February 11, 2014

Math 373

Fall 2013

Homework – Chapter 2

Chapter 2, Section 2

1. Brandon borrows 2000 at an interest rate of 7.2% compounded monthly. Brandon repays the

loan at the end of 6 years.

Determine the amount that Brandon repays.

Solution:

2000(1+

.072

12)12i6 = 3076.70

2. Guanling invests 600 in an account. After 6 years, Guanling has 1000.

Calculate the annual effective interest rate that Guanling earn on his investment.

Solution:

600(1+ i)6 = 1000

(1+ i)6 = 1.666666

(1+ i)66 = 1.6666666

1+ i = 1.0888669

i = 8.88669%

3. Kathy invests 10,000 in an account earning a nominal rate of interest of 8% compounded

quarterly.

Determine the amount of time in years that it will take for Kathy’s investment to grow to

25,000.

Solution:

10,000(1+.08

4)4iT = 25,000

(1+ .02)4iT = 2.5

ln(1.02)4iT = ln2.5

4T ln(1.02) = ln2.5

T =ln2.5

4 ln(1.02)= 11.5678

February 11, 2014

4. Avisek deposits 1000 in the bank earning an annual effective interest rate of i . Based on the

Rule of 72, Avisek believes that he will have 4000 at the end of 20 years.

Calculate the amount that Avisek will really have at the end of 20 years.

Solution:

Rule of 72: 0.72

i= amount of time for investment to double

In 20 years the investment would need to double twice in order to reach 4,000.

It would take the investment 10 years to double the first time, so we use this with the rule of 72.

.72

i= 10

i = .072

Now find the amount Chris will really have using i = 0.072.

(Initial Amount)(1 + i)T = (Final Amount)

1000(1.072)20 = 4016.94

5. Colleen invests 5250 in an account earning an annual effective interest rate of 7.2%. Colleen

wants to know when she will have 10,500. Her banker estimates that it will take X years using

the Rule of 72. Colleen calculates it exactly and gets Y years.

Calculate X Y .

Solution:

Banker:

X = 0.72 0.72

0.072i = 10 years

Colleen:

5,250 1.072 10,500

1.072 2

ln 1.072 ln 2

9.9696

Y

Y

Y

Y years

X – Y = 10 – 9.9696 = 0.0304

February 11, 2014

Chapter 2, Section 3

6. Camille borrows 1000 to buy a high definition television. She agrees to repay the loan with a

payment of P at the end of 2 years and an additional payment of P at the end of 3 years.

Camille is paying on the loan is an annual effective interest rate of 7% on the loan.

Determine P .

Solution:

1000(1.07)3 = P(1.07)+ P

1000(1.07)3 = P[(1.07)+1]

1225.04 = 2.07P

P = 591.81

7. Jose borrows 5000 from Yong. Jose repays the loan with a payment of 3000 at the end of 1 year

and 2500 at the end of 2 years.

Calculate the annual effective interest rate that Jose will pay on this loan.

Solution:

5000(1+ i)2 = 3000(1+ i)+ 2500

5000x2 - 3000x - 2500 = 0

Now use the quadratic formula and solve for x

3000 ± 30002 - 4(5000)(-2500)

2(5000)= 1.068115 = x

x = 1+ i

i = 1.068115 -1 = 6.8115%

February 11, 2014

8. Matt has 10,000 in his bank account today. Five years ago, he deposited 5000 into his account.

Additionally, two years ago, Matt deposited 4000 into his account.

Determine the annual effective interest rate that Matt has earned over the five year period.

Solution:

For this problem, use the financial calculator’s cash flow functionality.

CF0=5000

CO1=0

F01=1

CO2=0

FO2=1

CO3=4000

FO3=1

CO4=0

FO4=1

CO5= -10000

IRR, CPT

IRR=2.8899%

9. Devi deposits 1000 into a bank account on January 1, 2011. Devi also deposits 300 on October

1, 2011. On December 31, 2011, Devi has 1400.

Determine the annual effective interest rate earned by Devi during 2011.

Solution:

1000 300 -1400

Jan 1 Oct 1 Dec 1

Use the cash flow function on the calculator.

CF0=1000

C01 = 0

F01 = 8

C02= 300

F02 = 1

C03 = 0

F03 = 2

C04 = -1400

IRR CPT = 0.7456404612%

Annual Effective Rate = (1.007456404612)12 – 1 = 0.093239

February 11, 2014

10. Zach borrows 10,000 to be repaid with a payment of 6264.46 at time T and a payment of

6264.46 at time 2T . The annual effective rate on the loan is 7%.

Determine T in months.

Solution:

CF0 = 10,000

CO1 = -6264.46

FO1 = 2

IRRCPT = 16.443076%

(1.07)T = 1.16443076

T =ln1.16443076

ln1.07= 2.250008years*12 = 27months

Chapter 2, Section 4

11. Omer, Noah, and Dias enter into a financial arrangement. Under the arrangement, Omer will

pay Noah 1000 today. Noah will pay 500 to Omer at the end of 3 years as well as 650 to Dias at

the end of 4 years. Finally, Dias will make a payment of 800 to Omer at the end of 6 years.

Using the bottom line approach, determine the annual return for Omer on this transaction.

Solution:

Look at Omer’s cash flows.

-1000 500 800

0 3 6

1000 = 500(1 + i) -3 + 800(1 + i) -6

Let x = (1 + i) -3 and x2 = (1 + i) -6

800x2 + 500x – 1000 = 0

Use quadratic formula.

2

3

1

3

500 500 4 800 10000.848385976

2 800

1 0.848385976

0.848385976 1 0.05634

x

i

i

February 11, 2014

12. Cong, Erik, and Yifei enter into a financial arrangement. Cong agrees to pay Erik 3000 today. He

also agrees to pay Yifei 1000 at the end of one year. At the end of three years, Yifei will pay

Cong 4000. At the end of two years, Erik will pay X to Yifei and 1000 to Cong.

Using the bottom line approach, the annual yield rate or interest rate is the same for Cong and

Yifei.

Calculate X .

Solution:

Cong: -3000 -1000 +1000 +4000

Erik: +3000 -X-1000

Yifei: +1000 +X -4000

For Cong use the financial calculator:

CF0= -3000

CO1= -1000

CO2=1000

CO3=4000

CPT IRR= 9.14%

Yifei:

1 2 3

3 1

2

0 1000 1.0914 1.0914 4000 1.0914

4000 1.0914 1000 1.0914

1.0914

2573.56

X

X

X

February 11, 2014

13. Haoran, Jiayi, and Yinzhe are business partners. As part of their partnership, Haoran pays Jiayi

100,000 today. Additionally, at the end of 5 years, Jiayi agrees to pay 120,000 to Yinzhe. Finally,

at the end of T years, Yinzhe pays Haoran a total of 172,800.

Using the bottom line approach, all three partners have the same annual yield.

Determine T .

Solution:

Using Jiayi’s cash flows, we can find the interest rate using our financial calculators.

CF0=100000

C01=0

F01=4

C02= -120000

CPT IRR= 3.7137%

Now using this interest rate we can find T either using Haoran’s or Yinzhe’s cash flows. I will

use Haoran’s:

0 100000 172800 1.037137

0.5787 1.037137

ln(0.5787)15

ln(1.037137)

T

T

T years

February 11, 2014

Chapter 2, Section 5

14. Cody loans 1000 to Gabbie. Gabbie agrees to repay the loan with two payments of 600. One

payment will be made at the end of three years and the other payment will be made at the end

of six years.

a. Determine the annual effective interest rate on this loan from Gabbie’s viewpoint.

Solution:

It will be easiest to use the financial calculator here.

Remember that either the initial cash flow or the future cash flows need to be

negative. From Gabbie’s point of view it makes the most sense for the initial to

be positive (since he is receiving money) and the later cash flows to be negative

(since he is making payments).

CF0= 1000

C01=0, F01=2

C02= -600, F02=1

C03=0, F03=2

C04= -600, F04=1

IRR CPT = 4.1784%

b. Calculate Cody’s annual return on this loan over the six year period assuming that

he can reinvest repayments at an interest rate of 5%.

Solution:

Cody will be reinvesting the payments he receives. So at time six his investments will

be worth 3600(1.05) 600 1294.575 .

Setting this value equal to the amount he originally lent to Gabbie, we can find his

annual return.

6

6

1294.575 1000(1 )

1294.5751 4.39696%

1000

i

i

February 11, 2014

15. Collin makes a loan of 30,000 to John. John will repay the loan with payments of 10,000 at the

end of one year, 11,000 at the end of two years, and 12,000 at the end of three years.

Collin reinvests each payment at an interest rate of i .

Taking into account reinvestment, Collin’s annual yield is 9%.

Calculate i .

Solution:

We know that Collin’s annual yield is 9% when we take into account his reinvestments. This

gives us the equation: 330000(1.09) 38850.87 . We also know that this value must be equal to

the future value of the reinvested payments from John:

2

2

2

2

2

38850.87 10000(1 ) 11000(1 ) 12000

26650.87 10000(1 ) 11000(1 )

10000 11000 26650.87 0

1.12 2.665087

1.1 1.1 4( 2.665087)

2

1.17846 (1 )

17.846%

i i

i i

x x

x

x

x i

i

16. Derek lends 600 to a friend. The loan will be repaid with payments of P at the end of one year

and 300 at the end of 3 years. Derek reinvests the payment of P at an annual effective interest

rate of 8.25%.

Taking into account reinvestment, Derek earns an annual yield of 10% on this loan.

Determine P .

Solution:

Since Derek earns an annual yield of 10% taking into account reinvestment, we have 3600(1.1)

as the final amount of money Derek has. We know that this amount will be equal to the first

payment P reinvested for two years plus the final payment of 300.

3 2

3

2

600(1.1) (1.0825) 300

600(1.1) 300425.50

(1.0825)

P

P

February 11, 2014

Chapter 2, Section 6

17. Kylie has an account at Boing Brokerage House. She has made the following deposits and

withdrawals over the last two years.

Date Deposits Withdrawals Balance Before Cash Flow

January 1, 2009 5000 0 0

March 1, 2009 5000 0 4000

July 1, 2009 0 4000 10,000

May 1, 2010 3000 0 9,000

December 31, 2010 0 0 15,000

a. Estimate Kylie’s dollar weighted annual return using simple interest.

Solution:

A + Ct + I = BåA = 5000

B = 15000

Ct = 5000 - 4000 + 3000 = 4000å5000 + 4000 + I = 15000

I = 6000

j =I

A + Ct (1- t)å=

6000

5000 + 5000(1-1

12)- 4000(1-

1

4)+ 3000(1-

2

3)

=6000

7583.33= .7912

1+ i = (1.7912)1/2 ®1+ i = 1.33836® i = 33.836%

February 11, 2014

b. Calculate Kylie’s exact annual dollar weighted return using the cash flow functionality in

the BA II Plus.

Solution:

Treat Cash Flows as monthly:

0 5000

01 0; 01 1

02 5000; 02 1

03 0; 03 3

04 4000; 04 1

05 0; 05 9

06 3000; 06 1

07 0; 07 7

08 15,000; 08 1

2.478375127%

Since the periods are months, the IRR is the monthly

effective intere

CF

C F

C F

C F

C F

C F

C F

C F

C F

IRR CPT

(12)

12(12)

12

st rate which is 2.478375127%12

(1 ) 1 (1 0.02478375127) 34.149%12

i

ii i

February 11, 2014

18. Jing has a bank account balance of X on January 1, 2010. During the next two years Jing deposits 8000 into his account and withdraws 3000. On January 1, 2012, Jing has a bank account balance of 28,400. Assuming that all cash flows occur on January 1, 2011, Jing estimates his annual dollar weighted return assuming simple interest to be 4.5383664%. Calculate X.

Solution:

0.5

2

28400

0.5

8000 3000 5000

23400

(1 ) (1 ) 1.045383664

1.045383664 1 0.09287005

234000.092827005

(1 ) 5000(0.5)

0.092827005( 2500) 23400

21200

A X

B

k

C

I X

j i

j

I Xj

A C k X

X X

X

Chapter 2, Section 7

19. Wenxue invested 50,000 in an account three years ago. One year ago, the account had a

balance of 100,000 and Wenxue withdrew $50,000.

Today, Wenxue has 30,000 in the account.

Determine the annual effective time weighted return earned by Wenxue on this account.

Solution:

1+ j1 =B0

B1

=100,000

50,000= 2

1+ j2 =B2

B1 +C1

=30,000

100,000 - 50,000= .6

1+ jTW = (1+ j1)(1+ j2 ) = (2)(.6) = 1.2

1+ iTW = (1+ jTW )1/T

1+ iTW = (1.2)1/3 ® iTW = .06266

February 11, 2014

20. Kylie has an account at Boing Brokerage House. She has made the following deposits and

withdrawals over the last two years.

Date Deposits Withdrawals Balance Before Cash Flow

January 1, 2009 5000 0 0

March 1, 2009 5000 0 4000

July 1, 2009 0 4000 10,000

May 1, 2010 3000 0 9,000

December 31, 2010 0 0 15,000

Calculate Kylie’s annual time weighted return.

Solution:

1+ j1 =4000

5000= .8

1+ j2 =10000

4000 + 5000= 1.111111

1+ j3 =9000

10000 - 4000= 1.5

1+ j4 =15000

9000 + 3000= 1.25

1+ jTW = (.8)(1.1111)(1.5)(1.25) = 1.66667

1+ iTW = (1.66667)1/2 ® iTW = 29.099%

February 11, 2014

21. Evgeny invests in the Marano Mutual Fund. Over the next two years, Evgeny realizes an annual

time weighted yield of 12.5%.

Evgeny initially invests 100,000 with Marano and has 103,551.14 at the end of two years.

During the two year period, Evgeny also withdrew an amount of C to buy a new car. Before

the amount was withdrawn, the fund was worth 110,000.

Calculate C.

Solution:

First, let’s find the time weighted yield for the period of two years.

1/

2

(1 ) (1 )

(1.125) 1 0.265625

T

TW TW

TW

j i

j

Now we can solve for C

110000 103551.141 0.265625

100000 110000

110000 103551.14110000 20,000

100000 1.265625

c

c

February 11, 2014

Answers

1. 3076.60

2. 8.8867%

3. 11.5678 years

4. 4016.94

5. 0.0304

6. 591.81

7. 6.8115%

8. 2.8899%

9. 9.3239%

10. 27 months

11. 5.634%

12. 2573.56

13. 15 years

14.

a. 4.1784%

b. 4.3970%

15. 17.8464%

16. 425.50

17.

a. 33.836%

b. 34.149%

18. 21,200

19. 6.266%

20. 29.099%

21. 20,000