Illustrate the cashflow diagram and compute for the payback period for a project with the following characteristics, if the minimum attractive rate of return (MARR) is 10%? First Cost $20,000 Annual Benefits $8,000 Annual Maintenance $2,000 in year, then increasing by $500 per year Salvage Value $2,000 Useful Life 10 years

Illustrate the cashflow diagram and compute for the payback period for a project with the following characteristics, if the minimum attractive rate of return (MARR) is 10%? First Cost $20,000 Annual Benefits $8,000 Annual Maintenance $2,000 in year, then increasing by $500 per year Salvage Value $2,000 Useful Life 10 years

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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An investment proposal calls for $223,130 payment now and a second $259,450 7 years from now. The investment is for a project with perpetual life. The annual interest rate is 6%. What is the approximate capitalized cost?
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ces Consider two air-conditioning systems with the cash flow estimates as given below. Use AW analysis to dete economic decision to MARR values of 4%, 6%, and 8% per year. System Air-Conditioning System 1 Air-Conditioning System 2 First cost, $ -9,000 -13,000 AOC, $ per year -500 -175 Salvage value, $ -100 -300 New compressor and motor cost at midlife, $ -1,400 -2,900 Life, years 8 12 The annual worth of air-conditioning system 1 when the MARR is 4% is $ The annual worth of air-conditioning system 2 when the MARR is 4% is $ The annual worth of air-conditioning system 1 when the MARR is 6% is $[ The annual worth of air-conditioning system 2 when the MARR is 6% is $1 The annual worth of air-conditioning system 1 when the MARR is 8% is $ The annual worth of air-conditioning system 2 when the MARR is 8% is $
Using Incremental with EUAW analysis find the best alternative, MARR = %10. should use Excel and show your equations seperately, see below example: [A Benefit - [IC (A/P, i%, n) - Salvage (A/F, i, n)] + A Cost+ G Cost (A/G, i, n)] A B C $2,300,000 $2,780,000 $2,540,000 $82,000 $118,000 $97,000 $580,000 $670,000 $650,000 $65,000 $78,000 $71,000 $11,000 $15,000 $12,500 9 9 9 Data First Cost Salvage Value Annual Benefit M&O M&O Gradient Useful Life, Years You
The future worth of a project with initial cost P, positive annual cash flows of A, salvage value S, and interest rate of i over a life of n years can be calculated using which statement? (a) FW = −P(F/P, i%, n) + A(F/A, i%, n) + S(F/P, i%, n) (b) FW = P(F/P, i%, n) + A(F/A, i%, n) + S (c) FW = −P(P/F, i%, n) + A(F/A, i%, n) − A[(P/A, i%, n) + S (d) FW = −P(F/P, i%, n) + A(F/A, i%, n) + S?
Find the internal rate of return for the following investment (yes I want the actual rate). Is it a good idea if MARR=10%? Year %24 -160,400 75,000 -32,000 55,000 32,500 69,500 2 3 4 Attach File Browse Local Files
Most likely estimates for a project are as follows. MARR Useful life Initial investment Receipts - Expenses (R-E) 10% per year 9 years $5,000 $1,200/year Click the icon to view the relationship between the PW and the percent change in parameter. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. (b) To which variable is the PW most sensitive to? OA. Receipts - Expenses OB. Usefule life OC. Initial Investment (a) Determine whether the statement "An initial investment of $6,000 keeps the investment economical." is true or false. O False O True
Don't Use Excel function. Need Handwritten /Typed based solution using formula. Please Solve as fast as.
Q4
Given the following two alternatives and using the repeatability assumption and MARR=15%/year, the equation for computing the present worth of vendor B is Vendor A First Cost, $ -15,000 Annual cost, $ per year -3500 Salvage Value, $ 1000 Life, years 3 Vendor B -18000 -3100 2000 4 OPW=18000-18000(P/F,15%, 4)-18000(P/F,15%, 8)+2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000 (P/F, 15%, 12)-3100(P/A, 15%, 12) PW=18000-18000(P/F, 15%, 12)+2000(P/F, 15%, 4)+2000 (P/F, 15%, 8)+2000(P/F, 15%, 12)-3100(P/A, 15%, 12) O PW=-18000-18000(P/F,15%, 4) - 18000(P/F,15%, 8) +2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000 (P/F, 15%, 12)-3100(P/A, 15%, 12) O PW=-18000(P/F,15%, 4) -18000(P/F,15%, 8) +2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000(P/F, 15%, 12)-3100(P/A, 15%, 12)
QUESTION 8 For th below two machines and based on AW analysis which machine we should select? MARR=10% Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years 3 Answer the below question: B-the AW for machine B= 26,612 12,417 4,135 infinite 140,454 7,170