Consider this data with MARR = 10%. System A System B Initial Cost 65000 85000 S

Question

Consider this data with MARR = 10%. System A System B Initial Cost 65000 85000 Service Life yrs 5 9 Annual Revenue 19000 19000 1. Why would AW (Annual Worth) be a convenient method here? [0.3] 2. In order to use the AW approach there is something which must be planned for the future; what is it? [0.3] 3. What is the AW of each for the given MARR? [2.1] 4. Based on the result of 3 which is the better investment? [0.5] 5. Determine if the internal rate of return for System A is higher than 15%. [1.3] [4.5] Total Consider this data with MARR = 10%. System A System B Initial Cost 65000 85000 Service Life yrs 5 9 Annual Revenue 19000 19000 1. Why would AW (Annual Worth) be a convenient method here? [0.3] 2. In order to use the AW approach there is something which must be planned for the future; what is it? [0.3] 3. What is the AW of each for the given MARR? [2.1] 4. Based on the result of 3 which is the better investment? [0.5] 5. Determine if the internal rate of return for System A is higher than 15%. [1.3] [4.5] Total

Explanation / Answer

1. The Annual Worth Is convenient method because the we have to choose for the two alternatives which have unequal Service life and thus it is better to compare the Values with AW because it gives the results in $/time so will give results in same Units.

3. THe Annual WOrth = AW(i%)=R-E-CR(i%)

For System A-

R= Revenue= $19000

E= Annual Expense= $0 (not given )

MARR= 10%

Now Capital Recovery=CR(i%)=I(A/P, i%, N)-S(A/F, i%, N)

where,
I= initial investment for the project ($65000)
S=salvage(market) value at the end of the study period = $0 (not given )
N=project study period = 5 years

CR= 65000*(A/P, 10%, 5) = 65000*(0.2638) =17147

now AW(i%)=R-E-CR(i%)

19000-17147 = $1853 ------------------------ A)

Now For System B

R= Revenue= $19000

E= Annual Expense= $0 (not given )

MARR= 10%

Now Capital Recovery=CR(i%)=I(A/P, i%, N)-S(A/F, i%, N)

where,
I= initial investment for the project ($85000)
S=salvage(market) value at the end of the study period = $0 (not given )
N=project study period = 9 years

CR= 85000*(A/P, 10%, 9) = 85000*(0.1736) =14756

now AW(i%)=R-E-CR(i%)

19000-14756= $4244 ------------------------ B)

Now we can see from comparing equations A and B , that B>A so means System B has higher AW than System A

4- Investement in System B is better than A because it higher AW

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