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29 Nov 2019

Example: A $100 investment now in an account that pays compound interest annually will be worth $250 at a point exactly 31 years from now. What annual interest rate does this account pay?

Solving the equation for i:

250 = 100 (1 + i)³¹

1 + i = (250/100)^1/31= 1.0299

yields an answer of 3%.

Or, using interest tables, note that i = 3%, because (F/P,3%,31) = 2.500.

1. How many years are required for an investment to double in value at 10% interest?

a. n = 2 years, because (F/A,10%,2)>2 and (F/A,10%,1) < 2

b. n = 3 years, because (P/G,10%,3)>2 and (P/G,10%,2) < 2

c. n = 8 years, because (F/P,10%,8)>2 and (F/P,10%,7) < 2

d. n = 10 years, because interest is 10% per year

2. At what annual compound interest rate is $100 today equivalent to $370, seventeen years from now?

a. i = 6%

b. i = 7%

c. i = 8%

d. i = 9%

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Related questions

17) A $50,000 municipal bond has an interest rate of 6% per year, compounded semiannually. The bond will mature in 10 years. If the market interest rate is 8% per year compounded semiannually, a person purchasing the bond should be willing to pay:

a. An amount less than $50,000

b. $50,000

c. An amount greater than $50,000

d. The amount cannot be determined from the information provided

18) Two mutually exclusive alternatives, A and B, are to be evaluated by the ROR method. The initial investment for alternative B is greater than that of alternative A. If the overall ROR of both alternatives A, and B, is greater than the MARR, then:

a. Select alternative A because it has a lower initial investment cost.

b. Select alternative B because it has a higher initial investment cost.

c. Conduct an incremental analysis and select A if the ROR of the increment exceeds the MARR

d. Conduct an incremental analysis and select B if the ROR on the increment exceeds the MARR.

19) A person purchases a $5,000, 5% per year bond, with interest payable semiannually, for the amount of $4,000. The bond has a maturity date of 14 years from now. The equation for calculating how much the person must sell the bond for 6 years from now in order to make a rate of return of 12% per year, compounded semiannually, is:

a. 0 = 4,000 + 125(P/A,6%,12) + x(P/F,6%,12)

b. 0 = 4,000 + 100(P/A,6%,12) + x(P/F,6%,12)

c. 0 = 5,000 + 125(P/A,6%,12) + x(P/F,6%,12)

d. 0 = 4,000 + 125(P/A,12%,6) + x(P/F,12%,6)

20) How much must you deposit in your retirement account each year for 10 years starting now (i.e., years 0 through 9) if you want to be able to withdraw $50,000 per year forever beginning 30 years from now? Assume the account earns interest at 10% per year.

a. $4,239

b. $4,662

c. $4,974

d. $5,471

ceruleanrat918

1. The first cost of a bridge that is expected to have an infinite life is $10 million. The maintenance cost of the bridge will be $150,000 per year. At an interest rate of 10% per year, the annual worth of the bridge is equal to:

a. $3,200,000

b. $1,000,000

c. $1,150,000

d. $850,000

2. When calculating the i* value, all the net positive cash flow are assumed to be reinvested at:

a. The current market interest rate

b. The i* rate

c. The company's MARR

d. The company's cost of capital

3. If you have the present worth of an alternative with a 5-year life, you can obtain its annual worth by:

a. Multiplying the PW by i

b. Multiplying the PW by (A/F, i, 5)

c. Multiplying the PW by (P/A, i, 5)

d. Multiplying the PW by (A/P, i, 5)

4. If you have the capitalized cost of an alternative that has an infinite life, you can get its annual cost over a very long number of years by:

a. Multiplying the capitalized cost by i

b. Multiplying the capitalized cost by (A/F,i, n)

c. Dividing the capitalized cost by (P/A, i,n)

d. Dividing the capitalized cost by i

5. An investment of $85,000 resulted in uniforms income of $15,000 per year for 15 years. The rate of return of the investment is equal to:

a. 10.6%

b. 15.6%

c. 13.7%

d. 23.5%

plummanatee757

1. What is the capitalized value of an infinite series of $1,000 biennial (every two years) payments? MARR = 10%

		$4,950
		$4,025
	.	$4,350
		$4,762

2. Frank West Enterpises would like to choose one of the following plans for expansion. Using present worth analysis, find the best alternative among the three alternatives and the Do-Nothing.

	Alt. A	Alt. B	Alt. C
Initial Cost	$10,000	$21,000	$8,000
EUAB	$6,000	$7,000	$4,000
EUAC	$1,500	$1,000	$2,000
Salvage Value	$2,000	$8,000	$3,000
Life	2 Years	3 Years	2 Years

Use a MARR of 10%.

	.	Do-Nothing
	.	Alt. C
	.	Alt. B
	.	Alt. A

U. S. Engineering is investigating the possibility of acquiring new automated packaging equipment at a cost of $12,000. It is expected that the equipment will have a salvage value of $1,000 at the end of its useful life of 10 years. It is determined by the plant engineering department at the company that the operation and maintenance cost will be $500 in the first year and will gradually increase every year starting year 2 at the rate of $50 until the equipment is retired. Determine the equivalent uniform annual cost (EUAC) if MARR for the company is 10%.

		$2,575.95
		$2,150.50
	.	$2,176.60
	.	$2,184.40

The cash flows given in table below are for two different alternatives. MARR =10%

Data	C	D
Initial Cost	$20,000	$80,000
Uniform Annual Benefits	$6,000	$10,000
Salvage Value	$5,000	$20,000
Useful Life in years	5	infinity

The equivalent uniform annual worth (EUAW) of alternative N is ______________.

Hint: Assume the Salvage Value is never realized since N is infinite.

		$2,000
		$1,920
		$1,762
		$2,476

Compare the following plans using a MARR of 6%.

Data	Plan X	Plan Y
Equipment First Cost	$50,000	$75,000
Annual Operation & Maintenance Cost	$3,000	$2,500
Salvage Value	$10,000	$0
Service Life, Years)	25	50

		$6,728; $7,255
		$7,255; $9,000
	.	$6,728; $7,728
	.	$6,728; $8,200

LAEP, Inc. wants to evaluate two methods of shipping their products. The following cash flows are associated with each alternative:

Data	Method
	A	B
Life (Years)	10	10
First Cost	$700,000	$1,512,000
M&O Cost	$18,000	$9,000
M&O Cost Gradient	$900	$775
Annual Benefit	$154,000	$303,000
Salvage Value	$142,000	$210,000

Annual profits are based on amount of products which can ordinarily be shipped each year as a function of the amount of vehicles or service purchased with the first cost and the M&O costs.

Using a MARR of 15%, calculate the equivalent uniform annual cash flow (EUAB - EUAC) for each alternative.

Determine the most desirable alternative based on the results.

		$389.57, $445.90; Method A
		$445.90, $389.57; Method A
	.	$445.90, $389.57; Method B
	.	$389.57, $445.90; Method B

For the cash flow diagram shown, which of the following equations properly calculates the uniform equivalent (A)?

n	0	3	6	9	12	15
Dollars	$100	$100	$100	$100	$100	$100

		A=100(A/P,i,15)+ 100(A/F,i,3)
		A=100(A/P,i,15)
		A=100(A/F,i,3) + 100(A/F,i,15)
		A=100(A/P,i,3) + 100(A/F,i,3)

Three different alternatives shown in table below are being considered by U. S. Engineering systems.

Assume that alternatives X and Z are replaced at the end of their lives.

Data	Alternative X	Alternative Y	Alternative Z
Initial Cost	$6,000	$1,000	$1,500
Uniform Annual Benefits	$810	$125	$ 230
Useful Life in Years	20	infinite	10
MARR	12%

The NPW for alternative

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