Lös PuLP-modellen

Supply Chain Analytics i Python

Aaren Stubberfield

Supply Chain Analytics Mgr.

Vanlig modelleringsprocess i PuLP

~~Initiera modell~~
~~Definiera beslutsvariablar~~
~~Definiera målfunktionen~~
~~Definiera bivillkoren~~
Lös modell
- anropa metoden solve()
- kontrollera lösningens status
- skriv ut optimerade beslutsvariablar
- skriv ut optimerad målfunktion

Lös modell – solve-metoden

.solve(solver=None)

solver = Valfritt: den specifika lösaren som ska användas. Standardvärde är den förvalda lösaren.

# 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()

Lös modell – lösningens status

LpStatus[model.status]

Not Solved: Status innan problemet har lösts.
Optimal: En optimal lösning har hittats.
Infeasible: Det finns inga genomförbara lösningar (t.ex. om du anger bivillkoren x ≤ 1 och x ≥ 2).
Unbounded: Målfunktionen är inte begränsad – maximering eller minimering tenderar mot oändligheten (t.ex. om det enda bivillkoret var x ≥ 3).
Undefined: Den optimala lösningen kan existera men har eventuellt inte hittats.

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

Skriv ut variabler till standardutdata:

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

Pandas-datastruktur:

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))

loopa igenom modellens variabler
lagra värden i en pandas DataFrame

# 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))

Utdata:

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

Lös modell – optimerad målfunktion

Skriv ut värdet på den optimerade målfunktionen:

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

Sammanfattning

Lös modell

Anropa metoden solve()
Kontrollera lösningens status
Skriv ut beslutsvariablarnas värden
Skriv ut målfunktionens värde

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Supply Chain Analytics i Python