ECON 1020 Lecture 82: Lecture 82

62 views3 pages

erinchimpanzee657

17 May 2017

School

University of Manitoba

Department

Economics

Course

ECON 1020

Professor

Ryan A.Compton

For unlimited access to Class Notes, a Class+ subscription is required.

violetturtle181 and 7 others unlocked

ECON 1020 Full Course Notes

Verified Note

14 documents

Document Summary

Economic investment: refers either to paying for new additions to capital stock, or new replacements for capital stock that has worn out. Financial investment: refers to either buying or building an asset in the expectation of financial gain. The closer the payment is to you, the less it is discounted, and the more it is worth. Present value: the present-day value or worth of returns or costs that are expected to arrive in the future. X dollars today = x(1 + i)t dollars in t years at a compound interest rate of i. X dollars in t years = x / (1 + i)t dollars today. All investments share three common features: 1. Investors pay a price determined in the market to acquire them o 2. Owners have the chance to receive future payments: 3. Value of share will fall to nothing if corporation goes bankrupt. Limited liability rule: limits the risks (to the value of that share)

Related Questions

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

ECON 1020 Lecture 82: Lecture 82

ECON 1020 Full Course Notes

Document Summary

Get access

Related Documents

ECON 1020 Lecture 14: Macroeconomics Ch.14.docx

ECN 204 Lecture Notes - Lecture 9: Overnight Rate, Weighted Arithmetic Mean, Mutual Fund

ACC 1110 Lecture Notes - Lecture 1: Arbitrage, Time Preference, Risk Premium

Related Questions