optimum

TUGAS 3 OPERASI OPTIMUM SISTEM TENAGA LISTRIK

Oleh : Daniel Mahardhika (2213 105 052)

Diketahui data sebagai berikut:

Data Karakteristik Unit Pembangkit

Unit Max (MW) Min

(MW)

Incremental Heat Rate (Btu/kwh)

No-load Cost (R/h)

Full Load Cost

(R/mWh)

Min. Time Up

(jam)

Min. Time Down (jam)

1 80 25 10440 213,00 23,54 4 2 2 250 60 9000 585,62 20,34 5 3 3 300 75 8730 684,74 19,74 5 4 4 60 20 11900 252,00 28,00 1 1

Data Keadaan Awal

Unit Kondisi Awal Start-Up Cost Jam offline (-)/online (+) Hot (R) Cold (R) Cold Start (R) 1 -5 150 350 4 2 8 170 400 5 3 8 500 1100 5 4 -6 0 0 0

Data Pola beban

Beban Jam Beban (MW)

1 450 2 530 3 600 4 540 5 400 6 280 7 290 8 500

Data Tingkatan Kapasitas Unit Keadaan (State)

Kombinasi Unit

Maksimum Kapasitas Bersih

15 1111 690 14 1110 630 13 0111 610 12 0110 550 11 1011 440 10 1101 390 9 1010 380 8 0011 360 7 1100 330 6 0101 310 5 0010 300 4 1001 250 3 1001 140 2 1000 80 1 0001 60 0 0000 0

Fuel Cost 2,00 R/Mbtu

1. Kasus pertama, merupakan priority list schedule (rencana jadwal prioritas). (min up/down time : 1 jam untuk semua unit, start up cost : cold start cost)

Data Tingkat Prioritas Keadaan

(State) Kombinasi

Unit Maksimum Kapasitas Bersih

Untuk Tiap Kombinasi 15 1111 690 14 1110 630 12 0110 550 5 0010 300

A. J = 1; Jam ke-1, Beban = 450 MW

K = 15 (1,15) = min{} [(1,15) + (0,12; 1,15)]

F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (105) + 585,62 = 2475.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23,80 (20) + 252,00 = 728 PCost (1,15) = F1 + F2 + F3 + F4 = 9861.36 FCost (1,15) = 9861.36 + 350 = 10211.36

K = 14

(1,14) = min{} [(1,14) + (0,12; 1,14)] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (125) + 585,62 = 2835.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (1,14) = F1 + F2 + F3 = 9493.36 FCost (1,14) = 9493.36 + 350 = 9843.36

K = 12

(1,12) = min{} [(1,12) + (0,12; 1,12)] F2 = 18,00 (150) + 585,62 = 3285.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (1,12) = F2 + F3 = 9208.36 FCost (1,12) = 9208.36 + 0 = 9208.36

Biaya yang paling minimal pada stage 1 adalah FCost (1,12) = 9208.36

B. J = 2; Jam ke-2, Beban = 530 MW K = 15

(2,15) = min(12;14){} [(2,15) + (1, ; 2,15) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (185) + 585,62 = 3915.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23,80 (20) + 252,00 = 728 PCost (2,15) = F1 + F2 + F3 + F4 = 11301.36

FCost (2,15) = 11301.36 + (9208.36 + 350)(9843.36 + 0) = 20859.72

K = 14

(2,14) = min(12;14){} [(2,14) + (1, ; 2,14) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (205) + 585,62 = 4275.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (2,14) = F1 + F2 + F3 = 10933.36

FCost (2,14) = 10933.36 + (9208.36 + 350)(9843.36 + 0) = 20491.36

K = 12

(2,12) = min(12;14){} [(2,12) + (1, ; 2,12) + (1, )] F2 = 18,00 (230) + 585,62 = 4725.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (2,12) = F2 + F3 = 10648.36

FCost (2,12) = 10648.36 + (9208.36 + 0)(9843.36 + 0) = 19856.72


C. J = 3; Jam ke-3, Beban = 600 MW K = 15

(3,15) = min(12;14){} [(3,15) + (1, ; 3,15) + (1, )] F1 = 20,88 (30) + 213,00 = 839.4 F2 = 18,00 (250) + 585,62 = 5085.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23,80 (20) + 252,00 = 728 PCost (3,15) = F1 + F2 + F3 + F4 = 12575.76

FCost (3,15) = 12575.76 + (19856.72 + 350)(20491.36 + 0) = 32782.48

K = 14

(3,14) = min(12;14){} [(3,14) + (1, ; 3,14) + (1, )] F1 = 20,88 (50) + 213,00 = 1257 F2 = 18,00 (250) + 585,62 = 5085.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (3,14) = F1 + F2 + F3 = 12265.36

FCost (3,14) = 12265.36 + (19856.72 + 350)(20491.36 + 0) = 32472.08


D. J = 4; Jam ke-4, Beban = 540 MW K = 15


FCost (4,15) = 11481.36 + (32472.08 + 0)(32782.48 + 0) = 43953.44

K = 14

(4,14) = min(14;15){} [(4,14) + (1, ; 4,14) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (215) + 585,62 = 4455.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (4,14) = F1 + F2 + F3 = 11113.36

FCost (4,14) = 11113.36 + (32472.08 + 0)(32782.48 + 0) = 43585.44

K = 12

(4,12) = min(14;15){} [(4,12) + (1, ; 4,12) + (1, )] F2 = 18,00 (240) + 585,62 = 4905.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (4,12) = F2 + F3 = 10828.36

FCost (4,12) = 10828.36 + (32472.08 + 0)(32782.48 + 0) = 43300.44


E. J = 5; Jam ke-5, Beban = 400 MW K = 15


FCost (5,15) = 8967.06 + (43300.44 + 350)(43585.44 + 0) = 52552.5

K = 14

(5,14) = min(12;14){} [(5,14) + (1, ; 5,14) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (75) + 585,62 = 1935.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (5,14) = F1 + F2 + F3 = 8593.36

FCost (5,14) = 8593.36+ (43300.44 + 350)(43585.44 + 0) = 52178.8

K = 12

(5,12) = min(12;14){} [(5,12) + (1, ; 5,12) + (1, )] F2 = 18,00 (100) + 585,62 = 2385.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (5,12) = F2 + F3 = 8308.36

FCost (5,12) = 8308.36 + (43300.44 + 0)(43585.44 + 0) = 51608.8


F. J = 6; Jam ke-6, Beban = 280 MW K = 15


FCost (6,15) = 6868.86 + (51893.8 + 350)(52178.8 + 0) = 59047.66

K = 14

(6,14) = min(12;14){} [(6,14) + (1, ; 6,14) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (60) + 585,62 = 1665.62 F3 = 17,46 (195) + 684,74 = 4089.44 PCost (6,14) = F1 + F2 + F3 = 6490.06

FCost (6,14) = 6490.06 + (51893.8 + 350)(52178.8 + 0) = 58668.86

K = 12

(6,12) = min(12;14){} [(6,12) + (1, ; 6,12) + (1, )] F2 = 18,00 (60) + 585,62 = 1665.62 F3 = 17,46 (220) + 684,74 = 4525.94 PCost (6,12) = F2 + F3 = 6191.56

FCost (6,12) = 6191.56 + (51608.8 + 0)(52178.8 + 0) = 57800.36

K = 5

(6,5) = min(12;14){} [(6,5) + (1,; 6,5) + (1, )] F3 = 17,46 (280) + 684,74 = 5573.54 PCost (6,5) = F3 = 5573.54

FCost (6,5) = 5573.54 + (51608.8 + 0)(52178.8 + 0) = 57182.34


G. J = 7; Jam ke-7, Beban = 290 MW K = 15


FCost (7,15) = 7043.46 + (57182.34 + 750)(57800.36 + 350) = 64975.8

K = 14

(7,14) = min(5;12){} [(7,14) + (1,; 7,14) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (60) + 585,62 = 1665.62 F3 = 17,46 (205) + 684,74 = 4264.04 PCost (7,14) = F1 + F2 + F3 = 6664.66

FCost (7,14) = 6664.66 + (57182.34 + 750)(57800.36 + 350) = 64597.0

K = 12

(7,12) = min(5;12){} [(7,12) + (1,; 7,12) + (1, )] F2 = 18,00 (60) + 585,62 = 1665.62 F3 = 17,46 (230) + 684,74 = 4700.54 PCost (7,12) = F2 + F3 = 6366.16

FCost (7,12) = 6366.16 + (57182.34 + 400)(57800.36 + 0) = 63948.5

K = 5

(7,5) = min(5;12){} [(7,5) + (1, ; 7,5) + (1, )] F3 = 17,46 (290) + 684,74 = 5748.14 PCost (7,5) = F3 = 5748.14

FCost (7,5) = 5748.14 + (57182.34 + 0)(57800.36 + 0) = 62930.48


H. J = 8; Jam ke-8, Beban = 500 MW K = 15


FCost (8,15) = 20061.36 + (62930.48 + 750)(63948.5 + 350) = 83741.84

K = 14

(8,15) = min(5;12){} [(8,15) + (1,; 8,15) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (175) + 585,62 = 3735.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (8,14) = F1 + F2 + F3 = 10393.36

FCost (8,14) = 10393.36 + (62930.48 + 750)(63948.5 + 350) = 74073.84

K = 12

(8,15) = min(5;12){} [(8,15) + (1,; 8,15) + (1, )] F2 = 18,00 (200) + 585,62 = 4185.62 F3 = 17,46 (300) + 684,74 = 5922.74 PCost (8,12) = F2 + F3 = 10108.36

FCost (8,12) = 10108.36 + (62930.48 + 400)(63948.5 + 0) = 73438.84


2. Kasus kedua, sama dengan kasus pertama namun dengan melakukan complete enumeration (semua kombinasi diperhitungkan) dengan mengabaikan biaya hot start, minimum uptime dan minimum downtime. (min up/down time : 1 jam untuk semua unit, start up cost : cold start cost)

Data Kombinasi yang Memungkinkan Stage Load Kombinasi

1 450 15 ; 14 ; 13 ; 12 2 530 15 ; 14 ; 13 ; 12 3 600 15 ; 14 ; 13 4 540 15 ; 14 ; 13 ; 12 5 400 15 ; 14 ; 13 ; 12 ; 11 6 280 15 ; 14 ; 13 ; 12 ; 11 ; 10 ; 9 ; 8 ; 7 ; 6 ; 5 7 290 15 ; 14 ; 13 ; 12 ; 11 ; 10 ; 9 ; 8 ; 7 ; 6 ; 5 8 500 15 ; 14 ; 13 ; 12

A. J = 1; Jam ke-1, Beban = 450 MW

K = 15 FCost (1,15) = 9861.36 + 350 = 10211.36

K = 14

FCost (1,14) = 9493.36 + 350 = 9843.36

K = 13 (1,13) = min{} [(1,13) + (0,13; 1,13)]

F2 = 18,00 (130) + 585,62 = 2925.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (20) + 252.00 = 728 PCost (1,13) = F2 + F3 + F4 = 9576.62 FCost (1,13) = 9576.62 + 0 = 9576.62

K = 12

FCost (1,12) = 9208.36 + 0 = 9208.36 Biaya yang paling minimal pada stage 1 adalah FCost (1,12) = 9208.36

B. J = 2; Jam ke-2, Beban = 530 MW K = 15

FCost (2,15) = 11301.36 + (9208.36 + 350)(9576.62 + 0) = 20859.72

K = 14

FCost (2,14) = 10933.36 + (9208.36 + 350)(9576.62 + 0) = 20491.36

K = 13

(2,13) = min(12;13){} [(2,13) + (1, ; 2,13) + (1, )] F2 = 18,00 (210) + 585,62 = 4365.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (20) + 252.00 = 728 PCost (2,13) = F2 + F3 + F4 = 11016.36

