An engineering factory can produce five types of product (PROD 1, PROD 2,... PROD 5) by using two production processes: grinding and drilling. After deducting raw material costs, each unit of each product yields the following contributions to profit:

Each unit requires a certain time on each process. These are given below (in hours). A dash indicates when a process is not needed.

In addition, the final assembly of each unit of each product uses 20 hours of an employee’s time. The factory has three grinding machines and two drilling machines and works a six-day week with two shifts of 8 hours on each day. Eight workers are employed in assembly, each working one shift a day. The problem is to find how much of each product is to be manufactured so as to maximize the total profit contribution. The minimum demand for each product are as follows

The product mix problem constitutes of 5 design variables, 5 equality and 5 inequality linear constraints. In order to solve this problem the default termination criteria are used. This is an example for primal infeasible problem.

Mathematical Formulation:

.

1. Grinding available per week

2. Drilliing capacity available per week

3. Labour capacity available per week

4. Demand for each product

clc; 
nProducts = 5;
nMachines = [3 2];
nWorkers = 8;
nDays = 6;
nHours = 8;
nMachineShift = 2;
nWorkerShift = 1;
prodProfit = [550 600 350 400 200]';
ProcessingTime = [12 20 0 25 15; 10 8 16 0 0; 20*ones(1,nProducts)];

// Linear inequalities
A1 = ProcessingTime;
b1 = (nDays*nHours)*[nMachines*nMachineShift nWorkers*nWorkerShift]';
// Demand constraints
A2 = [-1 0 0 0 0;0 -1 0 0 0;0 0 -1 0 0; 0 0 0 -1 0; 0 0 0 0 -1];
b2 = [-5 -5 -2 -7 -2]';

A = [A1;A2];
b = [b1;b2];
lb = zeros(1,nProducts); 
Cost = -prodProfit;
[xopt,fopt,exitflag,output,lambda]=linprog(Cost, A, b, [], [], lb,[]);

// Result representation
if exitflag == 0 then
    disp(" Optimal Solution Found")
    P = [];
    for p = 1:nProducts
        P = [P strcat(["PROD",string(p)])];
    end

    disp([P;string(xopt')]);
    disp(["The optimal cost is ", string(fopt)])
elseif exitflag == 1 then
    nViolatedConstraints = sum(A*xopt-b>0)
    if nViolatedConstraints
        disp(strcat(["No of constraints violated : ", string(nViolatedConstraints)]));
    end
elseif exitflag == 2 then
    disp("Dual Infeasible")
else
    disp("Technical error encountered")
end

Expected Output:

Primal Infeasible.
  
 No of constraints violated : 2