit of Rs 7. How many units of each product is to be produced so as to give the maximum profit.

A Maximization Problem LP Formulation Max Z = 5x1 + 7x2 (Profit function)

s.t. x1 < 6 (Material A Constraint)

2x1 + 3x2 < 19 (Material B Constraint)

x1 + x2 < 8 (Labor Constraint)

x1, x2 > 0 (Non-negativity constraint)

Graphic Method of solving LPP Formulate linear programming problem. Each inequality in the constraints may be

written as equality. Plot the constraint lines considering them as

equations. Identify the feasible solution region. Locate the corner points of the feasible region. Calculate the value of objective function on the

corner points. Choose the point where the objective function

has optimal value.

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Graphical Solution Constraint #1 Graphed

x2

x1

x1 < 6

(6, 0)

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1

1 2 3 4 5 6 7 8 9 10

Graphical Solution Constraint #2 Graphed

2x1 + 3x2 < 19

x2

x1

(0, 6 1/3)

(9 1/2, 0)

Graphical Solution Constraint #3 Graphed

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5

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1 2 3 4 5 6 7 8 9 10

x2

x1

x1 + x2 < 8

(0, 8)

(8, 0)

Graphical Solution Combined-Constraint Graph

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1 2 3 4 5 6 7 8 9 10

2x1 + 3x2 < 19

x2

x1

x1 + x2 < 8

x1 < 6

Graphical Solution Feasible Solution Region

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1 2 3 4 5 6 7 8 9 10 x1

FeasibleRegion

x2

Example 1: Graphical Solution

The Five Extreme Points8

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1 2 3 4 5 6 7 8 9 10 x1

FeasibleRegion

E D

C

B

A

x2

Evaluation of objective function at feasible region corner points

Corner points

Coordinates Objective function Z = 5x1 + 7x2

Value

A (0,6.33) 5(0)+7(6.33) 44.31

B (5,3) 5(5)+7(3) 46

C (6,2) 5(6)+7(2) 44

D (6,0) 5(6)+7(0) 30

E (0,0) 5(0)+7(0) 0

Since Z has maximum value at point B (5,3), so

x1 = 5, x2 = 3Where profit is Rs 46.

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Alternate Solution by extreme points

Objective Function Line

x1

x2

(7, 0)

(0, 5)Objective Function5x1 + 7x2 = 35

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Alternate Graphical Solution Optimal Solution

x1

x2

Objective Function5x1 + 7x2 = 46

Optimal Solution(x1 = 5, x2 = 3)

Extreme Points and the Optimal Solution

The corners or vertices of the feasible region are referred to as the extreme points.

An optimal solution to an LP problem can be found at an extreme point of the feasible region.

When looking for the optimal solution, you do not have to evaluate all feasible solution points.

You have to consider only the extreme points of the feasible region.

Summary of the Graphical Solution Procedure for Maximization Problems

Prepare a graph of the feasible solutions for each of the constraints.

Determine the feasible region that satisfies all the constraints simultaneously.

Draw an objective function line. Move parallel objective function lines toward

larger objective function values without entirely leaving the feasible region.

Any feasible solution on the objective function line with the largest value is an optimal solution.

Minimization Problem

Min Z = 5x1 + 2x2

Subject to constraints 2x1 + 5x2 > 10

4x1 - x2 > 12 x1 + x2 > 4

x1, x2 > 0

Graphical Solution Graph the Constraints Constraint 1: When x1 = 0, then x2 = 2; when x2

= 0, then x1 = 5. Connect (5,0) and (0,2). The ">" side is above this line.

Constraint 2: When x2 = 0, then x1 = 3. But setting x1 to 0 will yield x2 = -12, which is not on the graph. Thus, to get a second point on this line, set x1 to any number larger than 3 and solve for x2: when x1 = 5, then x2 = 8. Connect (3,0) and (5,8). The ">" side is to the right.

Constraint 3: When x1 = 0, then x2 = 4; when x2 = 0, then x1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.

Graphical Solution Constraints Graphed

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x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

x1

Feasible Region

Graphical Solution Graph the Objective Function

Set the objective function equal to an arbitrary constant (say 20) and graph it. For 5x1 + 2x2 = 20, when x1 = 0, then x2 = 10; when x2= 0, then x1 = 4. Connect (4,0) and (0,10).

Move the Objective Function Line Toward Optimality

Move it in the direction which lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined by the last two constraints.

Graphical Solution Objective Function Graphed

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x2 Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

x1

Solve for the Extreme Point at the Intersection of the Two Binding Constraints

4x1 - x2 = 12 x1+ x2 = 4 Adding these two equations gives:

5x1 = 16 or x1 = 16/5. Substituting this into x1 + x2 = 4 gives: x2

= 4/5

Graphical Solution

Solve for the Optimal Value of the Objective FunctionSolve for z = 5x1 + 2x2 = 5(16/5) + 2(4/5) = 88/5.

Thus the optimal solution is x1 = 16/5; x2 = 4/5; z = 88/5

Graphical Solution Optimal Solution

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1 2 3 4 5 6

x2 Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

Optimal: x1 = 16/5 x2 = 4/5x1

Special Cases Multiple Optimal Solutions Infeasible problem Unbounded problem Redundant Constraint

Multiple optimal solutions

If objective function falls on more than one optimal point, the solution is said to be multiple optimal.

It exists when slope of the objective function and one of the constraint equation are same.

Infeasible problem

No feasible solution. It means there are no points that

simultaneously satisfy all constraints in the problem.

Common region do not develop in the first quadrant.

Objective function do not pass through any point of all constraints.

Example of Infeasible Problem

Max Z = 2x1 + 6x2

subject to constraints 4x1 + 3x2 < 12 2x1 + x2 > 8

x1, x2 > 0

Infeasible Problem There are no points that satisfy both

constraints, hence this problem has no feasible region, and no optimal solution.

x2

x1

4x1 + 3x2 < 12

2x1 + x2 > 8

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8

Unbounded Problem

When feasible solution space formed by the constraints is not confined to the first quadrants only.

If constraints passes through 2nd, 3rd or 4th quadrant.

No limits on the constraints. Common feasible region is not bounded in

any respect.

Example of Unbounded Problem

Max Z = 3x1 + 4x2

subject to x1 + x2 > 5 3x1 + x2 > 8

x1, x2 > 0

Unbounded Problem The feasible region is unbounded and the

objective function line can be moved parallel to itself without bound so that z can be increased infinitely. x2

x1

3x1 + x2 > 8

x1 + x2 > 5

Max 3x1 + 4x2

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5

8

2.67

Redundant Constraint

When one constraint is less restrictive than other constraint, it is said to be redundant constraint.

Advantages & DrawbacksS. No. Advantages  Drawbacks 1 Problem Insight Fraction Values2 Considers all possible Solution Co-efficients must be

Deterministic3 Better Decisions Function must be Linear

4 Better Tools Time & Un-certainty effect not considered

5 Highlights Bottlenecks Approximation of Complex problems

6 Clear specification of Management problems