Question

A company produces 3 widgets using two different machines. Every Widget A made requires 10 minutes of operation on machine 1 and 20 minutes of operation on machine 2. Widget A can be sold for $4 an item. Every Widget B requires 2 minutes of operation on machine 1 and 15 minutes of operation on machine 2. Widget B can be sold for $5 an item Each Widget C requires 30 minutes of operation on machine 1 and 8 minutes of operation on machine 2. Widget C can be sold for $2.

Each week 1500 minutes are available on machine A and 1500 hours are available on machine B. Demand for Widget B is extremely high, but demand for Widget A is a maximum of 35 items each week and demand for Widget C a maximum of 20 items each week.

Use the simplex method to solve for the amount of each widget is to be produced for a maximum profit each week.

Answer 1

Problem is Max Z = 4x + 5x + 2x subject to 10x + 2 X, + 30x, s 1500 20x + 15x, + 8 x 5 90000 $ 35 x 320 and X, X,X>0; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type'<' we should add slack variables 2. As the constraint-2 is of type '<' we should add slack variable S, 3. As the constraint-3 is of type's' we should add slack variables 4. As the constraint-4 is of type '<' we should add slack variable S. After introducing slack variables Max Z = 4x + 3x + 2x: +08, + OS, + OS + OS. subject to 10x + 2 x, + 30x, + S = 1500 20x + 15x, + 8 x + S = 90000 + S = 35 + S. = 20 and x , *.,*,5,5,5,5,20 Iteration-1 4 5 2 0 0 0 Min Ratio B X; x1 x2 3 s, S, Sg S4 X2 1500 S. 1500 10 (2) 30 1 0 0 0 = 750 2 $ 90000 20 15 8 0 1 0 0 90000 = 6000 15 S. 0 35 1 0 0 0 0 1 0 5. 0 20 0 1 0 0 1 Z= 0 0 0 0 0 0 0 0 2 2-c, -4 -51 0 0 0 0

Negative minimum Z-Cis-5 and its column index is 2. So, the entering variable is x, Minimum ratio is 750 and its row index is 1. So, the leaving basis variable is s. The pivot element is 2. Entering = x, Departing = S, Key Element = 2 + R (new) = R (old) = 2 + R (new) = R (old) - 15R, (new) + R (new) = R (old) + R (new) = R. (old) Iteration 2 4 5 2 0 0 0 0 B CE *[ X2 X3 S; MinRatio s 0.5 S2 0 54 0 5 5 1 15 0 X; 750 78750 35 0 -55 0 -217 -7.5 1 0 0 S. S S. 0 1 0 0 0 0 1 0 0 20 0 0 1 0 0 0 1 Z = 3750 25 5 75 0 0 0 2 2-0, 21 0 73 2.5 0 0 Since all Z -20 Hence, optimal solution is arrived with value of variables as: x = 0, x, = 750, x = 0 Max Z = 3750