Multi-objective knapsack
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This tutorial was generated using Literate.jl. Download the source as a .jl file.
The purpose of this tutorial is to demonstrate how to create and solve a multi-objective linear program. In addition, it demonstrates how to work with solvers which return multiple solutions.
Required packages
This tutorial requires the following packages:
using JuMP
import HiGHS
import MultiObjectiveAlgorithms as MOA
import PlotsMultiObjectiveAlgorithms.jl is a package which implements a variety of algorithms for solving multi-objective optimization problems. Because it is a long package name, we import it instead as MOA.
Formulation
The knapsack problem is a classic problem in mixed-integer programming. Given a collection of items , each of which has an associated weight, , and profit, , the knapsack problem determines which profit-maximizing subset of items to pack into a knapsack such that the total weight is less than a capacity . The mathematical formulation is:
where is if we pack item into the knapsack and otherwise.
For this tutorial, we extend the single-objective knapsack problem by adding another objective: given a desirability rating, , we wish to maximize the total desirability of the items in our knapsack. Thus, our mathematical formulation is now:
Data
The data for this example was taken from vOptGeneric, and the original author was @xgandibleux.
profit = [77, 94, 71, 63, 96, 82, 85, 75, 72, 91, 99, 63, 84, 87, 79, 94, 90]
desire = [65, 90, 90, 77, 95, 84, 70, 94, 66, 92, 74, 97, 60, 60, 65, 97, 93]
weight = [80, 87, 68, 72, 66, 77, 99, 85, 70, 93, 98, 72, 100, 89, 67, 86, 91]
capacity = 900
N = length(profit)17Comparing the capacity to the total weight of all the items:
capacity / sum(weight)0.6428571428571429shows that we can take approximately 64% of the items.
Plotting the items, we see that there are a range of items with different profits and desirability. Some items have a high profit and a high desirability, others have a low profit and a high desirability (and vice versa).
Plots.scatter(
profit,
desire;
xlabel = "Profit",
ylabel = "Desire",
legend = false,
)The goal of the bi-objective knapsack problem is to choose a subset which maximizes both objectives.
JuMP formulation
Our JuMP formulation is a direct translation of the mathematical formulation:
model = Model()
@variable(model, x[1:N], Bin)
@constraint(model, sum(weight[i] * x[i] for i in 1:N) <= capacity)
@expression(model, profit_expr, sum(profit[i] * x[i] for i in 1:N))
@expression(model, desire_expr, sum(desire[i] * x[i] for i in 1:N))
@objective(model, Max, [profit_expr, desire_expr])2-element Vector{AffExpr}:
77 x[1] + 94 x[2] + 71 x[3] + 63 x[4] + 96 x[5] + 82 x[6] + 85 x[7] + 75 x[8] + 72 x[9] + 91 x[10] + 99 x[11] + 63 x[12] + 84 x[13] + 87 x[14] + 79 x[15] + 94 x[16] + 90 x[17]
65 x[1] + 90 x[2] + 90 x[3] + 77 x[4] + 95 x[5] + 84 x[6] + 70 x[7] + 94 x[8] + 66 x[9] + 92 x[10] + 74 x[11] + 97 x[12] + 60 x[13] + 60 x[14] + 65 x[15] + 97 x[16] + 93 x[17]Note how we form a multi-objective program by passing a vector of scalar objective functions.
Solution
To solve our model, we need an optimizer which supports multi-objective linear programs. One option is to use the MultiObjectiveAlgorithms.jl package.
set_optimizer(model, () -> MOA.Optimizer(HiGHS.Optimizer))
set_silent(model)MultiObjectiveAlgorithms.jl supports many different algorithms for solving multiobjective optimization problems. One option is the epsilon-constraint method:
set_attribute(model, MOA.Algorithm(), MOA.EpsilonConstraint())Let’s solve the problem and see the solution
optimize!(model)
@assert termination_status(model) == OPTIMAL
solution_summary(model)* Solver : MOA[algorithm=MultiObjectiveAlgorithms.EpsilonConstraint, optimizer=HiGHS]
* Status
Result count : 9
Termination status : OPTIMAL
Message from the solver:
"Solve complete. Found 9 solution(s)"
* Candidate solution (result #1)
Primal status : FEASIBLE_POINT
Dual status : NO_SOLUTION
Objective value : [9.18000e+02,9.83000e+02]
Objective bound : [9.55000e+02,9.83000e+02]
Relative gap : 0.00000e+00
* Work counters
Solve time (sec) : 1.13743e-01
Simplex iterations : 0
Barrier iterations : 0
Node count : 0There are 9 solutions available. We can also use result_count to see how many solutions are available:
result_count(model)9Accessing multiple solutions
Access the nine different solutions in the model using the result keyword to solution_summary, value, and objective_value:
solution_summary(model; result = 5)* Solver : MOA[algorithm=MultiObjectiveAlgorithms.EpsilonConstraint, optimizer=HiGHS]
* Status
Result count : 9
Termination status : OPTIMAL
* Candidate solution (result #5)
Primal status : FEASIBLE_POINT
Dual status : NO_SOLUTION
Objective value : [9.36000e+02,9.42000e+02]@assert primal_status(model; result = 5) == FEASIBLE_POINT@assert is_solved_and_feasible(model; result = 5)objective_value(model; result = 5)2-element Vector{Float64}:
936.0
942.0Note that because we set a vector of two objective functions, the objective value is a vector with two elements. We can also query the value of each objective separately:
value(profit_expr; result = 5)936.0Visualizing objective space
Unlike single-objective optimization problems, multi-objective optimization problems do not have a single optimal solution. Instead, the solutions returned represent possible trade-offs that the decision maker can choose between the two objectives. A common way to visualize this is by plotting the objective values of each of the solutions:
plot = Plots.scatter(
[value(profit_expr; result = i) for i in 1:result_count(model)],
[value(desire_expr; result = i) for i in 1:result_count(model)];
xlabel = "Profit",
ylabel = "Desire",
title = "Objective space",
label = "",
xlims = (915, 960),
)
for i in 1:result_count(model)
y = objective_value(model; result = i)
Plots.annotate!(y[1] - 1, y[2], (i, 10))
end
ideal_point = objective_bound(model)
Plots.scatter!([ideal_point[1]], [ideal_point[2]]; label = "Ideal point")Visualizing the objective space lets the decision maker choose a solution that suits their personal preferences. For example, result #7 is close to the maximum value of profit, but offers significantly higher desirability compared with solutions #8 and #9.
The set of items that are chosen in solution #7 are:
items_chosen = [i for i in 1:N if value(x[i]; result = 7) > 0.9]11-element Vector{Int64}:
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