Definicion Del Problema De Investigacion

Chapter 4

Nominal and Effective Interest Rates


Solutions to Problems

4.1 (a) monthly (b) quarterly (c) semiannually

4.2 (a) quarterly (b) monthly (c) weekly

4.3 (a) 12(b) 4 (c) 2

4.4 (a) 1 (b) 4 (c) 12

4.5 (a) r/semi = 0.5*2 = 1% (b) 2% (c) 4%

4.6 (a) i = 0.12/6 = 2% per two months; r/4 months = 0.02*2 = 4%
(b) r/6 months =0.02*3 = 6%
(c) r/2 yrs = 0.02*12 = 24%

4.7 (a) 5% (b) 20%

4.8 (a) effective (b) effective (c) nominal (d) effective (e) nominal

4.9 i/6months = 0.14/2 = 7%4.10 i = (1 + 0.04)4 – 1
= 16.99%

4.11 0.16 = (1 + r/2)2 –1
r = 15.41%

4.12 Interest rate is stated as effective. Therefore, i = 18%

4.13 0.1881 = (1 + 0.18/m)m – 1Solve for m by trial and gives m = 2

4.14 i = (1 + 0.01)2 –1
i = 2.01%


4.15 i = 0.12/12 = 1% per month
Nominal per 6 months = 0.01(6) = 6%
Effective per 6 months = (1 + 0.06/6)6 – 1= 6.15%

4.16 (a) i/week = 0.068/26 = 0.262%
(b) effective

4.17 PP = weekly; CP = quarterly

4.18 PP = daily; CP = quarterly

4.19 From 2% table at n=12, F/P = 1.2682

4.20 Interest rate is effective
From 6% table at n = 5, P/G = 7.9345

4.21 P = 85(P/F,2%,12) = 85(0.7885)
= $67.02 million

4.22 F = 2.7(F/P,3%,60)
=2.7(5.8916)
= $15.91 billion

4.23 P = 5000(P/F,4%,16)
= 5000(0.5339)
= $2669.50

4.24 P = 1.2(P/F,5%,1) (in $million)
= 1.2(0.9524)
=$1,142,880

4.25 P = 1.3(P/A,1%,28)(P/F,1%,2) (in $million)
= 1.3(24.3164)(0.9803)
= $30,988,577

4.26 F = 3.9(F/P,0.5%,120) (in $billion)
= 3.9(1.8194)= $7,095,660,000

4.27 P = 3000(250 – 150)(P/A,4%,8) (in $million)
= 3000(100)(6.7327)
= $2,019,810
4.28 F = 50(20,000,000)(F/P,1.5%,9)
=...