Item Option 1 – Gravel Surface Option 2 – Paved Surface Initial Cost $10,753
ID: 1187673 • Letter: I
Question
Item
Option 1 – Gravel Surface
Option 2 – Paved Surface
Initial Cost
$10,753
$39,105
Maintenance
$3,386/yr (grading)
$3,804/yr (potholes)
Parking Fees Collected
$16,922/yr
$16,922/yr
Service Life
60 yrs
20 yrs
Demolition Cost
$6,473
Prepare cash flow diagrams for each option and an incremental cash flow diagram for option 1 – option 2. Assuming a cost of money or 3.2%, determine the following:
3a. What is the present worth of option 1?
3.b What is the present worth of option 2?
3c. Which is the better option, and why?
3d. Perform an annual equivalence analysis for the two options, and state which is the better option and why.
3e. Calculate the present worth of the incremental system (1 - 2), and state which option is better and why.
Please show your work. Thank you
Item
Option 1 – Gravel Surface
Option 2 – Paved Surface
Initial Cost
$10,753
$39,105
Maintenance
$3,386/yr (grading)
$3,804/yr (potholes)
Parking Fees Collected
$16,922/yr
$16,922/yr
Service Life
60 yrs
20 yrs
Demolition Cost
$6,473
Explanation / Answer
13536 13536 13536
PW1 0----- 1-----2-----......-------------60
10753
13118 13118 6645
PW2 0----- 1-----2-----......-----------20
39105
Present Worth of Option 1
= PW1 = -Initial Cost + (AINCOME - AEXPENSE)(P/A, 3.2%, 60)- FDemolition(P/F,3.2%,60)
SO PW1 = -10753 + (16922-3386)*26.5286 -0
ie PW1 = 348338 .....Ans (a)
Similarly PW2 = -39105+(16922-3804)*(P/A,3.2%,20) - 6473*(P/F,3.2%,20)
ie PW2 = -39105+(16922-3804)*14.6061- 6473*0.5326
ie PW2 = 149050 .....Ans (b)
c. As PW1 is hgher than PW2, Option 1 is better.
d. Annual equivalence analysis assuming 60 Yrs.
PW1 = 348338 ...No Change
PW2 = -39105+(16922-3804)*(P/A,3.2%,20) +(16922-3804)* (P/A,3.2%,40)(P/F,3.2%,20)- 6473*(P/F,3.2%,20)
ie PW2 = -39105+(16922-3804)*14.6061 +(16922-3804)*22.3853*0.5326 - 6473*0.5326
ie PW2 = 305448
So Option 1 is still better.
e. PW of Incremenatl sysyem = PW1-PW2 = 348338-305448
= 42890
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