PENELITIAN OPERASIONAL I (TIN 4109)

PENELITIAN OPERASIONAL I

Lecture 9

Lecture 9

• Outline:

– Analisa Sensitivitas Simplex

– Duality

• References:

– Frederick Hillier and Gerald J. Lieberman. Introduction

to Operations Research. 7th ed. The McGraw-Hill

Companies, Inc, 2001.

– Hamdy A. Taha. Operations Research: An Introduction.

8th Edition. Prentice-Hall, Inc

, 2007.

Sensitivity Analysis

– Objective Function –

• Contoh kasus:

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑧 = 3𝑥₁ + 2𝑥₂ + 5𝑥₃ 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑥₁ + 2𝑥₂ + 𝑥₃ ≤ 430 (𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 1) 𝑥₁ + 2𝑥₃ ≤ 460 (𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 2) 𝑥₁ + 4𝑥₂ ≤ 420 (𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 3) 𝑥₁, 𝑥₂, 𝑥₃ ≥ 0

• Hasil Optimal:

Basic 𝒙_𝟏 𝒙_𝟐 𝒙_𝟑 𝒙_𝟒 𝒙_𝟓 𝒙_𝟔 Solution 𝒛 4 0 0 1 2 0 1350 𝒙_𝟐 -1/4 _{1 0} 1/2 -1/4 _{0 100} 𝒙_𝟑 3/2 _{0 1 0} 1/2 _{0 230} 𝒙_𝟔 2 0 0 -2 1 1 20

𝑥

₁

= mainan – kereta

𝑥

₂

= mainan – truk

𝑥

₃

= mainan - mobil

Shadow Price

• It is often important for managers to determine how a

change in a constraint’s right-hand side changes the

LP’s optimal z-value.

• With this in mind, we define the shadow price for the

i

th

_{constraint of an LP to be the amount by which the}

optimal z-value is improved—increased in a max

problem and decreased in a min problem—if the

right-hand side of the i

th

_{constraint is increased by 1.}

• This definition applies only if the change in the

right-hand side of Constraint i leaves the current basis

6

Contoh: Shadow Price

Shadow price jika resource

b₁ bertambah 1 unit

b₁ ditambah 1 unit menjadi 9

Nilai z bertambah 4/5 (shadow price)  7 + 4/5 = 39/5  Max x₁ 3x₂ ST 2x₁3x₂  x₃ 8  x₁ x₂  x₄ 1 x₁ , x₂ , x₃, x₄ 0

D

U

A

L

I

T

A

S

Terima kasih kepada Prof.Dr.Ir. Abdullah Alkaff, M.Sc.

Linear Programming

Linear Programming

Kondisi keoptimalan:

sehingga persoalan LP dapat diintepretasikan sebagai

berikut:

cari y

₁

, y

₂

, …, y

_m

sedemikian hingga (1) dan (2)

terpenuhi

z

₀

= y b = y

₁

b

₁

+ y

₂

b

₂

+ … + y

_m

b

_m

(1)

Linear Programming

Dapat dilakukan dengan menyelesaikan LP sebagai

berikut:

Maka diperoleh problem LP yang baru yang disebut

DUAL dari problem semula atau disingkat PROBLEM

DUAL

Min z

₀

= y

₁

b

₁

+ y

₂

b

₂

+ … + y

_m

b

_m

subject to

y

₁

a

_1j

+y

₂

a

_2j

+ … + y

_m

a

_mj



c

_j

y

₁

, y

₂

, y

_m



0

Primal and Dual

The dual problem uses exactly the same parameters as the primal problem, but in different location.

Primal Problem

Dual Problem

Max

s.t.

Min

s.t.







n j j j

x

c

Z

1

,







m i i i

y

b

W

1

,



a

_ij

x

_j



b

_i

,

j1 n



_





m i j i ij

y

c

a

1 ,

for

i



1

,

2

,



,

m

.

for

j



1

,

2

,



,

n

.

for

i



1

,

2

,



,

m

.

for

j



1

,

2

,



,

n

.

,

0



j

x

y

_i



0

,

Primal Dual dalam Matriks

Where and are row

vectors but and are column vectors.

c

y





y

₁

,

y

₂

,



,

y

_m



b

x

Primal Problem

Dual Problem

Maximize

subject to

.

0



x

y



0

.

Minimize

subject to

b

Ax



yA



c

,

cx

Z



W



yb

,

Contoh: Primal – Dual

Max

s.t.

Min

s.t.

