EGG 307 HW 6

EGG 307 HW 6 - EGG 307: Engineering Economics: Solution to...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EGG 307: Engineering Economics: Solution to Homework No. 6 Spring 2009 Homework Problems: Chapter 5 Problem : 5-34 Chapter 6 Problems : 6-4; 6-47; 6-55. Notes: (1) For Problem 5-34, use the trial-and-error method to compute the IRR. (2) For Problem 6-4, use the AW method and confirm your results using the PW method. (3) For Problem 6-47, assume the useful lives for both designs is infinity (not 83 and 92 years). (4) For Problem 6-55, use the PW method (not the IRR method). Ignore the IRR information given. Solutions: 5-34(a) To estimate lower limit value of i , ignore the salvage value and find NPW = -10,000,000 + 2,800,000(P/A, i , 4) = 0 Therefore, (P/A, i , 4) = 10,000,000/2,800,000 = 3.5714 From tables, the nearest interest rate that gives (P/A, i , 4) close to 3.5714 is 5% To estimate upper limit value of i , use A = $2,800,000 + $5,000,000/4 = $4,050,000 Therefore, (P/A, i , 4) = 10,000,000/4,050,000 = 2.4691 From tables, the nearest interest rate that gives (P/A, i , 4) close to 2.4691 is 20%
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

EGG 307 HW 6 - EGG 307: Engineering Economics: Solution to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online