Pemrograman linier

Pengantar Optimasi di Python

Jasmin Ludolf

Content Developer

Pemrograman linier

Fungsi objektif dan kendala berhubungan linier
Berbeda dari kendala nonlinier: campuran unsur linier/nonlinier

Bauran produk

Tipe	Jumlah	Botol (j/kotak)	Cangkir (j/kotak)
$M_A$	4	1.5	1.3
$M_B$	3	0.8	2.1

Tiap mesin beroperasi 30 jam/minggu

Permintaan: botol untuk minuman, cangkir untuk yogurt

Maksimalkan laba

Laba per kotak

B ($/kotak)	C ($/kotak)
480	510

Fungsi objektif: $480B + 510C$

Kendala
- Mesin A: $1.5B+1.3C\leq 4 \times 30$
- Mesin B: $0.8B+ 2.1C \leq 3\times 30$
- Non-negatif: $B, C \geq 0$

Menggunakan PuLP

from pulp import *


model = LpProblem('MaxProfit', LpMaximize)


B = LpVariable('B', lowBound=0) 
C = LpVariable('C', lowBound=0)


model += 480*B + 510*C

model += 1.5*B + 1.3*C - 120, "M_A" 
model += 0.8*B + 2.1*C - 90, "M_B"

LpMaximize atau LpMinimize

Keluaran model PuLP

print(model)

ProductMix:
MAXIMIZE
480*B + 510*C + 0
SUBJECT TO
M_A: 1.5 B + 1.3 C <= 120

M_B: 0.8 B + 2.1 C <= 90

VARIABLES
B Continuous
C Continuous

Menyelesaikan dengan PuLP

status = model.solve()
print(status)

Welcome to the CBC MILP Solver 
...
1

print(f"Profit = {value(model.objective):.2f}")
print(f"Tons of bottles = {B.varValue:.2f}, tons of cups = {C.varValue:.2f}")

Profit = 40137.44
Boxes of bottles = 63.98, boxes of cups = 18.48

Menangani banyak variabel dan kendala

variables = LpVariable.dicts("Product", ['B', 'C'], lowBound=0)

variables = LpVariable.dicts("Product", range(2), 0)

A = LpVariable.matrix('A', (range(2), range(2)), 0)

box_profit = {'B': 480, 'C': 510}
model += lpSum([box_profit[i] * variables[i] for i in ['B', 'C']]) 

model += 1.5*variables['B'] + 1.3*variables['C'] <= 120, "M_A"
model += 0.8*variables['B'] + 2.1*variables['C'] <= 90, "M_B"

Ayo berlatih!

Pengantar Optimasi di Python