PENELITIAN OPERASIONAL I (TIN 4109)

Transkripsi

1 PENELITIAN OPERASIONAL I (TIN 4109)

2 Lecture 12 INTEGER PROGRAMMING

3 Lecture 12 Outline: Integer Programming: Binary References: Frederick Hillier and Gerald J. Lieberman. Introduction to Operations Research. 7th ed. The McGraw-Hill Companies, Inc, Hamdy A. Taha. Operations Research: An Introduction. 8th Edition. Prentice-Hall, Inc, Winston, Wayne L. Operations Research: Applications and Algorithms. 3rd edition. Wadsworth Inc Hartanto, Dody. Lecture PPT: Pemodelan dalam Pemrogaman Linier Integer (Modeling in Integer Linear Programming). ITS. 2012

4 INTEGER PROGRAMMING BINARY INTEGER PROGRAMMING

5 Algoritma Branch and Bound untuk BILP (Binary Integer Linear Programming) 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} Dody Hartanto

6 Algoritma Branch and Bound untuk BILP Dody Hartanto

7 Algoritma Branch and Bound untuk BILP(1) 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. Dody Hartanto

8 Algoritma Branch and Bound untuk BILP(2) Variabel keputusan X 1 merupakan variabel keputusan yang bernilai biner(binary variable) sehingga pencabangan dilakukan dengan menetapkan nilai X 1 = 0 pada salah satu cabang (sub masalah 1(SM 1)) dan X 1 = 1 pada cabang yang lain (sub masalah 2 (SM 2)). Dody Hartanto

9 Algoritma Branch and Bound untuk 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 X 2 yang bernilai 4/5 dan X 4 yang bernilai 4/5. Oleh karena itu, perlu dilakukan pencabangan pada sub masalah 2. Batas Bawah (Lower Bound) BILP(3) Dody Hartanto

10 Algoritma Branch and Bound untuk BILP(4) Pencabangan pada sub masalah 2 dilakukan dengan menetapkan X 2 = 0 pada salah satu cabang (sub masalah 3(SM 3)) dan X 2 = 1 pada cabang yang lain (sub masalah 4 (SM 4)). Dody Hartanto

11 Algoritma Branch and Bound untuk BILP(5) 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. Dody Hartanto

12 Algoritma Branch and Bound untuk BILP(6) Pencabangan pada sub masalah 4 dilakukan dengan menetapkan X 4 = 0 pada salah satu cabang (sub masalah 5(SM 5)) dan X 4 = 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. Dody Hartanto

13 Algoritma Branch and Bound untuk BILP(7) 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 Dody Hartanto

14 Algoritma Branch and Bound untuk BILP(8) 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 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 X 1 = 1, X 2 = 1, X 3 = 0, X 4 = 0, dan Z = 14 menjadi solusi optimal permasalahan ini. Dody Hartanto

15 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: PPT

16 Lecture 13 TRANSPORTATION

17 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, Hamdy A. Taha. Operations Research: An Introduction. 8th Edition. Prentice-Hall, Inc, 2007.

18 Transportation

19 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 d 1 c 11 c 12 s 1 1 c 13 c 21 c 22 2 d 2 s 2 2 c 23 3 d 3 SOURCES DESTINATIONS

20 LP Formulation Transportation Problem The linear programming formulation in terms of the amounts shipped from the origins to the destinations, x ij, can be written as: Min SSc ij x ij i j s.t. Sx ij < s i for each source i j Sx ij >= d j for each destination j i x ij > 0 for all i and j

21 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: delete the variable

22 Example: BBC Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as follows: Northwood tons, Westwood tons, and Eastwood 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 Plant

23 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)

24 Example: BBC LP Formulation Objective Function Minimize total shipping cost per week: Min 24x x x x x x 23 Constraints s.t. x 11 + x 12 + x 13 < 50 (Plant 1 capacity) x 21 + x 22 + x 23 < 50 (Plant 2 capacity) x 11 + x 21 >= 25 (Northwood demand) x 12 + x 22 >= 45 (Westwood demand) x 13 + x 23 >= 10 (Eastwood demand) all x ij > 0 (Non-negativity)

25 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)

