Introduction to Corporate Finance, Fifth Edition
Booth, Cleary, Rakita
Chapter 5: Time Value of Money
Multiple Choice Questions
1. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning Objective: 5.2; 5.3
Level of difficulty: Basic
Solution: C.
Simple interest rate: $1,000 + ($1,000)(8%)(6) = $1,480
Compound interest rate: $1,000(1+.08)6 = $1,586.87 = $1,587 rounded
Or using a financial calculator (TI BA II Plus),
N=6, I/Y=8, PV=-1,000, PMT = 0, CPT FV= 1,586.87
2. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning Objective: 5.2; 5.3
Level of difficulty: Intermediate
Solution: C
Simple interest: Total interest paid over three years: $6,200 - $5,000 = $1,200
Annual interest = $1,200/3 = $400
$400/$5,000 = 8%
Compound interest:
1
6,200 ⁄3
(
) − 1=7.43%
5,000
3. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning Objective: 5.2; 5.3
Level of difficulty: Intermediate
Solution: B
4. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning Objective: 5.2; 5.3
Level of difficulty: Basic
Solution: D.
A) $1,000 + ($1,000)(10%)(5) = $1,500
B) $1,000 + ($1,000)(8%)(10) = $1,800
C) $1,000(1.08)8 = $1,851
D) $1,000(1.07)10 = $1,967
Therefore, D is the largest.
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5. Section: 5.3 Compound Interest
Learning Objective: 5.3
Level of difficulty: Intermediate
Solution: B.
PV=$15,000,000/(1.05)25=$4,429,541.58 = $4,429,542 rounded
Or using a financial calculator (TI BA II Plus),
N=25, I/Y=5, FV=15,000,000, PMT = 0, CPT PV= –4,429,541.58
6. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning Objective: 5.2; 5.3
Level of difficulty: Intermediate
Solution: B. The greater the interest rate, the smaller the present value, given a $100 future value
and holding the time period constant.
7. Section: 5.3 Compound Interest
Learning Objective: 5.3
Level of difficulty: Intermediate
Solution: D.
FV=PV(1+k)n
16,000=10,000(1+ k)8
8ln(1+k)=ln(1.6), therefore k=6.05%
Or using a financial calculator (TI BA II Plus),
N=8, PV= –10,000, FV=16,000, PMT = 0, CPT I/Y=6.05%
8. Section: 5.3 Compound Interest
Learning Objective: 5.3
Level of difficulty: Intermediate
Solution: C.
FV=PV(1+k)n
Assume that the initial investment is $1.
(3)(1)=1 (1.09) n
ln(3)=(n)ln(1.09)
n=12.7 years
Or using a financial calculator (TI BA II Plus),
I/Y=9, PV= –1, FV=3, PMT = 0, CPT N=12.7
9. Section: 5.4 Annuities and Perpetuities
Learning objective: 5.4
Level of difficulty: Intermediate
Solution: D. An annuity due has a greater PV because it pays one year earlier than an ordinary
annuity.
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Introduction to Corporate Finance, Fifth Edition
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10. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Intermediate
Solution: C.
 (1  k )20  1
(1+.15)20 −1
=$2,000
[
] = $2,000(102.4436) = $204,887
FV20  PMT 

.15
k


Or using a financial calculator (TI BA II Plus),
N=20, I/Y=15, PMT= -2,000, PV = 0, CPT FV=204,887
11. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Intermediate
Solution: B.
1 

1  (1  k ) n 

PV0  PMT 
k




1
= $2,000 [
1 − (1.15)20
. 15
] = $2,000(6.25933) = $12,519
Or using a financial calculator (TI BA II Plus),
N=20, I/Y=15, PMT= –2,000, FV = 0, CPT PV=12,519
12. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Basic
Solution: D.
PV0=PMT/k=$1,500/.12=$12,500
13. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Basic
Solution: A.
PV0=PMT/k=$1,500/.12 + $1,500 =$14,000
14. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Intermediate
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Solution: B
PV of annuity of 120 remaining payments at 1% per month:
1 

1
1  (1  k ) n 
1−
(1.01)120
 = $3,303.26 [
] = $3,303.26(69.7005) = $230,238.95
PV0  PMT 
.01
k




Or using a financial calculator (TI BA II Plus),
N = 120, I/Y = 1, PMT = -3,303.26, FV = 0, CPT PV = 230,238.95
15. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Challenging
Solution: A.
The future value of a perpetuity cannot be computed as it is infinite.
Practice Problems
Basic
16. Section: 5.2 Simple Interest
Learning Objective: 5.2
Level of difficulty: Basic
Solution:
As this is simple interest, Dmitri will earn the same amount of interest each year. The annual
amount of interest is 8% x initial investment = .08 x $25,000 = $2,000
a. $2,000
b. $2,000
17. Section: 5.2 Simple Interest
Learning Objective: 5.2
Level of difficulty: Basic
Solution:
a. In one year, he will owe interest of P x k = $1,500 x 6% = $90
b. After three years, the total principal and interest paid will be: P + (n x P x k) = $1,500 + (3 x
$1,500 x 6%) = $1,770
18. Section: 5.2 Simple Interest
Learning Objective: 5.2
Level of difficulty: Basic
Solution:
As the exact amount of interest owing each year will be paid, there is no “compounding.” The
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amount of each annual payment will be P x k = $2,500 x 6% = $150.
19. Section: 5.2 Simple Interest
Learning Objective: 5.2
Level of difficulty: Basic
Solution:
Khalil will be paid interest each month for 12 months without compounding. The total interest
earned is n x P x k = 12 x $1,200 x 0.5% = $72
20. Section: 5.3 Compound Interest
Learning Objective: 5.3
Level of difficulty: Basic
Solution:
The payment of compound interest means that we must compound (or find the future value of) the
amount invested (the present value):
FV12 months  $1,200  (1  0.005)12  $1,274.01
Of this amount, $1,200 was the original amount invested, so $74.01 of interest will be earned.
21. Section: 5.2 Simple Interest; 5.3 Compound Interest
Learning outcome: 5.2; 5.3
Level of difficulty: Basic
Solution:
A. Value = P + (n x P x k) = $24 + (394 x $24 x 5%) = $497
B.𝐹𝑉394𝑦𝑒𝑎𝑟𝑠 = $24 × (1 + 0.05)394 = $5,355,438,979
22. Section: 5.3 Compound Interest
Learning outcome: 5.3
Level of difficulty: Basic
Solution:
The future value of the loan (the amount to be repaid) is $5,000. The amount that can be borrowed
is the present value amount, calculated as:
PV0  FV1 
1
1
 $5,000 
 $4,716.98
1
(1  k )
(1  .06)1
Or using a financial calculator (TI BA II Plus),
N=1, I/Y=6, FV= -5,000, PMT = 0, CPT PV=4,716.98
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23. Section: 5.3 Compound Interest
Learning Objective: 5.3
Level of difficulty: Basic
Solution:
a. FV1 year  $20,000  (1  0.10)1  $22,000.00
b. FV5 years  $20,000  (1  0.10)5  $32,210.20
c. FV10 years  $20,000  (1  0.10)10  $51,874.85
24. Section: 5.3 Compound Interest
Learning Objectives: 5.3
Level of difficulty: Basic
Solution:
Jon needs $800 in three years; that is the future value amount. The present value equivalent is:
PV0  FV3 
1
1
 $800 
 $691.07
3
(1  k )
(1  .05) 3
Or using a financial calculator (TI BA II Plus),
N=3, I/Y=5, FV= -800, PMT = 0, CPT PV=691.07
25. Section: 5.4 Annuities and Perpetuities
Learning outcome: 5.4
Level of difficulty: Basic
Solution:
Present value of the perpetual scholarship payment:
1
1
𝑃𝑉0 = 𝑃𝑀𝑇 [ ] = $5,000 × [
] = $166,667
𝑘
0.03
26. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Basic
Solution:
2
.
0725
0


1

1

7
.
38
%


For Bank A, k
2

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4
.
0720
0


1

1

7
.
40
%


For Bank B, k
4

12
.
0715
0


1

1

7
.
39
%


For Bank C, k
12


Bank B pays the highest effective annual rate.
27. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Basic
Solution:
a. For annual compounding, the effective annual rate will be the same as the quoted rate. To check
this:
m
1
 QR 
 9.5% 
k  1 
  1  1 
  1  9.5%
m 
1 


b. With quarterly compounding, set m=4,
4
 9.5% 
k  1 
  1  9.84%
4 

c. With monthly compounding, set m=12,
12
 9.5% 
k  1 
  1  9.92%
12 

28. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Basic
Solution:
a. 𝑘 = Quoted Rate = 6% ⇒ 𝐹𝑉1𝑦𝑒𝑎𝑟 = 𝑃𝑉0 (1 + 𝑘) = $50,000 × (1.06) = $53,000
12
 QR 
b. k  1 
  1  6.16778%  FV1 year  $50,000  (1.0616778)  $53,083.89
12 

QR 

c. k  1 

 365 
365
 1  6.18313%  FV1 year  $50,000  (1.0618313)  $53,091.57
29. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Basic
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Solution:
The value of any perpetual stream of payments can be valued as a perpetuity:
PMT
$2
PV0 

 $16.67
k
0.12
30. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Basic
Solution:
Because the fees are paid at the start of the year, this is an annuity due.
1


1  (1  0.06) 3 
PV0  $9,500  
  (1  0.06)  $26,917.23
0.06




Or using a financial calculator (TI BA II Plus),
Hit [2nd] [BGN] [2nd] [Set]
N=3, I/Y=6, PMT= -9,500, FV = 0, CPT PV=26,917.23
31. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Basic
Solution:
The future value amount is $60,000. The amount to be saved each year is really the payment on
an ordinary annuity:
 (1  0.07)8  1
$60,000  PMT 
  PMT  $5,848.07
0.07


Or using a financial calculator (TI BA II Plus),
N=8, I/Y=7, PV =0, FV= 60,000, CPT PMT= -5,848.07
32. Section: 5.5 Growing Annuities and Perpetuities
Learning Objective: 5.5
Level of difficulty: Basic
Solution:
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$100
𝑃𝑉 = .09−.03 = $1,666.67. The most I would be willing to pay for the investment is the present
value, therefore, $1,666.67.
33. Section: 5.8 Comprehensive Examples
Learning Objective: 5.8
Level of difficulty: Basic
Solution:
Annual investment = Annual income – Annual expenditure = $45,000 – $36,000 = $9,000
This is an annuity due.
 (1  k )n  1
FVn  PMT 
(1  k )
k


(1 + .126)35 − 1
= $9,000 [
] (1.126) = $(9,000)(497.2749)(1.126) = $5,039,384
. 126
Or using a financial calculator (TI BA II Plus),
Hit [2nd] [BGN] [2nd] [Set]
N=35, I/Y=12.6, PV = 0, PMT= -9,000, CPT FV=5,039,384
34. Section: 5.8 Comprehensive Examples
Learning Objective: 5.8
Level of difficulty: Basic
Solution:
This is an ordinary annuity.
 (1  0.10)15  1
FV15  $30,000  
  $953,174.45
0
.
10


No, Tommy will not achieve his goal before retirement.
Intermediate
35. Section: 5.1 Opportunity Cost
Learning Objective: 5.1
Level of difficulty: Intermediate
Solution:
Cost = tuition + textbook + loss of income = $800+$300+$900 (0.30 x $3,000) = $2,000
The rent and food are expenses that he will face regardless of taking the course.
We are assuming that the extra time he spends studying for the philosophy course will not have
any impact on his grades in his other courses and are not placing any value on his enjoyment of
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the subject.
36. Section: 5.4 Annuities and Perpetuities
Learning outcome: 5.4
Level of difficulty: Intermediate
Solution:
Present value of the perpetual scholarship payment at the end of 4 years:
1
1
𝑃𝑉4 = 𝑃𝑀𝑇 [ ] = $5,000 × [
] = $166,667
𝑘
0.03
The present value today is $166,667 / (1.03)4  $148,081 . Grace will need to endow $148,081
today for the scholarship to start in 5 years.
37. Section: 5.4 Annuities and Perpetuities
Learning outcome: 5.4
Level of difficulty: Intermediate
Solution:
Find the present value of the four-year annuity at year 3:
1 
1



1  (1  k ) n 
1  (1  0.05) 4 
  $6,000  
  $21,275.70
PV3  PMT 
k
0.05








Or using a financial calculator (TI BA II Plus),
N=4, I/Y=5, PMT= -6,000, FV = 0, CPT PV= 21,275.70
Now, find the present value of this amount today:
 1 
 1 
PV0  FV 
 $21,275.70  
 $18,378.75
3
3
 (1  k ) 
 (1.05) 
Or using a financial calculator (TI BAII Plus),
N=3, I/Y=5, PMT = 0, FV= 21,275.70, CPT PV= 18,378.75
38. Section: 5.4 Annuities and Perpetuities
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Learning outcome: 5.4
Level of difficulty: Intermediate
Solution:
To be indifferent between the two options means that the present value of the annuity must equal
$40 million (the immediate payout).
$40 = $5 (
1−
1
(1+𝑘)10
𝑘
). Solving this using the calculator is the easiest way. N=10, PMT = -5,
PV
= 40, FV = 0, CPT I/Y. We find an interest rate of 4.28%. If the interest rate is greater than
4.28%, I prefer the immediate payout of $40 million because the present value of the 10-year
annuity is less than $40 million. If the interest rate is less than 4.28%, I prefer the annuity
because the present value will be greater than $40 million.
39. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Intermediate
Solution:
Step 1: determine monthly effective rate
 .09  4 
k monthly  1 
 
4  

1 / 12
 1  0.7444%
Step 2: given the monthly effective rate, determine the quoted rate compounded monthly.
QR monthly = 12 x 0.7444
= 8.9333%
Therefore, 9% compounded quarterly is equivalent to 8.9333% compounded monthly.
40. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Intermediate.
Solution:
a. m = 365:
.24
𝑘 = (1 + 365)365 − 1 = 27.11%
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.24
b. m = 4:
𝑘 = (1 + 4 )4 − 1 = 26.25%
c. m = 3:
𝑘 = (1 + 3 )3 − 1 = 25.97%
d. m = 2:
𝑘 = (1 + 2 )2 − 1 = 25.44%
.24
.24
𝑘 = 𝑒 .24 − 1 = 27.12%
e. Continuous compounding:
f. The effective monthly rates for a. to d. are:
m
i. m=365, f=12
365
QR f
.24 12
)  1=2.02%
k  (1 
)  1= (1 
365
m
ii. m=4, f=12
4
QR f
.24 12
)  1 =1.96%
k  (1 
)  1= (1 
4
m
iii. m=3, f=12
k  (1 
iv. m=2, f=12
2
QR f
.24 12
)  1 =1.91%
k  (1 
)  1= (1 
2
m
m
m
3
QR f
.24 12
)  1 =1.94%
)  1= (1 
3
m
m
41. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Intermediate
Solution:
A. The future value of Jane’s account will be:
 (1  0.06)17  1
FV17  $1,000  
  $28,212.88
0.06


B. The grant has the effect of increasing the amount saved from $1,000 to $1,200. The future value
of the account will now be:
 (1  0.06)17  1
FV17  $1,200  
  $33,855.46
0.06


42. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Intermediate
Solution:
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Find the present value of the four-year annuity due:
1 
1



1  (1  k ) n 
1  (1  0.05) 4 
 (1  k )  $5,000  
  (1  0.05)  $18,616.24
PV5  PMT 
k
0.05








Or using a financial calculator (TI BA II Plus),
Hit [2nd] [BGN] [2nd] [Set]
N=4, I/Y=5, PMT= -5,000, FV = 0, CPT PV=18,616.24
Now, discount this amount back five years:
 1 
 1 
PV0  FV 
 $18,616.24  
 $14,586.31
5
5
 (1  k ) 
 (1.05) 
Or using a financial calculator (TI BA II Plus),
N=5, I/Y=5, PMT =0, FV= 18,616.24, CPT PV=-14,586.31
43. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Intermediate
Solution:
We have two separate annuities to consider: the tuition payments, and the savings amounts. First,
find the present value of the four annual tuition payments (at time 8, when Felix is due to begin
his university studies):
1
1 − (1+0.07)4
𝑃𝑉8 = $15,000 × (
) = $50,808.17
0.07
This is the amount of savings required at time 8. From the perspective of time 0, this is a future
value amount. Next, find the annual savings amount:
(1 + 0.07)8 − 1
$50,808.17 = 𝑃𝑀𝑇 [
] ⇒ 𝑃𝑀𝑇 = $4,952.16
0.07
44. Section: 5.5 Growing Annuities and Perpetuities
Learning Objective: 5.5
Level of difficulty: Intermediate
Solution:
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$100
Present value of Grow: 𝑃𝑉𝐺𝑅𝑂𝑊 = .05−.04 = $10,000
$1,000
Present value of Shrink:𝑃𝑉𝑆𝐻𝑅𝐼𝑁𝐾 = .05−(−.02) = $14,285.71
Grow’s present value exceeds the cost by $9,000, while Shrink exceeds the cost by $13,285.71.
Shrink is preferred, as it exceeds the investment cost more.
45. Section: 5.5 Growing Annuities and Perpetuities
Learning Objective: 5.5
Level of difficulty: Intermediate
Solution:
PV=
$100(1+.03)
0.09−0.3
+ $100
= $1,816.67
The most I would pay is $1,816.67, the present value of the investment. In this case, the cash
flows start immediately ($100) and then grow by 3% per year.
46. Section: 5.5 Growing Annuities and Perpetuities
Learning Objective: 5.5
Level of difficulty: Intermediate
Solution:
To solve this, we need to realize that the present value of a perpetuity (growing or otherwise)
occurs one period prior to the first cash flow. Hence, using the growing perpetuity formula will
give us the value of the cash flows in year 4. We need to discount those back to time 0.
= $1,180.71
The most I would be willing to pay for this investment is $1,180.71.
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47. Section: 5.4 Annuities and Perpetuities 5.6; Quoted versus Effective Rates
Learning Objective: 5.4; 5.6
Level of difficulty: Intermediate
Solution:
Solve the annuity equation to find k, the interest rate:
1 

1  (1  k ) 5 
k ?
$25,000.00  $6,935.24  
k




The calculations are most easily done with a financial calculator (TI BA II Plus),
PV = 25,000, PMT=-6,935.24, N= 5, FV = 0, CPT I/Y = 12%
The effective annual interest rate is 12 percent. With annual compounding, the nominal (annual)
rate will also be 12 percent per year.
48. Section: 5.4 Annuities and Perpetuities; 5.6 Quoted versus Effective Rates
Learning Objective: 5.4; 5.6
Level of difficulty: Intermediate
Solution:
a. There will be 5 x 12 = 60 monthly payments. The calculations are most easily done with a
financial calculator (TI BAII Plus),
PV = 25,000, PMT=-556.11, N= 60, CPT I/Y = 1.0%
Because we used monthly payments, and months as the time period, 1.0% is the effective
monthly rate.
b. The compounding period matches the payment frequency, so the nominal (annual) rate is:
QR  m  kmonthly  12  1.0%  12.0%
49. Section: 5.6 Quoted versus Effective Rates
Learning Objective: 5.6
Level of difficulty: Intermediate
Solution:
a. Scott will pay interest of ($800–$750) = $50 after one week. This implies a nominal interest rate
of $50/$750 = 6.67% per week. With 52 weeks in the year, the nominal rate per year is then 52 x
6.67% = 346.84%
b. The effective annual interest rate is k  (1  0.0667) 52  1  27.7210  2,772.10%
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50. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Intermediate
Solution:
a. In Canada, fixed-rate mortgages use semi-annual compounding of interest, so m=2. The
effective annual rate is therefore:
m
2
 QR 
 0.064 
k  1 
  1  1 
  1  6.5024%
m 
2 


b. With monthly payments, f=12. We can find the effective monthly interest rate from the effective
annual rate, k:
kmonthly  1  k  f  1  1  6.5024% 12  1  0.5264%
1
1
c. The amortization period is 20 years, or 20 x 12 = 240 months. Josephine’s monthly payments
can be computed as:
1


1  (1  0.005264) 240 
  PMT  $1,322.69
$180,000  PMT 
0.005264




Or using a financial calculator (TI BA II Plus),
N=240, I/Y=.5264, PV=180,000, FV = 0, CPT PMT = -1,322.69
d. With monthly compounding and payments, the effective monthly interest rate is:
m
12
 QR  f
 0.0636  12
kmonthly  1 

1


1 
  1  0.530%
m 
12 


The monthly payments can be computed using a financial calculator (TI BA II Plus),
N=240, I/Y=0.53, PV=180,000, FV = 0, CPT PMT = -1,327.24
Although the quoted rate is lower at the credit union than at the bank, the effective rate is higher.
Josephine should take the mortgage loan from Providence Bank. The monthly payment for the
credit union mortgage would be $1,327.24, which is higher than that at Providence Bank.
51. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Intermediate
Solution:
With semi-annual compounding (the norm in Canada) and monthly payments, m=2 and f=12.The
effective monthly rate is:
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m
Booth, Cleary, Rakita
2
 QR  f
 0.039  12
kmonthly  1 
  1  1 
  1  0.3224%
m 
2 


The present value of the mortgage payments over the amortization period (25 years x 12 = 300
months) is:
1


1

 (1  0.003224) 300 
  $576,236.50
PV0  $3,000.00  
0.003224




Or using a financial calculator (TI BA II Plus),
N=300, I/Y=.3224, PMT=-3,000, FV = 0, CPT PV = $576,236.50
In addition, Charlie has $130,000 available as a down payment. The most he can pay for the house
is $576,236.50 + $130,000 = $706,236.50.
52. Section: 5.8 Comprehensive Examples
Learning Objective: 5.8
Level of difficulty: Intermediate
Solution:
a. This is an annuity due. Timmy makes his first payment on his 21st birthday and the last
payment on his 60th birthday. When Timmy turns 61, the value of these 40 annuity payments is:
 (1  0.10) 40  1
FV40  $3,000  
  (1.10)  $1,460,555.43
0.10


Yes, Timmy will achieve his goal by a comfortable margin.
b. In the equation for part a, set FV = $1,000,000, and solve for the number of years, n. This is
easiest done with a financial calculator (TI BA II Plus),
Hit [2nd] [BGN] [2nd] [Set]FV = 1,000,000, I/Y = 10, PMT = -3,000, PV = 0, CPT N = 36.1
Timmy will hit the $1 million-dollar mark in just over 36 years, or shortly after his 57th birthday.
53. Section: 5.8 Comprehensive Examples
Learning Objective: 5.8
Level of difficulty: Intermediate
Solution:
a. 1st Calculate their yearly income available for investment:
Monthly income available = $9,000 – $3,000 – $850 –$1,450 = $3,700
Yearly available = $(3,700)(12) = $44,400
2nd Calculate the FV of their investment when they retire:
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𝐹𝑉30 = $44,400 [
Booth, Cleary, Rakita
(1+.1)30 −1
.1
]=$7,303,535
Or using a financial calculator (TI BA II Plus),
N=30, I/Y=10, PV = 0, PMT=- 44,400, CPT FV=7,303,535
3rd Calculate the amount they will have when they retire:
$7,303,535 + $150,000 = $7,453,535
b. This is an annuity due problem.
PV=7,453,535, k=10%, n=30
1
$7,453,535 = 𝑃𝑀𝑇 [
1 − (1+.1)30
.1
] (1 + .1)
PMT=$718,787
Or using a financial calculator (TI BA II Plus),
Hit [2nd] [BGN] [2nd] [Set]
N=30, I/Y=10, PV= 7,453,535, FV = 0, CPT PMT=-718,787
Challenging
54. Section: 5.1 Opportunity Cost; 5.3 Compound Interest
Learning Objective: 5.1; 5.3
Level of difficulty: Challenging
Solution:
Find the present value of the money paid back to Veda by each investment, using the interest rate
on the alternative (the bank account) as the discount rate.
For Investment A: PV0 
$500
$800

 $453.51  $691.07  $1,144.58
2
(1  0.05) (1  0.05)3
For Investment B:
PV0 
$200
$400
$700


 $190.48  $362.81  $604.69  $1,157.98
1
2
(1  0.05) (1  0.05)
(1  0.05)3
Veda would prefer Investment B, because it has the higher present value.
55. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Challenging
Solution:
The dividends for the first five years form an ordinary annuity. Starting in year 6, the reduced
dividend stream can be thought of as a perpetuity. However, the value of this perpetuity, as
determined by our formula, occurs at year 5 (one year before the first $2 dividend), and must be
discounted to the present:
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1
1 − (1+0.12)5
$2.00
1
𝑃𝑉0 = [$3.00 (
)] + [(
)
] = $10.81 + [$16.67 × 0.5674]
0.12
0.12 (1 + 0.12)5
= $20.27
56. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Challenging
Solution:
0.045
k monthly 
 0.375%
12
Rent payments are typically made at the start of each month (so this is an annuity due). Over
three years, we would expect 36 monthly rent payments. However, the last month’s rent must be
paid up front, so the annuity includes only 35 payments. The present value of the last month’s
rent is $950 because it will be paid today.
1


1

 (1  0.00375)35 
PV0  $950  
  (1  0.00375)  $950  $32,172.48
0.00375




57. Section: 5.4 Annuities and Perpetuities
Learning Objective: 5.4
Level of difficulty: Challenging
Solution:
It is tempting to view the first option as a perpetuity, but this would be incorrect as the man will
die at some time, and the payment will then cease. Thus, option one is an ordinary annuity, with
an uncertain number of payments. Option two is much easier to value; it includes exactly 240
monthly payments.
kmonthly 
0.06
 0.5%
12
Using a financial calculator (TI BA II Plus),
N = 240, PMT = 3,500, I/Y = 0.5, FV = 0, CPT PV = –488,532.70
For the first option to be a better deal, it must include enough payments so that its present value
is at least as great as for option two. Again, using the calculator,
PV = –488,532.70, PMT = 2,785, I/Y = 0.5, CPT N = 420.29
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So, option one must continue for over 420 monthly payments to equal the value of option two.
This is just over 35 years. Hence, the man must live to be at least 100 years old for option one to
be a better deal.
58. Section: 5.4 Annuities and Perpetuities
Learning outcome: 5.4
Level of difficulty: Challenging
Solution:
Step 1: determine Betty’s annual deposits:
 (1  0.05) 40  1
$1,000,000  PMT 
  PMT  $8,278.16
0.05


Or using a financial calculator (TI BA II Plus),
N=40, I/Y=5, FV= 1,000,000, PV = 0, CPT PMT= -8,278.16
Betty will have to make annual deposits of $8,278.16 per year for 40 years at 5% in order to have
$1 million.
Step 2: Abe will be making deposits of 2 x 8,278.16 = $16,556.32. How many annual deposits will
he need to make in order for the future value to be $1 million? (solve for N) The number of deposits
is: 28.52.
1,000,000×.05
𝑙𝑛 [ 16,556.32 + 1]
(1 + 0.05)𝑛 − 1
$1,000,000 = $16,556.32 [
]⇒𝑛=
= 28.52
0.05
𝑙𝑛( 1.05)
Or using a financial calculator (TI BA II Plus),
I/Y=5, PMT= -16,556.32, FV= 1,000,000, PV = 0, CPT N=28.52
Therefore, Abe can afford to wait 11 years (40 – 28.52 = 11.48) before he has to start making his
larger deposits.
59. Section: 5.1 Opportunity Cost; 5.2 Simple Interest; 5.3 Compound Interest; 5.4 Annuities and
Perpetuities
Learning outcome: 5.1; 5.2; 5.3; 5.4
Level of difficulty: Challenging
Solution:
The manager is confused. To make the choice between the two options you should consider the
present value of each set of payments, not the sum of the payments. Summing the payments
assumes that the opportunity cost is zero.
For example, if your opportunity cost is 10%, then the PV of the long option is $161,009. The
value of the house is $250,000 but the cost of the loan (to you) is only $161,009 – a net benefit
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of $88,991. The PV of the short option is $216,289 – in this case, with an opportunity cost of
10%, the short option costs you $55,280 more.
If instead, your opportunity cost is 1%, then the PV of the long option is $390,647, while the PV
of the short option is only $333,390. By taking the short option, you will save $57,257.
60. Section: 5.4 Annuities and Perpetuities; 5.6 Quoted versus Effective Rates
Learning Objective: 5.4; 5.6
Level of difficulty: Challenging
Solution:
Step 1: make the payment frequency match the compounding frequency. We need to convert the
6% compounded monthly to a quarterly effective rate.
12
 .06 
1  kannual   1 

 12 
1  kquarterly  1  kannual  4
1
1
 .06 12  4
  1 
 
 12  
12 4
 .06 
 1 

 12 
kquarterly  1.5075%
Step 2: Now we have an annuity of 5 x 4 = 20 quarterly payments, a present value of $50,000, and
an effective quarterly rate of 1.5075%. Solving for the payments, we get $2,914.44.
61. Section: 5.4 Annuities and Perpetuities; 5.6 Quoted versus Effective Rates
Learning Objective: 5.4; 5.6
Level of difficulty: Challenging
Solution:
Step 1: make the payment frequency match the compounding frequency. We need to convert the
6% compounded quarterly to a monthly effective rate.
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 .06 
1  kannual   1 

4 

Booth, Cleary, Rakita
4
1  kmonthly  1  kannual  12
1
1
 .06  4  12
  1 
 
4  

 .06 
 1 

4 

kmonthly  0.4975%
4 12
Step 2: Now we have an annuity of 10 x 12 = 120 monthly payments, a present value of
$250,000, and an effective monthly rate of 0.4975%. Solving for the payments, we get $2,771.75.
1


1  (1  0.004975)120 
  PMT  $2,771.75
$250,000  PMT  
0.004975




Or using a financial calculator (TI BA II Plus),
N=120, I/Y=.4975, PV= 250,000, FV = 0, CPT PMT= -2,771.75
62. Section: 5.3 Compound Interest; 5.4 Annuities and Perpetuities
Learning Objective: 5.3; 5.4
Level of difficulty: Challenging
Solution:
a. We know the future value and present value amounts, as well as the monthly interest rate.
Finding the number of time periods (months) is most easily done with a financial calculator (TI
BAII Plus),
PV = 15,000, FV = -20,000, I/Y = 0.5, PMT = 0, CPT N = 57.68
It will take approximately 58 months before Roger can afford to buy the car.
b. Solving the following equation for “n” we get:
$20,000 = $15,000(1.005)𝑛 + $250 [
(1.005)𝑛 −1
.005
]
n= 14.86.
Or using a financial calculator (TI BA II Plus),
I/Y=0.5, PV=15,000, FV= -20,000, PMT = 250, CPT N = 14.86
63. Section: 5.3 Compound Interest; 5.6 Quoted versus Effective Rates
Learning Objective: 5.3; 5.6
Level of difficulty: Challenging
Solution:
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Let’s assume the present value of the investment is $1. The future value, after doubling, is then
$2.
a. Annually: With annual compounding, the effective rate is the same as the quoted rate, 9%.
Using a financial calculator (TI BA II Plus),
PV = –1, FV = 2, I/Y = 9, PMT =0, CPT N = 8.04
So the investment will double in 8.04 years.
b. Quarterly: With quarterly compounding, the effective annual rate is:
4
 0.09 
k  1 
  1  9.3083% , and a financial calculator allows us to find:
4 

PV = -1, FV = 2, I/Y = 9.3083, PMT = 0, CPT N = 7.79
The higher effective rate means that only 7.79 years are needed to double the value of the
investment.
64. Section: 5.4 Annuities and Perpetuities
Learning Objectives: 5.4
Level of difficulty: Challenging
Solution:
a. The present value of the annual payments can be found with a financial calculator, (TI BAII
Plus): N=9, PMT = -6,000, I/Y = 5.0, FV = 0, CPT PV = 42,646.93
As this is less than $50,000, the immediate payment alternative is better.
b. This problem can be solved by trial and error, but the task is much easier with a financial
calculator, (TI BA II Plus), N=9, PMT = –6,000, PV = 50,000, FV = 0, CPT I/Y = 1.5675%. At
an interest rate below 1.5675% per year, the nine-year annuity would be preferable; above that
rate, the immediate payment is better.
65. Section: 5.5 Growing Annuities and Perpetuities
Learning Objective: 5.5
Level of difficulty: Challenging
Solution:
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n
PMT1   1  g  
PV0 
1 

k  g   1  k  
n
PMT1   1  g  
n
FVn  PV0 (1  k ) 
1 
 (1  k )

k  g  1 k  
n
$1,000,000 
=
PMT1   1  .04 
1 

.06  .04   1  .06 
PMT1
.02
25

(1  .06) 25  PMT1 1  .0625  1  .0425

.02



*1.62603439
The initial deposit is $12,299.86
 (1  0.06) 25  1
$1,000,000  PMT 
  PMT  $18,226.72
0.06


If Xiang made constant deposits (i.e., no growth), he would have to deposit $18,226.72 per year
for the next 25 years.
66. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Challenging
Solution:
a. The effective monthly interest rate is,
2
 0.051  12
kmonthly  1 
  1  0.4206%
2 

The amount of the mortgage loan will be ($380,000 – $150,000) = $230,000, and there will be
12 x 25 = 300 monthly payments, the value of which can be found with a financial calculator, (TI
BAII Plus), N=300, PV = –230,000, I/Y = 0.4206, FV = 0, CPT PMT = 1,350.89. Alysha’s two
friends will be paying 2 x $475 = $950 in rent, so she will need an additional $1,350.89 – $950 =
$400.89 to make the mortgage payments.
b. In two years, Alysha will have made 24 payments, leaving 276. The present value of these
payments is the outstanding value of the mortgage loan. Use the calculator again: N=276, I/Y =
0.4206, PMT = 1,350.89, FV = 0, CPT PV = 220,336.58. To pay off the loan, and recoup her
down payment, Alysha would have to sell the house for at least $220,336.58 + $150,000 =
$370,336.58.
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67. Section: 5.6 Quoted versus Effective Rates; 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.6; 5.7
Level of difficulty: Challenging
Solution:
a. First, find the effective interest corresponding to the frequency of Jimmie’s car payments
(f =12); with monthly compounding, set m=12,
 QR 
k monthly  1 

m 

m
f
12
 8.5% 
 1  1 

12 

12
 1  0.70833%
The 60 car payments form an “annuity” whose present value is the amount of the loan (the price
of the car):
1


1 
60 
(1  0.0070833) 
$29,000  PMT 
 PMT  $594.98
0.0070833






Or using a financial calculator (TI BA II Plus),
N=60, I/Y=.70833, PV= 29,000, FV = 0, CPT PMT= -594.98
b. Use the effective monthly interest rate from part a, k=0.70833%
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
...
35
36
(1) Principal
Outstanding
(2)
Payment
(3) Interest
=k*(1)
29,000.00
28,610.44
28,218.12
27,823.01
27,425.11
27,024.40
26,620.84
26,214.42
25,805.13
25,392.94
24,977.82
24,559.77
24,138.76
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
594.98
205.42
202.66
199.88
197.08
194.26
191.42
188.56
185.69
182.79
179.87
176.93
173.97
170.98
14,083.18
13,587.95
594.98
594.98
99.76
96.25
(4) Principal
Ending Principal
Repayment = (2)= (1)-(4)
(3)
389.56
28,610.44
392.32
28,218.12
395.10
27,823.01
397.90
27,425.11
400.72
27,024.40
403.56
26,620.84
406.42
26,214.42
409.29
25,805.13
412.19
25,392.94
415.11
24,977.82
418.05
24,559.77
421.01
24,138.76
424.00
23,714.76
495.22
498.73
13,587.95
13,089.22
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13,089.22
594.98
92.72
502.26
12,586.96
1,177.43
590.79
594.98
594.98
8.34
4.18
586.64
590.79
590.79
0.00
The first monthly payment repays $389.56 of the principal amount of the loan and the last payment
repays $590.79.
c. After three years, or 36 monthly payments, the principal outstanding is $13,089.22 (from the
amortization table). The present value of this amount is:


1
  $10,152.19
PV0  $13,089.22  
 (1  0.0070833) 36 


68. Section: 5.6 Quoted versus Effective Rates; 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.6, 5.7
Level of difficulty: Challenging
Solution:
The 60 monthly payments form an annuity whose present value is $30,000. Finding the interest
rate is most easily done with a financial calculator (TI BA II Plus),
N=60, PMT=-622.75, PV= 30,000, FV =0, CPT I/Y = 0.75%
Note that we used N=60 months, so the solution is a monthly interest rate, however, the problem
asks for the effective annual rate.
k  (1  k monthly)12  1  (1  0.0075)12  1  9.38%
The quoted rate would be:
1
1
QR  m  [(1  k ) 12  1]  12  [(1  0.0938) 12  1]  9.00%
Or simply:
QR  m  k monthly  12  0.0075  9.00%
69. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Challenging
Solution:
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Part 1: determine the principal outstanding after the 60th payment (i.e., How much will the next
mortgage be for?)
Step 1: determine effective monthly rate:
1
 .06 2  12
kmonthly  1 
   1  0.00493862
2  

Step 2: determine the monthly payments:
1


1  (1  0.00493862) 300 

$250,000  PMT  
0.00493862




PMT  $1,599.5162
Or using a financial calculator (TI BA II Plus),
N=300, I/Y=.493862, PV=250,000, FV =0, CPT PMT = -1,599.5162
Step 3: determine the present value of remaining (300 – 60) payments of $1,599.5162
1


1  (1  0.00493862) 30060 
  $224,591.7542
PV  $1,599.5162  
0.00493862




Or using a financial calculator (TI BA II Plus),
N=240, I/Y=.493862, PMT=-1,599.5162, FV = 0, CPT PV = $224,591.7542
Part 2: determine new monthly payment
Step 1: determine new effective monthly rate
1
 .08 2  12
kmonthly  1 
   1  0.00655820
2  

Step 2: determine the new monthly payment
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1


1  (1  0.00655820) 30060 

$224,591.7542  PMT  
0.00655820




PMT  $1,860.4231
Or using a financial calculator (TI BA II Plus),
N=240, I/Y=.65582, PV=224,591.7542, FV = 0, CPT PMT = -1,860.4231
Franklin’s new monthly payment is $1,860.42.
70. Section: 5.7 Loan or Mortgage Arrangements
Learning Objective: 5.7
Level of difficulty: Challenging
Solution:
a. PV=$200,000, monthly rate=12%/12=1%, N = (10)(12) =120 months
1
$200,000 = 𝑃𝑀𝑇 [
1 − (1+.01)120
. 01
]
1
𝑃𝑀𝑇 = $200,000/ [
1 − (1+.01)120
. 01
]
So, PMT=$2,869.42
Or using a financial calculator (TI BA II Plus),
N=120, I/Y=1, PV= 200,000, FV = 0, CPT PMT=-2,869.42
b. Remaining months to pay=120 – 18=102 months
1−
𝑃𝑉0 = $2,869.42 [
1
(1+.01)102
.01
]=$182,946.63
Or using a financial calculator (TI BA II Plus),
N=102, I/Y=1, PMT=- 2,869.42, FV = 0, CPT PV=182,946.63
2
c. kmonthly= (1 
.12 12
)  1 =.9759%
2
1
$200,000 = 𝑃𝑀𝑇 [
1 − (1+.009759)120
. 009759
]
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1
𝑃𝑀𝑇 = $200,000/ [
1 − (1+.009759)120
. 009759
]
So, PMT=$2,836.08
Or using a financial calculator (TI BA II Plus),
N=120, I/Y=.9759, PV=200,000, FV = 0, CPT PMT=-2,836.08
71. Section: 5.8 Comprehensive Examples
Learning Objective: 5.8
Level of difficulty: Challenging
Solution:
Investor A:
k=e.15 – 1=16.183424%.
1st, consider an ordinary annuity and the present value of the investment when A turns 25 years
old is:
1


1 
8 
(1  .16183424) 
=$23,749.19
PV25  $5,500


.16183424




Or using a financial calculator (TI BA II Plus),
N=8, I/Y=16.183424, PMT=-5,500, FV = 0, CPT PV= 23,749.19
2nd, discount this amount for five years back to today when she is 20.
PV0  FV5 
1
1
 $23,749.19 
 $11,218.3231
5
(1  k )
(1.16183424) 5
Or, N=5, I/Y=16.183424, PMT = 0, FV=- 23,749.19, CPT PV=11,218.3231
Investor B:
.16 4
k= (1 
)  1 =16.985856%
4
1


1 
10 
(1  .16985856) 
$11,218.3231  PMT 
(1.16985856)


.16985856




PMT=$2,057.38
Or using a financial calculator (TI BA II Plus),
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Hit [2nd] [BGN] [2nd] [Set]
N=10, I/Y=16.985856, PV=11,218.3231, FV = 0, CPT PMT= - 2,057.38
Therefore, Investor B has to make a yearly payment of $2,057.38 so that the present value of the
two investments is the same.
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Answers to Concept Review Questions
5.1 Opportunity Cost
Concept Review Questions
1. Why does money have a “time value”?
An investor can simply store dollars (tuck them under the bed!) and spend them in the future. In
this sense a dollar is always worth at least a dollar in the future. However, this ignores the fact
that the saver has other uses for that dollar, which in economics we call an “opportunity cost” or
simply an “alternative use.” This results in a “time value” of money.
2. What is an “opportunity cost”?
The opportunity cost of money is the interest rate you can earn by investing a dollar today.
5.2 Simple Interest
Concept Review Questions
1. Explain how simple interest payments are determined.
Simple interest payments are n × P × k, where n is the number of periods in years, P is the principal,
and k is the simple annual interest rate.
2. Does simple interest take into account the time value of money? Explain.
Simple interest can be used to calculate the future value of money assuming that only the
principal is reinvested. In practice, compound interest is more realistic.
5.3 Compound Interest
Concept Review Questions
1. Explain how to compute future values and present values when using compound interest.
FVn = PV0 (1 + 𝑘)𝑛 , where PV0 is the present value, k is the compound value interest factor, n
is the number of periods, and FVn is the future value in year n.
2. What is the relationship between FVIFs and PVIFs? Why does this make sense?
FVIF=1/PVIF. This relationship make sense because by definition, FVIF = (1 + 𝑘)𝑛 and PVIF
= 1/(1 + 𝑘)𝑛 .
3. Why does compound interest result in higher future values than simple interest?
Compound interest refers to a process whereby interest is earned on the invested principal
amount and on any accrued interest. However, simple interest is only earned on the principal
amount.
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5.4 Annuities and Perpetuities
Concept Review Questions
1. Explain how to calculate the present value and future value of an ordinary annuity and an
annuity due.
 (1  k )n  1
FVn  PMT 
 gives the future value of an ordinary annuity.
k


