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23 Nov 2019

Engineering Economics -EGR-226

If a Sum of $5,500.00 is deposited now and $7,500.00 is deposited three years from now, and $2,500.00 per year in years 4-10, the amount in year 10 at an intrest rate of 6% per year will be closest to:

What I got:

i=6%

**P= 5,500 * (P/A,6%,1) + 7,500 * (P/A,6%,3) + (2,500 * (P/G,6%,7))**

Is this the correct formula for the problem?

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Related questions

1 Two sets of cash flows are shown in the table below. If these sets are equal at 8% interest rate, determine the unknown value, X.

Year	0	1	2	3	4	5
Cash Flow, Series I	$125	$100	$75	$50	$25	$0
Cash Flow, Series II	$X	$2X	$3X	$4X	$5X	$7X

2 A deposit of $10,000 now at a interest rate of 10% per year will accumulate in 20 years to an amount equal to _______________.

	A.	$54,600
	B.	$61,500
	C.	$67,280
	D.	$51,900

3 Which of the following relationships between compound interest factors is not correct?

	A.	Single Payment Compound Amount Factor and Single Payment Present Worth Factor are reciprocals.
	B.	Sinking Fund Factor and Uniform Series Compound Amount Factor are reciprocals.
	C.	Capital Recovery Factor and the Sinking Fund Factor are reciprocals.
	D.	Capital Recovery Factor and Uniform Series Present Worth Factor are reciprocals.

4 What is the present worth of a series of ten deposits of $200 per year at 6% made at the end of each year?

	A.	$1,672
	B.	$1,072
	C.	$1,272
	D.	$1,472

5 Determine the net present worth (NPW) of the following cash flows using an interest rate of 7%.

Year	0	1	2	3	4	5	6	7	8
Cash Flow,$	-12,000	2,000	2,000	2,000	2,000	2,000	2,500	3,000	3,500

A.$2,771.05

B.$2,500
C.$1,500
D.$1,771.05

6 At Kal Tech Manufacturing Inc, the time required to make the first unit of an engineered product is 64.1 minutes. Determine the time required for 16^th unit if the learning curve percentage is 90%.

	A.	42.1 minutes
	B.	29.53 minutes
	C.	36.45 minutes
	D.	20.89 minutes

7 A sum of $994.62 deposited at time

Grandpa would like to deposit some money now for his grandson to have available to beginning withdrawing $18,000 per year for four years, starting 5 years from now. If the money earns 7% interest per year, the amount that Grandpa must deposit is closest to:

The number of time periods that should be used with the F/A factor is equal to:

To find the EUAW of a shifted gradient, find A/G.

The number of time periods that should be used with the P/A factor is equal to:

For a uniform annual series, the future worth is always located in the same period as the last A

To avoid errors in counting the number of periods, you should always:

About a uniform annual series, the present worth is always located:

To find the equivalent uniform annual worth of a cash flow that has both A's and F's convert everything to a P and then find the F.

The first cash flow of a uniform graduate will always be located in the first period.

greykangaroo414

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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