Question

Copy of Given the following information and options, calculate the optimal life of the project. Assume the cost of capital is 10% p.a. Maximum life is five years and replacement of like with like..

	Year 0	Year 1	Year 2	Year 3	Year 4	Year 5
Net Cash Flows ($)	(10,000)	2,200	3,000	3,500	2,500	2,000
Retirement Values ($)		6,000	5,000	4,800	3,000	1,000

Kindly please explain the formula used and the way to approach this type of question.

Thanks

Answer 1

i) Year 1 replacement :

Net present value (NPV) = Present value (P. V.) of cash inflows - Present value of cash outflow

Now,

NPV = ((Cash flow + Retirement value at year 1) / (1 + i)^n) - P. V of cash outflow

Here, i = Interest or cost of capital @ 10% or 0.10

n = no. Of years

NPV = ((2200 + 6000) / (1 + 0.10)^1) - 10000

NPV = (8200 / 1.10) - 10000 = - 2545.45

ii) Year 2 replacement :

NPV =((Cash flow year 1/(1+i)^n) +((Cash flow year 2 + Replacement value year 2)/(1+i)^n)) - P. V. Of cash outflow

NPV =((2200/(1+0.10)^1) + ((3000 + 5000)/ (1+0.10)^2)) - 10000

NPV = ((2200 / 1.10) + (8000 / 1.21)) - 10000

NPV = (2000 + 6611.57) - 10000 = - 1388.43

iii) Year 3 replacement :

NPV = ((Cash flow year 1/(1+i)^n) + (cash flow year 2 /(1+i)^n) + ((cash flow year 3 + replacement value at year 3)/(1+i)^n)) - P. V of cash outflow

NPV = ((2200 /(1+0.10)^1) + (3000 /(1+0.10)^2) + ((3500 + 4800) /(1+0.10)^3)) - 10000

NPV = ((2200 / 1.10) + (3000 / 1.21) + (8300 / 1.331)) - 10000

NPV = (2000 + 2479.34 + 6235.91) - 10000

NPV = 10715.25 - 10000 = 715.25

iv) Year 4 replacement :

NPV = ((Cash flow year 1 /(1+i)^n) + (Cash flow year 2 /(1+i)^n) + (Cash flow year 3 /(1+i)^n) + ((Cash flow year 4 + Replacement value at year 4) /(1+i)^n)) - P. V of cash outflow

NPV = ((2200 /(1+0.10)^1) + (3000 /(1+0.10)^2) + (3500 /(1+0.10)^3) + ((2500 + 3000) /(1+0.10)^4)) - 10000

NPV = ((2200 / 1.1) + (3000 / 1.21) + (3500 /1.331) + (5500 /1.4641)) - 10000

NPV = (2000 + 2479.34 + 2629.60 + 3756.57) - 10000

NPV = 10865.51- 10000 = 865.51

v) Year 5 replacement :

NPV = ((Cash flow year 1 /(1+i)^n) + (Cash flow year 2 /(1+i)^n) + (Cash flow year 3 /(1+i)^n) + (Cash flow year 4 /(1+i)^n) + ((Cash flow year 5 + Replacement value year 5) /(1+i)^n)) - P. V of cash outflow

NPV = ((2200 /(1+0.10)^1) + (3000 /(1+0.10)^2) + (3500 /(1+0.10)^3) + (2500 /(1+0.10)^4) + ((2000 + 1000) /(1+0.10)^5)) - 10000

NPV = ((2200 / 1.10) + (3000 /1.21) + (3500 /1.331) + (2500 /1.4641) + (3000 /1.6105)) - 10000

NPV = (2000 + 2479.34 + 2629.60 + 1707.53 + 1862.78) - 10000

NPV = 10679.25 - 10000 = 679.25

Now, calculation of equivalent annual cost each year

Equivalent annual cost (EAC) =NPV / ((1 - (1/(1+i)^n)) / i)

So,

Year 1 EAC = - 2545.45 / ((1 - (1/(1+0.10)^1)) / 0.10)

Year 1 EAC = -2545.45 / 0.9091 = - 2799.97

Year 2 EAC= - 1388.43/((1 - (1/(1+0.10)^2)) / 0.10)

Year 2 EAC = - 1388.43 / 1.7355 = - 800.02

Year 3 EAC = 715.25 / ((1 - (1/(1+0.10)^3)) / 0.10)

Year 3 EAC = 715.25 / 2.4869 = 287.61

Year 4 EAC = 865.51 / ($1 - (1/(1+0.10)^4)) / 0.10)

Year 4 EAC = 865.51 / 3.1699 = 273.04

Year 5 EAC = 679.25 / ((1 - (1/(1+0.10)^5)) / 0.10)

Year 5 EAC = 679.25 / 3.7908 = 179.18

Conclusion : As year 3 EAC is having higher and positive value of 287.61. Hence project should be replaced at the end of year 3.

Note : Figures are rounded off upto 4 decimals.