Mixed-Integer Linear Programming Basics: Problem-Based
This example shows how to solve a mixed-integer linear problem. Although not complex, the example shows the typical steps in formulating a problem using the problem-based approach. For a video showing this example, see Solve a Mixed-Integer Linear Programming Problem using Optimization Modeling.
For the solver-based approach to this problem, see Mixed-Integer Linear Programming Basics: Solver-Based.
You want to blend steels with various chemical compositions to obtain 25 tons of steel with a specific chemical composition. The result should have 5% carbon and 5% molybdenum by weight, meaning 25 tons*5% = 1.25 tons of carbon and 1.25 tons of molybdenum. The objective is to minimize the cost for blending the steel.
This problem is taken from Carl-Henrik Westerberg, Bengt Bjorklund, and Eskil Hultman, “An Application of Mixed Integer Programming in a Swedish Steel Mill.” Interfaces February 1977 Vol. 7, No. 2 pp. 39–43, whose abstract is at https://doi.org/10.1287/inte.7.2.39.
Four ingots of steel are available for purchase. Only one of each ingot is available.
Three grades of alloy steel and one grade of scrap steel are available for purchase. Alloy and scrap steels can be purchased in fractional amounts.
To formulate the problem, first decide on the control variables. Take variable
ingots(1) = 1 to mean that you purchase ingot 1, and
ingots(1) = 0 to mean that you do not purchase the ingot. Similarly, variables
ingots(4) are binary variables indicating whether you purchase ingots 2 through 4.
alloys(3) are the quantities in tons of alloys 1, 2, and 3 that you purchase.
scrap is the quantity of scrap steel that you purchase.
steelprob = optimproblem; ingots = optimvar('ingots',4,'Type','integer','LowerBound',0,'UpperBound',1); alloys = optimvar('alloys',3,'LowerBound',0); scrap = optimvar('scrap','LowerBound',0);
Create expressions for the costs associated with the variables.
weightIngots = [5,3,4,6]; costIngots = weightIngots.*[350,330,310,280]; costAlloys = [500,450,400]; costScrap = 100; cost = costIngots*ingots + costAlloys*alloys + costScrap*scrap;
Include the cost as the objective function in the problem.
steelprob.Objective = cost;
The problem has three equality constraints. The first constraint is that the total weight is 25 tons. Calculate the weight of the steel.
totalWeight = weightIngots*ingots + sum(alloys) + scrap;
The second constraint is that the weight of carbon is 5% of 25 tons, or 1.25 tons. Calculate the weight of the carbon in the steel.
carbonIngots = [5,4,5,3]/100; carbonAlloys = [8,7,6]/100; carbonScrap = 3/100; totalCarbon = (weightIngots.*carbonIngots)*ingots + carbonAlloys*alloys + carbonScrap*scrap;
The third constraint is that the weight of molybdenum is 1.25 tons. Calculate the weight of the molybdenum in the steel.
molybIngots = [3,3,4,4]/100; molybAlloys = [6,7,8]/100; molybScrap = 9/100; totalMolyb = (weightIngots.*molybIngots)*ingots + molybAlloys*alloys + molybScrap*scrap;
Include the constraints in the problem.
steelprob.Constraints.conswt = totalWeight == 25; steelprob.Constraints.conscarb = totalCarbon == 1.25; steelprob.Constraints.consmolyb = totalMolyb == 1.25;
Now that you have all the inputs, call the solver.
[sol,fval] = solve(steelprob);
Solving problem using intlinprog. LP: Optimal objective value is 8125.600000. Cut Generation: Applied 3 mir cuts. Lower bound is 8495.000000. Relative gap is 0.00%. Optimal solution found. Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within tolerance, options.IntegerTolerance = 1e-05 (the default value).
View the solution.
ans = 4×1 1.0000 1.0000 0 1.0000
ans = 3×1 7.2500 0 0.2500
ans = 3.5000
fval = 8.4950e+03
The optimal purchase costs $8,495. Buy ingots 1, 2, and 4, but not 3, and buy 7.25 tons of alloy 1, 0.25 ton of alloy 3, and 3.5 tons of scrap steel.