Solver-Based - MATLAB & Simulink (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.

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

To formulate the problem, first decide on the control variables. Take variable x(1)=1 to mean you purchase ingot 1, and x(1)=0 to mean you do not purchase the ingot. Similarly, variables x(2) through x(4) are binary variables indicating whether you purchase ingots 2 through 4.

Variables x(5) through x(7) are the quantities in tons of alloys 1, 2, and 3 that you purchase, and x(8) is the quantity of scrap steel that you purchase.

MATLAB® Formulation

Formulate the problem by specifying the inputs for intlinprog. The relevant intlinprog syntax is:

[x,fval] = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub)

Create the inputs for intlinprog from the first (f) through the last (ub).

f is the vector of cost coefficients. The coefficients representing the costs of ingots are the ingot weights times their cost per ton.

f = [350*5,330*3,310*4,280*6,500,450,400,100];

The integer variables are the first four.

intcon = 1:4;

Tip: To specify binary variables, set the variables to be integers in intcon, and give them a lower bound of 0 and an upper bound of 1.

The problem has no linear inequality constraints, so A and b are empty matrices ([]).

A = [];b = [];

The problem has three equality constraints. The first is that the total weight is 25 tons.

5*x(1) + 3*x(2) + 4*x(3) + 6*x(4) + x(5) + x(6) + x(7) + x(8) = 25

The second constraint is that the weight of carbon is 5% of 25 tons, or 1.25 tons.

5*0.05*x(1) + 3*0.04*x(2) + 4*0.05*x(3) + 6*0.03*x(4)

+ 0.08*x(5) + 0.07*x(6) + 0.06*x(7) + 0.03*x(8) = 1.25

The third constraint is that the weight of molybdenum is 1.25 tons.

5*0.03*x(1) + 3*0.03*x(2) + 4*0.04*x(3) + 6*0.04*x(4)

+ 0.06*x(5) + 0.07*x(6) + 0.08*x(7) + 0.09*x(8) = 1.25

Specify the constraints, which are Aeq*x = beq in matrix form.

Aeq = [5,3,4,6,1,1,1,1; 5*0.05,3*0.04,4*0.05,6*0.03,0.08,0.07,0.06,0.03; 5*0.03,3*0.03,4*0.04,6*0.04,0.06,0.07,0.08,0.09];beq = [25;1.25;1.25];

Each variable is bounded below by zero. The integer variables are bounded above by one.

lb = zeros(8,1);ub = ones(8,1);ub(5:end) = Inf; % No upper bound on noninteger variables

Solve Problem

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

[x,fval] = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);

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 0.00 (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.

x,fval

x = 8×1 1.0000 1.0000 0 1.0000 7.2500 0 0.2500 3.5000

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.

Set intcon = [] to see the effect of solving the problem without integer constraints. The solution is different, and is not realistic, because you cannot purchase a fraction of an ingot.