Primal Problem

in Algebraic Form

in Algebraic Form

Dual Problem

,

5

3

x

₁

x

₂

Z





,

18

12

4

y

₁

y

₂

y

₃

W







18

2

3

x

₁



x

₂



12

2

x

₂



4



1

x

0

x

,

0

x

₁



₂



5

2

2

y

₂



y

₃



3

3

₃





y

1

y

0

y

,

0

y

,

0

y

₁



₂



₃



Programa Dual

Hubungan antara PRIMAL dan DUAL adalah sebagai

berikut :

PRIMAL

DUAL

RHS

Fungsi Tujuan

MAX

MIN

Programa Dual

x

₁

x

₂



x

_n

RHS

y

₁

a

₁₁

a

₁₂



a

_1n



b

₁

y

₂

a

₂₁

a

₂₂



a

_2n



b

₂

y

_m

a

_m1

a

_m2

a

_mn



b

_m







c

₁

c

₂



c

_n











Koefisien Fungsi Objektif (Maksimisasi) Koefisie n Fu ng si Ob jek tif (Min imisa si)

PRIMAL

DUA

L

Contoh Programa Dual

PRIMAL : Max 3x₁ + 5x₂ s.t. x₁  4 2x₂  12 3x₁ + 2x₂  18 x₁, x₂  0

DUAL : Min 4y₁ + 12y₂ + 18y₃ s.t.

y₁ + 3y₃  3 2y₂ + 2y₃  5 y₁, y₂ , y₃  0

Primal of Diet problem

Diet Problem – Dual

PRIMAL – DUAL

Secara umum hubungan antara DUAL dan PRIMAL dapat

digambarkan seperti pada tabel di bawah ini

MINIMASI MAKSIMASI





_





_

Unrestricted



₌





_





_

=



Unrestricted

V

ariab

le

V

ariab

le

Co

nstr

ain

t

Constr

aint

Contoh

PRIMAL : Max 8x₁ + 3x₂ s.t. x₁ – 6x₂  4 5x₁ + 7x₂ = – 4 x₁  0 x₂  0 DUAL : Min 4w₁ – 4w₂ s.t. w₁ + 5w₂  8 – 6w₁ + 7w₂  3 w₁  0 w₂ unrestricted

Contoh 2

RI-1333 OR1/sew/2007/#6

Primal: Max. z = 3x₁ + 2x₂ (Obj. Func.)

subject to 2x₁ + x₂  100 (Finishing constraint) x₁ + x₂  80 (Carpentry constraint) x₁  40 (Bound on soldiers) x₁, x₂  0 Optimal Solution: z = 180, x₁ = 20, x₂ = 60

Dual : Min. w = 100y₁ + 80y₂ + 40y₃ (Obj. Func.)

subject to

2y₁ + y₂ + y₃  3 y₁ + y₂  2 y₁, y₂, y₃  0

Complementary Basic Solution

Problem Dual :

Constraint

z

_j

– c

_j



0 ; z

_j



c

_j

z

_j

– Surplus Var = c

_j

atau

Surplus Var. = z

_j

– c

_j

Dalam Tableau

Original Variables

Slack Variables

x

₁

x

₂



x

_n

x

_n+1

x

_n+2

 x

_n+m

Complementary Basic Solution

PRIMAL VARIABLES

DUAL VARIABLES

Original Variable : x

_j

z

_j

– c

_j

: Surplus Variable

Slack Variable : x

_n+i

y

_i

: Original Variable

Basic

Nonbasic

Nonbasic

Basic

Original Variables

Slack Variables

x

₁

x

₂



x

_n

x

_n+1

x

_n+2

 x

_n+m

Baris 0: z

₁

– c

₁

z

₂

– c

₂



z

_n

– c

_n

y

₁

y

₂



y

_m

Complementary Basic Solution

PRIMAL VARIABLES

DUAL VARIABLES

Original Variable : x

_j

z

_j

– c

_j

: Surplus Variable

Slack Variable : x

_n+i

y

_i

: Original Variable

Basic

Nonbasic

Nonbasic

Basic

1)

Bila x

_j

> 0, maka ………..….

2)

Bila x

_n+i

> 0, maka………….

3)

Bila y

_i

> 0, maka ……….……

Complementary Basic Solution

 Dapat diringkas sebagai berikut:

1. x

_n+i

y

_i

= 0

2. (z

_j

– c

_j

) x

_j

= 0

 Jadi constraint di satu problem adalah renggang

(non-binding), maka variabel yang berkaitan dengan constrain ini

dalam problem yang lain harus nol.

