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8/10/2019 Lecture 3 (Notes) (2)
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Lecture Day 3
Solving Linear Programming Problems
Using the Graphical Method Special Cases in Using the Graphical Method
Solving Linear Programming Problems
Using the Simplex Method
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(-4)(0.5X1+ X2= 81)
1.2X1+ 4X2= 240
X2= 81 0.5X1
1.2X1+ 4(81 0.5X1) = 240
1.2X1+ 324 - 2X1= 240
-0.8X1= -84
-0.8 -0.8
X1= 105
X2= 81 0.5X1
= 81 0.5(105)
= 81 52.5
X2= 28.5
-2X1- 4X2= -324
1.2X1+ 4X2= 240
-0.8X1 = -84
-0.8 -0.8
X1= 105
Z = 200X1+ 500X2
Point C:
= 200(105) + 500(28.5)
= 21,000 + 14,250
Z = 35,250
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X1
X2
50
50
100
100
150
150
2000
X2= 40
1.2X1+ 4X2= 240
0.5X1+ X2= 81
AB
C
D
Techniques to DetermineWhich Extreme Point
Gives the Optimal Solution
1. Use the objective function to single out theextreme point that is the optimum.
2. Find the coordinates of the extreme points andcompute the profit associated with each
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X1
X2
50
50
100
100
150
150
2000
X2= 40
1.2X1+ 4X2= 240
0.5X1+ X2= 81
AB
C
D
Point A
X1= 0
X2= 40
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
Z = 200X1+ 500X2
= 200(0) + 500(40)
= 20,000
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X1
X2
50
50
100
100
150
150
2000
X2= 40
1.2X1+ 4X2= 240
0.5X1+ X2= 81
AB
C
D
Point B
1.2X1+ 4X2= 240
X2= 40
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
Z = 200X1+ 500X2
= 33,333.33
1.2X1+ 4(40) = 240
1.2X1+ 160 = 240
1.2X1= 240 - 160
= 200(66.6) + 500(40)
1.2X1= 80
1.2 1.2
X1= 66.6
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X1
X2
50
50
100
100
150
150
2000
X2= 40
1.2X1+ 4X2= 240
0.5X1+ X2= 81
AB
C
D
Point D
0.5X1+ X2= 81
X2= 0
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
0.5X1= 81
0.5 0.5
X1= 162
Z = 200X1+ 500X2
= 32,400
= 200(162) + 500(0)
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Sample Problem # 2
The Reed Pump Co. manufactures pumping apparatus for thepetroleum industry. It is trying to determine an optimal production
and distribution plan for its new submersible pump. They have onemain plant and two regional warehouses. Warehouse 1 has demandfor at least 80 pumps this month, and warehouse 2 has demand forat least 100 units. The following are the combined costs of producingand shipping to the two warehouses:
To achieve economies of scale, the company wants to produce atleast 300 pumps this month. The only resource limitation involvesmanufacturing resource hours. Pumps sent to regional warehouse 1need an extra-fine filter and require five hours of manufacturing time.Those pumps sent to warehouse 2 require only four hours ofmanufacturing time. The company has 2,000 hours of manufacturing-
resource time available per month.
Warehouse 1 Warehouse 2
Cost of Producing & Shipping
P240 P280From Plant
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Let: X1= number of pumps to bedelivered to Warehouse 1
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2= number of pumps to be
delivered to Warehouse 2
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X2= 300
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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5X1+ 4X2= 2,000
if X2= 0
5X1= 2,000
5 5
X1= 400
if X1= 0
4X2= 2,000
4 4
X2= 500
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X2= 300
5X1+ 4X2= 2,000
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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240X1+ 280X2 = 100,000
240X1= 100,000
X1= 416.67
if X1= 0
280X2= 100,000
X2= 357.14
if X2= 0
240 240
280 280
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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240X1+ 280X2 = 50,000
240X1= 50,000
X1= 208.33
if X1= 0
280X2= 50,000
X2= 178.57
if X2= 0
240 240
280 280
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X
2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X
2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
Point D
X1+ X2 = 300
X2 = 100
X1+ 100= 300
X1= 300 - 100
X1= 200
Z = 240X1+ 280X2
= 240(200) + 280(100)
= 48,000 + 28,000
Z = 76,000
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X
2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
Point A
X1+ X2 = 300
X1 = 80
80 + X2 = 300
X2= 300 - 80
X2= 220
Z = 240X1+ 280X2
= 240(80) + 280(220)
= 19,200 + 61,600
Z = 80,800
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X
2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
Point B
5X1+ 4X2 = 2,000
X1 = 80
5(80) + 4X2 = 2,000
400 + 4X2 = 2,000
4X2 = 2,000 - 400
4X2 = 1,6004 4
X2 = 400
Z = 240X1+ 280X2
= 240(80) + 280(400)
= 19,200 + 112,000
Z = 131,200
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X1
X2
500
500
100
100
400
400
2000 300
200
300
X1= 80
X2= 100
X1+ X
2= 300
5X1+ 4X2= 2,000
A
B
C
D
Min. Z = 240X1+ 280X2
Subject to:
X1> 80
X2> 100
X1+ X2> 300
5X1+ 4X2< 2,000
Point C
5X1+ 4X2 = 2,000
X2 = 100
5X1+ 4(100)= 2,000
5X1 = 1,6005 5
X1 = 320
Z = 240X1+ 280X2
= 240(320) + 280(100)
= 76,800 + 28,000
Z = 104,800
5X1+ 400= 2,000
5X1= 2,000 - 400
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Special Cases in the Graphical Solutionto Linear Programming Problems
It is possible that an adjacent extreme point will yieldthe same value. Graphically, this happens wheneverthe slope of the objective function equals the slope
of a constraint equation that passes through anoptimal extreme point.
