Goal Programming n n Goal programming dapat digunakan

Goal Programming n n Goal programming dapat digunakan untuk memecahakan program linear dengan beberapa tujuan. Setiap tujuan disebut "goal". Di dalam goal programming, di+ and di- , variable deviasi, adalah jumlah i goal yang ditargetkan tercapai lebih atau kurang tercapai, secara berurutan Tujuan tujuan itu sendiri ditambahkan sebagai konstrain di lambangkan dgn di+ and di- yang dapat berfungsi sebagai variabel surplus and slack. Sebuah pendekatan untuk goal programming adalah mencapai goal dalam urutan prioritas. Goal kedua adalah dikejar tanpa mengurangi goal prioritas pertama. © 2003 Thomson. TM/South-Western Slide 1

Goal Programming n Untuk setiap level prioritas, fungsi tujuan nya adalah meminimalkan jumlah tertimbang dari deviasi goal. Pencapaian optimal sebelumnya dari goal ditambahkan ke konstrain (kendala) sehingga mereka tidak menurunkan goal prioritas sebelumnya. © 2003 Thomson. TM/South-Western Slide 2

Goal Programming Approach Step 1: Menetapkan level prioritas dari setiap goal. Step 2: Menetapkan timbangan di setiap goal. Jika sebuah level prioritas memiliki lebih dari sebuah goal, untuk setiap goal each i ditetapkan timbangan, wi , diletakkan di deviasi , di+ dan/atau di-, dari goal. Step 3: membuat linear program. Min w 1 d 1+ + w 2 d 2 s. t. Functional Constraints, and Goal Constraints Step 4: menyelesaikan linear program. jika ada level prioritas yang lebih rendah, silahkan ikuti step 5. jika tidak, solusi final telah dicapai. © 2003 Thomson. TM/South-Western Slide 3

Goal Programming Approach Step 5: Membuat program linear yang selanjutnya (bila ada tujuan sesudah penyelesaian pertama). Pertimbangkan gaol dengan prioritas lebih rendah dan buat formula sebuah fungsi tujuan baru berdasar tujuan tsb. Tambhakan sebuah konstrain yg mensyaratkan pencapaian goal level prioritas yang lebih tinggi berikutnya terjaga. Maka program linear baru menjadi: Min w 3 d 3+ + w 4 d 4 s. t. Functional Constraints, Goal Constraints, and w 1 d 1+ + w 2 d 2 - = k Go to step 4. (Repeat steps 4 and 5 until all priority levels have been examined. ) © 2003 Thomson. TM/South-Western Slide 4

Example: Conceptual Products adalah perusahaan komputer yang memproduksi komputer jenis CP 400 and the CP 500. komputer tersebut mengunakan mother board yang berbeda yang diproduksi secara besar oleh perusahaan, tetapi menggunakan kasing dan disk drives yang sama. CP 400 model menggunakan 2 floppy disk drives dan no zip disk drives sementara CP 500 model menggunakan satu floppy disk drive dan satu zip disk drive. © 2003 Thomson. TM/South-Western Slide 5

Example: Conceptual Products Disk drives dan kasing dibeli dari vendors. Tersedia 1000 floppy disk drives, 500 zip disk drives, and 600 cases perminggu. Hal ini memerlukan satu jam untuk membuat CP 400 dan keuntungannya adalah $200 and CP 500 memerlukan 1, 5 jam untuk diproduksi dengan profit $500. © 2003 Thomson. TM/South-Western Slide 6

Example: Conceptual Products The company has four goals which are given below: Priority 1: Meet a state contract of 200 CP 400 machines weekly. (Goal 1) Priority 2: Make at least 500 total computers weekly. (Goal 2) Priority 3: Make at least $250, 000 weekly. (Goal 3) Priority 4: Use no more than 400 man-hours per week. (Goal 4) © 2003 Thomson. TM/South-Western Slide 7

Example: Conceptual Products n n Variables x 1 = number of CP 400 computers produced weekly x 2 = number of CP 500 computers produced weekly di- = amount the right hand side of goal i is deficient di+ = amount the right hand side of goal i is exceeded Functional Constraints Availability of floppy disk drives: Availability of zip disk drives: Availability of cases: 600 © 2003 Thomson. TM/South-Western 2 x 1 + x 2 < 1000 x 2 < 500 x 1 + x 2 < Slide 8

Example: Conceptual Products n Goals (1) 200 CP 400 computers weekly: x 1 + d 1 - - d 1+ = 200 (2) 500 total computers weekly: x 1 + x 2 + d 2 - - d 2+ = 500 (3) $250(in thousands) profit: . 2 x 1 +. 5 x 2 + d 3 - - d 3+ = 250 (4) 400 total man-hours weekly: x 1 + 1. 5 x 2 + d 4 - - d 4+ = 400 Non-negativity: x 1, x 2, di-, di+ > 0 for all i © 2003 Thomson. TM/South-Western Slide 9

Example: Conceptual Products n Objective Functions Priority 1: Minimalkan jumlah state contract yang tidak terpenuhi: Min d 1 Priority 2: Minimize the number under 500 computers produced weekly: Min d 2 Priority 3: Minimize the amount under $250, 000 earned weekly: Min d 3 Priority 4: Minimize the man-hours over 400 used weekly: Min d 4+ © 2003 Thomson. TM/South-Western Slide 10

Example: Conceptual Products n Formulation Summary Min P 1(d 1 -) + P 2(d 2 -) + P 3(d 3 -) + P 4(d 4+) s. t. 1000 600 200 500 250 2 x 1 +x 2 x 1 < < 500 < +d 1 - -d 1+ +x 2 . 2 x 1+. 5 x 2 © 2003 Thomson. TM/South-Western = +d 2 - -d 2+ = +d 3 - -d 3+ = Slide 11

Example: Conceptual Products n Graphical Solution, Iteration 1 To solve graphically, first graph the functional constraints. Then graph the first goal: x 1 = 200. Note on the next slide that there is a set of points that exceed x 1 = 200 (where d 1 - = 0). © 2003 Thomson. TM/South-Western Slide 12

Example: Conceptual Products n Functional Constraints and Goal 1 Graphed x 2 1000 2 x 1 + x 2 < 1000 800 Goal 1: x 1 > 200 x 2 < 500 x 1 + x 2 < 600 400 Points Satisfying Goal 1 200 400 © 2003 Thomson. TM/South-Western 600 800 1000 1200 x 1 Slide 13

Example: Conceptual Products n Graphical Solution, Iteration 2 Now add Goal 1 as x 1 > 200 and graph Goal 2: x 1 + x 2 = 500. Note on the next slide that there is still a set of points satisfying the first goal that also satisfies this second goal (where d 2 - = 0). © 2003 Thomson. TM/South-Western Slide 14

Example: Conceptual Products n Goal 1 (Constraint) and Goal 2 Graphed x 2 1000 2 x 1 + x 2 < 1000 800 Goal 1: x 1 > 200 x 2 < 500 x 1 + x 2 < 600 400 Points Satisfying Both Goals 1 and 2 200 Goal 2: x 1 + x 2 > 500 200 400 © 2003 Thomson. TM/South-Western 600 800 1000 1200 x 1 Slide 15

Example: Conceptual Products n Graphical Solution, Iteration 3 Now add Goal 2 as x 1 + x 2 > 500 and Goal 3: . 2 x 1 +. 5 x 2 = 250. Note on the next slide that no points satisfy the previous functional constraints and goals and satisfy this constraint. Thus, to Min d 3 -, this minimum value is achieved when we Max. 2 x 1 +. 5 x 2. Note that this occurs at x 1 = 200 and x 2 = 400, so that. 2 x 1 +. 5 x 2 = 240 or d 3 - = 10. © 2003 Thomson. TM/South-Western Slide 16

Example: Conceptual Products n Goal 2 (Constraint) and Goal 3 Graphed x 2 1000 2 x 1 + x 2 < 1000 800 Goal 1: x 1 > 200 x 2 < 500 x 1 + x 2 < 600 (200, 400) Points Satisfying Both Goals 1 and 2 400 Goal 2: x 1 + x 2 > 500 200 Goal 3: . 2 x 1 +. 5 x 2 = 250 200 400 600 © 2003 Thomson. TM/South-Western 800 1000 1200 x 1 Slide 17