Contoh:

The Dakota Furniture Company manufactures desk, tables, and chairs. The manufacture of each type of furniture lumber and two types of skilled labor: finishing and carpentry. The amount of each resource needed to make each type of furniture is given in Table

At present, 48 bard feet of lumber, 20 finishing hours, and 8 carpentry hours are available. A desk sells for $60, and a table for $30, and a chair for $20. Dakota believes that demand for desks, chairs and tables is unlimited. Since available resource have already been purchased. Dakota wants to maximize total revenue

Resource Desk Table Chair

Lumber 8 board ft 6 board ft 1 board ft Finishing hours 4 hours 2 hours 1.5 hours Carpentry hours 2 hours 1.5 hours 0.5 hours

0

,

,

8

5

.

0

5

.

1

2

20

5

.

1

2

4

48

6

8

.

.

20

30

60

3 2 1 3 2 1 3 2 1 3 2 1 3 2 1



























x

x

x

x

x

x

x

x

x

x

x

x

t

s

x

x

x

z

Max

0

,

,

20

5

.

0

5

.

1

30

5

.

1

2

6

60

2

4

8

.

.

8

20

48

3 2 1 3 2 1 3 2 1 3 2 1 3 2 1



























y

y

y

y

y

y

y

y

y

y

y

y

t

s

y

y

y

w

Min

     





   

 





 

 

 



2 2 1.5 0 0.5 8



0 8 0 8 5 . 1 0 2 2 4 20 24 8 1 0 6 2 8 48 8 , 0 , 2 , 280 3 2 1 3 2 1                    s s s x x x z

     





   

 





 

 

 



1 0 1.510 0.510 20



0 5 30 10 5 . 1 10 2 0 6 0 60 10 2 10 4 0 8 10 , 10 , 0 , 280 3 2 1 3 2 1                    e e e y y y w

Contoh

0

,

1

8

3

2

Subject to

3

M ax

2 1 2 1 2 1 2 1















x

x

x

x

x

x

x

x

0

,

,

,

1

8

3

2

Subject to

3

M ax

4 3 2 1 4 2 1 3 2 1 2 1



















x

x

x

x

x

x

x

x

x

x

x

x

z x₁ x₂ x₃ x₄ RHS z 1 -1 -3 0 0 0 x₃ 0 2 3 1 0 8 x₄ 0 -1 1 0 1 1 z 1 -4 0 0 3 3 x₃ 0 5 0 1 -3 5 x₂ 0 -1 1 0 1 1 z 1 0 0 0.8 0.6 7 x₁ 0 1 0 0.2 -0.6 1 x₂ 0 0 1 0.2 0.4 2

Hubungan Primal - Dual

Primal

Dual

z x₁ x ₂ x ₃ x₄ RHS z 1 -1 -3 0 0 0 x ₃ 0 2 3 1 0 8 x ₄ 0 -1 1 0 1 1 z 1 -4 0 0 3 3 x ₃ 0 5 0 1 -3 5 x ₂ 0 -1 1 0 1 1 z 1 0 0 0.8 0.6 7 x ₁ 0 1 0 0.2 -0.6 1 x ₂ 0 0 1 0.2 0.4 2

Hubungan PRIMAL – DUAL

 Bila x adalah feasible terhadap PRIMAL dan y feasible

terhadap DUAL, maka cx



yb

 Nilai objektif problem Max



Nilai objektif problem

Min

DUAL Constraint

y A



c

x



0  y Ax



cx

Ax



b  y b



cx

Teorema Dualitas

● Bila x

*

_{adalah penyelesaian dari PRIMAL dan y}

*

_adalah

penyelesaian dari DUAL, maka cx

*

_{= y}

*

_b

● Bila x

₀

feasible terhadap PRIMAL dan y

₀

feasible

terhadap DUAL sedemikian hingga cx

₀

= y

₀

b

, maka x

₀

dan y

₀

adalah penyelesaian optimal

Menyelesaikan PRIMAL Menyelesaikan DUAL

z

DUAL FR PRIMAL FR Optimal

Teorema Dualitas

1. P optimal



D optimal

2. P tak terbatas

D tak terbatas





D tidak feasible

P tidak feasible

3. P tidak feasible

D tidak feasible





D tak terbatas/tidak feasible

P tak terbatas/tidak feasible

Lecture 10 – Preparation

• Materi: