ECON361 Lecture 3: Module 3 Use of Basic Tools

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tancockroach82

10 Dec 2017

School

University of Waterloo

Department

Economics

Course

ECON361

Professor

Fiona Rahman

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Document Summary

Module 3 use of basic tools (chapter 6) Multi-period investment projects tend to have costs and benefits occurring over many years. The same principles are used in private and public cba when analyzing cash flows. Common to all dcf analysis is he conceptualization of the investment project as a stream of net benefits. Although we use the term cash flow, the dollar values used might not be the same as the actual cash amounts. Dcf process: identification valuation comparison. Initial or present period is always year 0: year 1 amounts occur on the last day of the year. Future value analysis: where the future value in one year of an amount x invested at interest rate i is: Fv = x(1+i: or, fv = x(1+i) n for interest compounded annually, choose the project with the largest future value (fv) Present value analysis: where the present value of an amount y received in one year is.

Related Questions

The following are all True and False Questions:

1. Book Value represents the remaining underpreciated investment on the books after the depreciation charges to date have been subtracted

2. For the same interest rate, compounding interest does not always yield a larger future worth value than a simple interest rate

3. For any project, if its present worth is negative, the project will be rejected

4. The capatalized cost of a finite-life alternative can be obtained by calculating its annual worth over one life cycle and then multiplying by the interest rate

5. The depreciation amount is not directly included in CFBT (Cash Flow Before Tax), but depreciation is directly invlolved in the taxable calculation

6. Using incremental analysis of alternatives, the alternatice with the highest B/C ratio is selected.

7. Plotting PW vs i can be used to determine the IRR for a given project.

8. In conducting an incremental analysis, at any round, two alternatives can be compared and the alternative with the highest internal rate of return can be selected for comparison in the next round.

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

ECON361 Lecture 3: Module 3 Use of Basic Tools

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