PENELITIAN OPERASIONAL I (TIN 4109)

PENELITIAN OPERASIONAL I

Lecture 12

Lecture 12

• Outline:

– Integer Programming: Binary

• 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.

– Winston, Wayne L. Operations Research: Applications and

Algorithms. 3rd edition. Wadsworth Inc.1994.

– Hartanto, Dody. Lecture PPT: Pemodelan dalam Pemrogaman Linier Integer (Modeling in Integer Linear Programming). ITS. 2012

BINARY INTEGER PROGRAMMING

Algoritma Branch and Bound untuk

BILP (Binary Integer Linear

Programming)

Dody Hartanto

Maksimasi

Z = 9x

₁

+ 5x

₂

+ 6x

₃

+ 4x

₄

pembatas:

6x

₁

+ 3x

₂

+ 5x

₃

+ 2x

₄

 10

x

₃

+ x

₄



1

-x

₁

+ x

₃

 0

-x

₂

+ x

₄

 0

x

₁

, x

₂

, x

₃

, x

₄

= {0, 1}

Algoritma Branch and Bound untuk

BILP

Dody Hartanto

• Masalah asli (original problem) diselesaikan

dengan mengabaikan batasan yang

mengharuskan variabel keputusan bernilai

integer (ILP relaksasi).

• Solusi ILP relaksasi masalah asli memiliki satu

variabel keputusan yang tidak bernilai bulat

yaitu X

₁

sehingga perlu dilakukan

pencabangan.

Algoritma Branch and Bound untuk

BILP(1)

Dody Hartanto

Variabel keputusan X₁ merupakan variabel keputusan yang

bernilai biner(binary variable) sehingga pencabangan dilakukan dengan menetapkan nilai X₁ = 0 pada salah satu cabang (sub

masalah 1(SM 1)) dan X₁ = 1 pada cabang yang lain (sub masalah 2 (SM 2)).

Algoritma Branch and Bound untuk

BILP(2)

Dody Hartanto

• Nilai semua variabel keputusan pada solusi sub masalah 1 bernilai integer dengan nilai Z = 9 sehingga nilai ini menjadi batas bawah (lower bound).

• Terdapat dua variabel keputusan yang tidak bernilai integer pada solusi sub masalah 2 yaitu X₂ yang bernilai 4/5 dan X₄ yang bernilai 4/5. Oleh karena itu, perlu dilakukan

pencabangan pada sub masalah 2.

Batas Bawah (Lower Bound)

Algoritma Branch and Bound untuk

BILP(3)

Dody Hartanto

Pencabangan pada sub masalah 2 dilakukan dengan menetapkan X₂ = 0 pada salah satu cabang (sub masalah 3(SM 3)) dan X₂ = 1 pada cabang yang lain (sub masalah 4 (SM 4)).

Algoritma Branch and Bound untuk

BILP(4)

Dody Hartanto

• Tidak semua variabel keputusan pada solusi sub masalah 3 bernilai integer demikian pula dengan solusi pada sub masalah 4. Oleh karena itu perlu dilakukan pencabangan pada kedua sub masalah(sub masalah 3 dan sub masalah 4).

karena nilai Z pada sub masalah 4 yang bernilai 16 adalah lebih besar jika dibandingkan dengan nilai Z pada sub masalah 3 yang bernilai 13,8 maka pencabangan dilakukan terlebih dahulu pada sub masalah 4.

Algoritma Branch and Bound untuk

BILP(5)

Dody Hartanto

Pencabangan pada sub masalah 4 dilakukan dengan menetapkan X₄ = 0 pada salah satu cabang (sub masalah 5(SM 5)) dan X₄ = 1 pada cabang yang lain (sub masalah 6 (SM 6)). Semua variabel keputusan pada solusi sub masalah 5 bernilai integer dengan nilai fungsi tujuan Z= 14.

Algoritma Branch and Bound untuk

BILP(6)

Dody Hartanto

Nilai Z pada sub masalah 5 lebih besar jika dibandingkan dengan batas bawah

(lower bound) yang sekarang(solusi dari sub masalah 1) sehingga batas bawah

yang semula bernilai 9 diubah menjadi 14

Batas Bawah (Lower Bound) yang lama

Batas Bawah (Lower Bound) yang Baru

Algoritma Branch and Bound untuk

BILP(7)

Dody Hartanto

tidak ada sub masalah yang belum dicabangkan yang memiliki nilai batas atas (upper bound) yang lebih besar dari 14 sehingga proses pencarian solusi dapat dihentikan dan solusi sub masalah 5 dengan X₁ = 1, X₂ = 1, X₃ = 0, X₄ = 0, dan Z = 14 menjadi solusi

optimal permasalahan ini.

Solusi Sub Masalah 3 tidak mungkin bisa lebih baik dari solusi sub masalah 5 sehingga tidak perlu dicabangkan meskipun masih ada nilai variabel keputusan yang pecahan

Algoritma Branch and Bound untuk

BILP(8)

LP dan ILP dengan Software

• TUGAS KELOMPOK:

– Temukan satu permasalahan yang memerlukan solusi

optimal (permasalahan LP: 1 dan ILP: 1). Buat

formulasinya. Selesaikan dengan menggunakan

software: LINDO/LINGO dan Solver.

