Problem-Based - MATLAB & Simulink - MathWorks Deutschland (2024)

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.

$$\begin{array}{ccccc}Ingot& WeightinTons& \%Carbon& \%Molybdenum& \frac{Cost}{Ton}\\ 1& 5& 5& 3& \$350\\ 2& 3& 4& 3& \$330\\ 3& 4& 5& 4& \$310\\ 4& 6& 3& 4& \$280\end{array}$$

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.

$$\begin{array}{cccc}Alloy& \%Carbon& \%Molybdenum& \frac{Cost}{Ton}\\ 1& 8& 6& \$500\\ 2& 7& 7& \$450\\ 3& 6& 8& \$400\\ Scrap& 3& 9& \$100\end{array}$$

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 1.6.0: Copyright (c) 2023 HiGHS under MIT licence termsPresolving model3 rows, 8 cols, 24 nonzeros3 rows, 8 cols, 18 nonzerosSolving 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 0.0sSolving 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.