The following two mutually exclusive alternative investments with different service lives are being compared at MARR of 6%. Alt. Q 15 Service life (years) Net PW (over service life) Net AW (over service life) Based on the information given, which of the two investments is more economical? $1,834,000 $249,182 Alt P 20 O A. Alt. P because it has a higher net PW OB. Alt Q because it has a higher net AW $2,312,000 $238,050

The following two mutually exclusive alternative investments with different service lives are being compared at MARR of 6%. Alt. Q 15 Service life (years) Net PW (over service life) Net AW (over service life) Based on the information given, which of the two investments is more economical? $1,834,000 $249,182 Alt P 20 O A. Alt. P because it has a higher net PW OB. Alt Q because it has a higher net AW $2,312,000 $238,050

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

See similar textbooks

Similar questions

An equipment costing $57,500 is being considered for a production process at Dewey Chemicals. The expected benefits per year is $4,500 and estimated salvage value is $10,000. Determine the rate of return the company can get in this equipment proposal. Equipment life = 15 years.
I need solve this question in Engineering economics
We have the following K-T conditions: 1-10z + 3 – 6A 0 & x(10z+ 3 – 6) = 0 2-10x +1 – 31 0 & z(10x +1– 3) = 0 3-3 – 6x – 3z >0 & \>0 & \(3– 6x %D Which of the following best represents the problem associated with those K-T conditions? min(2.2) 10xz + 3x + z such that 3 6x + 3z Omin(r,2) 10æz + 3x + z such that 3 > 6x + 3z
The MARR is 15%. Three alternatives are available and the associated cash flow is as follow: Year First Cost A B C $1,700 $2,100 $3,750 Annual Benefit Useful Life 2 $1,000 $1,000 $1,000 3 6 Answer the following in this format: 1.23 1. The payback period for Alternative A is 2. The payback period for Alternative B is 3. The payback period for Alternative C is Based on Payback period analysis, Alternative should be selected. (Enter only the letter)
!
4-10 Improvement and expansion of a production facility are being given economic consideration. At the present time (Alternative A), profits are running at $450,000 per year. Assume a washout in operating profit in cach of years one through 10 so they will remain constant at $450,000 per year. Two mutually exclusive alternatives for improve- ment and improvement combined with expansion are being consid- ered with projected costs and revenues as shown on the following time diagrams with all dollar values expressed in thousands of dollars. 450 450 450 A) ..... .... 0. 2 10 ... -800 700 700 700 B) 2 ... 10 850 -900 C) -1,300 1,050 1,050 1 ..... 10 1) For a minimum rate of return of 15% determine whether the present project "A," improvement project "B," or improvement plus expansion project "C" is economically best using ROR, NPV, and PVR. 2) Then increase the minimum ROR to 25% from 15% over the entire 10-year evaluation life and re-evaluate the alternatives using both ROR and NPV…
am. 117.
4.3. Use ERR analysis to determine which alternative would be the best economical choice. Verify your decision with NPC analysis. Assume the MARR equals 10%. Alternative A: -500 -25 -25 -25 -25 L = 100 Alternative B: 1 3 -300 -50 -50 -50 -50 L= 25 Alternative C: 2 3 -250 -75 -60 -45 -30 L= 10 Alternative D: 2 450 -35 -35 -35 -35 L= 100
Two mutually exculsive alternative are being considered for the environmental protection equipment at a petroleum refinary. one of these alternatives must be selected. The estimated cash flow for each alternatives are as follows. Use MARR of 10% Note: Do not round in between solutions and round up to 2 decimal places for final asnwer. A. Which alternative is preferred, based on the repeatability assumptions? Use AW method.
A company is considering two machines. Machine X has a first cost of $30,000, AOC of $18.000, and 5 of $7000 after 4 years. Mact will cost $50,000 with an AOC of $16,000 and 5 of $9000 after 6 years. The AW equations for Machine X and Y at 12% interest rate Select one: O a AWX=-30,000(A/P12 % .4) -18,000 7,000(A/F.12%.4); AWY- 50,000(A/P12 %6) -16.000 9.000(A/F.12%,6) O b. AWX=-30,000(A/P12%24) -18,000 7,000(A/F.12%24) AWY- -50,000(A/P12%24)-16.000 9.000(A/F12%24) Oc AWX=-30.000(A/P12 % 6 ) -18,000 7,000(A/F.12%6) AWY 50.000(A/P12% 6)-16.000 9,000(A/F.12%,6) Od AWX-30,000(A/P12% 12)-18.000 7.000(A/F.12% 121 AWY- 50.000(A/P12% 12)-16,000 9,000(A/F.12%,12) .
7
The table below summarizes the comparison between four mutually exclusive alternatives using the IRR method. The MARR is 10%. Which alternative should be the base aftemative in the first step of the incremental analysis? Alternative Initial Cost Annual Income IRR P $110,000 $21.500 14.5% Q $130,000 $31,200 20.2% OA Atemative R because it has the lowest initial cost OB. Alternative 5 because it has the largest initial cost OC. Allamative Q because it has the highest IRR D. Alemative P because it has the lowest IRR R $95,000 $20,000 16.5% $ $354,000 $74,000 16.3%