1. A LINEAR PROGRAMMING PROBLEM HAS LINEAR OBJECTIVE FUNCTION AND LINEAR CONSTRAINT AND VARIABLES THAT ARE RESTRICTED TO NON-NEGATIVE VALUES. 1. -X 1 +2X 2 -X 3 <=70 2. 2X 1 -2X 3 =50 3. X 1 -2X 2 2 +4X3<=10 4. X1+X2+X3=6 5. 2X 1 +5X 2 +X 1 X 2 <=25 6. 3X 1/2 +2X 2 -X 3 >=15 Linear Programming

2. 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. Linear Programming

3. 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

4. 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

5. 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.

6. 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

7. 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 Problem

8. PRICE FOR PROFIT CONTRIBUTION STANDARD BAGS $10, PRICE FOR DELUXE BAG IS $9. Decision Making Standard Bag Deluxe Bag Max hrs Cutting &dyeing 7/10 hrs1hr630 Sewing½ hr5/6hr600 Finishing Dept 1hr2/3hr708 Inspection & Packaging 1/10hr1/4hr135

9. HOW MANY CONSTRAINTS ARE IN THIS PROBLEM. HOW MANY DECISION VARIABLE ARE IN THIS PROBLEM WHAT IS OBJECTIVE FUNCTION. Problem

10. 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 Problem formulation

11. . 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

12. 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

13. FIND THE POINTS LIE ABOVE THE FEASIBLE REGION Cutting and Dyeing Constrain

14. 1/2S+5/6D <=600 (SEWING) POINTS ???? 1S+2/3D<=708 (FINISHING) POINTS???? 1/10S+1/4D<=135 (INSPECTION & PACKAGING) POINTS????? Other Constraints

16.  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

17. 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.

18.  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

19.  Any unused capacity for a<=constraint is called as slack variables  Putting the value of the optimal solution in all constraint Slack Variables ConstraintHrs RequiredHrs available Unused Hrs C&D7/10(540)+1(252)=63 0 6300 Sewing½(540)+5/6(252)=4 80 600120 Finishing1(540)+2/3(252)=70 8 7080 I&P1/10(540)+1/4(252)= 117 135180

20. 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.

21.  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

22. 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

23.  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

24. 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

25. 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

26. Resources needed: Minimum Required Resources (hrs) Available Resource s Additional Resources needed Cutting&Dyeing7/10(500)+1(360)=7 10 63080 Sewing½(500)+5/6(360)=5 50 600None Finishing1(500)+2/3(360)=74 0 70832 Inspection&Pac king 1/10(500)+1/4(360) =140 1355

27. 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

28. Min 2A+3B 1A>=125 (demand for A) 1A+1B>=350 (Total Production) 2A+1B<=600 (Processing Time) A,B>=0 Minimization Problem

29. Feasible Region

30. 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

31.  Objective function:  2A+3B  2(125)+3(350)=1300  2(125)+3(225)=925  2(250)+3(100)=800 Optimal solution:

32. Example: Unbounded Problem Solve graphically for the optimal solution: Max 3x 1 + 4x 2 s.t. x 1 + x 2 > 5 3x 1 + x 2 > 8 x 1, x 2 > 0

33. 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. x2x2x2x2 x1x1x1x1 3x 1 + x 2 > 8 x 1 + x 2 > 5 Max 3x 1 + 4x 2 5 5 8 2.67

34. Surplus Variable

35. Model for Break-Even Analysis: