HW-Chap-3-2 - Engineering Economy 3-13 Double money at 4...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Engineering Economy GE 272 Chapter 3 3-13 Double money at 4% simple interest: 2P = P (1 + 0.04n) 2 = (1 + 0.04n) n = (2-1) /.04 = 25 years Double money at 4% compound interest: 2P = P (1 + 0.04) 2 log 2 = n log(1.04) n =log 2 / log 2.4 = 17.7 years 3-22 Year Cash Flow 1 −$2,000 2 −$4,000 3 −$3,625 4 −$3,250 5 −$2,875 3-26 (a) Future Worth $71 million = $165,000 (F/P, i%, 61) (F/P, i%, 61) = $71,000,000/$165,000 = 430.3 From interest tables: ( P/A, i %, 61) I 341.7 for 10% and 1,034.5 for 12% Performing linear interpolation:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/18/2011 for the course ENGINEER 300 taught by Professor Kashayar during the Spring '11 term at California State University Los Angeles .

Ask a homework question - tutors are online