PENELITIAN OPERASIONAL I

(TIN 4109)

Lecture 9

LINEAR PROGRAMMING

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 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒙𝟓 𝒙𝟔 Solution

𝒛 4 0 0 1 2 0 1350

𝒙𝟐 -1/4 1 0 1/2 -1/4 0 100

𝒙𝟑 3/2 0 1 0 1/2 0 230

𝒙𝟔 2 0 0 -2 1 1 20

𝑥1 = mainan – kereta 𝑥2 = mainan – truk 𝑥3 = 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 ith 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 ith constraint is increased by 1.

• This definition applies only if the change in the right-hand side of Constraint i leaves the current basis optimal.

6

Contoh: Shadow Price

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

DUALITAS

Terima kasih kepada Prof.Dr.Ir. Abdullah Alkaff, M.Sc. yang telah menyiapkan versi awal dalam tulisan tangan dari materi ini

Linear Programming

● LP dalam bentuk standard

Linear Programming

Linear Programming

Kondisi keoptimalan:

sehingga persoalan LP dapat diintepretasikan sebagai berikut:

cari y1, y2, …, ym sedemikian hingga (1) dan (2) terpenuhi

z0 = y b = y1b1 + y2b2 + … + ymbm (1)

zj = y aj = y1a1j + y2a2j + … + ymamj (2)

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 z0 = y1b1 + y2b2 + … + ymbm

subject to

y1a1j +y2a2j + … + ymamj cj

y1 , y2 , ym 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

jj xcZ1

,

m

i

ii ybW1

,

aijx j bi,j1

n

m

i

jiij cya1

,

for for .,,2,1 mi .,,2,1 nj

for .,,2,1 mi for .,,2,1 nj ,0jx ,0iy

Primal Dual dalam Matriks

Where and are row

vectors but and are column vectors.

c myyyy ,,, 21 b x

Primal Problem Dual Problem

Maximize

subject to

.0x .0y

Minimize

subject to

bAx cyA

,cxZ ,ybW

Contoh: Primal – Dual

Max

s.t.

Min

s.t.

Primal Problem in Algebraic Form

Dual Problem in Algebraic Form

,53 21 xxZ

,18124 321 yyyW

1823 21 xx

122 2 x41x

0x,0x 21

522 32 yy

33 3 y1y

0y,0y,0y 321

Programa Dual

Hubungan antara PRIMAL dan DUAL adalah sebagai berikut :

PRIMAL DUAL

RHS Fungsi Tujuan

MAX MIN

Constrain Variable

Programa Dual

x1 x2 xn RHS

y1 a11 a12 a1n b1

y2 a21 a22 a2n b2

ym am1 am2 amn bm

c1 c2 cn

Koefisien Fungsi Objektif

(Maksimisasi)

Ko

efi

sie

n F

un

gsi

Ob

jekti

f

(Min

imis

asi)

PRIMAL

DU

AL

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

DUAL dari DUAL adalah PRIMAL

Primal of Diet problem

RI-1333 OR1/sew/2007/#6

Diet Problem – Dual

RI-1333 OR1/sew/2007/#6

PRIMAL – DUAL

Secara umum hubungan antara DUAL dan PRIMAL dapat digambarkan seperti pada tabel di bawah ini

MINIMASI MAKSIMASI

Unrestricted =

= Unrestricted

Variable

Va

ria

ble

C

onstr

ain

t

Co

nstr

ain

t

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 zj – cj 0 ; zj cj

zj – Surplus Var = cj

atau

Surplus Var. = zj – cj

Dalam Tableau

Original Variables Slack Variables

x1 x2 xn xn+1 xn+2 xn+m

Baris 0: z1 – c1 z2 – c2 zn – cn y1 y2 ym

Complementary Basic Solution

PRIMAL VARIABLES DUAL VARIABLES

Original Variable : xj zj – cj : Surplus Variable

Slack Variable : xn+i yi : Original Variable

Basic Nonbasic

Nonbasic Basic

Original Variables Slack Variables

x1 x2 xn xn+1 xn+2 xn+m

Baris 0: z1 – c1 z2 – c2 zn – cn y1 y2 ym

Dalam Tableau

Complementary Basic Solution

PRIMAL VARIABLES DUAL VARIABLES

Original Variable : xj zj – cj : Surplus Variable

Slack Variable : xn+i yi : Original Variable

Basic Nonbasic

Nonbasic Basic

1) Bila xj > 0, maka ………..….

2) Bila xn+i > 0, maka………….

3) Bila yi > 0, maka ……….……

4) Bila zj–cj > 0, maka…….…...

Complementary Basic Solution

Dapat diringkas sebagai berikut:

1. xn+i yi = 0

2. (zj – cj) xj = 0

Jadi constraint di satu problem adalah renggang (non-binding), maka variabel yang berkaitan dengan constrain ini dalam problem yang lain harus nol.

Hasil ini biasa dikenal sebagai Complementary Slackness

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,,

85.05.12

205.1 24

48 6 8 ..

203060

321

321

321

321

321

xxx

xxx

xxx

xxxts

xxxzMax

0,,

205.05.1

305.1 26

60 2 4 8 ..

82048

321

321

321

321

321

yyy

yyy

yyy

yyyts

yyywMin

085.005.1228

085.1022420

2481062848

8 ,0 ,2 ,280

3

2

1

321

s

s

s

xxxz

020105.0105.101

530105.110206

06010210408

10 ,10 ,0 ,280

3

2

1

321

e

e

e

yyyw

Contoh

0 ,

1

832

Subject to

3 Max

21

21

21

21

xx

xx

xx

xx

0 , , ,

1

8 32

Subject to

3 Max

4321

421

321

21

xxxx

xxx

xxx

xx

z x 1 x 2 x 3 x 4 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

Pada Problem Dual

Hubungan Primal - Dual

Primal

Dual

z x 1 x 2 x 3 x 4 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 x0 feasible terhadap PRIMAL dan y0 feasible terhadap DUAL sedemikian hingga cx0 = y0b, maka x0 dan y0 adalah penyelesaian optimal

Menyelesaikan

PRIMAL

Menyelesaikan

DUAL

z DUAL FR

PRIMAL FR

Optimal

(PRIMAL – DUAL FEASIBLE)

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:

– Integer Linear Programming