solutions to end of chapter questions

Solutions to End of Chapter Questions

Calculator Settings Many of the solutions below utilize the BA II Plus calculator, and those solutions are based on the following assumptions:

The calculation method is set to the algebraic operating system (AOS). If the calculation method is currently set to the chain calculation method (Chn), then it can be changed to the AOS method with the following key strokes:

[2nd] [FORMAT] ↓↓↓↓ [2nd] [SET] [2nd] [QUIT] The payments per year (P/Y) and compounding periods per year (C/Y) are set to 1.

Changing the P/Y setting to 1 will automatically change the C/Y setting to 1: [2nd] [P/Y] 1 [ENTER] [2nd] [QUIT]

The relevant worksheet register has been cleared prior to data entry. The keystrokes to clear the registers are:

[2nd] [CLR TVM] to clear the time value of money worksheet [2nd] [CLR WORK] to clear the other worksheets

Unless stated otherwise, the calculator is set to treat payments as occurring at the end of each period. To toggle the calculator between treating payments as occurring at the beginning of each period or the end of each period, use the following key strokes: [2nd] [BGN] [2nd] [SET] [2nd] [QUIT]

If you take an actuarial exam, the exam administrator will reset your calculator prior to the exam. This means that it will then revert to using the chain calculation method, and it may change the settings for P/Y and C/Y. After it has been reset, you can easily change it back to the settings shown above. If you would like to practice, you can reset the calculator as follows: [2nd] [RESET] [ENTER] [2nd] [QUIT]

Chapter 1: Setting the Stage

Solution 1.01 C Section 1.04, Terminology Statement II is false. The current value could include the accumulated value of payments that have occurred in the past. Statement IV is false. A borrower receives funds now. The lender provides the funds.

Solution 1.02 E Section 1.05, Supply and Demand for Funds The supply and demand curves can change over time for a variety of reasons, including lifestyle decisions and demographic changes.

Chapter 2: Simple Interest and Discount Solutions to End of Chapter Questions

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Chapter 2: Simple Interest and Discount

Solution 2.01 D Section 2.01, Simple Interest Since the simple interest rate is expressed as an annual value, we use years as our unit of time. 18 months is equal to 1.5 years, so the funds accumulate from time 1.5 to time 5, which is 3.5 years. The accumulation function is based on the deposit being made at time 0, so we interpret t to be the time from the deposit until the valuation date:

0

3.5

3.5

(1 )10,000(1 0.07 3.5)

tAV PV itAVAV

12,450

Solution 2.02 D Section 2.02, Simple Interest The present value of $10,000 in 8 years is:

0

10,000 10,0001 1 0.07 8 1.56

tAVPV

it6,410.26

Solution 2.03 A Section 2.02, Simple Discount The present value of $10,000 in 8 years is: 0 (1 ) 10,000(1 0.07 8) 10,000(0.44)tPV AV dt 4,400

Solution 2.04 B Section 2.04, Simple Interest Let’s assume that the amount due in 45 days is X. Larry will choose option 2 if it results in having surplus funds after paying X at the end of 45 days. That is, he will choose option 2 if the accumulated value minus the amount due is greater than 0:

0.97 0301365

0.97 1 0301365

0.97 1301365

300.97 1365

30 1 0.97365

X Xd

d

d

d

d

d 0.3650

The minimum value of d that will cause Larry to choose option 2 is 36.50%.



Solution 2.05 E Section 2.02, Simple Interest and Discount The accumulated values are calculated below:

5

5

6

4

8

1,000(1 0.05 5) 1,250900 1,200

1 0.05 5900 1,184.21

1 0.04 61,000(1 0.06 4) 1,240

1,000 1,315.791 0.03 8

A AV

B AV

C AV

D AV

E AV

The highest accumulated value is Choice E.

Solution 2.06 C Section 2.03, Equivalent Simple Interest and Discount Rates The accumulated values are equal at the end of t years:

2

2

100100(1 0.05 )1 0.04

(1 0.05 )(1 0.04 ) 1

1 0.05 0.04 0.002 1

0.01 0.002 00.01 0.002 0

5

tt

t t

t t t

t tt

t

Eric’s accumulated value at the end of 5 years is: 100(1 0.05 ) 100(1 0.05 5)t 125

Solution 2.07 A Section 2.03, Equivalent Simple Interest and Discount Rates Let the amount of the initial loan be L. The accumulated values are equal at the end of t years:

2

2

(1 )111

1(1 )(1 ) 1

1 1

0

LL itdt

itdt

it dt

it dt idt

it dt idti d idt

i dtid

Equation I is the only valid expression for t.



Solution 2.08 B Section 2.04, Equations of Value under Simple Interest The amount of interest earned by Ann each year is: 800 0.06 48 Ann’s interest is twice the interest earned by Mike:

48 2 0.09XX

266.67

Solution 2.09 A Section 2.04, Equations of value under Simple Interest and Discount We begin by noting that 0 0.10x and 0 10T implies that 0 1xT . Let’s consider the possibility that Choice A is correct and Haley’s account value is certain to be greater than Tyler’s account value. If Choice A is correct, then:

2

2

2

11,000 1,000(1 )1

1 (1 )(1 )

1 1 ( )

0 ( )

( ) 0

xTxT

xT xT

xT

xT

xT

Since the square of xT is always greater than 0, the result above is consistent with Choice A being true. Working backwards, we can put this in the form of a proof:

2

2

2

( ) 0

0 ( )

1 1 ( )1 (1 )(1 )

11,000 1,000(1 )1

xT

xT

xTxT xT

xTxT

We’ve shown that Choice A is correct. Since Choice A precludes the other choices, the other choices must be false.

Solution 2.10 C Section 2.04, Equations of value under Simple Interest and Discount The accumulated value of the fund minus the accumulated value of the loan is:

11,000 1 0.06 1,000(1 0.05 )X t t

To find the point at which X is maximized, we take the derivative of X:

2

2

1,000(0.06) 1,000( 1)(1 0.05 ) ( 0.05)

5060(1 0.05 )

dX tdt

t



We set the derivative equal to zero to find the local maximum:

2

2

500 60(1 0.05 )

5(1 0.05 )6

51 0.056

50.05 16516

0.0538.2574 or 1.7426

t

t

t

t

t

t t

The question tells us that t must be between zero and 10, so we consider the value of 1.7426. To determine whether it is a local maximum or a local minimum, we take the second derivative of X and evaluate it at t = 1.7426:

2

2

2 3

2

2 3

3

5060(1 0.05 )

50( 2) ( 0.05)(1 0.05 )

5(1 0.05 )

5 6.5727(1 0.05 1.7426)

dXdt t

d Xdt t

d Xdt t

Since the second derivative is negative, there is a local maximum at 1.7426. The maximum value of X within 10 years is:

1

1

1,000 1 0.06 1,000(1 0.05 )

1,000 1 0.06 1.7426 1,000(1 0.05 1.7426)

1,104.55 1,095.45

X t t

9.11

Chapter 3: Compound Interest and Discount Solutions to End of Chapter Questions


Chapter 3: Compound Interest and Discount

Solution 3.01 D Section 3.01, Compound Interest The number of years until the 1-year-old, 4-year-old, 6-year-old, and 9-year-old reach the age of 17 are 16, 13, 11, and 8, respectively. The number of years until the 1-year-old, 4-year-old, 6-year-old, and 9-year-old reach the age of 22 are 21, 18, 16, and 13, respectively. The present value of the payments matches Choice D:

16 13 11 8 21 18 16 13

5 16 13 11 8

A v v v v B v v v v

A Bv v v v v

Solution 3.02 D Section 3.01, Compound Interest Let P be the purchase price. The two offers have the same present value:

9/12100

100

0.92 (1.07) 1

0.8745 1

X

X

P P

X

12.55

Solution 3.03 C Section 3.01, Compound Interest The equation of value at the outset can be used to solve for i:

2 2

2 2

2

100 200 300 604.42

100 200 0.7 300 0.7 604.42 0.7

0.9147

n n nv v v

v

vi

0.0456

Solution 3.04 E Section 3.01, Compound Interest The equation of value at the outset can be used to solve for i:

2 2

2 2

2

100 200 300 604.42

100 200 0.7 300 0.7 604.42 0.7

0.91470.045594

n n nv v v

v

vi



We now solve for n:

0.7

1 0.71.045594

ln(1.045594) ln(0.70)

n

nv

nn

8.0

Solution 3.05 B Section 3.01, Compound Interest Since the amount of interest earned in Wanda’s account during the 11th year is equal to the amount of interest earned in Claire’s account during the 15th year, the amount in Wanda’s account at the end of the 10th year must be equal to the amount in Claire’s account at the end of the 14th year:

10 14

4

1,000(1 ) 700(1 )10 (1 )7

0.09327

i i

i

i

The interest earned in Wanda’s account in the 11th year is:

10 101,000(1 ) 1,000(1.09327) 0.09327X i i 227.50

Solution 3.06 E Section 3.01, Simple Interest & Compound Interest The interest earned in Bonnie’s account is the product of the original deposit, the length of time elapsed, and the simple interest rate. The interest earned in Bonnie’s account during the 7th year is: 1,800 1 i

The interest in earned in Clyde’s account during the 7th year is the balance after 6 years times the annual effective interest rate:

61,000 1 i i

Setting Bonnie’s interest earned during the 7th year equal to Clyde’s interest earned during the 7th year allows us to solve for i:

6

6

6

1,800 1 1,000 1

1,800 1,000 1

1,800 1,000 1

i i i

i

i

i

10.29%



Solution 3.07 A Section 3.01, Compound Interest Equating the present value of the single payment with the present value of the set of three payments allows us to solve for T:

/12 1/12 1.5 2

/12

/12

1,553.50 300 500 7001.04 1.04 1.04 1.041,553.50 1,417.64341.041.09583 1.04

ln(1.09583) ln(1.04)12

T

T

T

T

T

27.9998

Solution 3.08 C Section 3.01, Compound Interest Let’s use 1B and 2B to denote the balances in the account as of July 1, 2007. We are given the balance in account #1 was three times the balance in account #2: 1 23B B

Nine years later the balance is 150,000:

9 91 2

9 92 2

2

(1.03) (1.05) 150,000

3 (1.03) (1.05) 150,00027,444.1395

B B

B BB

The sum of the two accounts on July 1, 2007 was: 1 2 2 2 23 4 4 27,444.1395B B B B B 109,776.56

Solution 3.09 A Section 3.01, Compound Interest There is no need to calculate the values of i or t. Using the equality of the first and third payment streams, we can find the value of v:

12

1/1212,000 8,000

812

0.9668

v

v

v

Using the equality of the second and third payment streams, we can find tv :

2

23,000 63,000 8,000

63 3 8 0

(21 8)(3 1) 08 1 or 21 3

t t

t t

t t

t t

v v

v v

v v

v v

Since the discount factor must be positive, we have:

13

tv



The present value of 6,000 at time (t +1) is:

1 16,000 6,000 6,000 0.96683

t tv v v 1,933.55

Solution 3.10 B Section 3.01, Compound Interest Jill’s accumulated amount is equal to Tom’s accumulated amount at the end of 15 years:

15 15 2

15 15 2

15 15 2

2

2

40 1 0.05 15 20 1 0.05 (15 6) 40(1.04) 20(1.04)

40 40 0.05 15 20 20 0.05 9 40(1.04) 20(1.04)

99 40(1.04) 20(1.04)

54.9712 40(1.04) 20(1.04)

54.9712(1.04) 40(1.04) 20

54.971

n n

n n

n n

n n

n n

22(1.04) 40(1.04) 20 0n n

Let’s use 1.04nx and use the quadratic formula to solve for x: 2

2

54.9712 40 20 0

40 ( 40) 4(54.9712)( 20)2(54.9712)

0.34059 or 1.06824

x x

x

x x

We use the positive value of x to solve for n:

1.06824

1.04 1.06824ln(1.04) ln(1.06824)

nx

nn

1.6831

Solution 3.11 C Section 3.03, Compound Discount The present value of the payments is:

3 50 10,000(1 0.065) 15,000(1 0.065)PV 18,892.88

Solution 3.12 E Section 3.03, Compound Discount The accumulated value of the payments is:

6 6 6 28,000 2,000

(1 0.07) (1 0.07)AV

15,038.56

Solution 3.13 B Section 3.03, Compound Interest and Discount The accumulated value of the deposit is:

4.54.5 2,000(1.08) 2,827.72AV



The present value of the payment is:

4.50 2,827.72(1 0.05)PV 2,244.88

Solution 3.14 B Section 3.03, Compound Discount The amount of interest earned in each fund is the accumulated value of the fund minus the original value of the fund:

1212

1212

12

1(1.03)3 1 3

1 1 1(1.03) 13 1 3

11.2773 11

11.070991

X XX Xd

d

d

dd

0.06628

Solution 3.15 B Section 3.03, Compound Discount There is no need to calculate the value of d. The present values of the two sets of payments are equal, allowing us to solve for the one-year discount factor:

2 3

2

2

169 169 196 196

169(1 ) 196 (1 )

169 196

169196

1314

v v v

v v v

v

v

v

The present value of the first set of payments is:

13169 169 169 16914

K v 325.93

Solution 3.16 E Section 3.04, Equivalent Compound Interest and Discount A is true:

1(1 ) (1 )1 1

d d di i id d v v vd

B is true:

2 2 2

221

d d did v v

C is true: (1 )id i v i iv i d



D is true:

2 21 1 11 v iv v ivi d i iv iv

v v v v v

E is false:

(1 )

(1 ) (1 )i d i iv i vi v i v

Solution 3.17 D Section 3.05, Interest Rate Conversion The annual effective interest rate is:

12

120.121 1 1.01 112

i

0.1268

Solution 3.18 E Section 3.05, Interest Rate Conversion The two-year effective interest rate is found below:

( ) ( )

1/2 12(1/2)

(1/2) 24

(1/2)

1 1

0.121 11 / 2 12

1 1.011 / 2

1 / 2

m pm pi im p

i

i

i

0.2697

Solution 3.19 A Section 3.05, Simple Interest & Compound Interest The interest earned in Patty’s account is the product of the original deposit, the length of time elapsed, and the simple interest rate. The interest earned in Patty’s account during the last 3 months of the 7th year is: 1,800 0.25 i

We would normally write the nominal interest rate compounded quarterly as (4)i , but since this question calls it i, we do likewise. The interest earned in Sally’s account is the balance after 6 years and 9 months, which is equal to 27 quarters, times the quarterly effective interest rate:

27

1,000 14 4i i



Setting Patty’s interest equal to Sally’s interest allows us to solve for i:

27

27

1,800 0.25 1,000 14 4

1,800 1,000 14

i ii

i

i

8.80%

Solution 3.20 C Section 3.05, Interest Rate Conversions The quarterly effective interest rate can be converted to the monthly effective interest rate:

(4) 0.12 0.034 4

i

There are 3 months in a quarter, so the monthly accumulation factor is the cube root of the quarterly accumulation factor:

13

(12)

(12)

(12)

1 1.0312

1 1.00990212

i

i

i

0.1188

Solution 3.21 D Section 3.05, Nominal Interest Rates Since the amount of interest earned in Wanda’s account during the 11th year is equal to the amount of interest earned in Claire’s account during the 15th year, the amount in Wanda’s account at the end of the 10th year must be equal to the amount in Claire’s account at the end of the 14th year:

10 12 14 12(12) (12)

4 12(12)

12(12)

1,000 1 700 112 12

10 17 12

1 1.0932712

i i

i

i


10 12 12(12) (12)101,000 1 1 1 1,000(1.09327) 0.09327

12 12i iX

227.50

Alternative Solution: Since the two accounts earn the same effective interest rate compounded monthly, they must also earn the same equivalent annual effective interest rate. Therefore, as shown below, we can answer the question without reference to the monthly effective interest rate.



Since the amount of interest earned in Wanda’s account during the 11th year is equal to the amount of interest earned in Claire’s account during the 15th year, the amount in Wanda’s account at the end of the 10th year must be equal to the amount in Claire’s account at the end of the 14th year:

10 14

4

1,000(1 ) 700(1 )10 (1 )7

0.09327

i i

i

i


10 101,000(1 ) 1,000(1.09327) 0.09327X i i 227.50

Solution 3.22 D Section 3.05, Nominal Interest Rates Although we usually use i to denote an annual effective interest rate, this question uses i to denote the annual interest rate that is compounded semiannually. We would usually use (2)i to denote this interest rate. Sam’s interest is based on the balance at the end of 13 6-month periods, and Dennis’ interest is based on the initial deposit:

13

13

1 22 2 2

1 22

i i iD D

i

i

10.95%

Solution 3.23 E Section 3.05, Nominal Interest Rates

Let (2)i be the annual interest rate charged to Adam. At the outset, the present value of Heidi’s loan is equal to the present value of Adam’s loan:

(2) (2)

(2)

(2)

(2)

(2)

40 40

2 2 2

1/40 4

2

41/40

2

1/40 1/40 (2) (2)

1/40 (2) 1/40

(2)

1,400 2,000

1 1

11,4002,000 1

10.7

1

0.7 0.7 0.5 1 0.25

0.7 0.5 0.25 1 0.7

0.245561

i i

i

i

i

i

i i

i

i

(2)0.008877

0.03615i



We can now find the present value of Heidi’s loan:

(2) 40 400.03615

44

1,400 1,400

11 iL

976.86

Solution 3.24 D Section 3.05, Nominal Interest Rates Interest is credited only at the end of each interest conversion period, so Michelle receives interest only every 3 months. Therefore, the moment at which Michelle’s account is at least double the amount in Lucy’s account will occur on some multiple of 3 months. Let t be the number of 3-month periods until Michelle’s account is at least double the amount in Lucy’s account. We need to find the minimum integer value of t that satisfies the following:

3

3

0.12 0.06100 1 100 2 14 12

1.03 2 1.005

ln(1.03) ln(2) 3 ln(1.005)47.4883

t t

t t

t tt

The smallest integer that satisfies the equation above is t = 48. The number of months in 48 three-month intervals is: 48 3 144

Solution 3.25 B Section 3.05, Nominal Interest Rates The monthly effective interest rate is:

0.09 0.007512

The equation of value that equates the value of the deposits with $5,900 six years from today is:

72 72 2

2 72

2

1,500 1.0075 3,000 1.0075 5,900

1,500 1.0075 3,000 1.0075 5,900 1.0075

3,000 1,500 3,445.1494 0 where: 1.0075

n n

n n

nX X X

We can use the quadratic formula to solve for X:

21,500 1,500 4(3,000)( 3,445.1494)2 3,000

1.3504 or 0.8504

X

X X

Using the positive value of X, we have the maximum possible value of n:

1.0075

0.8504 1.007521.6872

n

nX

n

Since n must be less than or equal to 21.6872, the maximum integral value of n is 21.



Solution 3.26 D Section 3.06, Nominal Interest and Discount Rates The accumulated value of the deposit is:

4.5 2

94.5

0.082,000 1 2,000 1.04 2,846.622

AV

The present value of the payment is:

4.5 12

540

0.052,846.62 1 2,846.62 0.995812

PV

2,272.01

Solution 3.27 D Section 3.06, Nominal Discount Rates The monthly effective discount rate is:

0.09 0.007512

The equation of value that equates the value of the deposits with $5,900 six years from today is:

72 72 2

72 2 72

2 72

2

1 11,500 3,100 5,9001 0.0075 1 0.0075

1,500 0.9925 3,100 0.9925 5,900

1,500 0.9925 3,100 0.9925 5,900 0.9925

3,100 1,500 3,431.2244 0 where: 0.9925

n n

n n

n n

nX X X

We can use the quadratic formula to solve for X:

21,500 1,500 4(3,100)( 3,431.2244)2 3,100

1.3215 or 0.8376

X

X X

Using the positive value of X, we have the maximum possible value of n:

0.9925

0.8376 0.992523.5412

n

nX

n

Since n must be less than or equal to 23.5412, the maximum integral value of n is 23.

Solution 3.28 C Section 3.07, Interest Rate Conversions The monthly accumulation factor is equal to the inverse of the monthly discount factor:

1(12) (12)

1(12)

(12)

1 112 12

0.221 112 12

i d

i

i

0.2241



Solution 3.29 C Section 3.07, Interest Rate Conversions The interest rate and the discount rate produce the same one-year accrual factor:

( ) ( )

412 (4)

3 (4)

(4)

1 1

0.141 112 4

0.141 112 4

m pm pi dm p

d

d

d

0.1368

Solution 3.30 B Section 3.07, Nominal Discount Rates We can use the ratio of A to B to find d:

16

4 416

48 2

2

2

4

5152

50 1 515250 1

1 51521

52 26 51 12.751 13.25

0.07457

d

d

d

d

AB

d dd

d

The value of d convertible quarterly is equivalent to an annual effective interest rate of i:

4

4

1 14

0.075471 14

d i

i

i

0.0792

Chapter 4: Constant Force of Interest Solutions to End of Chapter Questions


Chapter 4: Constant Force of Interest

Solution 4.01 B Section 4.02, Force of Interest The force of interest is: ln(1.080)r 0.07696

Solution 4.02 A Section 4.02, Force of Interest The force of interest is:

120.10ln 1

12r

0.09959

Solution 4.03 C Section 4.02, Force of Interest The force of interest is:

40.09ln 1

4r

0.09103

Solution 4.04 B Section 4.02, Force of Interest The force of interest is:

12ln 1 0.01r 0.1206

Solution 4.05 A Section 4.02, Force of Interest The force of interest is:

1/2

12

0.11ln 1r

0.09943

Solution 4.06 Section 4.02, Force of Interest a. The annual effective interest rate is:

0.06 1i e 0.06184 b. The monthly effective interest rate is:

(12)

0.06/12 112i e 0.00501



c. The annual interest rate compounded monthly is:

(12) 0.06/1212 1 12 0.00501i e 0.06015

d. The annual effective discount rate is:

0.061d e 0.05824 e. The quarterly effective discount rate is:

(4)

0.06/414

d e 0.01489

e. The annual discount rate convertible quarterly is:

(4) 0.06/44 1 4 0.01489d e 0.05955

Solution 4.07 B Section 4.02, Force of Interest The present value is:

0.07 2 0.07 5 0.07 73,000 8,000 10,0003,000 0.8694 8,000 0.7047 10,000 0.6126

e e e 14,371.84

Solution 4.08 D Section 4.02, Force of Interest The balance at the end of 10 years is:

0.07 10 0.07 6 0.07 42,000 1,500 8,0002,000 2.0138 1,500 1.5220 8,000 1.3231

e e e 12,329.60

Solution 4.09 E Section 4.02, Force of Interest The original force of interest is:

40.07ln 1 0.06939

4r

One half of the original force of interest is: 0.5 0.06939 0.03470 The new annual interest rate compounded quarterly is:

(4) 0.03470 0.254 ( 1)i e 0.0348



Solution 4.10 A Section 4.02, Force of Interest The equation of value at the outset can be used to solve for r:

2 2

2 2

2

100 200 300 604.42

100 200 0.7 300 0.7 604.42 0.7

0.91471 1.04559ln(1 )

n n nv v v

v

vi

i

0.0446

Solution 4.11 C Section 4.02, Force of Interest

The equation of value at the outset can be used to solve for nv :

2

2

2

100 200 300 600

300 400 100 0

3 4 1 0

(3 1)( 1) 011 or 3

n n n

n n

n n

n n

n n

v v v

v v

v v

v v

v v

The annual force of interest is greater than zero, so nv must be 1/3:

0.1221

13

13

10.1221 ln3

n

n

v

e

n

n

9.00

Solution 4.12 E Section 4.02, Force of Interest For a given force of interest, the equivalent nominal interest rate falls as its compounding frequency increases. The expression in Choice E can be rewritten as:

(1/2) (1)i i Increasing the compounding frequency from every other year to every year will result in a higher force of interest unless the interest rate with the higher compounding frequency is lower than the interest rate with the lower compounding frequency. Therefore, Choice E is false.



Solution 4.13 E Section 4.02, Force of Interest The number of quarters in 12 years and 1 month is:

1124 12 48.3333

The equation of value at the end of 12 years and 1 month can be used to solve for the force of interest:

112

48.3333 12

12.0833

12.0833

0.075200 1 2204

490.8684 220

2.2312ln(2.2312) 12.08330.8025 12.0833

e

e

e

0.06642

Solution 4.14 A Section 4.02, Force of Interest At the end of 5 years the amount in Suzie’s account is: 200(1 5 0.05) 250

The amount in both accounts is the same at the end of 5 years:

5250 220ln(250) ln(220) 5

e

0.02557

Solution 4.15 D Section 4.02, Force of Interest The derivative of the force of interest with respect to the annual effective interest rate is:

1ln(1 )1

d d i vdi di i

The derivative of the annual effective interest rate with respect to the annual effective discount rate is:

2 2 2 2(1 ) 1 ( 1) (1 ) 1 1

1 (1 ) (1 ) (1 )d d d d d d didd dd d d d d v

The product is:

21 1d d i v

di dd vv

1+ i

Chapter 5: Varying Rates Solutions to End of Chapter Questions


Chapter 5: Varying Rates

Solution 5.01 A Section 5.01, Varying Compound Interest The present value is:

1 4 1.5 12

1.50

0.08 0.06100 1.07 1 14 12

PV

76.3018

Solution 5.02 C Section 5.01, Varying Compound Interest The equation of value can be used to find the level equivalent interest rate:

8(2)1.5 4 18

8(2)1.5 4 18

(2)

(1.07) (1.02) (1.005) 12

(1.07) (1.02) (1.005) 12

iDeposit Deposit

i

i

0.0688

Solution 5.03 B Section 5.02, Varying Discount Rates The present value is:

1 4 1.5 12

1.50

0.08 0.06100 1 0.07 1 14 12

PV

75.5865

Solution 5.04 B Section 5.02, Varying Discount Rates The accumulated value is:

1.5 4 18 2 18100 0.93 0.98 0.995 50 0.98 0.995

132.2988 56.9774

189.2761

Solution 5.05 B Section 5.02, Varying Discount Rates The equation of value at time 5 can be used to solve for d:

36 2

62

2

1 12 100(1.10) 1.04 1500.99 1

1306.2072 215.38941

d

dd

0.1613



Solution 5.06 A Section 5.03, Varying Force of Interest The present value is:

1.5 0.07 0.08 1.5 0.06 0.2750 100 100PV e e e e 75.9572

Solution 5.07 D Section 5.03, Varying Force of Interest The accumulated value is:

1.5 0.07 0.08 1.5 0.06 0.5 0.08 1.5 0.06

0.275 0.13100 50

100 50

e e e e e

e e

188.5945

Solution 5.08 Section 5.03, Varying Force of Interest The accumulated value function and its derivative are:

(1 0.05 )

0.05

t

t

AV D t

d AVD

dt

The force of interest at time t is:

0.05 0.05(1 0.05 ) 1 0.05

t

tt

d AVDdtr

AV D t t

a. 112 1

12

0.051 0.05

r

0.04979

b. 10.05

1 0.05 1r

0.04762

c. 50.05

1 0.05 5r

0.0400

d. 100.05

1 0.05 10r

0.0333

Solution 5.09 Section 5.03, Varying Force of Interest The accumulated value function and its derivative are:

1

2 2

(1 0.05 )

(1 0.05 ) ( 0.05) 0.05 (1 0.05 )

t

t

AV D t

d AVD t D t

dt

The force of interest at time t is:

2

10.05 (1 0.05 ) 0.05

1 0.05(1 0.05 )

t

tt

d AVD tdtr

AV tD t

a. 112 1

12

0.051 0.05

r

0.05021



b. 10.05

1 0.05 1r

0.05263

c. 50.05

1 0.05 5r

0.0667

d. 100.05

1 0.05 10r

0.1000

Solution 5.10 C Section 5.03, Varying Force of Interest The present value at time 5 of the 12,000 payment is:

88 8155 2 5ln(2 )

5 8

ln(7) ln(10)

12,000 12,000712,000 12,000 8,40010

s s dsr ds sPV AV e e e

e

The equation of value at time 5 can be used to solve for X:

(5 0) (5 2)5,000(1.05) (1.05) 8,400X

X

1,743.74

Solution 5.11 B Section 5.03, Varying Force of Interest The force of interest at time t for Fund X is:

0.5

1 0.5

tXt

t

d AVdtrAV t

The force of interest at time t for Fund Y is:

21 0.5

tYt

t

d AVtdtr

AV t

Setting the two equal allows us to determine the time at which the two forces of interest are equal:

2

2 2

2

0.51 0.5 1 0.5

0.5 0.25 0.5

0 0.25 0.50

X Yt tr r

tt t

t t t

t t

We use the quadratic formula to solve for t:

21 1 4(0.25)( 0.5)2 0.25

4.4495 or 0.4495

t

t t

Discarding the negative solution, we have: t 0.4495



Solution 5.12 A Section 5.03, Varying Force of Interest Since X(0) and Y(0) are constants and X(0) = Y(0), the answer to this question does not depend on their value. For convenience, we assume that both are equal to 1: (0) (0) 1X Y

The accumulated value in Fund X at time t is:

00 0

11 lnln(1 ) 1ln(1 ) ln(1)1( ) 1tt t

st

dsr ds s tstX t AV e e e e e t

The accumulated value in Fund Y at time t is:

0 20

10

1 5( )tt

ss dsr ds s

tY t AV e e

Let’s use the following substitution:

21 5

10u sdu sds

The accumulated value in Fund Y at time t can now be written as:

22

20 2 0 0 0

1 510 1 lnln(1 5 ) 1ln( ) ln(1 5 ) ln(1)1 5

2( )

1 5

s tt s t s ts s s

ts ds sdu u ts uY t e e e e e e

t

We can find the maximum of ( )H t by setting its derivative equal to zero:

2( ) ( ) ( )

( ) 1 (1 5 )'( ) 1 10

1 10 0

H t X t Y t

H t t tH t t

tt

0.10

For the sake of thoroughness, we note that the second derivative of ( )H t is negative at 0.1t , indicating that (0.1)H is a maximum:

''( ) 10H t

Solution 5.13 D Section 5.03, Varying Force of Interest The equation of value is:

50

50 3

5

153(1 )

1,000 1,000(1 )

(1 )

sr ds

dss

e i

e i

To evaluate the integral, let’s use the following substitution:

(1 )u sdu ds



The integral is:

5 55 5 23

3 200 0 0

1 1 1 1 13 3 2 63(1 ) (1 )

1 1 1 0.162046 36

s ss

ss s

uds u dus s

We can now use the equation of value to solve for i:

50 3

153(1 )

0.16204 5

5

(1 )

(1 )

1.17590 (1 )

dsse i

e i

ii

0.032938

Solution 5.14 C Section 5.03, Varying Force of Interest The accumulated value in Fund B at time t is:

0t

sr dstAV e

Let’s evaluate the integral in the expression above:

3 2

4 2 2 2 20 0 0

2 6 2 ( 3) 26 9 ( 3) ( 3)

t t ts s s s sds ds ds

s s s s

Let’s use the following substitution:

2 32

u sdu sds

The integral is:

2

2 202 0

0 0

2 1 3ln( ) ln( 3) ln( 3) ln(3) ln3( 3)

t s t ts ts

s

s tds du u s tus

The accumulated value of Fund B at time t is:

02 33

tsr

ttAV e

At time 1, the accumulated value in Fund B is equal to the accumulated value in Fund A:

2

2

3 1 when 13

1 3 1 13

4 13

13

t it t

i

i

i



Let ( )H t be the difference between Fund A and Fund B. We can find the maximum of ( )H t by setting its derivative equal to zero:

2

2

3( ) 13

3( ) 13 3

1 2'( )3 3

1 2 03 31 2 02 1

tH t it

t tH t

tH t

t

tt

t

0.5

For the sake of thoroughness, we note that the second derivative of ( )H t is negative at 0.5t , indicating that (0.5)H is a maximum:

