a linear programming problem has linear objective function and linear constraint and variables that...
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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: