# 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.

### 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.

`$\begin{array}{ccccc}Ingot& Weight\phantom{\rule{0.2777777777777778em}{0ex}}in\phantom{\rule{0.2777777777777778em}{0ex}}Tons& %\phantom{\rule{0.2777777777777778em}{0ex}}Carbon& %\phantom{\rule{0.2777777777777778em}{0ex}}Molybdenum& \frac{Cost}{Ton}\\ 1& 5& 5& 3& 350\\ 2& 3& 4& 3& 330\\ 3& 4& 5& 4& 310\\ 4& 6& 3& 4& 280\end{array}$`

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.

`$\begin{array}{cccc}Alloy& %\phantom{\rule{0.2777777777777778em}{0ex}}Carbon& %\phantom{\rule{0.2777777777777778em}{0ex}}Molybdenum& \frac{Cost}{Ton}\\ 1& 8& 6& 500\\ 2& 7& 7& 450\\ 3& 6& 8& 400\\ Scrap& 3& 9& 100\end{array}$`

### 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. 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.

`sol.ingots`
```ans = 4×1 1.0000 1.0000 0 1.0000 ```
`sol.alloys`
```ans = 3×1 7.2500 0 0.2500 ```
`sol.scrap`
```ans = 3.5000 ```
`fval`
```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.