Pemrograman Linier.

Presentasi berjudul: "Pemrograman Linier."— Transcript presentasi:

1 Pemrograman Linier

2 Konsep Pemrograman Linear Solusi Grafis Bagi Persoalan
Metode Simpleks

3 Laba Biaya Tujuan Sumber Daya Terbatas Memaksimalkan Meminimalkan
Latar Belakang Laba Biaya

4 Macam Solusi Grafis Menggambarkan Batasan Secara Grafis
2. Metode Solusi Garis Iso-Profit 3. Metode Solusi Titik Sudut

5 CONTOH MEMINIMALKAN BIAYA
Perusahaan Peternakan Kalkun Holiday Meal sedang mempertimbankan untuk membeli dua jenis pakan kalkun yang berbeda. Setiap pakan mengandung, dalam proporsi yang berbeda, beberapa atau semua dari tiga kandungan gizi yang penting untuk menggemukkan kalkun. Merk Y seharga $ 2 per pon dan Merk Z seharga $3 per pon. Pengusaha peternakan bermaksud untuk menentukan menu pakan yang paling murah biayanya, yang tetap memenuhi persyaratan minimal bulanan setiap kandungan gizi. Tabel berikut berisi informasi yang berkaitan dengan komposisi pakan ternak merk Y dan Z, juga persyaratan minimal bulanan bagi setiap kandungan gizi untuk setiap ekor kalkun. Komposisi Setiap Pon Pakan Ternak Kandungan gizi Pakan Merk Y Pakan Merk Z Kebutuhan Minimal A 5 10 90 B 4 3 48 C 2 6 72 HARGA $2 $3

6 Formulating LP Problems
Walkman Watch-TVs Available Hours Department (X1) (X2) This Week Hours Required to Produce 1 Unit Electronic Assembly Profit per unit $7 $5 Table B.1 Decision Variables: X1 = number of Walkmans to be produced X2 = number of Watch-TVs to be produced

7 Formulating LP Problems
Objective Function: Maximize Profit = $7X1 + $5X2 There are three types of constraints Upper limits where the amount used is ≤ the amount of a resource Lower limits where the amount used is ≥ the amount of the resource Equalities where the amount used is = the amount of the resource

8 Formulating LP Problems
First Constraint: Electronic time available time used is ≤ 4X1 + 3X2 ≤ 240 (hours of electronic time) Second Constraint: Assembly time available time used is ≤ 2X1 + 1X2 ≤ 100 (hours of assembly time)

9 Assembly (constraint B) Electronics (constraint A)
Graphical Solution – 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch-TVs Number of Walkmans X1 X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3

10 Iso-Profit Line Solution Method
Graphical Solution Iso-Profit Line Solution Method – 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch TVs Number of Walkmans X1 X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3 Choose a possible value for the objective function $210 = 7X1 + 5X2 Solve for the axis intercepts of the function and plot the line X2 = X1 = 30

11 Graphical Solution $210 = $7X1 + $5X2 X2 Number of Watch-TVs (0, 42)
– 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch-TVs Number of Walkmans X1 X2 Figure B.4 $210 = $7X1 + $5X2 (0, 42) (30, 0)

12 Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2
– 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch-TVs Number of Walkmans X1 X2 $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 $210 = $7X1 + $5X2 $420 = $7X1 + $5X2 Figure B.5

13 Corner-Point Method 2 3 1 4 X2 Number of Watch-TVs X1
– 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch-TVs Number of Walkmans X1 X2 2 3 1 Figure B.7 4

14 The optimal value will always be at a corner point
Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410

15 Optimal solution point
Graphical Solution – 80 – 60 – 40 – 20 – | | | | | | | | | | | Number of Watch-TVs Number of Walkmans X1 X2 Maximum profit line Optimal solution point (X1 = 30, X2 = 40) $410 = $7X1 + $5X2 Figure B.6

16 Metode Simpleks Ketika permasalahan pemrograman linear memiliki lebih dari dua variabel dan rumit untuk diselesaikan dengan menggunakan grafik. Solusi terbaik : laba yg paling tinggi atau biaya yg paling rendah.

17 Terima Kasih