# Inen 509 - Economic Life of the Asset

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Homework Assignment 4 Solutions

1. Problem 5.1 (page 196) in textbook.

a. From equation 5.1

ib=0.15

tm=0.4

kb=(.15)(1-.4)=0.09 or 9%

b. from equation 5.2

1 + (ibtm/4)((P/A, kbp, 4) – (1 + ib)(P/F,kbp, 4) = 0

Try kbp = 9%

From table in Appendix A page 711

1 + (.15*.4/4)(3.2397199) – (1+ .15)(.7084252)= .2339068185

Try kbp =8%

From table in Appendix A page 710

1 + (.015)(3.3121268) – 1.15(.7350299) = .204398

Try kbp = 6%

From table in Appendix A page 708

1 + .015(3.4651056) – 1.15(.7920937) = 0.141071

Try kbp = 4%

From table in Appendix A page 706

1 + .015(3.6298952) – 1.15(.8548042) = 0.071427

Try kbp=2%

From table in Appendix A page 704

1 + .015(3.8077287) – 1.15(.9238454) = -0.0053063

kbp is therefore slightly more than 2% and you would need to increment it slightly and use 2.10 and 2.13 to calculate your factors for P/A and P/F

c. ibp = .15/4 = 0.0375

kb = [(1 + 0.0375(.6)]4 – 1 = 9.3%

d. ibp = (1+0.15).25 – 1

kb = [1 + 0.03556(.6)]4 – 1 = 0.08811 8.811%

2. Problem 5.5 (page 196) in textbook.

Source Amount x 106 % B/T cost A/T cost 3X5

Bank