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Mixed-Integer Linear Programming Basics: Problem-Based

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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.

Problem Description

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.

IngotWeightinTons%Carbon%MolybdenumCostTon1553$3502343$3303454$3104634$280

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.

Alloy%Carbon%MolybdenumCostTon186$500277$450368$400Scrap39$100

Formulate Problem

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(2) through ingots(4) are binary variables indicating whether you purchase ingots 2 through 4.

Variables alloys(1) through 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;

Solve Problem

Now that you have all the inputs, call the solver.

[sol,fval] = solve(steelprob);
Solving problem using intlinprog.
Running HiGHS 1.7.0: Copyright (c) 2024 HiGHS under MIT licence terms
Coefficient ranges:
  Matrix [3e-02, 6e+00]
  Cost   [1e+02, 2e+03]
  Bound  [1e+00, 1e+00]
  RHS    [1e+00, 2e+01]
Presolving model
3 rows, 8 cols, 24 nonzeros  0s
3 rows, 8 cols, 18 nonzeros  0s

Solving MIP model with:
   3 rows
   8 cols (4 binary, 0 integer, 0 implied int., 4 continuous)
   18 nonzeros

        Nodes      |    B&B Tree     |            Objective Bounds              |  Dynamic Constraints |       Work      
     Proc. InQueue |  Leaves   Expl. | BestBound       BestSol              Gap |   Cuts   InLp Confl. | LpIters     Time

         0       0         0   0.00%   0               inf                  inf        0      0      0         0     0.0s
         0       0         0   0.00%   8125.6          inf                  inf        0      0      0         4     0.0s
 R       0       0         0   0.00%   8495            8495               0.00%        5      0      0         5     0.0s

Solving report
  Status            Optimal
  Primal bound      8495
  Dual bound        8495
  Gap               0% (tolerance: 0.01%)
  Solution status   feasible
                    8495 (objective)
                    0 (bound viol.)
                    0 (int. viol.)
                    0 (row viol.)
  Timing            0.01 (total)
                    0.00 (presolve)
                    0.00 (postsolve)
  Nodes             1
  LP iterations     5 (total)
                    0 (strong br.)
                    1 (separation)
                    0 (heuristics)

Optimal solution found.

Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 1e-06. The intcon variables are integer within tolerance, options.ConstraintTolerance = 1e-06.

View the solution.

sol.ingots
ans = 4×1

     1
     1
     0
     1


sol.alloys
ans = 3×1

    7.2500
         0
    0.2500


sol.scrap
ans = 
3.5000

fval
fval = 
8495

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.

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