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Page 1: 4#Operations Research (2)

Operations Research

Industrial engineering

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DYNAMIC PROGRAMMING

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Characteristics Of Dynamic Programming Problem

• The problem can be divided into stages, with a policy decision required at each stage 1

• Each stage has a number of states associated with it 2

• The effect of the policy decision at each stage is to transform the current state into a state associated with the next stage (possibly according to a probability distribution)

3

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Characteristics Of Dynamic Programming Problem

• The solution procedure is designed to find an optimal policy for the overall problem, i.e., a prescription of the optimal policy decision at each stage for each of the possible states

4

• Given the current state, an optimal policy for the remaining stages is independent of the policy adopted in precious stages (this is the principle of the optimality for dynamic programming)

5

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Characteristics of dynamic programming problem: forward recursive

• The solution procedure begins by finding the optimal policy for the first stage 7

• A recursive relationship that identifies the optimal policy for stage n, given the optimal policy for stage (n - 1), is available

8

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Characteristics of dynamic programming problem: backward recursive

• The solution procedure begins by finding the optimal policy for the last stage 7

• A recursive relationship that identifies the optimal policy for stage n, given the optimal policy for stage (n + 1), is available

8

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• N = number of stages

• n = label for current stage (n = 1, 2, …, N)

• Sn = current state for stage n

• Xn = decision variable for stage n

• Xn* = optimal value of Xn (given Sn)

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nnSXX

n XfcSfn

n

*

1

* min

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• = contribution of stage n, n + 1, …, N to the objective function if the system starts in state Sn at stage n, the immediate decision id Xn, and optimal decisions are made thereafter

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nnn XSf ,

** , nnnnn XSfSf

nnSXX

n XfcSfn

n

*

1

* min

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• The recursive relationship will always be of the form

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nnnX

nn

nnnX

nn

XSfSf

or

XSfSf

n

n

,min

,max

*

*

nnSXX

n XfcSfn

n

*

1

* min

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Characteristics of dynamic programming problem

• When we use this recursive relationship, the solution procedure moves backward stage by stage – each time finding the optimal policy for that stage – until it finds the optimal policy starting at the initial stage

8.

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X1 X2 X3

S1

S2

S3

Xn

Sn

fn(Sn, Xn)

fn*(Sn) Xn*

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Contoh 1

Sebuah perusahaan mempunyai usulan dari ketiga pabriknya untuk kemungkinan mengembangkan sarana produksi. Perusahaan tersebut menyediakan anggaran $5 juta untuk alokasi pada ketiga pabrik. Setiap pabrik diminta untuk menyampaikan usulannya yang memberikan jumlah biaya (c) dan jumlah pendapatan (R) untuk setiap usulan.

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Contoh 1

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c1 R1 c2 R2 c3 R31 0 0 0 0 0 02 1 5 2 8 1 33 2 6 3 94 4 12

Pabrik 2 Pabrik 3Usulan

Pabrik 1

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Contoh 1: stage 1

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0 1 2

0 0 0 0

1 0 5 5 1

2 0 5 6 6 2

3 0 5 6 6 2

4 0 5 6 6 2

5 0 5 6 6 2

X1

S1

f1(S1, X1)

f1*(S1) X1*

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Contoh 1: Stage 2

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0 2 3 4

0 0 0 0

1 5 5 0

2 6 8 8 2

3 6 13 9 13 2

4 6 14 14 12 14 2 atau 3

5 6 14 15 17 17 4

X2*

X2

S2 f2*(S2)

f2(S2, X2)

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Contoh 1: Stage 3

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• Dana yang tersedia $5 juta dimanfaatkan semua • Alokasi dana pabrik 1 – pabrik 2 – pabrik 3

– 1 – 4 – 0 – 1 – 3 – 1 – 2 – 2 – 1

• Total pendapatan = $17 juta

0 1

5 17 17 17 0 atau 1

X3

S3 X3*

f3(S3, X3)

f3*(S3)

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Contoh 1: Rekursif Mundur

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0 1

0 0 0 0

1 0 3 3 1

2 0 3 3 1

3 0 3 3 1

4 0 3 3 1

5 0 3 3 1

X3

S3

f3(S3, X3)

f3*(S3) X3*

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Contoh 1: Rekursif Mundur

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0 2 3 4

0 0 0 0

1 3 3 0

2 3 8 8 2

3 3 11 9 11 2

4 3 11 12 12 12 3 atau 4

5 3 11 12 15 15 4

X2*

X2

S2

f2(S2, X2)

f2*(S2)

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Contoh 1: Rekursif Mundur

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• Dana yang tersedia $5 juta dimanfaatkan semua • Alokasi dana pabrik 1 – pabrik 2 – pabrik 3

– 1 – 3 – 1 – 1 – 4 – 0 – 2 – 2 – 1

• Total pendapatan = $17 juta

0 1 2

5 15 17 17 17 1 atau 2

X1

S1

f1(S1, X1)

f1*(S1) X1*

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Contoh 2

Suatu organisasi kesehatan dunia menyelenggarakan program peningkatan kepedulian pada kesehatan dan memberikan pendidikan kesehatan di beberapa negara terbelakang

Organisasi tersebut memiliki 5 tim medis yang siap ditugaskan di 3 negara

Satu negara paling tidak harus didatangi 1 tim medis

Performansi diukur dengan penambahan umur hidup

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Contoh 2

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1 2 3

1 45 20 50

2 70 45 70

3 90 75 80

4 105 110 100

5 120 150 130

Number of

Medical Teams

Thousands of Additional Person-

Years of Life

Country

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Contoh 2: Stage 3

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1 2 3

1 45 45 1

2 45 70 70 2

3 45 70 90 90 3

X3

S3

f3(S3, X3)

f3*(S3) X3*

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Contoh 2: Stage 2

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1 2 3

2 65 65 1

3 90 90 90 1 atau 2

4 110 115 120 120 3

X2

S2

f2(S2, X2)

f2*(S2) X2*

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Contoh 2: Stage 1

• Alokasi Tim Medis

– 1 – 3 – 1

– Total additional person-years of life = 170.000

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1 2 3

5 170 160 145 170 1

X1

S1

f1(S1, X1)

f1*(S1) X1*

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Contoh 2: asumsi suatu negara boleh tidak dikunjungi tim medis sama sekali

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0 1 2 3 4 5

0 0 0 0

1 0 45 45 1

2 0 45 70 70 2

3 0 45 70 90 90 3

4 0 45 70 90 105 105 4

5 0 45 70 90 105 120 120 5

f3*(S3) X3*

X3

S3

f3(S3, X3)

0 1 2 3 4 5

0 0 0 0

1 45 20 45 1

2 70 65 45 70 1

3 90 90 90 75 90 0 atau 1 atau 2

4 105 110 115 120 110 120 3

5 120 125 135 145 155 150 155 4

X2

S2

f2(S2, X2)

f2*(S2) X2*

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Contoh 2: asumsi suatu negara boleh tidak dikunjungi tim medis sama sekali

• Alokasi tim medis

– 1 – 3 – 1

– Total additional person-years of life = 170.000

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0 1 2 3 4 5

5 155 170 160 150 145 130 170 1

X1

S1

f1(S1, X1)

f1*(S1) X1*

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Soal 1

A college student has 7 days remaining before final examinations begin in her four courses, and she wants to allocate this study time as effectively as possible. She needs at least 1 day on each course, and she likes to concentrate on just one course each day, so she wants to allocate 1, 2, 3, or 4 days to each course. Having recently taken an operations research course, she decides to use dynamic programming to make these allocations to maximize the total grade points to be obtained from four courses. She estimates that the alternative allocations for each course would yield the number of grade points shown in the table. Solve this problem by dynamic programming. 01/04/2013 Operations Research 26

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Soal 1

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1 2 3 4

1 3 5 2 6

2 5 5 4 7

3 6 6 7 9

4 7 9 8 9

number of

study days

estimated grade points

courses

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Soal 2

Sebuah lembaga penelitian merencanakan suatu proyek penelitian. Terdapat tiga tim yang akan menyelesaikan proyek tersebut dengan pendekatan yang berbeda. Ketiga tim mempunyai kemungkinan gagal berturut-turut sebesar 0,4; 0,6; dan 0,8 sehingga kemungkinan proyek tersebut gagal adalah sebesar 0,4 x 0,6 x 0,8 = 0,192.

Untuk mengurangi tingkat kegagalan proyek, ada dua pakar yang ditugaskan untuk membantu kerja tim peneliti. Seorang pakar hanya bisa bergabung di satu tim. Tapi satu tim boleh dibantu oleh lebih dari satu pakar.

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Soal 2

• Data perkiraan tingkat kegagalan tim apabila dibantu oleh pakar:

• Tujuan: minimasi tingkat kegagalan

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Jumlah pakar yang bergabung

Tingkat kegagalan

Tim 1 Tim 2 Tim 3

0 0,40 0,60 0,80

1 0,20 0,40 0,50

2 0,15 0,20 0,30

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Soal 3

Selama satu horizon perencanaan produksi yang terdiri dari tiga bulan, permintaan akan produk A adalah satu unit produk per bulan

Jika inventory di awal dan akhir horizon perencanaan harus nol dan tidak diijinkan terjadi stok kosong pada saat permintaan datang, maka dalam tiap bulannya minimal harus tersedia satu unit produk. Pada bulan pertama harus diproduksi minimal satu unit produk. Stok pada bulan ketiga harus nol.

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Soal 3

Untuk memenuhi permintaan selama tiga bulan tersebut, produk dapat diproduksi sekaligus dalam satu bulan tertentu dengan resiko biaya simpan tinggi atau diproduksi dalam tiap bulan dengan resiko biaya setup tinggi

Tujuan: menentukan kebijakan produksi yang menghasilkan ongkos total minimum.

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Soal 3

• Rincian biaya

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Biaya Bulan ke-

1 2 3 Setup cost/run production 14 8 12 Production cost/unit 10 8 10 Holding cost/unit ending in inventory

4 2 6

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Soal 4

• Perusahaan X mempunyai data kebutuhan produksi dari produk A untuk 3 periode ke depan:

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Periode Kebutuhan Produksi

1 2 unit

2 2 unit

3 3 unit

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Soal 4

Setiap kali run produksi dihasilkan 1 unit produk A. Biaya setup untuk run pertama (1 unit pertama) adalah $20 dalam setiap periode dan naik sebesar $2 untuk setiap tambahan 1 unit yang diproduksi pada periode tersebut.

Biaya produksi + overhead untuk tiap periode konstan

Biaya simpan untuk setiap 1 unit kelebihan produk adalah $3 per periode

Kapasitas maksimum gudang adalah 3 unit per periode

Inventori awal periode 1 = 0

Tujuan: menentukan kebijakan produksi yang menghasilkan ongkos total minimum

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