Since each flow gets one extra period of compounding in an annuity due, the FV (annuity due) =
[FV (ordinary annuity)](1 + k). The present value of both an annuity and an annuity due
is PV0 = FVn /(1 + 𝑘)𝑛 .
2. Define “perpetuity.”
Perpetuities are special annuities in that they go on forever, so n goes to infinity in the annuity
equation.
3. Why is the present value of $1 million in 50 years’ time worth very little today?
If the required return is 12% a year, the present value of $1 million in 50 years’ time is only
$3,460. The small present value is caused by the discounting process.
5.5 Growing Perpetuities and Annuities
Concept Review Questions
1. Explain how to evaluate a growing perpetuity.
Estimate the payment (PMT), the required rate of return (k), and the expected growth rate to
infinity (g), and apply Equation 5.10.
2. Explain how to calculate the present value of a growing annuity.
Estimate the payment (PMT), the required rate of return (k), the number of years for the annuity
(n), the expected growth rate per period over a given period of time (g), and apply Equation 5.12.
5.6 Quoted Versus Effective Rates
Concept Review Questions
1. Why can effective rates often be very different from quoted rates?
If the quoted rates are not annually compounded, the effective rates are different from the quoted
rates because of the different number of compounding periods.
2. Explain how to calculate the effective rate for any period.
m
 QR 
The effective annual rate for any given compounding interval: k  1 
  1 , where k =
m 

effective annual rate, QR = quoted rate, and m = the number of compounding intervals per year.
Rates for payments that are other than annual payments require an effective period rate. The
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m
 QR  f
effective period rate for payments other than annual is given as k  1 
  1 , where k =
m 

effective period rate, QR is the nominal quoted rate, m = the number of compounding periods per
year, and f = the frequency of payments per year.
5.7 Loan or Mortgage Arrangements
Concept Review Questions
1. Explain how loan and mortgage payments can be determined using annuity concepts.
Since these loans involve equal payments at regular intervals based on one fixed interest rate
specified when the loan is taken out, the payments can be viewed as annuities.
2. What complications arise when dealing with mortgage loans in Canada?
In Canada, the interest rates are quoted semi-annually and the payments are made monthly.
Mortgages are amortized over long periods of time, but the rates are set for terms or periods that
may be shorter than the amortization period. When the term expires, a new rate of interest
needs to be negotiated and interest rates may have increased.
5.8 Comprehensive Examples
Concept Review Questions
1. Explain how timelines can be used to break a complicated time value of money problem into
manageable components.
You can visualize the problem and break complicated cash flows into its three constituent parts
since you should develop an understanding of what is approximately the right answer.
2. Demonstrate how to solve a typical retirement problem.
There are three steps. First, calculate the present value of retirement funds in the year of
retirement. Second, calculate the cash you need to raise through investment in the year of
retirement. Third, determine the required year-end payments to give you the future value of the
amount that is calculated in the second step.
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Case 5.1
1. a. Effective annual interest rate is:
(
k= 1+
QR
m
m
)
(
-1= 1+
0.045
2
2
) - 1 = 4.5506%
Effective monthly interest rate is:
kmonthly = (1 + k)1/f - 1 = (1 + 4.5506%)1/12 - 1 = 0.37153%
Shanna would pay an effective monthly interest rate of 0.37153%.
b. The mortgage principal is $325,000 – $48,750 = $276,250
There are 25 x 12 = 300 monthly payments.
Using a financial calculator (TI BA II Plus),
N = 300, I/Y = 0.37153, PV = 276,250, FV = 0, CPT PMT = –1,528.97
The monthly mortgage payment would be $1,528.97.
2. a. Shanna has $840 in savings each month over 48 months earning interest at 1.5%
annually, compounded monthly (1.5%/12 = 0.125%).
Using a financial calculator (TI BA II Plus),
N = 48, I/Y = 0.125%, PV = 0, PMT = 840, CPT FV = 41,527
Shanna will have $41,527 at the end of four years.
b. Shanna’s down payment will be the sum of her grandparents’ gift and her savings:
$48,750 + $41,527 = $90,277
c. The mortgage principal is $325,000 – $90,277 = $234,723
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Using a financial calculator (TI BA II Plus),
N = 300, I/Y = 0.37153, PV = 234,723, FV = 0, CPT PMT = –1,299.13
The monthly mortgage payment would be $1,299.13.
3. a. Shanna will make 48 monthly payments of $840 into her RRSP, earning interest at 6.5%
annually, compounded monthly (6.5%/12 = 0.5417%).
Using a financial calculator (TI BA II Plus),
N = 48, I/Y = 0.5417%, PV = 0, PMT = 840, CPT FV = 45,906
Shanna will have $45,906 at the end of four years.
b. Shanna will make a total of 43 x 12 = 516 payments of $840 each.
Using a financial calculator (TI BA II Plus),
N = 516, I/Y = 0.5417%, PV = 0, PMT = 840, CPT FV = 2,363,608
Shanna will have $2,363,608 at age 65.
4. a. Shanna will make 36 monthly payments of $840 into her RRSP and her opening balance
of $5,386 will grow as well at 0.5417% per month.
Using a financial calculator (TI BA II Plus),
N = 36, I/Y = 0.5417%, PV = 0, PMT = 840, CPT FV = 33,291
N = 36, I/Y = 0.5417%, PV = -5,386, PMT = 0, CPT FV = 6,542
Shanna’s account at age 26 will be $33,291 + $6,542 = $39,833
b. Without the Caribbean trip, Shanna had accumulated $45,906. She would have saved
$6,073 less ($45,906 – $39,833) by going on the trip. This is the correct answer because
the $5,000 would have grown to $6,073 over 36 months.
c.
Shanna will make a further 39 x 12 = 468 payments of $840 each.
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Using a financial calculator (TI BA II Plus),
N = 468, I/Y = 0.5417%, PV = 0, PMT = 840, CPT FV = 1,788,293
N = 468, I/Y = 0.5417%, PV = -39,833, PMT = 0, CPT FV = 499,202
Shanna will have $1,788,293 + $499,202 = $2,287,495 at age 65.
She will have $76,113 ($2,363,608 – $2,287,495) less money saved at age 65. This is
confirmed by calculating the future value of $5,000 over 504 months (42 years).
Shanna must weigh the advantages of enjoying a Caribbean vacation one year from now
versus allowing that $5,000 expenditure from either being invested and growing to $76,113
at age 65 or growing for a further three years to be used as a down payment. The mortgage
payment of $1,299 represents 32.5% of her gross monthly income of $4,000. This is a
higher percentage than is generally recommended. Her goal of acquiring a house in four
years seems to be too aggressive, even without the Caribbean vacation.
However, she may be able to acquire a house that lends itself to a rental opportunity,
possibly to university students. She needs to find a way to increase her gross income or
lower her expenditures, or both. She will benefit from the RRSP contributions for the first
four years, but then the RRSP one-time withdrawal of the accumulated amount will be
punitive for income tax purposes. The grandparents’ gift of $48,750 is an opportunity that
is significant, and she should strive to achieve her goal of home ownership. Her financial
picture would be much improved thanks to the time value of money if the Caribbean
vacation did not take place.
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