Question

JB Electronics

JB Electronics manufactures three components used to produce cell phones and other communication devices. In a given production period, demand for the three components may exceed JB's manufacturing capacity. In this case, the company meets demand by purchasing the components from another manufacturer at an increased cost per unit. JB's manufacturing cost per unit and purchasing cost per unit for the three components are as follows:

Source	Component 1	Component 2	Component 3
Manufacture	$          4.25	$            4.85	$          3.95
Purchase	$          6.50	$            7.80	$          6.25

The components that JB manufactures require time in their production, assembly and testing departments. The time required for each component and production department are shown in the table below. For instance, each unit of Component 1 that JB manufactures requires 2.50 minutes of production time, 1.25 minutes of assembly time, and 1 minute of testing time. For the next production period, JB has capacities of 300 hours in the production department, 240 hours in the assembly department, and 200 hours in the testing. JB must satisfy demand of 5000, 4500 and 3500 units for components 1, 2 and 3, respectively, while minimizing total manufacturing and purchasing costs.

Department	Component 1	Component 2	Component 3
Production	2.50	2.25	3.75
Assembly	1.25	1.25	3.25
Testing	1.00	1.50	2.00

1. In the JB electronics problem, how much would the company save if they could increase time in their production line to 20,000

2. In the JB electronics problem, according to the optimal solution, what is the total cost of manufacturing components?

Answer 1

Let the No. of Component 1 manufactured be C1m, purchased be C1p

Let the No. of Component 2 manufactured be C2m, purchased be C2p

Let the No. of Component 3 manufactured be C3m, purchased be C3p

Total Cost = 4.25*C1m + 4.85*C2m + 3.95*C3m + 6.50*C1p + 7.80*C2p + 6.25*C3p

We have to minimize this cost

Subject to constraints

2.50*C1m + 2.25*C2m + 3.75*C3m <= 300 hours * 60 minutes i.e. 18000..........Constraint for availability of production time

1.25*C1m + 1.25*C2m + 3.25*C3m <= 240 * 60 = 14,400 ..........Constraint for availability of assembly time

1.00*C1m + 1.50*C2m + 2.00*C3m <= 200*60 = 12000..........Constraint for availability of testing time

C1m + C1p >= 5000..........Constraint for demand of component 1

C2m + C2p >= 4500..........Constraint for demand of component 2

C3m + C3p >= 3500..........Constraint for demand of component 3

Hence, we get formulation as

Minimize Total Cost C = 4.25*C1m + 4.85*C2m + 3.95*C3m + 6.50*C1p + 7.80*C2p + 6.25*C3p

Subject to Constraints:

2.50*C1m + 2.25*C2m + 3.75*C3m <= 18000

1.25*C1m + 1.25*C2m + 3.25*C3m <= 14,400

1.00*C1m + 1.50*C2m + 2.00*C3m <= 12000

C1m + C1p >= 5000

C2m + C2p >= 4500

C3m + C3p >= 3500

C1m, C2m, C3m, C1p, C2p, C3p >=0

We solve in Excel using Excel solver as shown below:

The above table in the form of formulas along with Excel Solver extract is shown below for better understanding and reference:

As seen from above, total cost = $69,112.50

If the production time is increased to 20,000 minutes, we get

Total cost = $67,312.50

Saving in cost = 69,112.50 - 67,312.50 = $1,800

Hence, the company would save $1800 if they could increase time in their production line to 20,000 minutes

b. Total manufacturing cost of initial solution = C1m + cost per unit + C2m * Cost per unit + C3m * Cost per unit = 3150*4.25 + 4500*4.85 * 0*3.95 = 13,387.5 + 21,825‬ = $35,212.5‬

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