Dear students, despite the fact there is a lot of books or textbooks concerning Financial Mathematics, or Mathematics of Finance, I understand your
need of one comprehensive textbook dealing with all the topics which appear
in our course of Financial Mathematics taught by myself. I will try to avoid
so many mistakes, and errors as possible, nevertheless try to understand if
some appear and please let me know.
Enjoy each chapter by chapter.
Best regards Tatana Funiokova
–1–
–2–
Contents
1.
Simple Interest and Day–Count Conventions . . . . . . .
5
1.1.
Motivation and Basic Concepts
5
1.2.
Simple Interest Theorem . . . . . . . . . . . . .
1.2.1.
Application: Certificate of Deposit (CD)
. . . . .
1.2.2.
Application: Savings Account . . . . . . . . .
1.3.
Day-Count Conventions
1.4.
Discount . . . . . . . . . . . . . . . . . . .
1.5.
2.
. . . . . . . . . .
. . . . . . . . . . . . .
1.4.1.
Application: Treasury Bills (T–bills)
1.4.2.
Application: Bill of Exchange . . . . . . . . .
Summary
. . . . . . .
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . .
6
11
12
12
17
18
20
22
1.5.1.
Formulas to remember
1.5.2.
Questions
1.5.3.
Tasks/Problems . . . . . . . . . . . . . .
23
Compound interest . . . . . . . . . . . . . . . . .
27
2.1.
Compound Interest Theorem
27
2.2.
Effective Rate
2.3.
Summary
. . . . . . . . . . . . . . . .
. . . . . . . . . . .
22
22
. . . . . . . . . . . . . . . . .
32
. . . . . . . . . . . . . . . . . .
34
2.3.1.
Formulas to Remember
. . . . . . . . . . .
2.3.2.
Questions
2.3.3.
Tasks/Problems . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
–3–
34
34
35
3.
Cash Flows . . . . . . . . . . . . . . . . . . . .
37
3.1.
Cash Flows
37
3.2.
Net Present Value
3.3.
Internal Rate of Return
3.3.1.
3.4.
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . .
Application: Index Fund . . . . . . . . . . .
Summary
. . . . . . . . . . . . . . . . . .
3.4.1.
Formulas to Remember
. . . . . . . . . . .
3.4.2.
Questions
3.4.3.
Tasks/Problems . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
–4–
38
40
43
45
45
45
45
1
Chapter
Simple Interest and
Day–Count Conventions
1.1. Motivation and Basic Concepts
There are three crucial aspects that may affect the future value of your money
you keep under the mattress:
∙ Instead of keeping your money in all these „safe places“ in your house,
you may invest that amount, which means you potentially loose the
interest you can earn instead. Having $100 today – under some circumstances – is the same as having $102 in a year. And similarly, having
$100 today is not the same as having $100 in a year.
∙ Inflation – does it need any comments? 30 years ago 1 Czech roll was
for 0.4 CZK, today’s price is about 2.5 CZK? (Actually the range is 1.50
– 3.50. Moreover, the weight is lower - only 43g instead of formal 50g
...)
∙ Risk - what if the mattress would be stolen? What if lightning strikes
your house and devours everything including the mattress? This attitude may inspire you to to spend all your money immediately, but there
are better ways how you can handle your money.
It forces us to talk about the time value of money. In the following we will
discuss all these aspects in detail, and many other topics as well. Let me
introduce some basic concepts related to the time value of money:
∙ Present Value of an amount is its worth today;
∙ Future Value of an amount is its worth at a later time;
∙ Principal - an amount of money that is lent or invested to earn interest;
–5–
1. Simple Interest and Day–Count Conventions
∙ Interest - a compensation a borrower of capital pays to a lender of capital; lenders have to be compensated since they have temporarily lost
use of their capital;
∙ Financial Transaction - a principal is invested for a period of time; at
the end of the investment period, a total amount (Accumulated Value)
is returned;
∙ Interest Earned - the difference between the accumulated value and the
principal. (Interest and principal are almost always expressed in terms
of money.)
It is well known that the amount of interest depends on the interest rate,
the amount of money borrowed (principal) and the length of time that the
money is borrowed. In terms of Financial Mathematics we are also concerned
about the method of calculating the interest. On the one hand we need to
distinguish among simple interest, compound interest, and combination of
these two. On the other hand we need to know the exact time when the
interest is calculated and credited during the investment period. Also daycount conventions may have considerable impact.
1.2. Simple Interest Theorem
For a simplicity, let us focus on the annual interest rate at first, and the interest
is calculated at the end of the year. Let us invest $100 at 5% interest per year
for 4 years. It means the interest period is a year, and the investment period
is 4 years. Using simple interest for calculations means that the interest is
calculated from the initial principal during the whole investment period.
Table 1.1. Simple Interest (one interest period in a year)
Year
1
2
3
4
Year’s Beginning
Principal
$100
$100
$100
$100
Year’s End
Interest
$5
$5
$5
$5
Year’s End
Amount
$105
$110
$115
$120
For instance $115 is the future value of $100 after three years. Using the
following notation:
–6–
1.2. Simple Interest Theorem
∙ P0 . . . the initial principal (present value) invested
∙ n . . . the total number of years
∙ Pn . . . the future value of P0 at the end of the nth year
∙ i . . . the interest rate per year
we can easily verify that for the accumulated value after n years we have
Pn = P0 + P0 · n · i = P0 · (1 + n · i).
The annual interest rate is called the nominal rate, the quoted rate, or the
stated rate. But we do not need to restrict ourselves to annual rates. In general, each interest rate is associated with number (m) of interest periods per
year, i.e. how many times per year the interest is calculated. Interest period
may be a year, a month, or infinitely short time period. Therefore i(m) usually
denotes a nominal interest rate, where m is the number of interest periods in
a year.
Table 1.2. Nominal Interest Rates
number of
interest periods (m)
1
2
4
12
365 (366)
∞
nominal interest
rate i(m)
i(1)
i(2)
i(4)
i(12)
i(365)
i(∞)
the interest
is calculated
annually
semi-annually
quarterly
monthly
daily
continuously
The interest rate per interest period would be denoted by i (throughout
the whole textbook), and obviously we have
i=
i(m)
m
and
i(m) = i · m.
Note 1.1. For example, if the interest period is a month, it means we calculate the interest at the beginning or at the end of each month, and nominal
(annual) interest rate is i(12) = 6% then for the corresponding interest rate per
(12)
month i we have i = i12 = 0.5%.
–7–
1. Simple Interest and Day–Count Conventions
Table 1.3. Simple Interest (two interest periods in a year)
Semester
1
2
3
4
5
6
7
8
Semester’s Beginning
Principal
100
100
100
100
100
100
100
100
Semester’s End
Interest
2.5
2.5
2.5
2.5
2.5
2.5
2.5
2.5
Semester’s End
Amount
102.5
105.0
107.5
110.0
112.5
115.0
117.5
120.0
Let us invest $100 at 5% interest per year for 4 years, but the interest is
calculated twice a year, i.e. i(2) = 5% and i = 2.5%. It means the investment
period is 4 years again, but the interest period is a semester. Then accumulated values during the investment period are shown in Table 1.3
This example demonstrates, that number of interest periods has no effect
on the total interest earned. Using simple interest the accumulated value
depends just upon nominal interest rate, time, and principal invested. So if n
stands for number of semesters, and i for interest rate per interest period we
still can use the formula
Pn = P0 + P0 · n · i = P0 · (1 + n · i).
Finally, for general notation
∙ P0 . . . the initial principal (present value) invested
∙ n . . . the total number of interest periods
∙ Pn . . . the future value of P0 at the end of the nth interest period
∙ m . . . the number of interest periods per year
∙ i(m) . . . nominal rate (annual interest rate)
∙ i . . . the interest rate per interest period
we obtain the following theorem.
