CEIE 301 – Questions
Questions 1 through 10 are Multiple Choice Questions. Select the correct choice (50 Points).
About how many years will be required for $20,000 invested at 6% per year compounded annually to double in value?
a) 10
b) 12
c) 14
d) 16
A company that utilizes carbon fiber 3-D printing wants to have money available two years from now to add new equipment. The company currently has $650,000 in a capital account and it plans to deposit $200,000 now and another $200,000 one year from now. The total amount available in two years, provided it returns a compounded rate of 12% per year, is closest to:
a) $1,190,240
b) $1,290,240
c) $1,390,240
d) $1,490,240
What amount of money would have to be deposited in a bank at end of each year to accumulate a college fund of $60,000 at the end of 18 years? Interest on the account is 6%.
a) $1,560
b) $1,680
c) $1,820
d) $1,940
Sheryl is planning for her retirement. She expects to save $5000 in year 1, $6000 in year 2, and amounts increasing by $1000 each year through year 20. If the investments earn 10% per year, the amount Sheryl will have at the end of year 20 is closest to:
a) $242,568
b) $355,407
c) $597,975
d) $659,125
A self-employed civil engineer deposits $3,000 into a savings account at the end of every year for 10 years. He makes no deposits thereafter. How much will be in the account at the end of 20 years if interest rate is 6% per year.
a) $39,540
b) $68,660
c) $70,810
d) $76,320
A residential building lot is purchased for $500,000 cash. If the lot is held for 5 years and the return is 12% before taxes, what is the selling price close to if there is a 2% inflation rate?
a) $751,230
b) $881,150
c) $972,870
d) $1,433,210
The amount of money that you can spend now for a much safer car in lieu of spending $30,000 three years from now at an interest rate of 12% per year is closest to:
a) $15,710
b) $17,805
c) $19,300
d) $21,355
A flight company is considering the purchase of a helicopter to connect service between its base airport and a new field being built about 30 miles away. The choppers are assumed to be needed for only 6 years until rapid transit service is completed. Estimates for two choppers under consideration are as follows:
Chopper A Chopper B
First cost $100,000 $120,000
Annual maintenance $3,000 $7,000
Salvage value $15,000 $25,000
Selecting Life in years 3 6
Interest rate 8% 8%
What is the annual cost advantage of Chopper B?
a) $0
b) $5,888
c) $11,549
d) $7,631
Two alternative buildings are being considered. The first building has an initial cost of $600,000 and annual cost of $155,800 per year. The second building is estimated to cost $750,000 with annual cost of $125,000. With interest rate at 10% per year, what is the useful life of these two buildings to have the same equivalent value?
a) 7 years
b) 8 years
c) 9 years
d) 10 years
The following are the estimates for Present Worth for four independent projects.
The cost of money is 8% per year.
Project Initial
Investment, $
PW, $
1 −20,000 2,400
2 −35,000 9,200
3 −40,000 −7,300
4 −55,000 11,400
If only two projects may be selected with no more than $100,000 invested, the projects selected are:
a) 1 and 2
b) 3 and 4
c) 2 and 3
d) 2 and 4
For the following questions, you need to show your work to get the full credit.
(15 Points) A civil engineer starts investing his money when he graduates from college. He is able to afford investing $10,000.00 a year from the time he graduates in four years until the end of eight years (i.e. years 5 through 8). He also plans to increase his investment an additional $2,500.00 per year increasing by $2,500.00 every year until year eight. Use the interest rate of 10%.
a) Draw the cash flow diagram for the above cash flows.
b) How much will the civil engineer have saved by the end of year eight?
c) What is its present worth on the year he started the college?
(15 Points) A wastewater treatment process engineer is determining the most economical alternative by comparing three alternatives, as well as the do nothing alternative, for a new treatment process. He investigates the costs and the benefits for each of the alternatives and determines the net present worth of each alternative using a minimum attractive rate of return of 10%. All of the alternatives will have a life span of 20 years. The three potential alternatives being analyzed are listed below:
Alternative Total Investment Annual Benefit Salvage Value at end of 20 years
Process 1 $500,000 $51,000 $300,000
Process 2 $950,000 $105,000 $300,000
Process 3 $1,500,000 $150,000 $400,000
Do nothing $0 $0 $0
a) What is the net present worth value of each alternative?
b) What is the best alternative based on the net present worth value?
(20 Points) A civil engineer is asked to select a new scarper for his construction firm. He collects data from two equipment manufacturers and the data are shown in the Table below.
Cost or Disbursements Scraper 1 Scraper 2
Initial cost $1,600,000 $1,200,000
Cost to operate per hour $120 per hour $150 per hour
Number of hours used per year 2,000 hours 2,000 hours
Repairs $15,000 years 6 through 9 $20,000 years 6 through 9
Major overhaul $200,000 at year 5 $270,000 at year 5
Salvage value $260,000 $200,000
Life in years 12 10
a) Draw the cash flow diagrams for each of the scarper.
b) Determine which scraper he should recommend based on equivalent uniform annual worth analysis using an interest rate of 8%.
