1. It takes $25,000 every year to put on the Octagon Arts Festival here in Ames
ID: 2507311 • Letter: 1
Question
1. It takes $25,000 every year to put on the Octagon Arts Festival here in Ames each September. At the
beginning of September this year, an anonymous donor contributed $100,000 to be used for the annual arts
festival. If the committee uses the donation to pay for this years event and invests the rest in an account
paying 4% interest, how many more festivals (after this years) can be sponsored without raising more
money?
2. On Friday, February 21 a retired electronics technician from Cedar Rapids, Iowa won $200,000 in Thursday
evening drawing of the Iowa Lotterys Pick 4 game. (Source: www.usamega.com/lottery-news.asp) Lets say
the winner is offered a choice of a one-time payment of $200,000 or a guaranteed $22,000 per year for 10
years. If the value of money is 5%, which option should he choose?
3. George wants to buy a new snowmobile. Three purchase plans are available:
Plan A: $5000 cash immediately
Plan B: $1500 down and 36 monthly payments of $97.75
Plan C: $1000 down and 48 monthly payments of $96.50
If George expects to keep the snowmobile for 5 years, and his cost of money is 6% (compounded monthly),
which payment plan should he choose?
4. A bridge is being considered at a cost of $220M. The annual maintenance costs are estimated to be $150k. A
major renovation costing $50M is required every 25 years. What is the capitalized cost (analysis period =
infinite) of the bridge at 5% interest
5. The city council wants the municipal engineer to evaluate three alternatives for supplementing the city water
supply.
1. The first alternative is to continue deep-well pumping at an annual cost of $10,500.
2. The second alternative is to install an 18 pipeline from a surface reservoir. First cost is $25,000 and
annual pumping cost is $7000.
3. The third alternative is install a 24 pipeline from the reservoir at a first cost of $34,000 and annual
pumping cost of $5000.
The life of each alternative is 20 years. For the second and third alternatives, salvage value is 10% of first cost.
With interest at 8%, which alternative should be engineer recommend? Use present worth analysis.
6. An engineer is considering buying a life insurance policy for his family. He currently owes $77,500 in various
debts, and he would like his family to have an annual available income of $35,000 indefinitely (that is, to
ensure that the original capital does not decrease, the annual interest should amount to $35,000).
a. If the engineer assumes that any money from the insurance policy can be invested in an account paying a
guaranteed 4% annual interest, how much life insurance should be buy (to cover current debt and ensure
the annual income)?
b. If he now assumes that they money can be invested at 7% annual interest, how much life insurance
should be buy?
Explanation / Answer
1.)
remaing amount this year =$75000
the above amount will generate a cash flow of $25000 every year
75000 = 25000*PVIFA(4,n)
75000=25000*[1-(1.04^-n]/.04
1-3*.04 = 1.04^-n
1.13636 = 1.04^n
n = 3.25 year
so, 3 more years can be fully sponsored
2).
PV of future cash flow = 22000*PVIFA(5,10) = 22000*7.7217 = $169877.4
it should choose ist option i.e single payment of $200000
3).
interest rate per month = 6/12 = .5%
PV of cash flow of PLan B
PV = 97.75*PVIFA(.5,36) = 97.75*32.8710 = $3213.14
totala amount paid = 1500+3213.14 = $4713.14
PV of cash flow for paln C
PV = 96.5*PVIFA(.5,48) = 96.5*42.5803 = $4108.99
amount paid total = $4108.99+1000 = $5108.99
so, Plan be should be chosen by george
4).
let effective interest rate for 25 year = 1.05^25-1 = 2.3863 = 238.63%
tot calculate NPV of all future cash flow = 220000000+(150000/.05)+(50000000/2.3863) = 243952939.697
5).
for 1st alternative
total effective cost = 10500*PVIFA(8,20) = 10500*9.8181 = $103090.05
for 2nd alternativve = 25000+7000*PVIFA(8,20)-(2500/1.08^20) = 25000+7000*9.8181-(2500/1.08^20) = $93190.32
for 3rd alternative = 34000+5000*PVIFA(8,20)-(3400/1.08^20) = 34000+5000*9.8181-(3400/1.08^20) = $82361.036
alternative 3rd is best
6).
a) amount that he should buy @4%= 77500+(35000/.04) = $952500
b) amount that he should buy @7% = 77500+(35000/.07) = $577500
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