Chap09[2] LS Modif

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    Kuliah ke-3

    Linear Programming:

    The Simplex Method

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    2

    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