Theorem 1.2 (The Simple Interest Theorem). If we start with principal P0 ,
and invest it for n interest periods at a nominal rate of i(m) (expressed as a decimal)
–8–
1.2. Simple Interest Theorem
calculated m times a year using simple interest, then Pn , the future value of P0 at the
end of the nth interest period, is
Pn = P0 · (1 + n · i),
where i =
i(m)
m .
Accumulated Value of $100
500
10%
5%
400
300
200
100
0
0
5
10
15
Years
20
25
30
Figure 1.1. Accumulated Value of $100 at 5% and 10% simple interest
Note 1.3. There may be a need for defining some rates and index numbers
to evaluate the interest earned. We easily derive the interest earned during
the nth period
In = Pn − Pn−1 ,
obviously for simple interest we have In = P0 · i. For the whole investment period we can also derive the total interest earned (the principal appreciation)
IT = Pn − P0
where for simple interest we obtain IT = P0 · n · i. For every interest period
we can determine the rate of interest earned during the nth period:
in =
Pn − Pn−1
In
=
.
Pn−1
Pn−1
–9–
1. Simple Interest and Day–Count Conventions
For simple interest approach in is a decreasing function of n because we do
i
). Finally we
not use the interest earned to generate new interest ( in = 1+i·(n−1)
introduce the rate of return
Pn − P0
ROR =
P0
and the rate of return per period
RORP =
Pn − P0
.
n · P0
Again, for simple interest we obtain
ROR = n · i
RORP = i.
and
For our 4–year investment with annual interest rate of 5% the total interest
earned is $20, the rate of return 20%, the rate of return per period 5% and
interest and rate of interest earned during the nth period can be seen in the
Table 1.4
Table 1.4. Interest and rate for single periods
Year
1
2
3
4
Year’s Beginning
Principal
$100
$100
$100
$100
Year’s End
Interest
$5
$5
$5
$5
Year’s End
Amount
$105
$110
$115
$120
In
$5
$5
$5
$5
in
5%
4.76%
4.55%
4.35%
Note 1.4. Problems concerning simple interest approach may involve unknown accumulated value Pn , initial principal P0 , time n, total interest earned
IT , or interest rate i. For all these problems we can easily derive corresponding
formulas (the first formula is sometimes referred as “simple discounting”):
P0 = Pn
1
,
1+n·i
n=
Pn − P0
,
P0 · i
IT = P0 · n · i,
i=
Pn − P0
.
P0 · n
Example 1.5. What is better for us? To buy a parcel of land today for $82,000
or after 6 months for $83,500? Suppose that common stocks with the same
risk as this investment offer a 3,5% expected return.
Solution. We use simple approach because of short–time investment period. (For longer interest periods we usually use compound approach.) There
are three ways how we can solve the problem:
– 10 –
1.2. Simple Interest Theorem
based on P0 : we can compare $82,000 with value today of $83,500:
P0 =
$83, 500
= $82, 063.88.
1 + 0.5 · 3.5%
To pay $83,500 in six months is – under these circumstances – the same
as paying $82,063.88 today. So to pay $82,000 for the parcel of land is
better choice.
based on Pn : we can compare future value of $82,000 with $83,500:
Pn = $82, 000 · (1 + 0.5 · 3.5%) = $83, 435.
If we would not pay $82,000 now, we can invest it at 3.5%, but accumulated value of this principal would not reach the price of $83,500. So it
is better to pay $82,000 now.
based on i: we look for simple interest rate which accumulates $82,000 to
$83,500 in 6 months:
i=
$83, 500 − $82, 000
= 3.66%.
$82, 000 · 0.5
You need the opportunity cost to be 3.66% to increase $82,000 to $83,500,
but it is only 3,5%, so it is better to buy the parcel now.
1.2.1. Application: Certificate of Deposit (CD)
∙ issued by financial institutions
∙ the institution pays a fixed interest rate on the lender’s initial investment
for a specified term
∙ typical terms are: 6 months, 1 year, 2 years, 5 years
∙ the lender cannot withdraw the initial investment before the end of the
term without penalty, but the lender can withdraw the interest as it is
credited to the lender’s account, if desired
∙ payment frequency: at maturity for short–term; monthly for long–term
Example 1.6. Carla Bernoulli invests $10,000 in a CD at 6% a year for 4 years.
She withdraws the interest at the end of each year. What amount does she
have at the end of four years assuming that she does not spend or invest the
interest?
– 11 –
1. Simple Interest and Day–Count Conventions
Solution.
P4 = $10, 000 · (1 + 4 · 6%) = $12, 400
1.2.2. Application: Savings Account
∙ pays a stated annual interest rate
∙ the interest is often computed based on the daily balance
∙ the holder of the account may withdraw money at any time without
penalty
∙ frequently insured up to a maximum amount by the federal government
Example 1.7. Marie Toman has a savings account that pays interest at a nominal rate of 5%. Interest is calculated 365 times per year on the minimum daily
balance and credited to the account at the end of the month. Marie has an
opening balance of $1,000 at the beginning of April. On April 11th she deposits $200, and on April 21st she withdraws $300. How much interest does
she earn in April?
Solution. Because the initial principal varies over time, we need to determine number of days for these principals (see Table 1.5).
IT = $1, 000 · 11 ·
0.05
0.05
0.05
+ $1, 200 · 9 ·
+ $900 · 10 ·
= $4.22
365
365
365
1.3. Day-Count Conventions
Consider the following example concerning simple interest problem:
Example 1.8. Carl invests $1,000 on January 21st , 2012 at a nominal rate of
i(1) = 8%, and on April 16th , 2012 he withdraws the whole amount of money,
including interest earned. How much interest does he earn for this period?
This is the first time interest period is longer than investment period. In
fact, we are unable to express the investment period as a number (integer) of
investment periods. There are several methods how to treat the fraction representing incomplete part of interest period. We call them Day-Count Conventions.
– 12 –
1.3. Day-Count Conventions
Table 1.5. Miminum Daily Balance
day
1
..
.
daily balance
$1,000
..
.
minimum
$1,000
..
.
10
11
12
..
.
$1,000
$1,000 and $1,200
$1,200
..
.
$1,000
$1,000
$1,200
..
.
20
21
22
..
.
$1,200
$900 and $1,200
$900
..
.
$1,200
$900
$900
..
.
30
$900
$900
⇓
minimum daily balance
$1,000
$1,200
$900
number of days
11
9
10
The problem concerns the fraction n = Nk which represents proportion
of an investment period for which the interest is calculated. Let the interest
period be a year. We have three alternative how to treat denominator N:
“360” - despite a year has 365 or 366 days, we use 360 days for simplicity
“365” - despite a year has 365 or 366 days, we use 365 as a denominator
“ACT” - we use 365 for non–leap years and 366 days for leap years
Note 1.9. Are you sure you really know how to recognize a leap year? The
following algorithm may be helpful:
if year is divisible by 400 then
is_leap_year
else if year is divisible by 100 then
not_leap_year
else if year is divisible by 4 then
is_leap_year
else
– 13 –
1. Simple Interest and Day–Count Conventions
not_leap_year
On the other hand, we have three alternative how to treat numerator k as
well:
“ACT” - we use actual number of days of the investment period, the first
day1 is not taken into account2
“30E” - total number of days between D1 , M1 , Y1 and D2 , M2 , Y2 is
360 · (Y2 − Y1 ) + 30 · (M2 − M1 ) + (D2 − D1 )
if D2 or D1 equals 31 replace it by 30
“30” or “30A” - differs from “30E” in case D2 = 31 AND D1 ̸= 30 or 31, then
we use D2 = 31 in the formula
Note 1.10. For ACT approach the “Orders–of–Days Chart” table may be
helpful. Just simply subtract orders of corresponding days. (For a leap year
some orders must be increased by 1.)
Calculating number of days (k), we exclude one day using any of the three
methods above, not just ACT.
Combining all the approaches for numerator and denominator we can
obtain nine different methods, but only a few of them have some applications.
So here is the list of the most used day–count conventions:
“ACT/360” – money market basis, Banker’s Rule,commonly used for all
Euro Currency Libor rates, except sterling)
“ACT/365” – commonly used for all sterling interest rates, including Libor. It
is also used for money markets in Australia, Canada and New Zealand.
“ACT/ACT” – commonly used for all sterling bonds, US Treasury bonds,
Euro denominated bonds, and for some USD interest rate swaps.
“30E/360” - Eurobond basis,for calculating accrued interest on some legacy
currency pre Euro Eurobonds and on bonds in Sweden and Switzerland.
“30/360” – Bond basis, for calculating accrued interest on domestic US bonds
1
2
Omitting the last day instead of the first day may affect the final result.
The only exception is for interest period being a day.
– 14 –
1.3. Day-Count Conventions
Table 1.6. Orders–of–Days Chart
Jan
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
Feb
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
Mar
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
Apr
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
May
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
Jun
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
Jul
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
– 15 –
Aug
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
Sep
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
Oct
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
Nov
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
Dec
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
1. Simple Interest and Day–Count Conventions
Solution. [Example 1.8] We have P0 = $1, 000, i(1) = i = 8%, D1 = 21, M1 =
1, Y1 = 2012, D2 = 16, M2 = 4, Y2 = 2012, and we want to determine
IT = P0 · n · i = P0 ·
k
·i
N
For actual number of days let us assign orders to both of the dates. The order
of the 21st January is simply 21, and for the 16th April you can find 106 in the
table, but because 2012 is a leap year we need to add 1 to this order, so the
order which should be used is 107. 30E and 30A coincide here:
k = 360 · (2012 − 2012) + 30 · (4 − 1) + (16 − 21) = 90 − 5 = 85
Therefore, for the mentioned conventions we get:
“ACT/360” -
k
N
=
107−21
360
=
86
360
= 0.2389
“ACT/365” -
k
N
=
107−21
365
=
86
365
= 0.2356
“ACT/ACT” -
k
N
=
107−21
366
“30E/360” or “30/360” -
k
N
=
86
366
=
= 0.2350
85
360
= 0.2361
So we can reach 4 different results in general depending on day–count convention used:
“ACT/360” - IT = $1, 000 ·
86
360
· 0.08 = $19.11
“ACT/365” - IT = $1, 000 ·
86
365
· 0.08 = $18.85
“ACT/ACT” - IT = $1, 000 ·
86
366
· 0.08 = $18.80
“30E/360” or “30/360” - IT = $1, 000 ·
85
360
· 0.08 = $18.89
Note 1.11. Number of days (k) between two dates date1 and date2 can be also
determined using Excel:
“ACT” - just subtract corresponding dates date2 − date1
“30E” - DAYS360(date1 ;date2 ;TRUE)
“30” or “30A” - DAYS360(date1 ;date2 ;FALSE) or the last parameter
may be omitted, i.e. DAYS360(date1 ;date2 )
– 16 –
1.4. Discount
Example 1.12. How many days are there between January 13th , 2004 and
May 31st , 2007? If the interest period is one year, calculate n for these day–
count conventions: ACT/ACT, 30E/ACT, and 30A/ACT.
Solution. For ACT we have 1234 days, for 30E 1217, and for 30A 1218 days.
Because the first year (2004) is a leap year we need to split these counts according to particular years for further calculations:
ACT/ACT n =
353
366
+
365
365
+
365
365
+
= 3.378
151
365
30E/ACT n =
348
366
+
360
365
+
360
365
+
150
365
= 3.334
30A/ACT n =
347
366
+
360
365
+
360
365
+
150
365
= 3.332
1.4. Discount
What if the simple interest would be calculated at the beginning of the interest period? If we start with $100 then future value at the end of the year
at 10% simple interest would be $110 using Simple Interest Theorem. Now
the question is what amount of money should I start with if I want the future value in one year to be $100, and the interest earned would be 10% of
that future amount? Well the starting amount is $100 − 10% · $100 = $90. In
other words, we have already defined the rate of interest earned during the
nth period as a percentage of the balance at the beginning of the period
in =
Pn − Pn−1
In
=
Pn−1
Pn−1
similarly, we can define the rate of discount based on the balance at the end
of the period
In
Pn − Pn−1
=
dn =
Pn
Pn
for n = 1 and dn = d we obtain
P0 = P1 (1 − d)
generally
P0 = Pn (1 − d · n).
If accumulated value of $90 is $100 at discount rate d = 10%, obviously we
can find corresponding interest rate id at which we accumulate $90 to $100
$100 = $90 · (1 + id )
⇒
– 17 –
id = 11.11%
1. Simple Interest and Day–Count Conventions
in general we have (for positive values of i and d) these relationships between
i and d:
d
i
id =
>d
and
di =
< i,
1−d
1+i
and for arbitrary n we have:
id =
d
>d
1−d·n
di =
and
i
< i,
1+i·n
1.4.1. Application: Treasury Bills (T–bills)
∙ treasury securities that have a maturity of a year or less when issued
(typical terms: 4, 13, 26, 52 weeks)
∙ pay no coupons, they make a single payment of par value (face value)
at maturity
∙ they are issued at a discount to par, so an investor who holds a T-bill
to maturity earns the difference between the par value and discounted
value at which the instrument was purchased (or the difference between subsequent sale price and the discounted value)
∙ are quoted at a discount from face value, with the discount expressed
as an annual rate based on a 360–day year (ACT/360)
∙ prices are quoted for $100 of par value
∙ issued in increment of $5,000 above a minimum amount of $10,000
You can see an example of Treasury Bill Quotation (from the Wall Street Journal, Friday, February 6th , 2015) in Table 1.7.
Table 1.7. Treasury Bill Quotations
Maturity
10/15/2015
Bid
0.115
Asked
0.105
Chg.
0.020
Asked Yield
0.107
The first number refer to the bill’s maturity date, October 15th , 2015, so
the current date (February 6th , 2015) is 251 days before maturity. The Bid and
Asked are discount yields (in percentages). The Bid 0.115% is the interest rate
the dealer proposes as a buyer of this bill.
– 18 –
1.4. Discount
The formula for discounting (P0 = Pn (1 − d · n)) can be transformed to the
more revealing formula:
P = F(1 − yd ·
k
),
360
where P is either the purchase price from a dealer (asked price) or the selling
price to a dealer (bid price)3 , F is the face value of the T–bill, yd is the bid or
asked rate (the discount yield), and k is the number of days to maturity (the
number of actual calendar days between the settlement date and maturity
date).
According to our example the dealer is offering to pay
PBid = $10, 000(1 − 0.115% ·
251
) = $9, 991.98
360
for a T–bill with a face value of $10,000, and the dealer proposes as a seller of
this bill
PAsked = $10, 000(1 − 0.105% ·
251
) = $9, 992.68.
360
The Chg. of 0.020 in the quotation is the difference between the present
day’s listed bid and the preceding day’s bid in hundredths of a percentage
point. Thus, the change in our example means the discount rate of return on
the previous day’s bid was 0.095. The Asked Yield yi (or B.Y.E. as the bond
yield equivalent) is based on the asked rate and is the annualized rate of return if held to maturity:
P=
F
1+
k
365
⇒
· yi
yi =
F−P
P·
k
365
=
$10, 000 − $9, 992.68
$9, 992.68 ·
251
365
= 0.107%
(Note that asked discount yield will always be lower than the asked yield yi .)
Note 1.13. There are some useful function in Excel, which we can use to solve
T–bill problems:
3
Bid quotes on bills are always higher than the asked
– 19 –
1. Simple Interest and Day–Count Conventions
∙ function PRICEDISC( ) can be used to calculate PBid and PAsked :
=PRICEDISC(s–date;m–date;discount-rate;face-value;2),
where s–date stand for settlement date, and m–date for maturity date
∙ on the other hand, if we know price we can determine face value
using function RECEIVED
=RECEIVED(s–date;m–date;price;discount-rate;2).
∙ the discount rate can be found using DISC():
=DISC(s–date;m–date;price;face-value;2).
∙ finally, there are two ways how to find the asked yield:
=YIELDDISC(s–date;m–date;price;face-value;3).
=INTRATE(s–date;m–date;price;face-value;3).
1.4.2. Application: Bill of Exchange
∙ A written, unconditional order by one party (the drawer) to another
(the drawee) to pay a certain sum, either immediately (a sight bill) or
on a fixed date (a term bill) for payment of goods and/or services received. The drawee accepts the bill by signing it, thus converting it into
a postdated check and a binding contract.
Assume merchant A purchases goods worth $10,000 from another merchant B at a credit of 3 months. Then, B prepares a bill, called the bill of
exchange. A signs this bill and allows B to withdraw the amount from A’s
bank account after exactly 3 months. The date exactly after 3 months is called
nominally due date. Three days (known as grace days) are usually added to
this date to get a date known as legally due date. The amount given on the
bill ($10,000) is called the face value (F).
Suppose B wants to have the money before the legally due date. Then he
can have the money (Price P) from the banker or broker, who deducts simple
interest (using discount rate r) on the face value for the period from the date
– 20 –
1.4. Discount
on which the bill was discounted (i.e., paid by the banker) and the legally due
date. This amount is known as a Banker’s discount (BD).
The present value of Banker’s discount, using simple interest rate r is
called the True Discount (TD). The Banker’s gain (BG) is the difference between banker’s discount and the true discount for the unexpired time. The
present value PW is such amount which, if placed at interest rate r for a specified period will amount to F at the end of the specified period. The interest
on the present value is nothing but TD.
If n is time expressed in years, then we have these formulas for Bill of
Exchange calculations:
BD = F · n · r
P = F − BD = F(1 − n · r)
PW =
TD =
F
1+n·r
BD
= PW · n · r
1+n·r
BG = BD − TD = TD · n · r
Example 1.14. A Bill of Exchange for $4,000 was drawn on January 1st , 2008
for six months and discounted on February 5th , 2008. Calculate the Banker’s
Discount at 12% per annum. Determine Banker’s gain as well. (Allow three
days of grace, and use ACT/360.)
Solution. If January 1st , 2008 is the date on which the bill was drawn, then
the nominal due date would be July 1st , 2008. For further calculations we need
to know legally due date, so adding 3 grace days we obtain July 4th , 2008. We
use the actual number of days to determine number of days between the date
on which the bill was discounted and legally due date. Using Excel we obtain
150 days. So the Banker’s discount is
BD = F · n · r = $4, 000 ·
150
· 0.12 = $200
360
so the price is $3,800. The present value of F is
PW =
$4, 000
1+
150
360
· 0.12
= $3, 809.52
and so the True discount is
TD = $3, 809.52 ·
150
· 0.12 = $190.48
360
– 21 –
1. Simple Interest and Day–Count Conventions
or
$200
= $190.48
1 + 150
360 · 0.12
and so the Banker’s gain is $200-$190.48 = $9.52.
TD =
1.5. Summary
1.5.1. Formulas to remember
Pn = P0 · (1 + n · i)
In = Pn − Pn−1
ROR =
Pn − P0
P0
BD = F · n · r
P0 = Pn (1 − d · n)
Pn − Pn−1
Pn − Pn−1
IT = Pn − P0
in =
dn =
Pn−1
Pn
Pn − P0
d
i
RORP =
id =
di =
n · P0
1−d·n
1+i·n
k
P = F(1 − yd ·
),
360
F
BD
PW =
TD =
BG = BD − TD
1+n·r
1+n·r
1.5.2. Questions
∙ What does the amount of interest depend on?
∙ What is the difference between investment period and interest period?
∙ Why is there ambiguity in counting the number of days between two
dates?
∙ How do you count the number of days between two dates? (day-count
conventions)
∙ How can we evaluate (measure) short–term investments?
∙ Do you know any applications of Simple Interest Theorem?
∙ Do you know any applications of Discount?
∙ Describe a T–bill quotation Table
∙ How are the discount and investment yields for a T–bill computed?
∙ How are the Banker’s discount, true discount and Banker’s gain calculated?
∙ If you want to invest, which day–count convention would you prefer?
ACT/360 or 30E/ACT?
– 22 –
1.5. Summary
1.5.3. Tasks/Problems
Example 1.15. Paul Simon invests $4,000 in a CD at 3.8% a year for 5 years.
He withdraws the interest at the end of each year. What amount does he have
at the end of five years assuming that he does not spend or invest the interest?
[$4,760]
Example 1.16. Determine number of days using 30E, 30A, and ACT for investment periods given in the table below. Before you start, where can you
expect differences between 30E and 30A method?
date1
25.5.2015
25.5.2015
30.5.2015
30.5.2015
31.5.2015
31.5.2015
date2
16.8.2015
31.8.2015
16.8.2015
31.8.2015
16.8.2015
31.8.2015
30E
30A
ACT
Example 1.17. Carl invests $10,000 on May 25th , 2012 at a nominal rate of
i(1) = 12%, and on August 31st , 2012 he withdraws the whole amount of
money, including interest earned. How much interest does he earn for this
period?
[$311.48–$326.67]
Example 1.18. You have two possibilities how to invest $30,000 between
February 4th , 2015 and March, 5th , 2015.
(A) at interest rate 3% using 30E/360
(B) at interest rate 3.2% using ACT/365
Which alternative would you prefer?
[(A)]
Example 1.19. How many days are there between April 25th , 2011 and
February 11st , 2013? Calculate n for given day–count conventions: ACT/ACT,
30E/ACT, ACT/360, and 30E/360.
[1.8;1.767;1.828;1.794]
Example 1.20. Paul Simon invests $500,000 on February 5th , 2012 at a nominal annual rate 2.5%. When can he withdraw the whole amount of money,
including interest earned, to be sure he has $515,000? Use the simple interest
approach. Use ACT/ACT, and 30E/360 day–count conventions. [April 19th ,
2013; April 17th , 2013]
Example 1.21. Max Envit invests $1,000 at a nominal rate
(a) of i(1) , and he withdraws the interest at the end of each year
– 23 –
1. Simple Interest and Day–Count Conventions
(b) of i(2) , and he withdraws the interest at the end of each semester
At the end of the fourth year he has earned $300 in total interest. What nominal interest rate does he earn?
[7.5%]
Example 1.22. Helen Kendrick has a savings account that pays interest at a
nominal rate of 5%. Interest is calculated 365 times per year on the minimum
daily balance and credited to the account at the end of the month. Helen
has an opening balance of $1,000 at the beginning of April. On April 11 she
deposits $200, and on April 21 she withdraws $300. How much interest does
she earn in April ? (The interest is calculated for the whole month, including
the 1st day.)
[$1.51+$1.48+$1.23=$4.22]
Example 1.23. Suppose you could by a 91–day T–bill at an asked price of $98
per $100 face value and you could sell to the dealer at a bid price of $97.95
per $100 face value. What are the bid and asked discount yields? What is the
asked yield?
[7.91%; 8.11%; 8.19%]
Example 1.24. Consider a 13–week $10,000 face value T–bill with an asked
discount yield of 8.88%. What is the cost of this T–bill? What is the asked
yield of this T–bill?
[$9,775.53; 9.21% ]
Example 1.25. A drawer has a bill for $10,000. He discounted this bill with
his bank two months before its due date at 15% p.a. rate of discount. What
amount of money will the drawer receive from a bank? (When the date of
the bill is not given, grace days are not to be added.) What is Banker’s gain?
[$9,750; $6.1]
Example 1.26. An exporter draws a Bill of Exchange on his overseas customer on July 1st for $10,000 for three months. He discounts the Bill with his
bankers on July 23rd at 8% per annum
(a) how much does the exporter receive for the discounted Bill?
(b) If the sum received from the bank had been $40 more, what would have
been the rate of discount (Allow three days of grace, and use ACT/360.)
[$9,837.78; $6.027]
Example 1.27. Find the difference between true discount and banker’s discount on a 90–day bill for $800, drawn on March 8th and discounted on March
28th at 7%. Use ACT/365.
[$0.155]
Example 1.28. An exporter draws a Bill of Exchange on his customer on
January 1st , 2014 for $16,500 for five months. He sells the bill to a trading
company on March 8th , 2014 at discount rate 9.5% per annum. Later, on April
5th , 2014, the trading company discounts that bill with bankers at 9.3% per
annum. What interest rate did the trading company earned? (Allow three
– 24 –
1.5. Summary
days of grace, use ACT/360 for discounting, and ACT/365 for the simple
interest.)
[10.306%]
– 25 –
1. Simple Interest and Day–Count Conventions
– 26 –
2
Chapter
Compound interest
2.1. Compound Interest Theorem
Simple interest is generally charged for borrowing money for short time periods. For long periods we usually apply compound interest. Compound
interest is similar, but the total amount due at the end of each period is calculated and further interest is charged against both the original principal but
also the interest that was earned during that period. So the compound interest generates interest on interest, whereas simple interest does not.
Again, starting with the annual interest rate, let us invest $100 at 5% interest per year for 4 years.
Table 2.1. Compound Interest (one interest period in a year)
Year
1
2
3
4
Year’s Beginning
Principal
$100.00
$105.00
$110.25
$115.76
Year’s End
Interest
$5.00
$5.25
$5.51
$5.79
Year’s End
Amount
$105.00
$110.25
$115.76
$121.55
For the 5% interest compounded semi–annually, i.e. i(2) = 5% and i = 2.5%
different accumulated values come out (see Table 2.2).
Comparing accumulated values at the end of all four years, we can see the
number of interest periods in a year is important. It means compound interest
depends upon nominal interest rate, time, principal invested, and how many
times a year the interest is calculated and credited.
We now state the general formula for compounding the accumulated value
(Pn ) of P0 at the end of the nth interest period. So we let
– 27 –
2. Compound interest
Table 2.2. Compound Interest (two interest periods in a year)
Semester
1
2
3
4
5
6
7
8
Semester’s Beginning
Principal
$100.00
$102.50
$105.06
$107.69
$110.38
$113.14
$115.97
$118.87
Semester’s End
Interest
$2.50
$2.56
$2.63
$2.69
$2.76
$2.83
$2.90
$2.97
Semester’s End
Amount
$102.50
$105.06
$107.69
$110.38
$113.14
$115.97
$118.87
$121.84
∙ P0 . . . the initial principal (present value) invested,
∙ n . . . the total number of interest periods,
∙ Pn . . . the future value of P0 (accumulated principal) at the end of the nth
interest period,
∙ m . . . the number of interest periods per year,
∙ i(m) . . . nominal rate (annual interest rate),
∙ i . . . the interest rate per interest period (i =
i(m)
m ).
Theorem 2.1 (The Compound Interest Theorem). If we start with principal
P0 , and invest it for n interest periods at nominal rate of i(m) (expressed as a decimal)
compounded m times a year using simple interest, then Pn , the future value of P0 at
the end of nth interest period, is
Pn = P0 · (1 + i)n
where i =
i(m)
m .
Note 2.2.
∙ Unlike simple interest, accumulated value is non–linear in i
and n for compound interest. See Fig. 2.1 and 2.1.
∙ In the last chapter (see Note 1.3) we have defined some measures to
evaluate the interest earned. Now using the compound interest we obtain:
In = P0 ·i·(1+i)n−1
IT = P0 ·((1+i)n −1)
– 28 –
in = i
ROR = (1+i)n −1
2.1. Compound Interest Theorem
∙ As for simple approach, we can solve various problems connected with
compounding involving unknown Pn , P0 , n, or i. Derived formulas are
as follows:
√︂
ln Pn − ln P0
Pn
Pn
,
n=
,
i= n
− 1.
P0 =
n
(1 + i)
ln(1 + i)
P0
Accumulated Value of $100
2000
10%
5%
1600
1200
800
400
0
0
5
10
15
Years
20
25
30
Figure 2.1. Accumulated Value of $100 at 5% and 10% compound interest
Example 2.3. Find the future value of $1,000 invested at a nominal rate of
8% at the end of the 5th year, if interest is compounded
(a) annually,
(b) semi–annually,
(c) monthly,
(d) daily (365 × a year),
(e) continuously, that is, the number of interest periods per year grows
without bound.
– 29 –
2. Compound interest
Accumulated Value of $100
1 600,00 Kč
10 years
1 200,00 Kč
5 years
800,00 Kč
400,00 Kč
0,00 Kč
0%
5%
10%
15%
Interest rate
20%
25%
30%
Figure 2.2. Accumulated Value of $100 after 5 and 10 years using compound
interest
Solution.
(a) P5 = $1, 000 · 1.085 = $1, 469.33,
(b) P10 = $1, 000 · (1 +
0.08 10
2 )
= $1, 480.24,
(c) P60 = $1, 000 · (1 +
0.08 60
12 )
= $1, 489.85,
(d) P1825 = $1, 000 · (1 +
0.08 1825
365 )
(e) P5·∞ = limm→∞ $1, 000 · (1 +
= $1, 491.76,
0.08 5·m
m )
= 1, 000 · e5·0.08 = $1, 491.82.
Note 2.4. For calculations of future value the Excel function FV can be used.
Following the Example above, we have
(a) =FV(0,08;5;0;-1000;0)
(b) =FV(0,08/12;12*5;0;-1000;0) . . .
Function FV is meant primary for saving–calculations. The syntax is
– 30 –
2.1. Compound Interest Theorem
=FV(rate,nper,pmt,[pv],[type]),
where rate is the interest rate per interest period, nper is the total number
of interest periods, pmt is regular payment (which is omitted in this case),
[pv] is present value (but because the idea is based on terms of cash–flows
it should be negative to reach positive result), and the last parameter [type]
specify whether the interest is calculated at the end (FALSE) or at the beginning of the interest period (TRUE). Similarly we can use function PV to
determine P0 , RATE for i, or NPER for n.
=PV(rate,nper,pmt,[pv],[type]),
=RATE(nper;pmt;pv;[fv];[type];[guess]), where guess may be omitted if you expect just one solution for the problem
=NPER(rate;pmt;pv;[fv];[type]).
Day–count conventions can be also used for compound interest:
Example 2.5. Carl invests $1,000 on January 4th , 2015 at a nominal rate of
i(1) = 8%, and on April 21st , 2015 he withdraws the whole amount of money,
including interest earned. How much interest does he earn for this period using compound interest approach? What if the interest is compounded semi–
annually? Use ACT/365.
Solution. For ACT/365 we have:
107
I = P0 · (1 + i)n − P0 = $1, 000 · (1 + 0.08) 365 − $1, 000 = $22.82
If the interest is compounded semi–annually we just need to adjust our formulas:
I = P0 · (1 + i)n − P0 = $1, 000 · (1 +
0.08 2· 107
) 365 − $1, 000 = $23.26
2
(For all these results function FV can be used.)
Consider the constant number of investment periods in a year (m). We already know, that it has no effect on the interest earned using simple interest
approach. Now the question is: do we reach the higher interest using simple interest approach or the compound interest approach? From examples,
where we invest $100 for four years it seems the compound interest produces
the higher interest earned. But have a look at the following example.
– 31 –
2. Compound interest
Example 2.6. You can invest $1,000 at a nominal interest rate of 4% for 6
months. Would you prefer simple or compound interest to be used, if
(a) m=1,
(b) m=2,
(c) m=4.
Solution. For the simple interest we have
(a) P 1 = $1, 000 · (1 +
2
1
2
· 0.04) = $1, 020,
(b) P1 = $1, 000 · (1 + 1 · 0.02) = $1, 020,
(c) P2 = $1, 000 · (1 + 2 · 0.01) = $1, 020.
For the compound interest we have
1
(a) P 1 = $1, 000 · (1 + 0.04) 2 = $1, 019.80 ⇒ simple approach generates the
2
higher interest earned,
(b) P1 = $1, 000 · (1 +
0.04 1
2 )
= $1, 020.0 ⇒ the same result,
2
(c) P2 = $1, 000 · (1 + 0.04
4 ) = $1, 020.10 ⇒ compound approach generates
the higher interest earned.
Well, if the investment period is shorter than the interest period, we would
prefer simple interest approach for our investment to be used, and for investment period longer than interest period, compound interest approach is
preferred because it generates the higher interest earned. You can see the
difference between compound and simple interest for m=1 in Fig. 2.1
2.2. Effective Rate
In order to compare investments with different nominal interest rates and
frequencies of compounding, it is common to calculate equivalent annual interest rate for each investment so we do not need to calculate future values
to compare given alternatives. We are looking for an annual interest rate ie f f
such that
i(m) m
(1 + i)m = (1 +
) = (1 + ie f f )
m
– 32 –
2.2. Effective Rate
Accumulated Value of $100
400
compound
300
simple
200
100
0
0
0,5
1
Years
1,5
2
Figure 2.3. Accumulated Value of $100 at 100% interest rate
The effective rate (EFF), ie f f , is the annual rate of return i(1) that is equivalent to the nominal rate i(m) (compounded m times a year), or the nominal
rate i(∞) (compounded continuously).
ie f f = (1 + i)m − 1
or
ie f f = e i
(∞)
− 1.
Example 2.7. What are the EFFs for (a)-(e) from Example 2.3?
Solution.
(a) ie f f = (1 + 0.08)1 − 1 = 8%,
(b) ie f f = (1 +
0.08 2
2 )
− 1 = 8.16%,
...
(e) ie f f = limm→∞ (1 + 0.08)m − 1 = e0.08 = 8.329%.
Note 2.8. An Excel function for calculations of the effective interest rate is
=EFFECT(nominal_rate, npery).
– 33 –
2. Compound interest
It has just two parameters: nominal interest rate, and number of interest
periods in a year (m). So we can reach 8.300 % from the Example above using
=EFFECT(0,08;12).
On the other hand, sometimes we want to find a nominal interest rate corresponding to the given effective rate, for this purpose function NOMINAL
can be used.
=NOMINAL(effect_rate,npery).
It means these functions are inverse and so we have
NOMINAL(EFFECT(0, 08; 12), 12) = 0, 08,
or
EFFECT(NOMINAL(0, 678; 14); 14) = 0, 678.
2.3. Summary
2.3.1. Formulas to Remember
Pn = P0 · (1 + i)n
ie f f = (1 + i)m − 1
ie f f = e i
(∞)
−1
2.3.2. Questions
∙ If you want to invest . . .
– would you prefer simple or compound interest to be used?
– and simple interest approach is used, would you prefer interest
period to be a month or a year?
– and compound interest approach is used, would you prefer interest period to be a month or a year?
∙ What does compounded continuously mean?
∙ What is the EFF?
– 34 –
2.3. Summary
2.3.3. Tasks/Problems
Example 2.9. Find the future value of $600 invested at a nominal rate of 10%
at the end of the 3rd year, if interest is compounded
(a) semi-annually,
(b) quarterly,
(c) monthly,
(d) weekly,
(e) daily,
(f) continuously, that is, the number of interest periods per year grows
without bound.
[$804.06; $806.93; $808.91; $809.68; $809.88; $809.92]
Example 2.10. What are the EFFs for (a)-(f) from the previous example.
[10.250%; 10.381%; 10.471%; 10.506%; 10.516%; 10.517%]
Example 2.11. You can invest $10,000 at an annual interest rate of 10% for
(a) 4 years
(b) 4 months
Do you prefer simple or compound interest to be used? [(a) compound; (b)
simple]
Example 2.12. Carl invests $10,000 on April 16th , 2014 at a nominal rate of
12%, compounded
(a) annually
(b) semi-annually
(c) quarterly
(d) continuously
and on August 7th , 2015 he withdraws the whole amount of money, including interest earned. How much interest does he earn for this period? Use
ACT/365 convention.
[ $11,599.93; $11,648.78; $11,674.71; $11,701.72]
Example 2.13. Do you prefer to invest at 26% compounded quarterly, at 27%
compounded at the end of every even month, or at 28% compounded semiannually?
[27%]
– 35 –
2. Compound interest
Example 2.14. Carl invests $1,000 at a nominal rate of i(1) , and when the
interest is credited at the end of each year, he leaves it in his account. At the
end of the fourth year he has earned $300 in total interest. Determine i(1) .
[6.779%]
Example 2.15. Carl invests $1,000 at a nominal rate of i(2) , and when the
interest is credited at the end of each six months, he leaves it in his account.
At the end of the fourth year he has earned $300 in total interest. Determine
i(2) . Would you expect it to be higher or lower than the answer to previous
example?
[6.668%]
Example 2.16. At what nominal interest rate, compounded daily, will your
money double in 7 years?
[9.903%]
Example 2.17. How long will it take for an investment of $1,000 to increase
to $1,500 at a nominal interest rate of 7% compounded semi-annually?
[n=11.786 → 6 years]
Example 2.18. Peter invests $10,000 on February 1st , 2013 at a nominal interest rate 3%. When can he withdraw the whole amount of money, including
interest earned, to be sure he has $12,000? The interest is calculated semi–
annually using compound interest. For counting the days use Banker’s rule
ACT/360.
[February 15th ,2019]
Example 2.19. Mark invests $70,000 on October 13th , 2008 at a nominal annual rate 6%. What amount of money can he withdraw on February 14th ,
2011? When can he withdraw the whole amount of money, including interest earned, to be sure he has $80,000? The interest is calculated annually using
compound interest. For counting the days use 30E/ACT.
[$79 940.83;
th
February 19 ,2011]
Example 2.20. If the EFF of an investment is 8%, then what is the nominal
interest rate, if interest is compounded monthly?
[7.721%]
– 36 –
3
Chapter
Cash Flows
3.1. Cash Flows
In the following we will represent a cash flow (the flow of cash in and out of
an investment) using tables consisting of two rows
∙ the 1st row represents time increasing from the present (denoted by 0)
∙ the 2nd row represents cash coming in (a positive cash flow), and cash
going out (a negative cash flow)
Time
Cash Flow
0
C0
1
C1
2
C2
...
...
n-1
Cn−1
n
Cn
C0 , . . . , Cn may be positive, negative, or zero.
For example, suppose that we invest $1,000 at 5% compounded annually
for three years. At the end of the third year this $1,000 grows to the future
value of $1, 000(1 + 5%)3 = $1,157.625. The cash flows are represented as follows: At year zero (at the beginning of the first year) we invested $1,000 (so
Time
Cash Flow
0
-$1,000
1
0
2
0
3
$1,157.625
the cash went out, and hence the negative sign), and at the end of the third
year we received $1,157.625 (so the cash came in, and hence the positive sign,
which we normally omit).
In general, cash flows for compounding and for the opposite problem of
compounding, called discounting (i.e. we know a future value of the initial
principal which we want to determine) are represented in Table 3.1.
– 37 –
3. Cash Flows
Table 3.1. CF for Compounding and Discounting
Time
Cash Flow
Time
Cash Flow
0
−P
1
0
2
0
...
...
0
P
− (1+i)
n
1
0
2
0
n-1
0
...
...
n
P · (1 + i)n
n-1
0
n
P
3.2. Net Present Value
The net present value (NPV), of an investment is the difference between the
present value of the cash inflows and the present value of the cash outflows,
that is,
NPV =
n
∑︁
k=0
C1
C2
Cn
Ck
= C0 +
+
+ ... +
,
2
k
(1
+
i)
(1
+ i)n
(1 + i)
(1 + i)
where i is prevailing interest rate1 . This measure, NPV, represents one of the
ways how to make decisions among investments with the same risk.
Example 3.1. Jane is considering two investments with annual cash flows:
Years
Cash Flow (Investment 1)
Cash Flow (Investment 2)
0
-$8,000
-$8,000
1
$2,500
$3,500
2
$3,000
$2,050
3
$3,500
$3,400
∙ Which is the better investment if the prevailing annual interest rate is
4%?
∙ Which is the better investment if the prevailing annual interest rate is
5%?
Solution. For the interest 4% we have
NPV1 = −$8, 000 +
NPV2 = −$8, 000 +
$2, 500 $3, 000 $3, 500
+
+
= $289,
1.04
1.042
1.043
$3, 500 $2, 050 $3, 400
+
+
= $283.31,
1.04
1.042
1.043
1
the most common at particular time; the general interest rate level for a given type of loan
across numerous lenders
– 38 –
3.2. Net Present Value
thus the first investment is better for her. And for 5% we have
NPV1 = $125.47,
NPV2 = $129.79,
and thus the second investment is better for her.
Note 3.2.
∙ We can see that cash flow, on its own, is not sufficient for
decision making between two investments. The interest rate which we
should take under consideration is necessary.
∙ We can use Excel function NPV for this purpose. Unfortunately the
output of this function is not exactly NPV as we understand it, but only
the present value of cash flows for future incomes (outcomes). So if we
want to reach NPV as defined, we need to add (or deduct) the initial
principal. For the CF from Example 3.1 we have
= −8000 + NPV(0, 04; 2500; 3000; 3500).
=NPV(rate,value1,[value2],...)
The idea of NPV can be used generally for semesters, months, days etc.
Example 3.3. Jane is considering two investments with monthly cash flows:
Months
Cash Flow (Investment 1)
Cash Flow (Investment 2)
0
-$8,000
-$8,000
1
$2,500
$3,500
2
$3,000
$2,050
3
$3,500
$3,400
∙ Which is the better investment if the prevailing annual interest rate is
4%?
Solution.
NPV1 = −$8, 000 +
$2, 500
NPV2 = −$8, 000 +
$2, 500
1.04
1
12
+
$3, 000
+
$3, 000
1.04
2
12
+
$3, 500
+
$3, 500
3
1.04 12
= $938.15
= $892.05
3
1.05
1.05
1.05 12
If you want to use NPV function for this problem you need to find appropriate
interest rate which you can use instead of “rate”. You are looking for a rate
which satisfies
1
1.04
1
12
=
1
1+i
1
12
⇒
– 39 –
2
12
(1 + 4%)1 = (1 + i)12 ,
3. Cash Flows
but we are able to find i(12) using NOMINAL, which can be applied this way:
(1 + 4%)1 = (1 +
i(12) 12
) ,
12
and using
i(12)
12
we obtain i. So the final application of Excel would be
i=
= −8000 + NPV(NOMINAL(0, 04; 12)/12; 2500; 3000; 3500).
Sometimes we know exact dates for our cash flow:
Example 3.4. Jane is considering an investments with these cash flows:
Dates
CF
Jan 2nd , 2014
-$8,000
March 4th , 2014
$2,500
July 5th , 2014
$3,000
Sep. 6th , 2014
$3,500
Calculate NPV if the prevailing annual interest rate is 4%.
Solution. Using actual number of days we obtain:
NPV = −$8, 000 +
$2, 500
1.04
61
365
+
$3, 000
1.04
184
365
+
$3, 500
247
1.04 365
= $938.15
Note 3.5. Even for problems concerning dates function NPV can be used,
but it is better to apply XNPV. This function is based on ACT/ACT convention and it is really the net present value as we know it, because it has three
parameters: annual rate, values from our CF – involving the first C0 as well –
and all these dates.
=XNPV(rate, values, dates)
3.3. Internal Rate of Return
We introduce concept of IRR using two basic problems:
Example 3.6. If we invest $3,000 at 4% per annum, and then a year later
invest $5,000 at 2% per annum. What rate have we really been earning over
the two years?
– 40 –
3.3. Internal Rate of Return
Solution. The future value of the entire investment after a total of two years
is:
$3, 000 · (1 + 4%)2 + $5, 000 · (1 + 2%) = $8, 344.8.
We need to remember that both principals were deposited at different times.
The corresponding CF is as follows:
Years
Cash Flow
0
-$3,000
1
-$5,000
1
$8,344.8
and taking time value of money into account we are looking for an interest
rate i such that:
$3, 000 · (1 + i)2 + $5, 000 · (1 + i)1 = $8, 344.8,
if we focus on future value of our investment, or
$3, 000 +
$5, 000 $8, 344.8
=
,
(1 + i)1
(1 + i)2
if we focus on present value of our investment. To find the i is to find the
solution of this quadratic equation.
i1,2
√︀
−5, 000 ± 5, 0002 − 4 * 3, 000 * 8, 344.8
− 1 = 3.108% or − 369.775%
=
2 * 3, 000
however we expect i to be positive, so i = 3.108%. This annual interest rate is
known as the IRR.
We can meet another kind of problem:
Example 3.7. We invest $20,000 at 2% compounded 365 days a year. At the
end of the year we move the proceeds to another account with nominal interest rate of 3% compounded 4 times a year. How much interest do we have at
the end of the second year? What rate have we really been earning over these
two years?
Solution. At the end of the first year we have
$20, 000 · (1 +
2% 365
) = $20, 404.02.
365
At the end of the second year we can expect
$20, 000 · (1 +
2% 365
3% 4
) · (1 +
) = $21, 023.06,
365
4
– 41 –
3. Cash Flows
so we earned $1,023.06 in interest.
To find appropriate interest rate we have earned, we want to find the rate
i such that
$20, 000 · (1 + i)2 = $21, 023.06.
so
√︂
i = ± (1 +
3% 4
2% 365
) · (1 +
) − 1 = 2.526% or − 202.526%.
365
4
Thus the asked rate is 2.526%.
The internal rate of return (IRR), iirr , for an investment is the interest rate
that is equal to the annually compounded rate earned on a savings account
with the same cash flows. (It is the rate that makes the present value of the
expected future cash flows equal to the initial cost of the investment.) So for
iirr we have:
C0 +
C1
C2
Cn
+
+ ... +
= 0.
2
(1 + iirr ) (1 + iirr )
(1 + iirr )n
Example 3.8. Find the IRR for these CFs:
Years
Cash Flow (Investment 1)
Cash Flow (Investment 2)
0
-$1,000
-$1,000
1
$2,000
$2,150
2
-$1,500
-$1,155
Solution. For the first investment we obtain quadratic equation
−$1, 000 · (1 + iirr )2 + $2, 000(1 + iirr )1 − $1, 500 = 0,
i.e.
(1 + iirr )2 − 2 · (1 + iirr )1 + 1.5 = 0
which leads to the complex solution
iirr;1,2
√
√
2 ± 22 − 4 * 1 * 1.5
2±i· 2
=
−1=
− 1,
2
2
which is not applicable to our CF. So there is not such an interest rate which
fulfills attributes of IRR.
For the second one we have
−$1, 000 · (1 + iirr )2 + $2, 150(1 + iirr )1 − $1, 155 = 0,
i.e.
(1 + iirr )2 − 2.15 · (1 + iirr )1 + 1.155 = 0,
– 42 –
3.3. Internal Rate of Return
for which we obtain two solutions:
√
2.15 ± 2.152 − 4 * 1 * 1.155
iirr;1,2 =
− 1 = 5% or 10%.
2
So there comes the question, how to recognize how many positive roots
can we expect? If we found one, should we go on looking for another? Next
theorem gives us partial answer to our question.
