# Manufacturing Constraints

Linear Programming - Manufacturing

A manufacturing company produces plastic plates and cups.

Both products require time on two different machines.

Producing a run of plastic plates requires one hour on Machine A and two hours on Machine B.

Producing a run of cups requires three hours on Machine A and one hour on Machine B.

Each machine is operated a maximum of 15 hours per day.

It costs the company \$30 on each run of plastic plates and \$22 on each run of cups.

How many runs of each should be produced in a day in order to minimize costs?

What are the constraints for this problem?

 May 27, 2013 Linear Programming by: Staff Answer Part I A manufacturing company produces two products: plastic plates and cups. x = number of runs of plastic plates manufactured y = number of runs of cups manufactured Both products require time on two different machines. m₁ = total time used on Machine A m₂ = total time used on Machine B T = total machine time used = m₁ + m₂ The manufacturing cost is for each product is different c₁ = cost for each run of plastic plates = \$30 c₂ = cost for each run of cups = \$22 P = c₁x + c₂y P = 30x + 22y -----------------------------------------------------------------

 May 27, 2013 Linear Programming by: Staff -----------------------------------------------------------------Part IIConstraints - run of plastic plates requires: 1 hour on Machine A 2 hours on machine B - run of cups requires: 3 hours on Machine A 1 hour on Machine B • Machine Time: - maximum machine time available 15 hours available on machine A (1 hour) * (number of runs of plastic plates) + (3 hours) * (number of runs of cups) ≤ 15 hours 1x + 3y ≤ 15 15 hours available on machine B (2 hours) * (number of runs of plastic plates) + (1 hour) * (number of runs of cups) ≤ 15 hours 2x + 1y ≤ 15 The objective function: • maximize machine time utilized: - Machine A: m₁ = 1x + 3y - Machine B: m₂ = 2x + 1y T = total machine time used = m₁ + m₂ T = (1x + 3y) + (2x + 1y) T = 3x + 4y The system of linear equations to maximize the machine time used: • Solve for the values of x, and y which maximize the following objective function (maximize the total machine time): T = 3x + 4y • Subject to the following constraints: 1x + 3y ≤ 15 2x + 1y ≤ 15 x ≥ 0 y ≥ 0 -----------------------------------------------------------------

 May 27, 2013 Linear Programming by: Staff ----------------------------------------------------------------- Part III The system of linear equations to minimize the cost: • Solve for the values of x, and y which minimize the following objective function (minimize the total manufacturing cost): P = 30x + 22y • Subject to the following constraints: 1x + 3y ≤ 15 2x + 1y ≤ 15 x ≥ 0 y ≥ 0 Since there are only two independent variables to solve for (x and y), a graphical solution is the best way. This is shown below. If there were more than two independent variables (for example: x, y, z, and q), a graphical solution is not possible. In that case, you should use the Gaussian Elimination method. A graphical solution using corner points Plot the restrictions: 1x + 3y ≤ 15 (the red line and all the area below the red line) 2x + 1y ≤ 15 (the green line and all the area below the green line) x ≥ 0 (all the area to the right of the vertical blue line, which is the “y-axis”) y ≥ 0 (all the area above the horizontal pink line, which is the “x-axis”) • When these boundaries are plotted, the area inside the boundaries forms the Bounded Feasible Region: Bounded Feasible Region (shaded yellow area) -----------------------------------------------------------------

 May 27, 2013 Linear Programming by: Staff ----------------------------------------------------------------- Part IV • The corner points are (in x,y format): (0,0), (0,5), (6,3), (7.5,0) • Find the minimum value of P by computing the values of P at the corner points. P = 30x + 22y Corner point: (0,0) P = 30 * 0 + 22 * 0 P = 0 + 0 P = 0, minimum value of P Corner point: (0,5) P = 30 * 0 + 22 * 5 P = 0 + 110 P = 110 Corner point: (6,3) P = 30 * 6 + 22 * 3 P = 180 + 66 P = 246, maximum value of P Corner point: (7.5,0) P = 30 * 7.5 + 22 * 0 P = 225 + 0 P = 225 • Find the maximum value of T by computing the values of T at the corner points. T = 3x + 4y Corner point: (0,0) T = 3 * 0 + 4 * 0 T = 0 + 0 T = 0, minimum value of T Corner point: (0,5) T = 3 * 0 + 4 * 5 T = 0 + 20 T = 20 Corner point: (6,3) T = 3 * 6 + 4 * 3 T = 18 + 12 T = 30 , maximum value of T Corner point: (7.5,0) T = 3 * 7.5 + 4 * 0 T = 22.5 + 0 T = 22.5 Final Answer Minimum cost occurs when the machines are not used: x = 0, y = 0 Maximum cost of \$246 per day occurs when: x = 6, y = 3 Minimum machine utilization time occurs when the machines are not used: x = 0, y = 0 Maximum machine utilization time of 30 hours per day occurs when: x = 6, y = 3 Thanks for writing. Staff www.solving-math-problems.com