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Linear Programming

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Linear Programming

A linear programming problem has linear objective function and linear constraint and variables that are restricted to non-negative values.

-X1+2X2-X3<=70

2X1-2X3=50

X1-2X22+4X3<=10

X1+X2+X3=6

2X1+5X2+X1X2<=25

3X1/2 +2X2-X3>=15

Linear Programming

Feasible solution:

The solution of the problem that satisfy every constraint is called as feasible solution.

For constraint <=0 feasible solution lie under the constraint

For constraint >=0 feasible solution above under the constraint

For constraint ==0 feasible solution lie on constraint

.

Graphical Method

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

.Determine the feasible region by identifying the solutions that satisfy all constrain simultaneously.

. Draw an objective function line showing the values of the decision variable yield a specified value of objective function.

A Linear programming problem involving two variables can be solved using the graphical solution

Graphical Method

Move parallel objective function lines toward larger objective function values until further movement would take the line completely out of feasible region

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

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.

Problem

A small manufacturing company decided to move in market for standard and deluxe golf bags. Initial analyses showed that each bag produced will require following operation

1.cutting and dyeing material

2.sweing

3.Finishing

4.Inspection and packing

Problem

For a standard bag each bag will require:

7/10 hr in cutting & dyeing,

1/2 hr in sewing

1 hr in finishing department

1/10 hr in inspection

For a high -priced bag each bag will require :

1 hr in cutting & dyeing,

5/6 hr in sewing

2/3 hr in finishing department

1/4 hr in inspection

Decision Making

Price for profit contribution standard bags $10, Price for Deluxe bag is $9.

Problem

How many constraints are in this problem.

How many decision variable are in this problem

what is objective function.

Problem formulation

Max 10S +9D

s.t

7/10S+1D<=630 (Cutting & dyeing)

1/2S+5/6D <=600 Sewing

1S+2/3D<=708 Finishing

1/10S+1/4D<=135 Inspection & Packaging

Solutions

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

How?

Take S on x-axis and D on Y-axis

For each constrain put S= 0 Get D to obtain a point (0,D) then put D=0 Get S to obtain (S,0) , join these two point to get the line.

Cutting and Dyeing Constrain

Constrain Equation:

7/10S+1D<=630

Solve for equality first by determining two points as described above:

7/10S+1D=630

S=0 -> D=630, D=0 -> S=900

Points (0,630) and (900,0)

Cutting and Dyeing Constrain

Find the points lie above the feasible region

Other Constraints

1/2S+5/6D <=600 (Sewing)

Points ????

1S+2/3D<=708 (Finishing)

Points????

1/10S+1/4D<=135 (Inspection & Packaging)

Points?????

Feasible Solutions as Profit increases

Let us take any arbitrary profit and draw it on feasible solution

10S+ 9D=1800; (putting S, D 0) find points

10S+ 9D=3600

10S+9D=5400

Feasible Solutions V/S Profit

Profit lines are parallel to each other, higher value of the objective for higher profit lines, However at some value line would be outside the feasible region. The point in the feasible region that lies on highest profit line is optimal solution.

Feasible Solutions V/S Profit

Intersection of cutting & Dyeing and finishing constraint gives an optimal solution.

7/10S+1D<=630 (Cutting & dyeing)

1S+2/3D<=708 Finishing

Find point of intersection and check the profit for the optimal solution.

10s+9d=???

Slack Variables

Any unused capacity for a<=constraint is called as slack variables

Putting the value of the optimal solution in all constraint

A linear program in which all the variables are non-negative and all the constraints are equalities is said to be in standard form.

Standard form is attained by adding slack variables to "less than or equal to" constraints, and by subtracting surplus variables from "greater than or equal to" constraints.

Slack and surplus variables represent the difference between the left and right sides of the constraints.

Slack and surplus variables have objective function coefficients equal to 0.

Standard Form

Add four slack variables to constraint having zero coefficient for unused capacity.

Max 10S+9D+0S1+0S2+0S3+0S4

7/4S + 1D + 1S1=630

1/2S+5/6D+1S2=600

1S+2/3D+1S3=708

1/10S+1/4D+1S4=135

S1=0,S2=120,S3=0,S4=18

Extreme Points

The vertices of feasible region is called as vertices of the feasible region. It has 5 extreme points.

Optimal solutions occur at one of vertices of the feasible solutions, one producing highest value of objective function is the required optimal solution.

II. Objective function: Max 5S+ 9D

What about constraints

Check the objective function at the extreme points.

S=300,D=420 find objective function

S=540,D=252 find objective function

Which one produce max objective function..

Alternative Optimal Solution

III. Max 6.3S +9D

Check the objective function at the extreme points.

S=300,D=420 find objective function

S=540,D=252 find objective function

Infeasibility

Suppose if management want to produce at least S=500 and D=360 with the same given constraints

Then there would be no feasible region

Resources needed:

Minimization Problem

A company sales two products A and B. The combined production for production A and B must be at least 350 gallons.

Additionally a major customer require 125 gallon of Product A.

Product A requires 2 hrs and product B requires 1 hr of processing time per gallon respectively.

Total 600 processing hrs is available

Production cost is $2 gallon and $3 gallon for A and B respectively.

Objective function & constraint

Minimization Problem

Min 2A+3B

1A>=125 (demand for A)

1A+1B>=350 (Total Production)

2A+1B<=600 (Processing Time)

A,B>=0

Feasible Region

Feasible Region

Find Extreme point using intersection of three lines

1.(A,B)=(125,225)

2.(A,B)=(125,350)

3.(A,B)=(250,100)

Optimal solution:

Objective function:

2A+3B

2(125)+3(350)=1300

2(125)+3(225)=925

2(250)+3(100)=800

Solve graphically for the optimal solution:

Max 3x1 + 4x2

s.t. x1 + x2> 5

3x1 + x2> 8

x1, x2> 0

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

3x1 + x2> 8

8

Max 3x1 + 4x2

5

x1 + x2> 5

x1

2.67

5

Surplus Variable

Model for Break-Even Analysis: