Chap09[2] LS Modif
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Transcript of Chap09[2] LS Modif
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Kuliah ke-3
Linear Programming:
The Simplex Method
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Learning Objectives
Students will be able to
Convert LP constraints to equalities with slack, surplus,
and artificial variables.
Set up and solve both maximization and minimization
LP problems with simplex tableaus.
Interpret the meaning of every number in a simplextableau.
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Learning Objectives - continued
Students will be able to
Recognize cases of infeasibility,
unboundedness, degeneracy, and multiple
optimal solutions in a simplex output.
Understand the relationship between the primal
and dual and when to formulate and use the
dual.
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Chapter Outline
9.1 Introduction
9.2 How to Set Up the Initial Solution
9.3 Simplex Solution Procedures
9.4 The Second Simplex Tableau
9.5 Developing the Third Simplex Tableau
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Chapter Outline - continued
9.6 Review of Procedures for Solving LPMaximization Problems
9.7 Surplus and Artificial Variables
9.8 Solving Minimization Problems9.9 Review of Procedures for Solving LP
Minimization Problems
9.10 Special Cases
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Flair Furniture Company
Maximize:Objective: 21 57 XX +
Hours Required to Produce One Unit
DepartmentX1
Tables
X2
Chairs
Available
Hours This
Week
CarpentryPainting/Varnishing
42
31
240100
Profit/unit
Constraints:
$7 $5
)varnishing&(painting10012 21 + XX
)(carpentry24034 21 + XX
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Flair Furniture Company'sFeasible Region & Corner Points
Numbero
fChairs100
80
60
40
20
0 20 40 60 80 100 X
X2
Number of Tables
B = (0,80)
C = (30,40)
D = (50,0)
Feasible
Region
24034 21
+
XX
10012 11 +XX
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Flair Furniture - Adding Slack Variables
(carpentry)24034 21 + XX
(painting&varnishing10012 21 + XX
Constraints:
Constraints with Slack Variables
24034
10012
221
121
=++
=++
SXX
SXX
21 57 XX +
Objective Function
Objective Function with Slack Variables
2121 0057 SSXX +++
(carpentry)
(painting&varnishing
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Flair Furnitures Initial Simplex Tableau
Profit
per
Unit
Column
Prod.
Mix
Column
Real Variables
Columns Slack
Variables
Columns
Constant
Column
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$7 $5 $0 $0
Profit
perunit row
2 1 1 0
4 3 0 1
$0 $0 $0 $0
$7 $5 $0 $0
$0
$0
S1
S2
Zj
Cj -Zj
100
240
$0
$0
Constraintequation
rowsGrossProfit
rowNet
Profitrow
Si l S f
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Simplex Steps forMaximization
1. Choose the variable with the greatest positiveCj - Zj to enter the solution.
2. Determine the row to be replaced by selecting thatone with the smallest (non-negative) quantity-to-
pivot-column ratio.3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the Cj and Cj - Zj values for this tableau.If there are any Cj - Zj values greater than zero,return to Step 1.
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Calculate the Zj and Cj-Zj rows
Zj(for X1 column) = ($0)(2) + ($0)(4) = $0
Zj(for X2 column) = ($0)(1) + ($0)(3) = $0Zj(for S1 column) = ($0)(1) + ($0)(0) = $0
Zj(for S2 column) = ($0)(0) + ($0)(1) = $0
Zj(for Total Profit) = ($0)(100) + ($0)(240) = $0
Cj Sol Mix X1 X2 S1 S2 Quantity Rasio
$0 S1 2 1 1 0 100 100/2=50
$0 S2 4 3 0 1 240 240/4=60
X1 X2 S1 S2Cj for column $7 $5 $0 $0
Zj for column $0 $0 $0 $0
CjZj for column $7 $5 $0 $0
Pivot column (the greatest positive)
Pivot numberPivot row (the smallest
/non-negative)
Step 1:
Step 2:Step 2:
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Pivot Row, Pivot Number Identified inthe Initial Simplex Tableau
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$7 $5 $0 $0
2 1 1 04 3 0 1
$0 $0 $0 $0
$7 $5 $0 $0
$0$0
S1
S2
Zj
Cj -Zj
100240
$0
$0
Pivot
row
Pivot number
Pivot column
C l l i N Pi R (X ) &
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Calculating New Pivot Row (X1) &Number in New S2 Row
= - x43
0
1
240
(4(4
(4
(4
(4
1)1/2)
1/2)
0)
50)
= - x
= - x
= - x
= - x
-
=
rowX
newin
number
ingCorrespond
number
pivot
below
Number
rowS
oldin
Number
RowS
Newin
Number
1
22
Cj Sol Mix X1 X2 S1 S2 Quantity
$7 X1 2/2=1 1/2=1/2 1/2=1/2 0/2=0 100/2=50
(0)
(1)
(-2)
(1)
(40)
Step 3:
Step 4:
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Calculate the Zj and Cj-Zj rows
Zj(for X1 column) = ($7)(1) + ($0)(0) = $7
Zj(for X2 column) = ($7)(1/2) + ($0)(1) = $7/2Zj(for S1 column) = ($7)(1/2) + ($0)(-2) = $7/2
Zj(for S2 column) = ($7)(0) + ($0)(1) = $0
Zj(for Total Profit) = ($7)(50) + ($0)(40) = $350
Cj Sol Mix X1 X2 S1 S2 Quantity Rasio
$7 X1 1 1/2 1/2 0 50 50/(1/2)=100
$0 S2 0 1 -2 1 40 40/1=40
X1 X2 S1 S2Cj for column $7 $5 $0 $0
Zj for column $7 $7/2 $7/2 $0
CjZj for column $0 $3/2 -$7/2 $0
Pivot column (the greatest positive)
Pivot numberPivot row (the smallest
/non-negative)
Step 1:
Step 2:Step 2:Step 5:
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Pivot Row, Column, and NumberIdentified in Second Simplex Tableau
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$7 $5 $0 $0
1 1/2 1/2 0
0 1 -2 1
$7 $7/2 $7/2 $0
$0 $3/2 -$7/2 $0
$7
$0
X1
S2
Zj
Cj -Zj
50
40
$350(Total
Profit)
Pivotrow
Pivot number
Pivot column
C i
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Calculating the NewX1 Rowfor Flairs Third Tableau
= - x10
3/2
-1/2
30
11/2
1/2
0
50
(1/2)(1/2)
(1/2)
(1/2)
(1/2)
(0)(1)
(-2)
(1)
(40)
= - x
= - x
= - x
= - x
-
=
rowX
newin
number
ingCorrespond
number
pivot
above
Number
rowX
oldin
Number
RowX
Newin
Number
ii
Cj Sol Mix X1 X2 S1 S2 Quantity$5 X2 0/1=0 1/1=1 -2/2=-2 1/1=1 40/1=40
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Fi l Si l T bl f
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Final Simplex Tableau forthe Flair Furniture Problem
CjSolution
Mix X1 X2 S1 S2 Quantity
$7 $5 $0 $0
1 0 3/2 -1/2
0 1 -2 1
$7 $5 $1/2 $3/2
$0 $0 -$1/2 -$3/2
$7
$5
X1
X2
Zj
Cj -Zj
30
40
$410
Si l St f
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Simplex Steps forMinimization
1. Choose the variable with the greatest negative Cj - Zj
to enter the solution.
2. Determine the row to be replaced by selecting that one
with the smallest (non-negative) quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).5. Calculate the Cj and Cj - Zj values for this tableau. If
there are any Cj - Zj values less than zero, return to
Step 1.
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Surplus & Artificial VariablesConstraints
Constraints-Surplus & Artificial Variables9003025
2108105
21
321
=+
++
XX
XXX
9003025
2108105
221
11321
=++
=+-++
AXX
ASXXX
Objective Function
21 65 XX +:Min
21121 065 MAMASXX ++++:Min
Objective Function-Surplus & Artificial Variables
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The Muddy River Chemical Company
Muddy River Chemical Companymemproduksi 1.000 pound campuran
phosphate dan potassium.
Biaya phosphate $5 per pound dan potassium$6 per pound. Tidak lebih dari sama dengan
300 pound phosphate, dan paling sedikit 150
pound potassium.
Masalahnya adalah berapa biaya terendah
campuran dua bahan kimia tersebut?
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Model Matematika Muddy R.C.
Constraints
Objective Function65 + 21 XX:Min Z =
300
1,000
2
1
21
=+
X
X
lbXX
lb
150 lb
2
X 01,X
X1 = jumlah pound phosphat
X2 = jumlah pound potassium
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Surplus & Artificial Variables
Constraints
Objective Function
X1 + X2 + A1 = 1.000
X1 + S1 = 300
X2 - S2 + A2 = 150
X1, X2, S1, S2, A1, A2 0
Min Z = 5X1 + 6X2 +0S1+0S2+MA1+MA2
Objective Function-Surplus & Artificial Variables
3001,000
2
1
21
=+
XX
lbXXlb
150 lb
2X 01,X
Constraints Function-Surplus & Artificial Variables
Min Z = 5X1 + 6X2
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Initial Simplex Tableau
Cj - Zj
Zj
15010-1010A2M300000101S10
1000010011A1M
QtyA2A1S2S1X2X1Sol
Mix
MM0065Cj
Total Cost
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Calculate the Zj and Cj-Zj rows
Zj(for X1 column) = ($M)(1) + ($0)(1) + ($M)(0) = $M
Zj(for X2 column) = ($M)(1) + ($0)(0) + ($M)(1) = $2MZj(for S1 column) = ($M)(0) + ($0)(1) + ($M)(0) = $0
Zj(for S2 column) = ($M)(0) + ($0)(0) + ($M)(-1) = -$M
Zj(for A1 column) = ($M)(1) + ($0)(0) + ($M)(0) = $M
Zj(for A2 column) = ($M)(0) + ($0)(0) + ($M)(1) = $M
Zj(for Total Profit) = ($M)(1.000) + ($0)(300) + ($M)(150) = $1.150M
Cj Sol
Mix
X1 X2 S1 S2 A1 A2 Quantity Ratio
$M A1 1 1 0 0 1 0 1.000 1.000/1=1.000
$0 S1 1 0 1 0 0 0 300 300/0 = ~
$M A2 0 1 0 -1 0 1 150 150/1 = 150
X1 X2 S1 S2 A1 A2
Cj for column $5 $6 $0 $0 $M $M
Zj for column $M $2M $0 -$M $M $M
CjZj for column -$M+5 -$2M+6 $0 $M $0 $0Pivot Column
Pivot
Row
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Ratio
Step 2:
Baris A1 = 1.000/1 = 1.000
Baris A2 = 150/1 =150 (rasio terkecil, maka dipilih sebagai pivot row)
Baris S1 = 300/0 = ~ (tidak terdefinisi, maka pivot row ditolak)
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Initial Simplex Tableau
00M0-2M
+6
-M+5Cj - Zj
1.150MMM-M02MMZj
15010-1010A2M300000101S10
1000010011A1M
QtyA2A1S2S1X2X1Sol
Mix
MM0065Cj
Pivot Column
Total Cost
Pivot
Row
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Developing a Second Tableau, Othernew rows are calculated with formula
Calculating the NewA1Row
1 = 1(1 x 0)
0 = 1 - (1 x 1)0 = 0 - (1 x 0)
1 = 0 - (1 x -1)
1 = 1 - (1 x 0)
-1 = 0 - (1 x 1)
850 = 1.000(1 x 150)
-
=
rowreplaced
newlyin
number
ingCorrespond
number
pivot
beloworabove
Number
row
oldinNumbers
numbersrow
New
Step 4:
Calculating the NewS1 Row
1 = 1(0 x 0)
0 = 0 - (0 x 1)1 = 1 - (0 x 0)
0 = 0 - (0 x -1)
0 = 0 - (0 x 0)
0 = 0 - (0 x 1)
300 = 300(0 x 150)
St 5 Fi ll Z d C Z
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Step 5: Finally Zj and Cj-Zj rows arerecomputed
Zj(for X1 column) = ($M)(1) + ($0)(1) + ($6)(0) = $M
Zj(for X2 column) = ($M)(0) + ($0)(0) + ($6)(1) = $6Zj(for S1 column) = ($M)(0) + ($0)(1) + ($6)(0) = $0
Zj(for S2 column) = ($M)(1) + ($0)(0) + ($6)(-1) = $M-6
Zj(for A1 column) = ($M)(1) + ($0)(0) + ($6)(0) = $M
Zj(for A2 column) = ($M)(-1) + ($0)(0) + ($6)(1) = -$M+6
Zj(for Total Profit) = ($M)(850) + ($0)(300) + ($6)(150) = $850M+900
Cj Sol Mix X1 X2 S1 S2 A1 A2 Quantity
$M A1 1 0 0 1 1 -1 850
$0 S1 1 0 1 0 0 0 300
$6 X2 0 1 0 -1 0 1 150
X1 X2 S1 S2 A1 A2
Cj for column $5 $6 $0 $0 $M $M
Zj for column $M $6 $0 $M-6 $M -$M+6
CjZj for column -$M+5 $0 $0 -$M+6 $0 $2M-6
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Ratio
Step 2:
Baris A1 = 850/1 = 850
Baris X2 = 150/0 = ~ (tidak terdefinisi, maka pivot row ditolak)
Baris S1 = 300/1 = 300 (rasio terkecil, maka dipilih sebagai pivot row)
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D l i Thi d T bl
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Developing a Third Tableau, Othernew rows are calculated with formula
Calculating the NewA1 Row
0 = 1(1 x 1)
0 = 1 - (1 x 0)-1 = 0 - (1 x 1)
1 = 0 - (1 x 0)
1 = 1 - (1 x 0)
-1 = 1 - (1 x 0)
550 = 8500(1 x 300)
-
=
rowreplaced
newlyin
number
ingCorrespond
number
pivot
beloworabove
Number
row
oldinNumbers
numbersrow
New
Step 4:
Calculating the NewS1 Row
0 = 0(0 x 1)
1 = 1 - (0 x 0)0 = 0 - (0 x 1)
-1 = -1 - (0 x 0)
0 = 0 - (0 x 0)
1 = 1 - (0 x 0)
150 = 150(0 x 300)
Step 5: Finally Z and C Z rows are
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Step 5: Finally Zj and Cj-Zj rows arerecomputed
Zj(for X1 column) = ($M)(0) + ($5)(1) + ($6)(0) = $5
Zj(for X2 column) = ($M)(0) + ($5)(0) + ($6)(1) = $6Zj(for S1 column) = ($M)(-1) + ($5)(1) + ($6)(0) = -$M+5
Zj(for S2 column) = ($M)(1) + ($5)(0) + ($6)(-1) = $M-6
Zj(for A1 column) = ($M)(1) + ($5)(0) + ($6)(0) = $M
Zj(for A2 column) = ($M)(-1) + ($5)(0) + ($6)(1) = -$M+6
Zj(for Total Profit) = ($M)(550) + ($5)(300) + ($6)(150) = $$550M+2.400
Cj Sol Mix X1 X2 S1 S2 A1 A2 Quantity
$M A1 0 0 -1 1 1 -1 550
$5 X1 1 0 1 0 0 0 300
$6 X2 0 1 0 -1 0 1 150
X1 X2 S1 S2 A1 A2
Cj for column $5 $6 $0 $0 $M $M
Zj for column $5 $6 -$M+5 -$M-6 $M -$M+6
CjZj for column $0 $0 $M-5 -$M+6 $0 $2M-6Pivot Column
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Ratio
Step 2:
Baris A1 = 550/1 = 550 (baris yang diganti/row to be replaced)
Baris X2 = 150/-1 = -150 (tidak memenuhi syarat, karena negatif)
Baris X1 = 300/0 = ~ (tidak terdefinisi, maka pivot row ditolak)
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Developing a Fourth Tableau, Othernew rows are calculated with formula
Calculating the NewX1 Row
1 = 1(0 x 0)
0 = 0 - (0 x 0)1 = 1 - (0 x -1)
0 = 0 - (0 x 1)
0 = 0 - (0 x 1)
0 = 0 - (0 x -1)
300 = 300(0 x 550)
-
=
rowreplaced
newlyin
number
ingCorrespond
number
pivot
beloworabove
Number
row
oldinNumbers
numbersrow
New
Step 4:
Calculating the NewX2 Row
0 = 0(-1 x 0)
1 = 1 - (-1 x 0)-1 = 0 - (-1 x -1)
0 = -1 - (-1 x 1)
1 = 0 - (-1 x 1)
0 = 1 - (-1 x -1)
700 = 150(-1 x 550)
Step 5: Finally Z and C Z rows are
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Step 5: Finally Zj and Cj-Zj rows arerecomputed
Zj(for X1 column) = ($0)(0) + ($5)(1) + ($6)(0) = $5
Zj(for X2 column) = ($0)(0) + ($5)(0) + ($6)(1) = $6Zj(for S1 column) = ($0)(-1) + ($5)(1) + ($6)(-1) = -$1
Zj(for S2 column) = ($0)(1) + ($5)(0) + ($6)(0) = $0
Zj(for A1 column) = ($0)(1) + ($5)(0) + ($6)(1) = $6
Zj(for A2 column) = ($0)(-1) + ($5)(0) + ($6)(0) = $0
Zj(for Total Profit) = ($0)(550) + ($5)(300) + ($6)(700) = $5.700
Cj Sol Mix X1 X2 S1 S2 A1 A2 Quantity
$0 A1 0 0 -1 1 1 -1 550
$5 X1 1 0 1 0 0 0 300
$6 X2 0 1 -1 0 1 0 700
X1 X2 S1 S2 A1 A2
Cj for column $5 $6 $0 $0 $M $M
Zj for column $5 $6 -$1 $0 $6 $0
CjZj for column $0 $0 -$1 $0 $M-6 $MPivot Column
F th & O ti l S l ti t th M dd
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Fourth & Optimal Solution to the Muddy
River Chemical Corporation Problem
$M$M-6$0$1$0$0Cj
- Zj
$5.700$0$6$0-$1$6$5Zj
700010-110X2$630000010X1$5
550-111-100S2$0
QtyA2A1S2S1X2X1Sol
Mix
MM0065Cj
Total Cost
1
Solusi Optimal X1=300, X2=700, S2=550, Artificial Variable = 0
Jadi dalam keputusan manajemen perusahaan Muddy River Chemical Corp.
harus mencampurkan 300 pounds phosphate (X1) dengan 700 pounds
potassium (X2). Surplus potassium (S2) = 550 pounds potassium lebih daripada batasan X2 150.
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Special Cases Infeasibility
02M+21M-31200Cj - Zj
1800+2M0-21-M31-M-285Zj
201-1-1000A2M1000-12110X28
2000-13-201X15
QtyA2A1S2S1X2X1Sol
Mix
MM0085Cj
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Special Cases Unboundedness
Pivot Column
0-18015Cj - Zj
2700189-9Zj
101-10-2S1
30021-1X1
QtyS2S1X2X1Sol
Mix
0096Cj
Rasio 0
atauNegatif
tidak boleh
30/-1=-30
10/-2=-5
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Special Cases Degeneracy
Pivot Column
Cj 5 8 2 0 0 0
Solution
Mix
X1 X2 X3 S1 S2 S3 Qty
8 X2 1/4 1 1 -2 0 0 10
0 S2 4 0 1/3 -1 1 0 20
0 S3 2 0 2 2/5 0 1 10
Zj 2 8 8 16 0 0 80Cj-Zj 3 0 -6 -16 0 0
Rasio
10/(1/4)=40
20/4=5
10/2=5
Rasio
terkecil ini
indikasi
degeneracy
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Special Cases Multiple Optima
Cj 3 2 0 0Sol
Mix
X1 X2 S1 S2 Qty
2 X1 3/2 1 1 0 60 S2 1 0 1/2 1 3
Zj 3 2 2 0 12
Cj - Zj 0 0 -2 0
Multiple Optima atau Alternatif solusi optimal mungkin terjadi jika
nilai Cj-Zj = 0
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Kasus
High Note Sound Company memproduksi Compact Disk (CD)
players dan Stereo Receivers (SR). Membuat 1 unit CD player
jam kerja yang dibutuhkan untuk pekerja listrik 2 jam-kerja dan
pekerja teknik audio 3 jam-kerja, sedangkan 1 unit SR pekerja
listrik membutuhkan 4 jam-kerja dan pekerja teknik audio 1 jam-
kerja. Masing-masing pekerja memiliki waktu yang tersediauntuk menyelesaikan pekerjaannya, pekerja listrik tersedia waktu
sebanyak-banyaknya 80 jam-kerja dan pekerja audio sebanyak-
banyaknya 60 jam-kerja.
Pimpinan perusahaan menghendaki setiap CD player
mendapatkan profit $50 dan SR $120. Berapa banyak CD player
dan SR harus diproduksi untuk memperoleh profit maksimum?
Model Matematis
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Model MatematisHigh Note Sound Company
XX
XX
XX
:toSubject
:Max
X1, X2 > 0
Sensitivity Analysis
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Sensitivity Analysis
High Note Sound Company
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Simplex Steps for Maximization1. Choose the variable with the greatest positive
Cj - Zj to enter the solution.
2. Determine the row to be replaced by selecting thatone with the smallest (non-negative) quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the Cj and Cj - Zj values for this tableau.
If there are any Cj - Zj values greater than zero,return to Step 1.
High Note Sound Coy - Adding
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High Note Sound Coy. - AddingSlack Variables
(hours of audio technicians time available)6013 21 + XX
(hours of electricians time available)8042 21 + XX
Constraints:
Constraints with Slack Variables
21 12050 XX +
Objective Function
Objective Function with Slack Variables
21 00 SS ++
6013 21 + XX
8042 21 =+ XX + 1S
+ 2S =
21 12050 XX +
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Initial Simplex TableauProfit
perUnit
Column
Prod.
Mix
Column
Real Variables
Columns Slack
Variables
Columns
Constant
Column
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$50 $120 $0 $0Profit
perunit row
2 4 1 0
3 1 0 1
$0 $0 $0 $0
$50 $120 $0 $0
$0
$0
S1
S2
Zj
Cj -Zj
80
60
$0
$0
Constraintequation
rowsGrossProfit
rowNet
Profitrow
Pivot Row Pivot Number Identified in the
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Pivot Row, Pivot Number Identified in theInitial Simplex Tableau (Iteration 1)
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$50 $120 $0 $0
2 4 1 0
3 1 0 1
$0 $0 $0 $0$50 $120 $0 $0
$0
$0
S1
S2
ZjCj -Zj
80
60
$0$0
Pivot
rowPivot number
Pivot column
Step 1: Variable entering the solution has the largest positive Cj-Z
j
Step 2: Variable leaving the solution is
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Step 2: Variable leaving the solution isdetermined by a ratio we must compute
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$50 $120 $0 $0
2/4 4/4 1/4 0/4
3/1 1/1 0/1 1/1
$0 $0 $0 $0
$50 $120 $0 $0
$0
$0
S1
S2
ZjCj -Zj
80/4=20
60/1=60
$0
$0
Pivot
rowPivot number
Pivot column
Step 3: New pivot-row calculations are
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Step 3: New pivot row calculations aredone next
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$50 $120 $0 $0
1/2 1 1/4 03 1 0 1
$60 $120 $30 $0
$10 $0 -$30 $0
$120$0
X2S2
Zj
Cj -Zj
2060
$240
$0
Pivot
rowPivot number
Pivot column
Step 4: Other new rows are
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Step 4: Other new rows arecalculated with formula
= - x5/2
0
-1/4
1
40
3
1
0
1
60
(1)
(1)
(1)
(1)
(1)
(1/2)
(1)
(1/4)
(0)
(20)
= - x
= - x
= - x
= - x
-
=
rowX
newin
number
ingCorrespond
number
pivot
below
Number
rowS
oldin
Number
RowS
Newin
Number
2
22
Calculating the New S2 Row
Step 4: Other new rows are calculated
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Step 4: Other new rows are calculatedwith formula
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$50 $120 $0 $0
1/2 1 1/4 05/2 0 -1/4 1
$0 $0 $0 $0
$50 $120 $0 $0
$120$0
X2S2
Zj
Cj -Zj
2040
$0
$0
Pivot
rowPivot number
Pivot column
Step 5: Finally Zj and Cj-Zj rows are
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Step 5: Finally Zj and Cj Zj rows arerecomputed
Zj(for X1 column) = ($120)(1/2) + ($0)(5/2) = $60
Zj(for X2 column) = ($120)(1) + ($0)(0) = $120Zj(for S1 column) = ($120)(1/4) + ($0)(-1/4) = $30
Zj(for S2 column) = ($120)(0) + ($0)(1) = $0
Zj(for Total Profit) = ($120)(20) + ($0)(40) = $2,400
Cj Solution
Mix
X1 X2 S1 S2 Quantity
$120 X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
X1 X2 S1 S2
Cj for column $50 $120 $0 $0
Zj for column $60 $120 $30 $0
CjZj for column -$10 $0 -$30 $0
Simplex Optimal Solution
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Simplex Optimal Solution
High Note Sound Company
Cj 50 120 0 0
Sol
Mix
X1 X2 S1 S2 Qty
120 X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj 60 120 30 0 2.400
Cj - Zj -10 0 -30 0
N b i Obj ti F ti C ffi i t
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Nonbasic Objective Function Coefficients
Cj 50 120 0 0
Sol Mix X1 X2 S1 S2 Qty
120 X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj 60 120 30 0 2400CjZj -10 0 -30 0
CjZj 0Cj Zj;
X1 nilai Cj = $50 dan Zj = $ 60
- Cj (untuk X1) $60
- Cj (untuk S1) $30
Basic Objective Function Coefficients
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j
Perubahan pada Profit Contribution Sterio Receivers
Cj 50 120 + 0 0
Sol Mix X1 X2 S1 S2 Qty
120+ X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj
60+ /2 120+ 30+ /4 0 2400+20
Cj - Zj -10- /2 0 -30- /4 0
CjZj 0Cj = Zj;
X1 nilai -101/2 0 -10 1/2 -20 atau -20
S1 nilai -301/4 0 -30 1/4 -120 atau -120
- $100 Cj (untuk X2)
Simplex Solution
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Simplex Solution
High Note Sound Company
Objective increases by (shadow price) 30 if 1 additional hour of electricians
time is available.Jadi shadow price 30 jam kerja sah (valid) karena berada
antara 0 s.d. 240 jam kerja.
Cj 50 120 0 0
Sol Mix X1 X2 S1 S2 Qty
X1 1 1/4 0 20
S2 5/2 0 -1/4 1 40
Zj
60 120 30 0 40
Cj - Zj 0 0 -30 0 2400
Quantity S1 Ratio Meaning
20 1/4 20/(1/4) = 80 Pengurangan 80 jam-kerja s.d. 0 jam-kerja
40 -1/4 40/(-1/4) = -160 Meningkatkan 160 jam-kerja s.d. 240
jam-kerja
Change in Resource
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Change in Resource(Shadow Prices)
Shadow Price: Value of One Additional Unit of aScarce Resource
Found in Final Simplex Tableau in C-Z Row
Negatives of Numbers in Slack Variable Column
Hiring electrician, cost-nya $22 per hour in wages,shadow price = $30, jadi cost jadi bertambah sebesar
$30 - $22 = $8
Hiring audio jadi $14 per hour tdk mungkin karena
shadow price $0
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Change in RHS
ORIGINALQUANTITY S1 NEW QUANTITY
20 1/4 20+{(1/4)(12)} = 23
40 -1/4 40+{(-1/4)(12)} = 37
New quantity = original quantity + (substitution rate)(change in RHS)
Example : 12 electrician hours
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Steps to Form the Dual
To form the Dual: If the primal is max., the dual is min., and vice versa.
The right-hand-side values of the primal constraints
become the objective coefficients of the dual.
The primal objective function coefficients become the
right-hand-side of the dual constraints.
The transpose of the primal constraint coefficients
become the dual constraint coefficients. Constraint inequality signs are reversed.
P i l & D l
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Primal & Dual
Primal: Dual
6013
8042
21
21
+
+
XX
XX
Subject to:
12014
5032
21
21
+
+
UU
UU
Subject to:
12050 21 + XX:Max 6080 21 + UU:Min
Comparison of the Primal and Dual
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pOptimal Tableaus
P
rimalsOptima
lSolution
DualsOptimalSolution
Cj
SolutionMix
Quantity
$7
$5
X2
S2
Zj
Cj -Zj
20
40
$2,400
X1 X2 S1 S2
$50 $120 $0 $0
1/2 1 1/4 0
5/2 0 -1/4 1
60 120 30 0
-10 0 -30 0
CjSolution
Mix
Quantity
$7
$5
U1
S1Zj
Cj -Zj
30
10
$2,400
X1 X2 S1 S2
80 60 $0 $0
1 1/4 0 -1/4
0 -5/2 1 -1/2
80 20 0 -20
$0 40 0 20
A1 A2
M M
0 1/2
-1 1/2
0 40
M M-40
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Problem 9 2 Bentuk Konversi
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Problem 9-2 Bentuk Konversi
2X1 + 1X2 + 1S1 + 0S2 = 40X1 + 3X2 + 0S1 + 1S2 = 30
X1, X2, S1, S2 > 0
Max : $9X1 + $7X2 + $0S1 + $0S2
Subject to :
Initial Tableau
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Initial Tableau
Cj $9 $7 $0 $0Sol
Mix
X1 X2 S1 S2 Qty
$0 S1 2 1 1 0 40$0 S2 1 3 0 1 30
Zj $0 $0 $0 $0 $0
Cj - Zj 9 7 0 0
Second Tableau
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Second Tableau
Cj $9 $7 $0 $0
Sol
Mix
X1 X2 S1 S2 Qty
$9 X1 1 1/2 1/2 0 20
$0 S2 0 5/2 -1/2 1 10
Zj $9 $9/2 $9/2 $0 $180Cj - Zj 0 5/2 0 0
Third Tableau
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Third Tableau
Cj $9 $7 $0 $0
Sol
Mix
X1 X2 S1 S2 Qty
$9 X1 1 0 3/5 -1/5 18
$7 S2 0 1 -1/5 2/5 4
Zj $9 $7 $4 $1 $190
Cj - Zj 0 0 -4 -1
Solusi Optimal diperoleh pada tableau ketiga,
yaitu X1
= 18, X2
= 4, Profit = $190
Analisis SensitivitasP bl 9 2
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Problem 9-2
Berapa shadow price kedua pembatas? Berapa RHS yang baru untuk pembatas 1?
Jika pembatas 1 meningkat sebesar 10,
berapa kemungkinan profit maksimum? Tentukan interval optimalitas untuk profit
X1!
Shadow Price = - (Cj-Zj)
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Shadow Price = - (Cj-Zj)
Pembatas 1: shadow price= -(-4) = 4 Pembatas 2: shadow price= -(-1) = 1
Ch i RHS
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Change in RHS
ORIGINAL
QUANTITY S1 RATIO
18 3/5 18/(3/5) = 30
4 -1/5 4/(-1/5) = -20
Rasio terkecil = 30, sehingga pengurangan RHS untuk pembatas 1
sebesar 30 unit (dengan batas bawah 4030 = 10 unit). Rasio -20
meningkatkan RHS pembatas 1 sebanyak 20 unit (dengan batasatas 40 + 20 = 60)
Ch i P fi X
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Change in Profit X1:
ORIGINAL
QUANTITYS1 NEW QUANTITY
18 3/5 18+(3/5)(10) = 24
4 -1/5 4+(-1/5)(10) = 2
X1 = 24, X2 = 2, S1 = 0, S2 = 0 (KEDUA SLACK NON BASIS
VARIABLE
PROFIT = 9(24) + 7(2) = 230
Kemungkinan Profit maksimum = original profit + 10 (shadow price)
= 190 + 10(4) = 230
Basic Objective Function Coefficients
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Perubahan pada Profit Contribution X1
Cj 9 + 7 0 0
Sol Mix X1 X2 S1 S2 Qty
9 + X1 1 0 3/5 -1/5 18
7 X2 0 1 -1/5 2/5 4
Zj 9 + 7 4 + (3/5) 1 -
(1/5)
190+18
Cj - Zj 0 0 -4 - (3/5) -1 -
(1/5)
Nilai solusi optimal CjZj 0Cj = Zj ;
-4(3/5) 0 -4 (3/5) -20/3 dan
-1(1/5) 0 (1/5) 1 5
Jadi perubahan profit antara -20/3 dan 5. Profit asal 9 dan solusi optimal profit X1