2''( )3

H t

Solution 5.15 E Section 5.03, Varying Force of Interest The integral of the force of interest from time 4 to time 8 is:

83 2 3 2 3 28 8 2

4 44

8 8 4 40.001( ) 0.001 0.0013 2 3 2 3 2

0.001 138.6667 13.3333 0.1253

tt tdt t t dt

The accumulated value at time 8 is:

840.05 4 0.20 0.1253100 100tdt

e e e e 138.45

Solution 5.16 C Section 5.03, Varying Force of Interest The integral of the force of interest from time 6 to time 10 is:

103 210 10 2

6 66

3 2 3 2

0.002( ) 0.0023 2

10 10 6 60.002 0.002 383.3333 903 2 3 2

0.5867

tt tdt t t dt

The present value at time 2 is:

106 0.02 (6 2) 0.5867 0.08100 100tdt

e e e e 51.3417



Solution 5.17 D Section 5.03, Varying Force of Interest The interest accumulation factors must be the same for Marcia and Jan over the course of 5 years. The interest accumulation factor for Jan is:

55

0 0

5

1 ln( 0.20 )exp exp exp 5 ln( 1) 5 ln( )0.20 0.20

1 1exp 5 ln

K tdt K KK t

K KK K

The interest accumulation factor for Jan is equal to the interest accumulation factor for Marcia:

5 5

2

2

1 136

1 136

136

1366

K KK

K KK

KK K

K

K

Jan’s accumulated value at the end of 5 years is:

5 51 7100 100

6K

K

216.14

Solution 5.18 B Section 5.03, Varying Force of Interest At time 4, before the deposit of X, the value of the fund is:

44348,0000 2,000 0 0.032100 100 100

tt dte e e

The accumulation factor from time 4 to time 8 is:

84388,0004 2,000 4 0.512 0.032 0.48

tt dte e e e

The interest earned from time 4 to time 8 is equal to X:

0.032 0.48

0.032

100 1

100 0.6161

63.6108 0.3839

e X e X

e X X

XX

165.6851



Solution 5.19 D Section 5.03, Varying Force of Interest The equation of value is:

0

23

23

0 4

0 4

3

exp 3

ln3

tsds

t ss

t ss

De D

ds

ds


3

24

3

u s

du s ds

The integral is:

23 3

2 3100 0 04 4 0

33

1 1 1 1 13 ln ln 43 3 3 3

1 1 4ln 4 ln(4) ln3 3 4

s tt t s t s tssss s s

ds s ds du u su

tt

We can now solve for t:

230 4

313

1/33

ln3

4ln ln34

4 34

t ss

ds

t

t

t

4.7027

Solution 5.20 E Section 5.03, Varying Force of Interest The present value is:

80800 sds

e


3

2

4180

60

su

sdu ds

The integral is:

22

3 3

3

5 512060 00 0 0

180 1805

125180 180

0

1 10.5 0.5 0.5 ln4 4

0.5 ln 4 0.5 ln 4 ln(4) ln 1.0833

st t s sssss s

ss

s

ds ds du uu



The present value is:

80 ln(1.0833) 800800 800

1.0833sds

e e 738.46

Solution 5.21 E Section 5.04, Mix of Varying Rates The accumulated value is:

12 4

0.10 12 4 0.100.06 0.07100 1 1 100(0.995) (1.0175)12 4

e e

125.80

Solution 5.22 A Section 5.04, Mix of Varying Rates The present value is:

12 4

0.10 12 4 0.100.06 0.07100 1 1 100(0.995) (1.0175)12 4

e e

79.49

Solution 5.23 C Section 5.04, Mix of Varying Rates The equation of value at the end of 6 years can be used to find d:

2 24

44

500 1 1 7672

1 1.534 12

1 0.5 1.1129 1

1.6129 0.1129

dd

d d

d d

dd

0.07

Solution 5.24 C Section 5.04, Mix of Varying Rates The equation of value at the end of 20 years can be used to find d:

2 7 13 12 10 12

14 13 12

14

0.07 0.07100 1 1 50 1 5002 12 12

0.07100 1 1 399.51692 12

1 1.61242

d

d

d

d

0.0671



Solution 5.25 C Section 5.05, Accumulation Function The accumulation function at the end of 15 years is:

2(15) 1 0.04 15 0.002 15 2.05a

The accumulated value of 1 at the end of 15 years is equal to the accumulated value found using the equivalent discount rate:

15

15

1(15)(1 )

12.05(1 )

0.9533 1

ad

dd

d 0.04673

solutions to end of chapter questions

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