Question

The machines shown below are under consideration for an improvement to an automated candy bar wrapping! process. First cost, $ Annual cost, $/year Salvage value, $ Life, years (Source: Blank and Tarquin) Machine C 40,000 -15,000 12.000 Machine D -75,000 -10,000 25.000 Based on the data provided and using an interest rate of 5% per year, the Annual Worth “AW" of Machine C is closest to (All the alternatives presented below were calculated using compound interest factor tables including all decimal places) L ANVC =$25 882 AW c = -$86,098 AWO --$21.117 ANC= -$16,390

Answer 1

Problem 1

Annual worth of machine C is given by

AW_C=-40000*(A/P,5%,3)-15000+12000*(A/F,5%,3)

Let us calculate interest factors\

$(A/P,i,n)=\frac{i}{1-\frac{1}{(1+i)^{n}}}$

$(A/P,0.05,3)=\frac{0.05}{1-\frac{1}{(1+0.05)^{3}}}=0.367209$

$(A/F,i,n)=\frac{i}{(1+i)^{n}-1}$

$(A/F,0.05,3)=\frac{0.05}{(1+0.05)^{3}-1}=0.317209$

AW_C=-40000*0.367209-15000+12000*0.317209=-$25881.85

Correct Option is

AWC=-$25882

Problem 2

Initial Cost=I=-$75000

Salvage=S=$25000

Time period=n=6

Interest rate=i=5%

Capital Recovery for alternative D is given by

CR_D=I*(A/P,5%,6)+S*(A/F,5%,6)

CR_D=-75000*(A/P,5%,6)+25000*(A/F,5%,6)

Correct Option is

CR_D=-75000*(A/P,5%,6)+25000*(A/F,5%,6)

Problem 3

We know that

CR_D=-75000*(A/P,5%,6)+25000*(A/F,5%,6)

Let us calculate interest factors

$(A/P,i,n)=\frac{i}{1-\frac{1}{(1+i)^{n}}}$

$(A/P,0.05,6)=\frac{0.05}{1-\frac{1}{(1+0.05)^{6}}}=0.197017$

$(A/F,i,n)=\frac{i}{(1+i)^{n}-1}$

$(A/F,0.05,6)=\frac{0.05}{(1+0.05)^{6}-1}=0.147017$

CR_D=-75000*0.197017+25000*0.147017=-$11100.85

Correct Option is

CR_D=-$11,101