– Isi laporan:

• Deskripsi Permasalahan • Formulasi

• Langkah-langkah pengerjaan dengan software • Solusi dan Kesimpulan

– Format laporan:

Lecture 13

Lecture 13

• Outline:

– Transportation: starting basic feasible solution

• 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

Description

A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n

demand points using the capacities of m supply points.

1 2 3 1 2 c₁₁ c₁₂ c₁₃ c_{21 c22} c₂₃ d₁ d₂ d₃ s₁ s₂ SOURCES DESTINATIONS

Transportation Problem

• LP Formulation

The linear programming formulation in terms of the

amounts shipped from the origins to the destinations, x_ij, can be written as:

Min SSc_ijx_ij

i j

s.t. Sx_ij < s_i for each source i

j

Sx_ij >= d_j for each destination j

i

Transportation Problem

• LP Formulation Special Cases

The following special-case modifications to

the linear programming formulation can be

made:

– Minimum shipping guarantees from i to j:

x

_ij

> L

_ij

– Maximum route capacity from i to j:

x

_ij

< L

_ij

– Unacceptable routes:

Example: BBC

Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. BBC has two plants, each of which can produce 50 tons per week.

How should end of week shipments be made to fill the above orders given the following delivery cost per ton:

Northwood Westwood Eastwood

Plant 1 24 30 40

Example: BBC

• LP Formulation

– Decision Variables Defined

x_ij = amount shipped from plant i to suburb j where i = 1 (Plant 1) and 2 (Plant 2)

j = 1 (Northwood), 2 (Westwood), and 3 (Eastwood)

Example: BBC

• LP Formulation

– Objective Function

Minimize total shipping cost per week:

Min 24x₁₁ + 30x₁₂ + 40x₁₃ + 30x₂₁ + 40x₂₂ + 42x₂₃

– Constraints

s.t. x₁₁ + x₁₂ + x₁₃ < 50 (Plant 1 capacity) x₂₁ + x₂₂ + x₂₃ < 50 (Plant 2 capacity) x₁₁ + x₂₁ >= 25 (Northwood demand) x₁₂ + x₂₂ >= 45 (Westwood demand) x₁₃ + x₂₃ >= 10 (Eastwood demand) all x_ij > 0 (Non-negativity)

Exercise….

Powerco has three electric power plants that supply the electric needs of four cities.

• The associated supply of each plant and demand of each city is given in the table as follows:

• The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.

From To

City 1 City 2 City 3 City 4 Supply (Million kwh) Plant 1 $8 $6 $10 $9 35 Plant 2 $9 $12 $13 $7 50 Plant 3 $14 $9 $16 $5 40 Demand (Million kwh) 45 20 30 30

Solution

1. Decision Variable:

Since we have to determine how much electricity is sent from each plant to each city;

X_ij = Amount of electricity produced at plant i and sent to city j X₁₄ = Amount of electricity produced at plant 1 and sent to city 4

2. Objective Function:

Since we want to minimize the total cost of shipping from plants to cities;

Minimize Z = 8X₁₁+6X₁₂+10X₁₃+9X₁₄+9X₂₁+12X₂₂+13X₂₃+7X₂₄ +14X₃₁+9X₃₂+16X₃₃+5X₃₄

3. Supply Constraints X₁₁+X₁₂+X₁₃+X₁₄ <= 35 X₂₁+X₂₂+X₂₃+X₂₄ <= 50 X₃₁+X₃₂+X₃₃+X₃₄ <= 40 4. Demand Constraints

Solution

X₁₁+X₂₁+X₃₁ >= 45 X₁₂+X₂₂+X₃₂ >= 20 X₁₃+X₂₃+X₃₃ >= 30 X₁₄+X₂₄+X₃₄ >= 30

Since each supply point has a limited production capacity;

Since each supply point has a limited production capacity;

5. Sign Constraints

LP Formulation of

Powerco’s Problem

Min Z = 8X₁₁+6X₁₂+10X₁₃+9X₁₄+9X₂₁+12X₂₂+13X₂₃+7X₂₄ +14X₃₁+9X₃₂+16X₃₃+5X₃₄ S.T.: X₁₁+X₁₂+X₁₃+X₁₄ <= 35 (Supply Constraints) X₂₁+X₂₂+X₂₃+X₂₄ <= 50 X₃₁+X₃₂+X₃₃+X₃₄ <= 40 X₁₁+X₂₁+X₃₁ >= 45 (Demand Constraints) X₁₂+X₂₂+X₃₂ >= 20 X₁₃+X₂₃+X₃₃ >= 30 X₁₄+X₂₄+X₃₄ >= 30 Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

Balanced Transportation Problem

If Total supply equals to total demand, the problem

is said to be a balanced transportation problem:















j

n

j

j

m

i

i

i

d

s

1

1

Finding Basic Feasible Solution for

Transportation Problem

Unlike other Linear Programming problems,

a balanced TP with m supply points and n demand points is easier to solve, although it has m + n equality constraints.

