Menyelesaikan model PuLP

Analitik Rantai Pasok dengan Python

Aaren Stubberfield

Supply Chain Analytics Mgr.

Proses pemodelan umum di PuLP

~~Inisialisasi model~~
~~Definisikan variabel keputusan~~
~~Definisikan fungsi objektif~~
~~Definisikan kendala~~
Selesaikan model
- panggil metode solve()
- periksa status solusi
- cetak variabel keputusan optimal
- cetak nilai fungsi objektif

Selesaikan model - metode solve

.solve(solver=None)

solver = Opsional: solver spesifik yang akan dipakai, default ke solver bawaan.

# Initialize, Define Decision Vars., Objective Function, and Constraints
from pulp import *
import pandas as pd
model = LpProblem("Minimize Transportation Costs", LpMinimize)
cust = ['A','B','C']
warehouse = ['W1','W2']
demand = {'A': 1500, 'B': 900, 'C': 800}
costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264, 
         ('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250}
ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust], 
                         lowBound=0, cat='Integer')
model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust])
for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c]

# Solve Model
model.solve()

Selesaikan model - status solusi

LpStatus[model.status]

Not Solved: Status sebelum pemecahan masalah.
Optimal: Solusi optimal ditemukan.
Infeasible: Tidak ada solusi layak (mis. jika kendala x ≤ 1 dan x ≥ 2).
Unbounded: Fungsi objektif tak berbatas; memaksimalkan atau meminimalkan menuju tak hingga (mis. jika satu-satunya kendala x ≥ 3).
Undefined: Solusi optimal mungkin ada tetapi belum ditemukan.

¹ Keen, Ben Alex. “Linear Programming with Python and PuLP ² Part 2.” _Ben Alex Keen_, 1 Apr. 2016, benalexkeen.com/linear-programming-with-python-and-pulp-part-2/._{{5}}

# Initialize, Define Decision Vars., Objective Function, and Constraints
from pulp import *
import pandas as pd
model = LpProblem("Minimize Transportation Costs", LpMinimize)
cust = ['A','B','C']
warehouse = ['W1','W2']
demand = {'A': 1500, 'B': 900, 'C': 800}
costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264,
         ('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250}
ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust], lowBound=0, cat='Integer')
model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust])
for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c]
# Solve Model
model.solve()
print("Status:", LpStatus[model.status])

Status: Optimal

Cetak variabel ke output standar:

for v in model.variables():
    print(v.name, "=", v.varValue)

Struktur data pandas:

o = [{A:ship[(w,'A')].varValue, B:ship[(w,'B')].varValue, C:ship[(w,'C')].varValue}
     for w in warehouse]
print(pd.DataFrame(o, index=warehouse))

loop variabel model
simpan nilai ke DataFrame pandas

# Solve Model
model.solve()
print(LpStatus[model.status])
o = [{A:ship[(w,'A')].varValue, B:ship[(w,'B')].varValue, C:ship[(w,'C')].varValue}
     for w in warehouse]
print(pd.DataFrame(o, index=warehouse))

Output:

Status: Optimal
|       |A      |B      |C      |
|:------|:------|:------|:------|
|W1     |1500.0 |0.0    |0.0    |
|W2     |0.0    |900.0  |800.0  |

Selesaikan model - fungsi objektif optimal

Cetak nilai fungsi objektif optimal:

print("Objective = ", value(model.objective))

# Solve Model
model.solve()
print(LpStatus[model.status])
output = []
for w in warehouse: t = [ship[(w,c)].varValue for c in cust] output.append(t)
opd = pd.DataFrame.from_records(output, index=warehouse, columns=cust)
print(opd)
print("Objective = ", value(model.objective))

Status: Optimal
|       |A      |B      |C      |
|:------|:------|:------|:------|
|W1     |1500.0 |0.0    |0.0    |
|W2     |0.0    |900.0  |800.0  |
Objective = 757700.0

Ringkasan

Selesaikan model

Panggil metode solve()
Periksa status solusi
Cetak nilai variabel keputusan
Cetak nilai fungsi objektif

Ayo berlatih!

Analitik Rantai Pasok dengan Python