Kiofor
commented
1 year ago I have the same issue. Running a very simple LP:
import pulp
prob = pulp.LpProblem("test080", pulp.LpMinimize)
x = pulp.LpVariable("x", 0, 5)
y = pulp.LpVariable("y", -1, 1)
z = pulp.LpVariable("z", 0)
c1 = x+y <= 5
c2 = x+z >= 10
c3 = -y+z == 7
prob += x + 4*y + 9*z, "obj"
prob += c1, "c1"
prob += c2,"c2"
prob += c3,"c3"
prob.solve()
print(c1.pi, c2.pi, c3.pi)I get None None None as result.
Python 3.9.13
pulp 2.6.0
However, when I downgrade to Python 3.7.7 and pulp 1.6.8, this MWE works with no issues.
What is your question
Hello there,
I was trying to run the same code that my teacher uploaded since i was expecting the same result but the sensitivity analysis that i run gave me "none" output as slack variable and so. Is there any thing that i missed?
Import PuLP modeler functions
from pulp import *
Creates a list of the Ingredients
Ingredients = ['CHICKEN', 'BEEF', 'MUTTON', 'RICE', 'WHEAT', 'GEL']
A dictionary of the costs of each of the Ingredients is created
costs = {'CHICKEN': 0.013, 'BEEF': 0.008, 'MUTTON': 0.010, 'RICE': 0.002, 'WHEAT': 0.005, 'GEL': 0.001}
A dictionary of the protein percent in each of the Ingredients is created
proteinPercent = {'CHICKEN': 0.100, 'BEEF': 0.200, 'MUTTON': 0.150, 'RICE': 0.000, 'WHEAT': 0.040, 'GEL': 0.000}
A dictionary of the fat percent in each of the Ingredients is created
fatPercent = {'CHICKEN': 0.080, 'BEEF': 0.100, 'MUTTON': 0.110, 'RICE': 0.010, 'WHEAT': 0.010, 'GEL': 0.000}
A dictionary of the fibre percent in each of the Ingredients is created
fibrePercent = {'CHICKEN': 0.001, 'BEEF': 0.005, 'MUTTON': 0.003, 'RICE': 0.100, 'WHEAT': 0.150, 'GEL': 0.000}
A dictionary of the salt percent in each of the Ingredients is created
saltPercent = {'CHICKEN': 0.002, 'BEEF': 0.005, 'MUTTON': 0.007, 'RICE': 0.002, 'WHEAT': 0.008, 'GEL': 0.000}
Create the 'prob' variable to contain the problem data
prob = LpProblem("Blending_Model", LpMinimize)
A dictionary called 'ingredient_vars' is created to contain the referenced Variables
ingredient_vars = LpVariable.dicts("Ingr",Ingredients,0)
The objective function is added to 'prob' first
ingredient_vars
The objective function is added to 'prob' first
prob += lpSum([costs[i]*ingredient_vars[i] for i in Ingredients]), "Total Cost of Ingredients per 100 gr can"
The five constraints are added to 'prob'
prob += lpSum([ingredient_vars[i] for i in Ingredients]) == 100, "GramsSum" prob += lpSum([proteinPercent[i] ingredient_vars[i] for i in Ingredients]) >= 8.0, "ProteinRequirement" prob += lpSum([fatPercent[i] ingredient_vars[i] for i in Ingredients]) >= 6.0, "FatRequirement" prob += lpSum([fibrePercent[i] ingredient_vars[i] for i in Ingredients]) <= 2.0, "FibreRequirement" prob += lpSum([saltPercent[i] ingredient_vars[i] for i in Ingredients]) <= 0.4, "SaltRequirement"
prob
The problem is solved using PuLP's choice of Solver
prob.solve()
Each of the variables is printed with it's resolved optimum value
for v in prob.variables(): print(v.name, "=", v.varValue)
The optimised objective function value is printed to the screen
print("Total Cost the Cat Food = ", value(prob.objective))
import pandas as pd
Report Sensitivity Analysis
print("\nSensitivity Analysis") Cons_Sensitivity_Report=[{'Constraint_Name':name,'Slack':c.slack,"Shadow_Price":c.pi} for name, c in prob.constraints.items()] print(pd.DataFrame(Cons_Sensitivity_Report)) print("\n") Variable_Sensitivity_Report = [{'Variable_Name': v.name, 'Value':v.varValue,'Reduced_Cost': v.dj} for v in prob.variables()]
print(pd.DataFrame(Variable_Sensitivity_Report)[['Variable_Name','Value','Reduced_Cost']])
Sensitivity analysis report output is looks like below: Sensitivity Analysis Constraint_Name Slack Shadow_Price 0 GramsSum None None 1 ProteinRequirement None None 2 FatRequirement None None 3 FibreRequirement None None 4 SaltRequirement None None
Variable_Name Value Reduced_Cost 0 Ingr_BEEF 60.0 None 1 Ingr_CHICKEN 0.0 None 2 Ingr_GEL 40.0 None 3 Ingr_MUTTON 0.0 None 4 Ingr_RICE 0.0 None 5 Ingr_WHEAT 0.0 None