Question

Question 12 For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine A Machine B 240,000 First cost (AED) 20,000 Annual maintenance cost per year, AED 5,000 2.300 Periodic cost every 10 years, AED 10,000 Salvage cost 2000 Life. vears Match the closest correct answers for the below questions: Calculate the present value of the maintenance costs for Alternative B A. Alternative Al B. -44,483.50] C. I-269,275] D. Alternative B] E. I-627.5] F -1,300] G. I-253,627.50] H. [-23,000] . -108,784) Calculate the Anuual value of the periodic costs for Alternative B . # | Calculate CC of Alternative B Calculate the CC of Alternative A # which alternative should be selected?

Answer 1

Ans 1)

We need to consider only Maintenance cost parameter of Machine B therefore,

Present Worth of Alternative B = PW of Annual Maintenance Cost

=-2300(P/A,10%,infinity)=

Formula for (P/A, r%, infinity)=1/r

=-2300(P/A,10%,infinity)=-2300/0.1=-23000

Hence PW of Maintenance cost of Alternative B=-$23000

Option H is correct

Ans 2)

Annual Value of Periodic Cost=PW of Periodic Cost/Sum of discounting factors

PW of Periodic Cost=-(10000/(1.1)^10+10000/(1.1)^20+...infinity)

=-10000/(1.1)^10*(1/(1-(1/1.1)^10))=-6274.9

Annual worth of Periodic Cost=-6274.9/(1/0.1)=-6274.9*0.1=-$627.49

Hence Option E is correct

Ans 3)

CC of Alternative B=Initial Cost+(Annual Maintenance Cost/0.1)+Present Worth of Periodic Cost

=-(240000+2300/0.1+6274.5)=-$269,274.5

Hence option C is correct

Ans 4)

CC of Alternative A=-20000-18000/(1.1)^4-18000/(1.1)^8-...infinity-5000/(0.1)

=-20000-18000/(1.1)^4(1+(1/1.1^4)+(1/1.1^8)+...infinity)-50000

=-20000-(18000/((1.1)^4-1))-50000

=-70000-38784.75=-$108,784.75

Option I is correct

Ans 5)

Capitalized Cost for Alternative A is less than that of Alternative B therefore Alternative A to be chosen over Alternative B