In my most recent post on linear programming I applied PuLP for solving below linear optimization problem, using two approaches. Approach #1 was based on solving a sub-problem with one objective only first, then adding the optimal outcome of that problem to a second sub-problem that considered objective two only. Approach #2 is based on combining all objectives into one overall objective function, combining single objectives with scalar weights.
In this post I want to show an alternative approach for solving below multi-objective linear optimization problem: I will use approach #1, but apply weights instead:
The approach I now use is to solve the problem in two steps, where a sub-problem with one objective only is solved first and its optimal outcome is added to a second sub-problem with only the the second objective as a constraint. This approach has been demonstrated by before, but this time I will apply a scalar weight-factor to that constraint, considering the optimal outcome of sub-problem with 0-100%.
Starting with the first objective, the first sub-problem to solve would be the following:
The optimal objective value to above sub-problem is 30.
After having solved the first sub-problem, the second objective would be considered by a second sub-problem adding the optimal outcome to above problem as a weighted constraint:
Here is the implementation in Python, using the PuLP module and applying a step-size of 0.01:
# import PuLP for modelling and solving problems import pulp # import matplotlib.pyplot for visualization import matplotlib.pyplot as plt # import pandas and numpy for being able to store solutions in DataFrame import numpy as np import pandas as pd # define step-size stepSize = 0.01 # initialize empty DataFrame for storing optimization outcomes solutionTable = pd.DataFrame(columns=["beta","x1_opt","x2_opt","obj_value"]) # declare optimization variables using PuLP and LpVariable x1 = pulp.LpVariable("x1",lowBound=0) x2 = pulp.LpVariable("x2",lowBound=0) # model and solve sub-problem no. 1 linearProblem = pulp.LpProblem("First sub-problem",pulp.LpMaximize) linearProblem += 2*x1 + 3*x2 # add objective no. 1 linearProblem += x1 + x2 <= 10 # add contraints from original problem statement linearProblem += 2*x1 + x2 <= 15 solution = linearProblem.solve() # store optimal outcome of sub-problem no. 1 into variable optimalObj1 = pulp.value(linearProblem.objective) # iterate through beta values from 0 to 1 with stepSize, and write PuLP solutions into solutionTable for i in range(0,101,int(stepSize*100)): # declare the problem again linearProblem = pulp.LpProblem("Multi-objective linear maximization",pulp.LpMaximize) # add the second objective as objective function to this sub-problem linearProblem += 4*x1-2*x2 # add the constraints from original problem statement linearProblem += x1 + x2 <= 10 linearProblem += 2*x1 + x2 <= 15 # add additional constraint at level beta, considering optimal outcome of sub-problem no. 1 linearProblem += 2*x1 + 3*x2 >= (i/100)*optimalObj1 # solve the problem solution = linearProblem.solve() # write solutions into DataFrame solutionTable.loc[int(i/(stepSize*100))] = [i/100, pulp.value(x1), pulp.value(x2), pulp.value(linearProblem.objective)] # visualize optimization outcome, using matplotlib.pyplot # -- set figure size plt.figure(figsize=(20,10)) # -- create line plot plt.plot(solutionTable["beta"],solutionTable["obj_value"],color="red") # -- add axis labels plt.xlabel("beta",size=20) plt.ylabel("obj_value",size=20) # -- add plot title plt.title("Optimal outcome of sub-problem 2, depending on beta",size=32) # -- show plot plt.show()
The results indicate that there is some range for beta where an increase in beta will not affect the optimal outcome of sub-problem 2, i.e. will not affect objective 2.
Data scientist focusing on simulation, optimization and modeling in R, SQL, VBA and Python