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"""
Author: Louis Luangkesorn <lugerpitt@gmail.com> 2019
https://github.com/lluang
Title: Gemstone Optimization problem
Problem taken from Data, Models, and Decisions by Bertsimas and Freund, 4th Edition
DMD 7.2
## 2 stage problem
- **Scenarios:** $s \in S = (1, 2, 3, 4)$
- **Probability scenario occuring:** $p^s$
- **Cost of steel:** $c$
- **Total steel:** $cap_{steel}$
- **Total molding and assembly hours:** $cap_{molding}, cap_{assembly}^s$
- **Wrench and plier earnings by scenario:** $w^s, p^s$
- **Max demand wrenches and pliers:** $UB_w, UB_p$
- **Decision variables**
- $(W_{t+1}^s, P_{t+1}^s)$
- **Objective** $Max \sum_s (p^s * (w^s W_{t+1}^s + p^s P_{t+1}^s) - c$
- **Steel Constraint:** $1.5W_{t+1}^1 + P_{t+1}^1 - C \le 0$
- **Molding Constraint:** $W_{t+1}^1 + P_{t+1}^1 \le cap_{molding}$
- **Assembly Constraint:** $0.3 W_{t+1}^1 + 0.5 P_{t+1}^1 \le cap_{molding}^s$
- **Demand Limit W:** $W \le UB_w$
- **Demand Limit P:** $P \le UB_p$
- **Nonnegativity:** $W, P \ge 0$
"""
import pulp
# parameters
products = ["wrenches", "pliers"]
price = [130, 100]
steel = [1.5, 1]
molding = [1, 1]
assembly = [0.3, 0.5]
capsteel = 27
capmolding = 21
LB = [0, 0]
capacity_ub = [15, 16]
steelprice = 58
scenarios = [0, 1, 2, 3]
pscenario = [0.25, 0.25, 0.25, 0.25]
wrenchearnings = [160, 160, 90, 90]
plierearnings = [100, 100, 100, 100]
capassembly = [8, 10, 8, 10]
production = [(j, i) for j in scenarios for i in products]
pricescenario = [[wrenchearnings[j], plierearnings[j]] for j in scenarios]
priceitems = [item for sublist in pricescenario for item in sublist]
# create dictionaries for the parameters
price_dict = dict(zip(production, priceitems))
capacity_dict = dict(zip(products, capacity_ub * 4))
steel_dict = dict(zip(products, steel))
molding_dict = dict(zip(products, molding))
assembly_dict = dict(zip(products, assembly))
# Create variables and parameters as dictionaries
production_vars = pulp.LpVariable.dicts(
"production", (scenarios, products), lowBound=0, cat="Continuous"
)
steelpurchase = pulp.LpVariable("steelpurchase", lowBound=0, cat="Continuous")
# Create the 'gemstoneprob' variable to specify
gemstoneprob = pulp.LpProblem("The Gemstone Tool Problem", pulp.LpMaximize)
# The objective function is added to 'gemstoneprob' first
gemstoneprob += (
pulp.lpSum(
[
pscenario[j] * (price_dict[(j, i)] * production_vars[j][i])
for (j, i) in production
]
- steelpurchase * steelprice
),
"Total cost",
)
for j in scenarios:
gemstoneprob += pulp.lpSum(
[steel_dict[i] * production_vars[j][i] for i in products]
) - steelpurchase <= 0, ("Steel capacity" + str(j))
gemstoneprob += pulp.lpSum(
[molding_dict[i] * production_vars[j][i] for i in products]
) <= capmolding, ("molding capacity" + str(j))
gemstoneprob += pulp.lpSum(
[assembly_dict[i] * production_vars[j][i] for i in products]
) <= capassembly[j], ("assembly capacity" + str(j))
for i in products:
gemstoneprob += production_vars[j][i] <= capacity_dict[i], (
"capacity " + str(i) + str(j)
)
# Print problem
print(gemstoneprob)
# The problem data is written to an .lp file
gemstoneprob.writeLP("gemstoneprob.lp")
# The problem is solved using PuLP's choice of Solver
gemstoneprob.solve()
# The status of the solution is printed to the screen
print("Status:", pulp.LpStatus[gemstoneprob.status])
# OUTPUT
# Each of the variables is printed with it's resolved optimum value
for v in gemstoneprob.variables():
print(v.name, "=", v.varValue)
production = [v.varValue for v in gemstoneprob.variables()]
# The optimised objective function value is printed to the console
print("Total price = ", pulp.value(gemstoneprob.objective))
|
