Question

A company produces and sells three types of products A, B, and C. The table below shows the selling price per unit and the total demand for each product type.

	$/unit	Demand
A	5	300
B	6	250
C	7	200

For this purpose, the company can commission any of 7 plants with varying costs. Furthermore, each plant has a fixed capacity (in units) regardless of the type of product made. Production costs and capacities for each of the six plants are shown in the table below

Plant	$/unit			Capacity
	A	B	C
1	2.5	1.5	2.1	200
2	2.1	2.5	1.4	250
3	2.2	2.2	1.6	300
4	1.9	2.1	2.3	400
5	1.8	1.6	1.5	500
6	2.1	1.5	2.1	750

• Write an optimization model that solves the company's problem maximizing its profit while meeting the demand.

Consider now a fixed rent for each plant that is producing as shown in the table below

Plant	1	2	3	4	5	6
Rent, $	120	240	270	450	420	620

• What would be the new optimal solution? (make the necessary adjustments to the model)

Furthermore, consider each plant must produce a single product, but products can be commissioned from as many as necessary to meet demand.

• Make the necessary adjustments and give the new optimal solution.

Finally, if the plant can produce as many as two different products paying a $50 fee to change configurations between them,

• what would be the new optimal solution? (make the necessary adjustments to the model).

Answer 1

PRODUCT	$/UNIT	DEMAND
A	5	300
B	6	250
C	7	200
TOTAL REVENUE	4400
		(5*300+6*250+7*200)

PLANT	A	B	C	CAPACITY
1	2.5	1.5	2.1	200
2	2.1	2.5	1.4	250
3	2.2	2.2	1.6	300
4	1.9	2.1	2.3	400
5	1.8	1.6	1.5	500
6	2.1	1.5	2.1	750
MINIMUM	1.8	1.5	1.4

A should be produced in plant 5. Cost to produce 300 is 300*1.8 = 480

B should be produced in plant 1 and 6. Cost to product 250 is 250*1.5 = 375

C should be produced in plant 2. Cost to produce 200 is 200*1.4 = 280

Total costs = 480 + 375 + 280 = 1135

Profit = 4400 - 1135 = 3265

(NEXT CASE)

CAPACITY	RENT	RENT/UNIT
200	120	0.60
250	240	0.96
300	270	0.90
400	450	1.13
500	420	0.84
750	620	0.83

ABOVE RENT PER UNIT IS CALCULATED CONSIDERING THAT THE PLANT IS RUNNING AT FULL CAPACITY.

ADDING THE SAME TO PER UNIT VARIABLE COSTS GIVES THE FOLLOWING TABLE:

	300	250	200
PLANT	A	B	C	CAPACITY
1	3.10	2.10	2.70	200
2	3.06	3.46	2.36	250
3	3.10	3.10	2.50	300
4	3.03	3.23	3.43	400
5	2.64	2.44	2.34	500
6	2.93	2.33	2.93	750
MINIMUM	2.64	2.1	2.34

This table suggests that to fulfill, 750 capacity different cases can be made.

Plant 5 is suitable to manufacture both A and C, as the demand can be met as well as the cost is least. But, plant 1 is not sufficient to meet the demand for product B as its demand is 250 and capacity for plant A is 200 only.

So, following cases can be made.

1. Use Plant 6 to manufacture all 3.

2. Use Plant 5 to manufacture A and C , along with some other plant to manufacture B.

1. Use Plant 6 to manufacture all 3.

FC(Plant 6) = 620

VCA=2.1*300=630

VCB=1.5*250=375

VCC=2.1*200=420

TVC=630+375+420=1425

TC = 1425+620 = 2045

2. Use Plant 5 to manufacture A and C , along with some other plant to manufacture B.

FC(Plant 5) = 420

VCA=1.8*300=480

VCC=1.5*200=300

B can be manufactured using Plant 2,3,4

If plant 2 is used TCB = VCB + FC = 2.5*250+240 = 865

If plant 3 is used TCB = VCB + FC = 2.2*250+270 = 820

If plant 4 is used TCB = VCB + FC = 2.1*250+450 = 975

So, B should be manufactured in Plant 3.

TC = 420(FC)+480(VCA)+300(VCC)+820(TCB) = 2080

Comparing the 2 approaches, it can be said that first approach is better as 2045<2080.

So, the optimal approach is to manufacture A,B and C in plant 6.

(NEXT CASE)

Now, if each plant can manufacture only one product then above optimal solution will change.

PLANT	RENT	TVCC	TCC
1	120	420	540
2	240	280	520
3	270	320	590
4	450	460	910
5	420	300	720
6	620	420	1040
	MIN		520

It is optimal to manufacture C in Plant 2.

PLANT	RENT	TVCB	TCB
1
2	240	625	865
3	270	550	820
4	450	525	975
5	420	400	820
6	620	375	995
	MIN		820

It is optimal to manufacture B in Plant 3 and Plant 5.

PLANT	RENT	TVCA	TCA
1
2
3	270	660	930
4	450	570	1020
5	420	540	960
6	620	630	1250
	MIN		930

It is optimal to manufacture A in plant 3.

So in conclusion, A (plant 3), B(plant 5) and C(Plant 2).

(NEXT CASE)

Plants now can manufacture 2 products with extra fee of 50.

We can use the second case here. (from above)

Without any extra fee following were the total cost

Use Plant 5 to manufacture A and C , along with plant 3 to manufacture B. (Total Cost = 2080)

In second one, there is extra fee of 50 to change configurations once. (New total cost = 2130)

So, the optimal solution here is to manufacture A and C in plant 5 and B in plant 3.