A LINEAR PROGRAMMING PROBLEM HAS LINEAR OBJECTIVE FUNCTION AND LINEAR CONSTRAINT AND VARIABLES THAT ARE RESTRICTED TO NON-NEGATIVE VALUES. 1. -X 1 +2X.

A LINEAR PROGRAMMING PROBLEM HAS LINEAR OBJECTIVE FUNCTION AND LINEAR CONSTRAINT AND VARIABLES THAT ARE RESTRICTED TO NON-NEGATIVE VALUES.1. -X1+2X2-X3<=70

2. 2X1-2X3=50

3. X1-2X22+4X3<=10

4. X1+X2+X3=65. 2X1+5X2+X1X2<=25

6. 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 CONSTRAINTFOR CONSTRAINT >=0 FEASIBLE SOLUTION ABOVE UNDER THE CONSTRAINTFOR CONSTRAINT ==0 FEASIBLE SOLUTION LIE ON CONSTRAINT

•.

Linear Programming

•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

Graphical Method

Summary of the Graphical Solution Procedurefor 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.

•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 DEPARTMENT1/10 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

Problem

PRICE FOR PROFIT CONTRIBUTION STANDARD BAGS $10, PRICE FOR DELUXE BAG IS $9.

Decision MakingStandardBag

DeluxeBag

Max hrs

Cutting &dyeing

7/10 hrs 1hr 630

Sewing ½ hr 5/6hr 600

Finishing Dept

1hr 2/3hr 708

Inspection &Packaging

1/10hr 1/4hr 135

• HOW MANY CONSTRAINTS ARE IN THIS PROBLEM.

• HOW MANY DECISION VARIABLE ARE IN THIS PROBLEM

• WHAT IS OBJECTIVE FUNCTION.

Problem

MAX 10S +9DS.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

Problem formulation

. 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.

Solutions

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

Cutting and Dyeing Constrain

900,0

0, 630

0

200

400

600

800

1000

1200

0 200 400 600 800 1000 1200 1400

Cutting & Dyeing

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

POINTS ????

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

POINTS????

1/10S+1/4D<=135 (INSPECTION & PACKAGING)

POINTS?????

Other Constraints

900,0

0, 630

1200,0

0,720

0

200

400

600

800

1000

1200

0 200 400 600 800 1000 1200 1400

Cutting & Dyeing Sewing

900,0

0, 630

1200,0

0,720

708,0

0,1062

0

200

400

600

800

1000

1200

0 200 400 600 800 1000 1200 1400

Cutting & Dyeing Sewing FinishingLine

900,0

0, 630

1200,0

0,720

708,0

0,1062

0,540

1350,0

0

200

400

600

800

1000

1200

0 200 400 600 800 1000 1200 1400 1600

Cutting & Dyeing Sewing FinishingLine I&P fissibleregion

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 as Profit increases

Feasible Solutions V/S Profit

0 100 200 300 400 500 600 700 8000

100

200

300

400

500

600

700

530

425

252

0

Feasible region

18003600

5400

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.

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=???

Feasible Solutions V/S Profit

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

Putting the value of the optimal solution in all constraint

Slack Variables

Constraint Hrs Required Hrs available

UnusedHrs

C&D 7/10(540)+1(252)=630

630 0

Sewing ½(540)+5/6(252)=480

600 120

Finishing 1(540)+2/3(252)=708

708 0

I&P 1/10(540)+1/4(252)=117

135 180

Slack and Surplus Variables

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.

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

Standard Form

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.

Extreme Points

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..

II. Objective function: Max 5S+ 9D

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

Alternative Optimal Solution

0 100 200 300 400 500 600 700 8000

100

200

300

400

500

600

Feasible Region

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

Infeasibility

0 100 200 300 400 500 600 700 800 9000

100

200

300

400

500

600

700

360

500,600

500,360 360

In Feasibilty

Points Satisfying Depart-mental Costraints

Points Saisfying Maximum Pro-duction

Resources needed:Minimum Required Resources (hrs)

Available Resources

Additional Resources needed

Cutting&Dyeing

7/10(500)+1(360)=710

630 80

Sewing ½(500)+5/6(360)=550

600 None

Finishing 1(500)+2/3(360)=740

708 32

Inspection&Packing

1/10(500)+1/4(360)=140

135 5

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

Minimization Problem

Feasible Region

0 50 100 150 200 250 300 350 4000

100

200

300

400

500

600

700

A= 1251A+1B2A+B=600

Find Extreme point using intersection of three lines

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

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

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

Feasible Region

Objective function: 2A+3B 2(125)+3(350)=1300 2(125)+3(225)=925 2(250)+3(100)=800

Optimal solution:

Example: Unbounded Problem

Solve graphically for the optimal solution:

Max 3x1 + 4x2

s.t. x1 + x2 > 5

3x1 + x2 > 8

x1, x2 > 0

Example: 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

5

5

8

2.67

Surplus Variable

Model for Break-Even Analysis: