Pemrograman Linier

Konsep Pemrograman Linear Solusi Grafis Bagi Persoalan Metode Simpleks

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

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

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

Formulating LP Problems Walkman Watch-TVs Available Hours Department (X1) (X2) This Week Hours Required to Produce 1 Unit Electronic 4 3 240 Assembly 2 1 100 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

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

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)

Assembly (constraint B) Electronics (constraint A) Graphical Solution – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch-TVs Number of Walkmans X1 X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3

Iso-Profit Line Solution Method Graphical Solution Iso-Profit Line Solution Method – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 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 = 42 X1 = 30

Graphical Solution $210 = $7X1 + $5X2 X2 Number of Watch-TVs (0, 42) – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch-TVs Number of Walkmans X1 X2 Figure B.4 $210 = $7X1 + $5X2 (0, 42) (30, 0)

Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch-TVs Number of Walkmans X1 X2 $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 $210 = $7X1 + $5X2 $420 = $7X1 + $5X2 Figure B.5

Corner-Point Method 2 3 1 4 X2 Number of Watch-TVs X1 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch-TVs Number of Walkmans X1 X2 2 3 1 Figure B.7 4

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

Optimal solution point Graphical Solution – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 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

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.

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