FCost (2,13) = 11016.36 + (9208.36 + 0)(9576.62 + 0) = 20224.72

K = 12

FCost (2,12) = 10648.36 + (9208.36 + 0)(9576.62 + 0) = 19856.72


C. J = 3; Jam ke-3, Beban = 600 MW K = 15

FCost (3,15) = 12575.76 + (19856.72 + 350)(20224.72 + 0) = 32782.48

K = 14

FCost (3,14) = 12265.36 + (19856.72 + 350)(20224.72 + 0) = 32472.08

K = 13

(3,13) = min(12;13){} [(3,13) + (1, ; 3,13) + (1, )] F2 = 18,00 (250) + 585,62 = 5085.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (50) + 252.00 = 1442 PCost (3,13) = F2 + F3 + F4 = 12450.36

FCost (3,13) = 12450.36 + (19856.72 + 0)(20224.72 + 0) = 32307.08


D. J = 4; Jam ke-4, Beban = 540 MW K = 15

FCost (4,15) = 11481.36 + (32307.08 + 350)(32472.08 + 0) = 43953.44

K = 14

FCost (4,14) = 11113.36 + (32307.08 + 350)(32472.08 + 0) = 43585.44

K = 13

(4,13) = min(13;14){} [(4,13) + (1, ; 4,13) + (1, )] F2 = 18,00 (220) + 585,62 = 4545.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (20) + 252.00 = 728 PCost (4,13) = F2 + F3 + F4 = 11196.36

FCost (4,13) = 11196.36 + (32307.08 + 0)(32472.08 + 0) = 43503.44

K = 12

FCost (4,12) = 10828.36 + (32307.08 + 0)(32472.08 + 0) = 43135.44


E. J = 5; Jam ke-5, Beban = 400 MW K = 15

FCost (5,15) = 8967.06 + (43135.44 + 350)(43503.44 + 0) = 52452.5

K = 14

FCost (5,14) = 8593.36+ (43135.44 + 350)(43503.44 + 0) = 52078.8

K = 13

(5,13) = min(12;13){} [(5,13) + (1, ; 5,13) + (1, )] F2 = 18,00 (80) + 585,62 = 2025.62 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (20) + 252.00 = 728 PCost (5,13) = F2 + F3 + F4 = 8676.36

FCost (5,13) = 8676.36 + (43135.44 + 0)(43503.44 + 0) = 51811.8

K = 12

FCost (5,12) = 8308.36 + (43135.44 + 0)(43503.44 + 0) = 51443.8

K = 11

(5,11) = min(12;13){} [(5,13) + (1, ; 5,13) + (1, )] F1 = 20.88 (80) + 213.00 = 1883.4 F3 = 17,46 (300) + 684,74 = 5922.74 F4 = 23.8 (20) + 252.00 = 728 PCost (5,11) = F1 + F3 + F4 = 8534.14

FCost (5,11) = 8534.14 + (43135.44 + 350)(43503.44 + 0) = 52019.58


F. J = 6; Jam ke-6, Beban = 280 MW K = 15

FCost (6,15) = 6868.86 + (51443.8 + 350)(51811.8 + 0) = 58662.66

K = 14

FCost (6,14) = 6490.06 + (51443.8 + 350)(51811.8 + 0) = 58283.86

K = 13

(6,13) = min(12;13){} [(6,13) + (1, ; 6,13) + (1, )] F2 = 18,00 (25) + 585,62 = 1035.62 F3 = 17,46 (235) + 684,74 = 4787.84 F4 = 23.8 (20) + 252.00 = 728 PCost (6,13) = F2 + F3 + F4 = 6551.46

FCost (6,13) = 6551.46 + (51443.8 + 0)(51811.8 + 0) = 57995.26

K = 12

FCost (6,12) = 6191.56 + (51443.8 + 0)(51811.8 + 0) = 57635.36

K = 11

(6,11) = min(12;13){} [(6,11) + (1, ; 6,11) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (235) + 684,74 = 4787.84 F4 = 23.8 (20) + 252.00 = 728 PCost (6,11) = F1 + F3 + F4 = 6250.84

FCost (6,11) = 6250.84 + (51443.8 + 350)(51811.8 + 0) = 58044.64

K = 10

(6,10) = min(12;13){} [(6,10) + (1, ; 6,10) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (235) + 585,62 = 4815.62 F4 = 23.8 (20) + 252.00 = 728 PCost (6,10) = F1 + F2 + F4 = 6278.62

FCost (6,10) = 6278.62 + (51443.8 + 350)(51811.8 + 0) = 58072.42

K = 9

(6,9) = min(12;13){} [(6,9) + (1,; 6,9) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (255) + 684,74 = 5137.04 PCost (6,9) = F1 + F3 = 5872.04

FCost (6,9) = 5872.04 + (51443.8 + 350)(51811.8 + 0) = 57665.84

K = 8 (6,8) = min(12;13){} [(6,8) + (1,; 6,8) + (1, )]

F3 = 17,46 (260) + 684,74 = 5224,34 F4 = 23,88 (20) + 252,00 = 728 PCost (6,8) = F3 + F4 = 5952.34

FCost (6,8) = 5952.34 + (51443.8 + 0)(51811.8 + 0) = 57396.84

K = 7

(6,7) = min(12;13){} [(6,7) + (1,; 6,7) + (1, )] F1 = 20,88 (30) + 213,00 = 839,4 F2 = 18,00 (250) + 585,62 = 5085,62 PCost (6,7) = F1 + F2 = 5925.02

FCost (6,7) = 5925.02 + (51443.8 + 350)(51811.8 + 0) = 57718.82

K = 6

(6,6) = min(12;13){} [(6,6) + (1,; 6,6) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (255) + 684,74 = 5137.04 PCost (6,6) = F1 + F3 = 5872.04

FCost (6,6) = 5872.04 + (51443.8 + 350)(51811.8 + 0) = 57315.84

K = 5

FCost (6,12) = 5573.54 + (51443.8 + 0)(51811.8 + 0) = 57017.34


G. J = 7; Jam ke-7, Beban = 290 MW K = 15

FCost (7,15) = 7043.46 + (57017.34 + 750)(57315.84 + 1450) = 64810,8

K = 14

FCost (7,14) = 6664.66 + (57017.34 + 750)(57315.84 + 1450) = 64432,0

K = 13

(7,13) = min(5;6){} [(7,13) + (1, ; 7,13) + (1, )] F2 = 18,00 (25) + 585,62 = 1035.62 F3 = 17,46 (245) + 684,74 = 4962,44 F4 = 23.8 (20) + 252.00 = 728 PCost (7,13) = F2 + F3 + F4 = 6726.06

FCost (7,13) = 6726.06 + (57017.34 + 400)(57315.84 + 1100) = 64143,4

K = 12

FCost (7,12) = 6366.16 + (57017.34 + 400)(57315.84 + 1100) = 63783,5

K = 11

(7,11) = min(5;6){} [(7,11) + (1, ; 7,11) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (245) + 684,74 = 4962.84 F4 = 23.8 (20) + 252.00 = 728 PCost (7,11) = F1 + F3 + F4 = 6425.84

FCost (7,11) = 6425.84 + (57017.34 + 350)(57315.84 + 1450) = 63793,18

K = 10

(7,10) = min(5;6){} [(7,10) + (1, ; 7,10) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F2 = 18,00 (245) + 585,62 = 4995.62 F4 = 23.8 (20) + 252.00 = 728 PCost (7,10) = F1 + F2 + F4 = 6278.62

FCost (7,10) = 6458.62 + (57017.34 + 750)(57315.84 + 350) = 64124.46

K = 9

(7,9) = min(5;6){} [(7,9) + (1,; 7,9) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (265) + 684,74 = 5311,64 PCost (7,9) = F1 + F3 = 6046,64

FCost (7,9) = 6046,64 + (57017.34 + 350)(57315.84 + 1450) = 63413,98

K = 8 (7,8) = min(5;6){} [(7,8) + (1,; 7,8) + (1, )]

F3 = 17,46 (270) + 684,74 = 5398,94 F4 = 23,88 (20) + 252,00 = 728 PCost (7,8) = F3 + F4 = 6126,94

FCost (7,8) = 6126,94 + (57017.34 + 0)(57315.84 + 0) = 63144.28

K = 7

(7,7) = min(5;6){} [(7,7) + (1,; 7,7) + (1, )] F1 = 20,88 (40) + 213,00 = 1048,2 F2 = 18,00 (250) + 585,62 = 5085,62 PCost (7,7) = F1 + F2 = 6133,82

FCost (7,7) = 6133,82 + (57017.34 + 750)(57315.84 + 350) = 63799.66

K = 6

(7,6) = min(5;6){} [(7,6) + (1,; 7,6) + (1, )] F1 = 20,88 (25) + 213,00 = 735 F3 = 17,46 (265) + 684,74 = 5311.64 PCost (7,6) = F1 + F3 = 6046.64

FCost (7,6) = 6046.64 + (57017.34 + 400)(57315.84 + 0) = 63362.48

K = 5

(7,5) = min(5;6){} [(7,5) + (1,; 7,5) + (1, )] F3 = 17,46 (290) + 684,74 = 5748.14 PCost (7,5) = F3 = 5748.14

FCost (7,5) = 5748.14 + (57017.34 + 0)(57315.84 + 0) = 62765.04


H. J = 8; Jam ke-8, Beban = 500 MW K = 15

FCost (8,15) = 20061.36 + (62765.04 + 750)(63144.28 + 750) = 83576.4

K = 14

FCost (8,14) = 10393.36 + (62765.04 + 750)(63144.28 + 750) = 73908.4

K = 13

(8,13) = min(5;8){} [(8,13) + (1, ; 8,13) + (1, )] F2 = 18,00 (180) + 585,62 = 3825.62 F3 = 17,46 (300) + 684,74 = 5922,74 F4 = 23.8 (20) + 252.00 = 728 PCost (8,13) = F2 + F3 + F4 = 10476,36

FCost (8,13) = 10476,36 + (62765.04 + 400)(63144.28 + 400) = 73641,4

K = 12

FCost (8,12) = 10108.36 + (62765.04 + 400)(63144.28 + 400) = 73273.4


3. Kasus ketiga, melibatkan hot/cold start, minimum uptime dan minimum downtime. Menggunakan PowerGen dengan melakukan complete enumeration (semua kombinasi diperhitungkan). PowerGen memiliki kombinasi sebagai berikut:

Terdapat perbedaan urutan kombinasi dengan kombinasi berdasarkan urutan kapasitas unit (konsdisi awal)

Data Perbandingan Kombinasi PowerGen dengan Kombinasi urutan Kapasitas

PowrGen Kombinasi Kondisi Awal 1 2 3 4 1 1 1 1 1 15 2 1 1 1 0 14 3 1 1 0 1 10 4 1 1 0 0 7 5 1 0 1 1 11 6 1 0 1 0 9 7 1 0 0 1 4 8 1 0 0 0 2 9 0 1 1 1 13 10 0 1 1 0 12 11 0 1 0 1 6 12 0 1 0 0 3 13 0 0 1 1 8 14 0 0 1 0 5 15 0 0 0 1 1

Dengan inputan data sebagai berikut:

Didapatkan hasil:

Perbandingan Kasus 1, Kasus 2, dan Kasus 3:

Data Kombinasi Perjam Tiap Kasus

Jam Kombinasi

Kasus 1

Kasus 2

Kasus 3

1 12 12 12 2 12 12 12 3 14 13 13 4 12 12 12 5 12 12 12 6 5 5 12 7 5 5 12 8 12 12 12

Grafik Kombinasi Perjam Tiap Kasus

optimum

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