26 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 X 14 = 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 = 8X 11 +6X X 13 +9X 14 +9X X X 23 +7X X 31 +9X X 33 +5X 34

27 3. Supply Constraints X 11 +X 12 +X 13 +X 14 <= 35 X 21 +X 22 +X 23 +X 24 <= 50 X 31 +X 32 +X 33 +X 34 <= 40 Solution Since each supply point has a limited production capacity; 4. Demand Constraints X 11 +X 21 +X 31 >= 45 X 12 +X 22 +X 32 >= 20 X 13 +X 23 +X 33 >= 30 X 14 +X 24 +X 34 >= 30 Since each supply point has a limited production capacity; 5. Sign Constraints Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

28 LP Formulation of Powerco s Problem Min Z = 8X 11 +6X X 13 +9X 14 +9X X X 23 +7X X 31 +9X X 33 +5X 34 S.T.: X 11 +X 12 +X 13 +X 14 <= 35 (Supply Constraints) X 21 +X 22 +X 23 +X 24 <= 50 X 31 +X 32 +X 33 +X 34 <= 40 X 11 +X 21 +X 31 >= 45 (Demand Constraints) X 12 +X 22 +X 32 >= 20 X 13 +X 23 +X 33 >= 30 X 14 +X 24 +X 34 >= 30 Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

29 Balanced Transportation Problem If Total supply equals to total demand, the problem is said to be a balanced transportation problem: i m i 1 j n si dj j 1

30 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 that remaining constraint automatically.

31 Methods to find the BFS for a balanced Transportation Problem There are three basic methods: 1. North West Corner Method 2. Minimum Cost Method 3. Vogel s Method

32 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 x 11 as large as possible (here the limitations for setting x 11 to a larger number, will be the demand of demand point 1 and the supply of supply point 1. Your x 11 value can not be greater than minimum of this 2 values).

33 According to the explanations in the previous slide we can set x 11 =3 (meaning demand of demand point 1 is satisfied by supply point 1) X

34 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 X X 3 3 X X 2 3 2

35 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 X X X X X X X X 2 2

36 Finally, we will have the following bfs, which is: x 11 =3, x 12 =2, x 22 =3, x 23 =2, x 24 =1, x 34 =2 3 2 X X X X X X 2 X

37 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.

38 2. Minimum Cost Method (Cont d) The Steps: 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.

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

40 Step 2: Cross-out column X 4 6

41 Step 3: Find the new cell with minimum shipping cost and crossout row X X 4 6

42 Step 4: Find the new cell with minimum shipping cost and crossout row X X 15 5 X 4 6

43 Step 5: Find the new cell with minimum shipping cost and crossout column X X X X

44 Step 6: Find the new cell with minimum shipping cost and crossout column X X 6 X X X 6

45 Step 7: Finally assign 6 to last cell. The bfs is found as: X 11 =5, X 21 =2, X 22 =8, X 31 =5, X 33 =4 and X 34 = X X X X X X X

46 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 crossout the row or column as in the previous methods. Compute new penalties and use the same procedure.

47 An example for Vogel s Method Step 1: Compute the penalties Supply 10 Row Penalty 7-6= =63 Demand Column Penalty 15-6=9 80-7= =70

48 Step 2: Identify the largest penalty and assign the highest possible value to the variable Supply 5 15 Row Penalty 8-6= =63 Demand 15 X 5 Column Penalty 15-6=9 _ 78-8=70

49 Step 3: Identify the largest penalty and assign the highest possible value to the variable Supply 0 15 Row Penalty Demand 15 X X Column Penalty 15-6=9

50 Step 4: Identify the largest penalty and assign the highest possible value to the variable Supply X 15 Row Penalty Demand 15 X X Column Penalty _

51 Step 5: Finally the bfs is found as X 11 =0, X 12 =5, X 13 =5, and X 21 = Supply X Row Penalty _ X _ Demand X X X Column Penalty _

52 Exercise Bazaraa Chapter 10: Solve the following transportation problem: Destination Origin si dj cij matrix

53 Exercise Bazaraa Chapter 10: Solve the following transportation problem: si dj cij matrix

54 Lecture 14 Preparation Materi: Transportasi: Optimum Solution

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