Case study
A firm operates at full capacity to manufacture 10,000 items a year of a product to meet market demand. It is anticipated the demand will increase in the following few years. The firm is considering two alternatives to meet the anticipated increase in demand. The planning horizon is 4 years, and the MARR is 10% p.a.
Alternative 1: Use the existing machine and cater for the increase in demand by using overtime. The costing data for this option are shown below.
Present value of the existing machine: $10,000
Fixed annual operating cost (regardless of the quantity manufacture): $10,000
Added value per item up 10,000 items $5
Added value per item for all items above 10,000 (due to overtime costs) $4
Salvage value of the machine after 4 years $0
Note: Added value per item is the contribution per item, i.e. Gross profit which is the difference between selling price and direct cost.
Alternative 2: Sell the existing machine and replace it with a new automatic machine with higher capacity. Due to productivity improvements, the added value per item will be $8. The costing data for this option are shown below.
Initial cost of the new machine: $50,000
Fixed annual operating cost (regardless of the quantity manufacture): $30,000
Added value per item $8
Salvage value of the machine after 4 years $0
The firm feels confident about the costing data except the fixed annual operating cost for the new machine. It feels that the estimated fixed annual operating cost for the new machine may be subject to error.
In addition, the estimated increase in demand is also subject to error.
Advise the firm the option that it should adopt under different scenarios, i.e. changes in fixed operating cost of the new machine and changes in annual demand for the product.
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Answer:
Given that
A firm operates at full capacity to manufacture 10,000 items
The planning horizon is N= 4 years
MARR is 10% p.a.
therefore MARR i= 10/100
= 0.1
we have to check the difference in present value for 10000 items
plus one item after those 10000 units for both the alternates.
Alternate 1
= -P-P(P/A,i,N)+P*ITEM(P/A,i,N)+OVERTIME(P/A,i,N)
= -P-P((1+i)^N -1)/(i.(1+i)^N)+P*(PER ITEM)((1+i)^N
-1)/(i.(1+i)^N)+OVER TIME((1+i)^N -1)/(i.(1+i)^N)
= -10000- 10000( (1.1)4-1)/0.1(1.1)4 +50000( (1.1)4-1)/0.1(1.1)4 +4( (1.1)4-1)/0.1(1.1)4
= -10000- 31698.65+158493+12.64
= 116807
ALternate
= -P-F(F/A,i,N)(P/A,i,N) + P*ITEM(P/A,i,N)+OVERTIME(P/A,i,N)
= -P-F((1+i)^N -1)/(i.(1+i)^N)+P*(PER ITEM)((1+i)^N
-1)/(i.(1+i)^N)+OVER TIME((1+i)^N -1)/(i.(1+i)^N)
= -50000- 30000( (1.1)4-1)/0.1(1.1)4 +80000( (1.1)4-1)/0.1(1.1)4 +8 ( (1.1)4-1)/0.1(1.1)4
=-50000-95095.96+253589.23+25.28
=108518
Hence alternative1 is greater than alternative 2
therefore choose alternative1.
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