ch 7 solutions final

7/28/2019 Ch 7 Solutions Final

1/19

Chapter 7

Rate of Return Analysis: Single Alternative

Solutions to Problems

7.1 A rate of return of 100% means that the entire investment is lost.

7.2 Balance = 10,000(1.50) 5(2638)= $1810

7.3 (a) Annual payment = [10,000/4 + 10,000(0.10)]= $3500

(b) A = 10,000(A/P,10%,4)= 10,000(0.31547)= $3154.70

7.4 Monthly pmt = 100,000(A/P,0.5%,360)= 100,000(0.00600)= $600

Balloon pmt = 100,000(F/P,0.5%,60) 600(F/A,0.5%,60)= 100,000(1.3489) 600(69.7700)

= $93,028

7.5 0 = -150,000 + (33,000 27,000)(P/A,i,30)(P/A,i,30) = 25.0000

i = 1.2% per month (interpolation or Excel)

7.6 0 = -400,000 + [(10(200) + 25(50) +70(100)](P/A,i%,48)(P/A,i%,48) = 39.0240

Solve by trial and error or Exceli = 0.88% per month (Excel)

7.7 0 = -30,000 + (27,000 18,000)(P/A,i%,5) + 4000(P/F,i%,5)

Solve by trial and error or Exceli = 17.9 % (Excel)

Chapter 7 1

PROPRIETARY MATERIAL. The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may bedisplayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used

beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you area student using this Manual, you are using it without permission.


2/19

7.8 0 = -130,000 49,000(P/A,i%,8) + 78,000(P/A,i%,8) + 1000(P/G,i%,8)+ 23,000(P/F,i%,8)

Solve by trial and error or Exceli = 19.2% (Excel)

7.9 (100,000 10,000)i = 10,000i = 11.1%

7.10 0 = -10 4(P/A,i%,3) - 3(P/A,i%,3)(P/F,i%,3) + 2(P/F,i%,1) + 3(P/F,i%,2)+ 9(P/A,i%,4)(P/F,i%,2)

Solve by trial and error or Exceli = 14.6% (Excel)

7.11 (a) 0 = -(220,000 + 15,000 + 76,000)(A/P,i%,36) + 12,000(2.00 1.05) + 2000

+ 100,000(A/F,i%,36)0 = -(311,000)(A/P,i%,36) + 13,400 + 100,000(A/F,i%,36)


(b) Nominal per year = 3.3(12)= 39.6% per year

Effective per year = (1 + 0.396/12)12 1= 47.6% per year

7.12 0 = -40 28(P/A, i%,3) + 5(P/F,i%,4) + 15(P/F,i%,5) + 30(P/A,i%,5)(P/F,i%,5

Solve by trial and error or Exceli = 5.2% per year (Excel)

7.13 (a) 0 = -41,000,000 + 55,000(60)(P/A,i%,30)


(b) 0 = -41,000,000 + [55,000(60) + 12,000(90)](P/A,i%,30)0 = -41,000,000 + (4,380,000)(P/A,i%,30)


Chapter 7 2




3/19

7.14 Cash flow tabulation is below. Note that both series endat the end of month 11.

Month M3Turbo M3Power Difference0 -7.99 -(14.99 + 10.99) -17.99

1 -7.99 0 +7.992 -7.99 -10.99 - 3.003 -7.99 0 + 7.994 -7.99 -10.99 - 3.005 -7.99 0 +7.996 -7.99 -10.99 -3.007 -7.99 0 +7.998 -7.99 -10.99 -3.009 -7.99 0 +7.9910 -7.99 -10.99 - 3.0011 -7.99 0 +7.99

(a) 0 = -17.99 + 7.99(P/F,i%,1) 3.00(P/F,i%,2) + 7.99(P/F,i%,3) 3.00(P/F,i%,4)+ ... + 7.99(P/F,i%,9) 3.00(P/F,i%,10) + 7.99(P/F,i%,11)


(b) Nominal per year = 12.2(12) = 146.4% per year

Effective = (1 + 0.122)12 1= 298% per year

7.15 0 = -90,000(A/P,i%,24) 0.014(6000) + 0.015(6000)(150)0 = -90,000(A/P,i%,24) + 13,416


7.16 0 = -110,000 + 4800(P/A,i%,60)(P/A,i%,60) = 22.9167

Use tables or Exceli = 3.93% per month (Excel)

7.17 0 = -210 150(P/F,i%,1) + [100(P/A,i%,4) + 60(P/G,i%,4)](P/F,i%,1)


Chapter 7 3




4/19

7.18 0 = -450,000 [50,000(P/A,i%,5) 10,000(P/G,i%,5)] + 10,000(P/A,i%,5)+10,000(P/G,i%,5) + 80,000(P/A,i%,7)(P/F,i%,5)

Solve by trial and error or Excel

i = 2.36% per quarter (Excel)= 2.36(4)= 9.44% per year (nominal)

7.19 0 = -950,000 + [450,000(P/A,i%,5) + 50,000(P/G,i%,5)] )(P/F,i%,10)


7.20 10,000,000(F/P,i%,4)(i) = 100(10,000)Solve by trial and error

i = 7.49%

7.21 [(5,000,000 200,000)(F/P, i%,5) 200,000(F/A,i%,5)](i) = 1,000,000Solve by trial and error

i = 13.2%

7.22 In a conventional cash flow series, there is only one sign change in the net cashflow. A nonconventional series has more than one sign change.

7.23 Descartes rule uses net cash flows while Norstroms criterion is based oncumulative cash flows.

7.24 (a) three; (b) one; (c) five

7.25 Tabulate net cash flows and cumulative cash flows.

Quarter Expenses Revenue Net Cash Flow Cumulative0 -20 0 -20 -201 -20 5 -15 -352 -10 10 0 -353 -10 25 15 -204 -10 26 16 -45 -10 20 10 +66 -15 17 2 +87 -12 15 3 +118 -15 2 -13 -2

(a) From net cash flow column, there are two possible i* values

Chapter 7 4




5/19

(b) In cumulative cash flow column, sign starts negative but it changes twice.Therefore, Norstroms criterion is not satisfied. Thus, there may be up to twoi* values. However, in this case, since the cumulative cash flow is negative,there is no positive rate of return value.

7.26 (a) There are two sign changes, indicating that there may be two real-number rate ofreturn values.

(b) 0 = -30,000 + 20,000(P/F,i%,1) + 15,000(P/F,i%,2) - 2000(P/F,i%,3)Solve by trial and error or Exceli = 7.43% per year (Excel)

7.27 (a) There are three sign changes, Therefore, there are three possible i* values.

(b) 0 = -17,000 + 20,000(P/F,i%,1) - 5000(P/F,i%,2) + 8000(P/F,i%,3)Solve by trial and error or Excel

i = 24.4% per year (Excel)

7.28 The net cash flow and cumulative cash flow are shown below.

Year Expenses, $ Savings, $ Net Cash Flow, $ Cumulative, $0 -33,000 0 -33,000 -33,0001 -15,000 18,000 +3,000 -30,0002 -40,000 38,000 -2000 -32,0003 -20,000 55,000 +35,000 +30004 -13,000 12,000 -1000 +2000

(a) There are four sign changes in net cash flow, so, there are four possiblei* values.

(b) Cumulative cash flow starts negative and changes only once. Therefore, there isonly one positive, real solution.

0 = -33,000 + 3000(P/F,i%,1) - 2000(P/F,i%,2) + 35,000(P/F,i%,3)-1000(P/F,i%,4)


7.29 Cumulative cash flow starts negative and changes only once, so, there isonly one positive real solution.

0 = -5000 + 4000(P/F,i%,) + 20,000(P/F,i%,4) - 15,000(P/F,i%,5)


Chapter 7 5




6/19

7.30 Reinvestment rate refers to the interest rate that is used for funds that arereleasedfrom a project before the project is over.

7.31 Tabulate net cash flow and cumulative cash flow values.

Year Cash Flow, $ Cumulative, $1 -5000 -5,0002 -5000 -10,0003 -5000 -15,0004 -5000 -20,0005 -5000 -25,0006 -5000 -30,0007 +9000 -21,000

8 -5000 -26,0009 -5000 -31,00010 -5000 + 50,000 +14,000

(a) There are three changes in sign in the net cash flow series, so there are threepossible ROR values. However, according to Norstroms criterion regardingcumulative cash flow, there is only one ROR value.

(b) Move all cash flows to year 10.

0 = -5000(F/A,i,10) + 14,000(F/P,i,3) + 50,000

Solve for i by trial and error or Excel

i = 6.3% (Excel)

(c) If Equation [7.6] is applied, all F values are negative except the last one. Therefore,i is used in all equations. The composite ROR (i) is the same as the internalROR value (i*) of 6.3% per year.

7.32 First, calculate the cumulative CF.

Year Cash Flow, $1000 Cumulative CF, $10000 -65 -651 30 -352 84 +493 -10 +394 -12 +27

(a) The cumulative cash flow starts out negatively and changes sign only once,

Chapter 7 6




7/19

indicating there is only one root to the equation.

0 = -65 + 30(P/F,i,1) + 84(P/F,i,2) 10(P/F,i,3) 12(P/F,i,4)

Solve for i by trial and error or Excel.


(b) Apply net reinvestment procedure because reinvestment rate, c, is not equalto i* rate of 28.6% per year:

F0 = -65 F0 < 0; use iF1 = -65(1 + i) + 30 F1 < 0; use iF2 = F1(1 + i) + 84 F2 > 0; use c(F2 must be > 0 because last two terms

are negative)F3 = F2(1 + 0.15) -10 F3 > 0; use c(F3 must be > 0 because last term is

negative)F4 = F3(1 + 0.15) 12

Set F4 = 0 and solve for i by trial and error:

F1 = -65 65i + 30F2 = (-65 65i + 30)(1 + i) + 84

= -65 - 65i + 30 65i 65i 2 + 30i +84= -65i2 100i + 49

F3 = (-65i2 100i + 49)(1.15) 10

= -74.8 i2 115i + 56.4 10= -74.8 i2 115i + 46.4

F4 = (-74.8 i2 115i + 46.4)(1.15) 12

= -86 i2 132.3i + 53.3 12= -86 i2 132.3i + 41.3

Solve by quadratic equation, trial and error, or spreadsheet.


7.33 Apply net reinvestment procedure.

F0 = 3000 F0 > 0; use cF1 = 3000(1 + 0.14) - 2000

= 1420 F1 > 0; use cF2 = 1420(1 + 0.14) + 1000

= 2618.80 F2 > 0; use cF3 = 2618.80(1 + 0.14) 6000

= -3014.57 F3 < 0; use i

Chapter 7 7




8/19

F4 = -3014.57(1 + i) + 3800

Set F4 = 0 and solve for i.0 = -3014.57(1 + i) + 3800i = 26.1%

7.34 Apply net reinvestment procedure because reinvestment rate, c, is not equalto i* rate of 44.1% per year (from problem 7.29):

F0 = -5000 F0 < 0; use iF1 = -5000(1 + i) + 4000

= -5000 5000i + 4000= -1000 5000i F1 < 0; use i

F2 = (-1000 5000i)(1 + i)= -1000 5000i 1000i 5000i2

= -1000 6000i 5000i2 F2 < 0; use i

F3 = (-1000 6000i 5000i

2

)(1 + i)= -1000 6000i 5000i2 1000i 6000i2 5000i3

= -1000 7000i 11,000i2 5000i3 F3 < 0; use iF4 = (-1000 7000i 11,000i

2 5000i3)(1 + i) + 20,000= 19,000 8000i 18,000i2 16,000i3 - 5,000i4 F4 > 0; use c

F5 = (19,000 8000i 18,000i2 16,000i3 - 5,000i4)(1.15) 15,000

= 6850 9200i 20,700i2 18,400i3 - 5,750i4

Set F5 = 0 and solve for i by trial and error or spreadsheet.

i = 35.7% per year7.35 (a) i = 25,000(0.06)/2

= $750 every six months

(b) The bond is due in 22 years, so, n = 22(2) = 44

7.36 i = 10,000(0.08)/4= $200 per quarter

0 = -9200 + 200(P/A,i%,28) + 10,000(P/F,i%,28)

Solve for i by trial and error or Exceli = 2.4% per quarter (Excel)

Nominal i/yr = 2.4(4) = 9.6% per year

Chapter 7 8




9/19

7.37 (a) i = 5,000,000(0.06)/4 = $75,000 per quarter

After brokerage fees, the City got $4,500,000. However, before brokeragefees, the ROR equation from the Citys standpoint is:

0 = 4,600,000 75,000(P/A,i%,120) - 5,000,000(P/F,i%,120)


i = 1.65% per quarter (Excel)

(b) Nominal i per year = 1.65(4)= 6.6% per year

Effective i per year = (1 + 0.066/4)4 1

= 6.77% per year

7.38 i = 5000(0.04)/2= $100 per six months

0 = -4100 + 100(P/A,i%,22) + 5,000(P/F,i%,22)


i = 3.15% per six months (Excel)

7.39 0 = -9250 + 50,000(P/F,i%,18)(P/F,i%,18) = 0.1850

Solve directly or use Excel


7.40 i = 5000(0.10)/2= $250 per six months

0 = -5000 + 250(P/A,i%,8) + 5,500(P/F,i%,8)


i = 6.0% per six months (Excel)

Chapter 7 9




10/19

7.41 (a) i = 10,000,000(0.12)/4= $300,000 per quarter

By spending $11 million, the company will save $300,000 every three months

for 25 years and will save $10,000,000 at that time. The ROR will be:

0 = -11,000,000 + 300,000(P/A,i%,100) + 10,000,000(P/F,i%,100)i = 2.71% per quarter (Excel)

(b) Nominal i per year = 2.71(4)= 10.84% per year

FE Review Solutions

7.42 Answer is (d)

7.43 Answer is (c)

7.44 0 = 1,000,000 20,000(P/A,i,24) 1,000,000(P/F,i,24)Solve for i by trial and error or Exceli = 2% per month (Excel)Answer is (b)

7.45 Answer is (b)

7.46 0 = -60,000 + 10,000(P/A,i,10)(P/A,i,10) = 6.0000

From tables, i is between 10% and 11%Answer is (a)

7.47 Answer is (c)

7.48 0 = -50,000 + (7500 5000)(P/A,i,24) + 11,000(P/F,i,24)

Solve for i by trial and error or Exceli = 2.6% per month (Excel)Answer is (a)

7.49 0 = -100,000 + (10,000/i) (P/F,i,4)

Solve for i by trial and error or Exceli = 9.99%% per year (Excel)Answer is (a)

Chapter 7 10




11/19

7.50 i = 4500/50,000= 9% per yearAnswer is (c)

7.51 Answer is (d)

7.52 250 = (10,000)(b)/2b = 5% per year payable semiannuallyAnswer is (c)

7.53 Since the bond is purchased for its face value, the interest rate received by thepurchaser is the bond interest rate of 8% per year payable quarterly. This is anominal interest rate per year. The effective rate per quarter is 2%Answer is (a)

7.54 Answer is (a)

7.55 Since the bond was purchased for its face value, the interest rate received by thepurchaser is the bond interest rate of 10% per year payable quarterly. Answers (a)and (b) are correct. Therefore, the best answer is (c).

Chapter 7 11




12/19

Extended Exercise 1 Solution

Solution by hand

1. Charles payment = 5000(A/P,10%,3)

= 5000(0.402115) (by formula)= $2010.57

Beginning Total Endingunrecovered amount unrecovered

Year balance Interest owed Paymentbalance

(1) (2) (3) = 0.1(2) (4)=(2)+(3) (5) (6)=(4)+(5)0 $-5000.00 $-5000.001 $-5000.00 $-500.00 -5500.00 $2010.57 -3489.432 -3489.43 -348.94 -3838.37 2010.57 -1827.80

3 -1827.80 -182.78 -2010.58 2010.57 -0.01*$-1031.72 $6031.71

*round-offJeremys payment = $2166.67

Beginning Total Endingunrecovered amount unrecovered

Year balance Interest owed Paymentbalance

(1) (2) (3) = 0.1(5000) (4)=(2)+(3) (5) (6)=(4)+(5)0 $-5000.00 $-5000.001 $-5000.00 $-500 -5500.00 $2166.67 -3333.332 -3333.33 -500 -3833.33 2166.67 -1666.673 -1666.67 -500 -2166.67 2166.67 0.00

$-1500 $6500.01

Plot year versus column (4) in the form of Figure 71 in the text.

2. More forJeremy Charles Jeremy

Interest $1500.00 $1031.72 $468.28Total 6500.01 6031.71 468.30

Solution by computer

1. The following spreadsheets have the same information as the two tables above.The x-y scatter charts are year (column A) versus total owed (column B). (Theindicator lines and curves were drawn separately.)

Chapter 7 12




13/19

2. The second spreadsheet shows that $468.28 more is paid by Jeremy.

Chapter 7 13




14/19

Chapter 7 14




15/19

Extended Exercise 2 Solution

The spreadsheet above summarizes all answers in $1000. Some cells must be changed toobtain the rate of return values shown in column H. These are described below.

1. Use IRR function in year 4 and add $225 in cell C8 for year 4.i* (sell after 4) = 38.09%

With no sale, IRR results in:i* (no sale after 4) = 3.04%

2. Use IRR function with $60 added into cell C11 for year 7.i* (sell after 7) = 17.74%i* (no sale after 7) = 10.95

3. i* (no sale after 10) = 4.84%

4. Use IRR function with $25 added into cell C14 for year 10.i* (charity after 10) = 9.5%

Chapter 7 15




16/19

Case Study Solution

Chapter 7 16




17/19

Chapter 7 17




18/19

Chapter 7 18




19/19

Chapter 7 19


beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are

ch 7 solutions final

Documents