Alternative Optima
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X1
X2
50
50
100
100
150
150
2000
X2= 40
1.2X1+ 4X2= 240
0.5X1+ X2= 81
A
D
C
B
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
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Special Cases in the Graphical Solutionto Linear Programming Problems
It is possible for an LP problem to have a non-emptyset of feasible solutions and yet have no finite optimalsolution.
This can occur whenever the feasible region extendsinfinitely in the direction of improvement for theobjective function.
Having this scenario implies that the model has beenformulated incorrectly. Usually, a constraint has beenomitted or the signs on some of the coefficients havebeen reversed.
Unbounded Solution
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Max. Z = X1+ X2
Subject to:
5X1 - X2> 10
3X1- 2X2< 9
X1, X2> 0
X1
X2
5
5
1
1
4
4
20 3
23
-1-2-3-4-5-6-7-8-9
-10
5X1X2 = 10
3X12X2= 9
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Special Cases in the Graphical Solutionto Linear Programming Problems
It is also possible to have an LP problem in which nofeasible solution exists.
This situation corresponds to a problem that is
formulated incorrectly or has conflicting restrictionswithin the constraint set.
No Feasible Solution
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Min. Z = X1+ X2
Subject to:
X1 - X2> 1
-X1+ X2> 1
X1, X2> 0
X1
X2
1
1
2 3
23
-1-2-3
-2-3
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SolvingLinear Programming Problems
Using the Simplex Method
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GeorgeDantzig developed the simplex methodin 1947. It is the most widely used method for
solving LP problems.
Thesimplex method is a solution algorithm. Analgorithmis an iterative procedure withspecific computational rules that solves aproblem in a finite number of steps.
Thesimplex algorithm is a systematic algebraicprocedure for examining the extreme points ofthe LP feasible region. The extreme points areexamined in a sequence such that eachsuccessive point yields a solution that is at leastas effective as the previous point.
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Thesimplex method requires that all constraints be
expressed as equations, or that the problem betransformed in standard form.
That is, there is a need to convert less-than-or-equal-to and greater-than-or-equal-to
constraints into equations. This can be done by adding dummy
variablesto the inequalities.
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Sample Problem In Standard Form
0X1 +1X2 +1S1 = 40
1.2X1 +4X2 +0S1+1S2 = 240
Let : X1= number of
power amplifiersX2= number of
preamplifiers
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
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Let : X1= number of
power amplifiersX2= number of
preamplifiers
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
Sample Problem In Standard Form
0X1 +1X2 +1S1 = 40
1.2X1 +4X2 +0S1
+
1S2 = 240
0S2
+
0.5X1 +1X2 +0S1+0S2+1S3 = 81
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Let : X1= number of
power amplifiersX2= number of
preamplifiers
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
Sample Problem In Standard Form
0X1 +1X2 +1S1 = 40
1.2X1 +4X2 +0S1
+
1S2 = 240
0S2
+
0.5X1 +1X2 +0S1+0S2+1S3 = 81
+0S3
+0S3
200X1 +500X2 +0S1 +0S2 +0S3
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Canonical Form of the Simplex Tableau
The table consisting of the system equations and other relevantinformation of an LP problem is called a Simplex Tableau.
00
.
.
.
0
S1S2.
.
.
Sm
a11a21.
.
.
am1
a12a22.
.
.
am2
a1na2n.
.
.
amn
10
.
.
.
0
01
.
.
.
0
00
.
.
.
1
b1b2.
.
.
bm
SolutionValues
. . .
. . .
. . .
BasicVariables
X1 X2 Xn
. . .
. . .
. . .
CB . . . S2 Sm. . .
Cj C1 C2 Cn. . . 0 0 0. . .
Decision Variables Slack Variables
Zj
Cj- Zj
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Let : X1= number of
power amplifiersX2= number of
preamplifiers
Max. Z = 200X1+ 500X2
Subject to:
X2< 40
1.2X1+ 4X2< 240
0.5X1+ X2< 81
X1, X2> 0
Sample Problem In Standard Form
0X1 +1X2 +1S1 = 40
1.2X1 +4X2 +0S1
+
1S2 = 240
0S2
+
0.5X1 +1X2 +0S1+0S2+1S3 = 81
+0S3
+0S3
200X1 +500X2 +0S1 +0S2 +0S3
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Cj
BasicVariables
CB SolutionValues
X1 X2 S1 S2 S3
Zj
Cj- Zj
S1
S2
S3
200 500 0 0 0
0 1 1 0 0 40
1.2 4 0 1 0 240
0.5 1 0 0 1 81
0
0
0
0 0 0 0 0 0
200 500 0 0 0
Tableau 1
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End of Lecture Day 3
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