This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1 = 0.0539 5.39%/year. Problem 3 Solution: (a) A = P[A/P, i , N], where P = 90% of $180,000 = $162,000; I = %; N = 120 months; i.e., A = 162,000(A/P, %, 120) = 162,000x0.0127 = $2,057 (factor obtained from tables) (b) Balance after 24 months = PW of remaining (12024 = 96) payments, i.e., P B = A[P/A, %, 96] = 2,057x68.259 = $140,436 (factor obtained from tables) Total new loan amount, P N = 140,436 + 1,500 = $141,936 New monthly payment, A = P N (A/P, %, 36) = 141,936x0.0304 = $4,315 Problem 4 Solution: Present worth of the revenues, P R = $1,200,000(P/A, 20%, 9)  $80,000(P/G, 20%, 9) = $1,200,000x4.031 80,000x11.434 = $3,922,480 Present worth of the costs, P C = $1,000,000 + 300,000(P/A, 20%, 9) = $1,000,000 + 300,000x4.031 = $2,209,290 Since PW(revenues) > PW(costs) This is an economical investment....
View
Full
Document
This note was uploaded on 07/13/2009 for the course EGG a taught by Professor Dr.kaseko during the Spring '09 term at University of Nevada, Las Vegas.
 Spring '09
 Dr.Kaseko

Click to edit the document details