Solutions presented on engineering graph paper Only neatly written, bgible solto
ID: 2779847 • Letter: S
Question
Solutions presented on engineering graph paper Only neatly written, bgible soltons w be graded A correct answer without an outline of work will carry no grade . Where more than one solution is presented, the first solution will be graded . No presumptions wil be made about numbers by the grader: identfy all numbers 1. A county parks department is considering building entry deciks to monuments located at 30 yrs $1,800 $5/yr 7% Wood 10 yrs $450 $20lyr 7% Initial Cost Determine the best option. 2. A loan shark lends money to desperate people who are in need of immediate cash Hs terms for a recent loan are ..give you S50 today Monday, you owe me$60 Monday" rate per year? What are the nominal interest rate per year and the effective initerest 3. A project has a first cost of $10,000. net annual benefits of $2.000, and a salvage value of $3,000 after 10 years. The project will be replaced identicaly at the end of 10 years The MARR is 15%. a) Draw a cash Sow diagranm b) is the system economically viable? c) What change in annual benefits will make the system d) What is the B/C ratio? break even?Explanation / Answer
1)
CF diagram : ( Draw a straight line, up arrow on line for positive cashflow, down arrow for negative cashflow)
Year 0 1 2 .... ..........................10
CF -10000 2000 2000 ..........................5000 (=2000+3000)
2)
need to find NPV (Present Worth)
PW of 10 years = -10000 + 2000(P/A, 15%, 10) + 3000(P/F, 15%, 10)
= -10000 + 10037.54 + 741.55
= 779.09
3)
To breakeven PW = 0
Assume benefits = X
-10000 + X*(P/A, 15%, 10) + 3000(P/F, 15%, 10) = 0
X*(P/A, 15%, 10) = 10000 - 741.55
X*5.02 = 9258.45
X = 1844.76 (break even annual benefits)
4)
B/C = PW of benefits / PW costs
PW costs = 10000
PW of benefits = 2000(P/A, 15%, 10) + 3000(P/F, 15%, 10)
= 10037.54 + 741.55
= 10779.09
B/C = 10779.09 / 10000 = 1.0779
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