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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?

Comments for Manufacturing Constraints

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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


math - variables x and y





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

math - manufacturing - linear equation - total machine time used





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

math - manufacturing - linear equation - Cost Function





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May 27, 2013
Linear Programming
by: Staff

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Part II


Constraints

- 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

math - manufacturing - linear equation - machine time function










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May 27, 2013
Linear Programming
by: Staff


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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

math - manufacturing - linear equation - machine cost function - minimize





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)


math - manufacturing - linear equation - plot constraints









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May 27, 2013
Linear Programming
by: Staff


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Part IV


• The corner points are (in x,y format):

(0,0), (0,5), (6,3), (7.5,0)


math - manufacturing - linear equations - corner points





• 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


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