The reason for that is, if a set of decision variables (x_ij’s) satisfy all but one constraint, the values for x_ij’s will satisfy

Methods to find the BFS for a balanced

Transportation Problem

There are three basic methods:

1. North West Corner Method

2. Minimum Cost Method

1. Northwest Corner Method

To find the bfs by the NWC method:

Begin in the upper left (northwest) corner of the

transportation tableau and set x₁₁ as large as possible (here the limitations for setting x₁₁ to a larger number, will be the demand of demand point 1 and the supply of supply point 1. Your x₁₁ value can not be greater than minimum of this 2 values).

According to the explanations in the previous slide we can set x₁₁=3 (meaning demand of demand point 1 is satisfied

by supply point 1). 5 6 2 3 5 2 3 3 2 6 2 X 5 2 3

After we check the east and south cells, we saw that we can go east (meaning supply point 1 still has capacity to fulfill some demand).

3 2 X 6 2 X 3 2 3 3 2 X 3 3 2 X X 2 3

After applying the same procedure, we saw that we can go south this time (meaning demand point 2 needs more supply

by supply point 2). 3 2 X 3 2 1 2 X X X 3 3 2 X 3 2 1 X 2 X X X 2

Finally, we will have the following bfs, which is: x₁₁=3, x₁₂=2, x₂₂=3, x₂₃=2, x₂₄=1, x₃₄=2 3 2 X 3 2 1 X 2 X X X X X

2. Minimum Cost Method

The Northwest Corner Method dos not utilize shipping costs. It can yield an initial BFS easily but the total shipping cost may

be very high.

The minimum cost method uses shipping costs in order come up with a BFS that has a lower cost.

1. First, We look for the cell with the minimum cost of shipping in the overall transportation tableau.

2. We should cross out row i and column j and reduce the supply or demand of the noncrossed-out row or column by the value of Xij.

3. Then we will choose the cell with the minimum cost of shipping from the cells that do not lie in a crossed-out row or column and we will repeat the procedure.

2. Minimum Cost Method (Cont’d)

An example for Minimum Cost Method Step 1: Select the cell with minimum cost.

2 3 5 6 2 1 3 5 3 8 4 6 5 10 15 12 8 4 6

Step 2: Cross-out column 2 2 3 5 6 2 1 3 5 8 3 8 4 6 12 X 4 6 5 2 15

Step 3: Find the new cell with minimum shipping cost and cross-out row 2 2 3 5 6 2 1 3 5 2 8 3 8 4 6 5 X 15 10 X 4 6

Step 4: Find the new cell with minimum shipping cost and cross-out row 1 2 3 5 6 5 2 1 3 5 2 8 3 8 4 6 X X 15 5 X 4 6

Step 5: Find the new cell with minimum shipping cost and cross-out column 1 2 3 5 6 5 2 1 3 5 2 8 3 8 4 6 5 X X 10 X X 4 6

Step 6: Find the new cell with minimum shipping cost and cross-out column 3 2 3 5 6 5 2 1 3 5 2 8 3 8 4 6 5 4 X X 6 X X X 6

Step 7: Finally assign 6 to last cell. The bfs is found as: X₁₁=5, X₂₁=2, X₂₂=8, X₃₁=5, X₃₃=4 and X₃₄=6 2 3 5 6 5 2 1 3 5 2 8 3 8 4 6 5 4 6 X X X X X X X

3. Vogel’s Method

1. Begin with computing each row and column a penalty. The

penalty will be equal to the difference between the two smallest shipping costs in the row or column.

2. Identify the row or column with the largest penalty.

3. Find the first basic variable which has the smallest shipping cost in that row or column.

4. Then assign the highest possible value to that variable, and cross-out the row or column as in the previous methods. Compute new penalties and use the same procedure.

An example for Vogel’s Method Step 1: Compute the penalties.

Supply Row Penalty

6 7 8 15 80 78 Demand Column Penalty 15-6=9 80-7=73 78-8=70 7-6=1 78-15=63 15 5 5 10 15

Step 2: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8 5 15 80 78 Demand Column Penalty 15-6=9 _ 78-8=70 8-6=2 78-15=63 15 X 5 5 15

Step 3: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8 5 5 15 80 78 Demand Column Penalty 15-6=9 _ _ _ _ 15 X X 0 15

Step 4: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8 0 5 5 15 80 78 Demand Column Penalty _ _ _ _ _ 15 X X X 15

Step 5: Finally the bfs is found as X₁₁=0, X₁₂=5, X₁₃=5, and X₂₁=15

Supply Row Penalty

6 7 8 0 5 5 15 80 78 15 Demand Column Penalty _ _ _ _ _ X X X X X

Exercise

Bazaraa Chapter 10:

Solve the following transportation problem:

1

2

3

s

i

1

5

4

1

3

2

1

7

5

7

d

j

2

5

3

Origin Destination cij matrix

Exercise

Bazaraa Chapter 10:

1

2

3

4

s

i

1

7

2

-1

0

10

2

4

3

2

3

30

3

2

1

3

4

25

d

j

10

15

25

15

c

ij

matrix

Lecture 14 – Preparation

• Materi: