Given the data for two different alternatives in the table below, determine the best alternative usingthe incremental rate of return (AROR) analysis. MARR =9%.ABFirst cost$15,000 $20,000O&M Cost/year 1,600900Benefit/year8,0009,000Salvage value3,0004,600Life in years44

Calculate the rate of return (ROR) for each alternative individually.Compare the ROR of each…

Answered: Given the data for two different…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Three independent alternatives are given below. If MARR is 18%, what is your decision? A B C Initial Cost $4.50 $1.90 $1.20 Annual Revenues $4.00 $2.50 $3.00 Salvage Value $0.50 $0.90 $0.00 Annual Operating & Maintenance Costs $1.20 $1.90 $2.70 Estimated life, in years 3 Infinite
Q1 Table Q1 shows the engineering project with minimum attractive rate of return (MARR) at 15% per year. Table Q1 Proposal A Investment Cost RM 10,000 Expected Life 5 years Market (Salvage Value) -RM 1,000 Annual Receipts RM 8,000 Annual Expenses RM 4,000 (a) Determine the present worth (PW), future worth (FW) and annual worth (AW) based on details given in Table Q1.
Please no written by hand solution Two projects are shown below. Interest rate is 10% per year with yearly compounding. Project A Project B Project Life 10 years 10 years Benefits $500 at year = 0 and $100 per year (at year = 1,2,3,…,10) $15 per year starting year 2 (at year = 2,3,4,…,10) Disbenefits $40 per year (at year = 1,2,3,…,10) $1 per year starting year 2 (at year = 2,3,…,10) Costs $600 at year = 0, $100 at year = 1 $50 at year = 0 (1) Find the conventional B/C ratio for Project A. (2) Find the conventional B/C ratio for Project B. (3) Based on B/C analysis (using conventional B/C ratio), which between A and B is more attractive?
Given the following pertinent data for four alternatives, what is the best alternative using the incremental ROR method given MARR=10% Initial Cost Annual CF, $ Life, years 30 O Alt. A Alt. B O Alt. C A O Alt. D B +22,000 +35,000 с -200,000 -275,000 -190,000 -331,350 30 D +19500 +42,000 30 30
please answer all with complete solution and draw the cash flow diagram,,, do not use an excel please. thank you....
I need the answer as soon as possible
A deposit of $20,000 now at an interest rate of 10% per year will accumulate in 10 years to an amount closest to, A 51,900 B $61,500 (c) s54,600 D $67,300
Consider the comparison between alternatives A and B. Using present worth method of analysis, and at an interest rate of 10% per year, the values of n that you should use in the uniform series factors to make a correct comparison are: A B First Cost ₱500,000 ₱900,000 Annual Operating Cost ₱100,000 ₱40,000 Salvage Value ₱130,000 ₱150,000 Life 3 years 6 years a. n=3 years for A and n=3 years for B b. n=3 years for A and n=6 years for B c. n=6 years for A and n=6 years for B d. n=6 years for A and n=3 years for B
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?
Given the data for three different alternatives in the table below, determine the best alternative using the incremental rate of return (∆RoR) analysis. MARR =9%. A B C First cost $15,000 $25,000 $20,000 O &M Cost/ year 1,600 400 900 Benefit/year 8,000 13,000 9,000 Salvage value 3,000 6,000 4,600 Life in years 4 4 4 The better alternative between the second increment (B – C) is:. Group of answer choices Alt. B Alt. C Alt. B or Alt. C Alt. A
Q4

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Given the data for two different alternatives in the table below, determine the best alternative using the incremental rate of return (AROR) analysis. MARR =9%. A B First cost $15,000 $20,000 O&M Cost/year 1,600 900 Benefit/year 8,000 9,000 Salvage value 3,000 4,600 Life in years 4 4