Theorem 3.9 (The IRR Uniqueness Theorem I.). If there exists an integer p for
which C0 , C1 ,. . . , Cp are of the same sign or zero (but not all zero) and for which
Cp+1 , Cp+2 , . . . , Cn are all of the opposite sign or zero (but not all zero), then
C0 +
C1
C2
Cn
=0
+
+ ... +
(1 + i) (1 + i)2
(1 + i)n
has at most one positive solution 1+i.
Note 3.10. This theorem guarantees that there is at most one positive solution, 1 + i. However, that does not guarantee that i is positive.
There is another theorem which can be used if assumptions of the last one
are not fulfilled.
Theorem 3.11 (The IRR Uniqueness Theorem II.). If there exists an i for which:
(a) 1 + i > 0,
∑︀p
(b) k=0 Ck · (1 + i)p−k > 0 for all integers p satisfying 0 ≤ p ≤ n − 1 (that is, the
future value of all the cash flows up to period p are positive), and
∑︀
(c) nk=0 Ck · (1 + i)n−k = 0,
then i is unique.
3.3.1. Application: Index Fund
An Index Fund (or Tracker Fund) is a fund whose collection of assets is designed to mirror the performance of a particular broad–based stock market
index. Usually index funds are managed so that decisions are automated
and transactions are infrequent. There is usually a minimum opening balance, unless the investor invests a regular amount each month. Investors are
usually discouraged from frequent buying and selling, and they may be penalized for this.
– 43 –
3. Cash Flows
Example 3.12. At the beginning of every month for 12 months, we buy $100
worth of shares in an index fund. At the end of the twelfth month our shares
are worth $1,300. What is the internal rate of return, of this investment?
Solution. Our investment can be represented by this CF:
Months
Cash Flow
0
-$100
1
-$100
...
...
11
-$100
12
+$1,300
Focus on future value of this investment at an annual interest rate of iirr :
12
11
1
$100 · (1 + iirr ) 12 + $100 · (1 + iirr ) 12 + . . . + $100 · (1 + iirr ) 12 = $1, 300
1
Using substitution 1 + i = (1 + iirr ) 12 we obtain
$100 · (1 + i)12 + $100 · (1 + i)11 + . . . + $100 · (1 + i)1 = $1, 300
which leads to equation
(1 + i)12 + (1 + i)11 + . . . + (1 + i)1 = 13
This is a polynomial equation of degree 12, which have 12 roots (some of
which may be complex). In general we are unable to solve equation of this
type analytically, but it can be solved numerically (bisection method or Newton’s method), or graphically (see Fig. 3.3.1). In Table 3.3.1 net present value
of our investment is tabulated for various values of i ∈ {0%, 1%, . . . 10%}. As
you can see there is one change in sign, so we can expect one solution to
be between 1% and 2%. Adding another digit we restrict our solution to
< 1.2%; 1.3% >, and similarly to < 1.22%; 1.23% >, etc. (1.225279%), thus i
is about 1.225% and due the IRR uniqueness theorem, we know that there is
just one reasonable solution:
iirr = (1 + 1.225%)12 − 1 = 15.736%.
Note 3.13. Fortunately (similarly as for NPV) we can use Excel function for
this concept. The output of IRR() is nothing but internal rate of return, and
our problem concerns exact days we can you XIRR(). Because we already
know there may be number of outputs, the last parameter of these functions
(which can be omitted), the guess, can be used as first approximation of expected result. Contrary to NPV, the input includes the initial cash flow CF0
as well.
IRR(values, [guess])
XIRR(values, dates, [guess])
– 44 –
3.4. Summary
i
0%
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
Table 3.2. Bisection method for IRR problem
NPVi
i
NPVi
i
NPVi
-$100.00 1.0% -$16.92 1.20% -$1.87
-$16.92 1.1%
-$9.33 1.21% -$1.13
$53.64 1.2%
-$1.87 1.22% -$0.39
$113.47 1.3%
$5.48 1.23% $0.35
$164.07 1.4% $12.70 1.24% $1.08
$206.75 1.5% $19.81 1.25% $1.82
$242.63 1.6% $26.80 1.26% $2.55
$272.65 1.7% $33.68 1.27% $3.29
$297.65 1.8% $40.44 1.28% $4.02
$318.32 1.9% $47.10 1.29% $4.75
$335.29 2.0% $53.64 1.30% $5.48
3.4. Summary
3.4.1. Formulas to Remember
NPV =
n
∑︁
k=0
C0 +
Ck
Cn
C1
C2
+ ... +
= C0 +
+
,
2
k
(1 + i) (1 + i)
(1 + i)n
(1 + i)
Cn
C1
C2
= 0.
+
+ ... +
(1 + iirr ) (1 + iirr )2
(1 + iirr )n
3.4.2. Questions
∙ What is the NPV?
∙ Why is the IRR Uniqueness Theorem(I.) important?
∙ What does the IRR Uniqueness Theorem (I.) says?
3.4.3. Tasks/Problems
Example 3.14. Marie buys shares in MSFT for $52. Three months later they
are worth $75. What is Marie’s IRR?
[332.745%]
Example 3.15. What is the IRR for investment paying 4.5% compounded
continuously?
[4.603%]
Example 3.16. To be able to withdraw
– 45 –
3. Cash Flows
$300
$200
$100
$0
0,0%
1,0%
2,0%
3,0%
4,0%
5,0%
-$100
-$200
Figure 3.1. Graphical solution for IRR problem
(a) $2,700 in a year and $2,800 in two years
(b) $2,700 in two years and $2,800 in three years
(c) $2,700 in a month and $2,800 in two months
what amount should I invest assuming 4% nominal interest compounded annually?
Example 3.17. If I borrow $5,000 now I should be able to pay
(a) $2,700 in a year and $2,800 in two years. What is the IRR?
(b) $2,700 in two years and $2,800 in three years. What is the IRR?
(c) $2,700 in a month and $2,800 in two months. What is the IRR?
[6.555%; 3.879%;114.233%]
Example 3.18. An initial amount of $10,000 is invested for 2 years at successive annual interest rates of 5% and 4% compounded annually. Do you
think the future value of this investment is different from the future value of
– 46 –
3.4. Summary
$10,000 invested for 2 years at successive annual interest rates of 4% and 5%
compounded annually? Justify your answer.
Example 3.19. Cindy can buy a parcel of land for $75,000 now, or for $40,000
in one year, and $40,000 in two years. For which annual interest rate are these
investments comparable? (i.e. there is no difference in PVs of these investments).
[4.413%]
Example 3.20. You invested $10,000 on February 8th ,2011. On April 6th you
received the first payment of $5,500. On May 8th you had to invest additional
amount of $4,000 to obtain the final amount of $9,000 on October, 6th . Do you
consider such an investment to be profitable for you if market interest rate for
investments with the same risk was 6%?
[9.28%, $168.31, OK]
Example 3.21. Find the IRR for these CFs:
Years
Cash Flow (Investment 1)
Cash Flow (Investment 2)
Cash Flow (Investment 3)
0
-$10,000
$10,000
$10,000
1
$400
-$10,800
-$10,800
2
-$100
-$9,950
-$9,990
3
$15,300
$10,850
$10,800
[16.87%, does not exist, 7.29% or 0.69 %]
Example 3.22. Find the IRR for these CF:
Dates
CF
Janurary 6th , 2011
-$10,000
June 12th , 2011
$5,050
June 18th , 2011
$5,050
[29.6%]
Example 3.23. Peter’s business can buy a piece of equipment for $200,000
now, or for $70,000 now, $70,000 in one year, and $80,000 in two years.
(a) Which option is better if money can be invested at a nominal rate of
4.8% compounded annually?
(b) Which option is better if money can be invested at a nominal rate of
4.8% compounded semi–annually?
(c) For which nominal interest rate compounded annually are these investments comparable?
(d) For which nominal interest rate compounded quarterly are these investments comparable?
[now, payments, 4.81%, 4.72 %]
Example 3.24. Peter’s business can buy a piece of equipment for $80,000
now, or for $5,000 now, $40,000 in one year, and $40,000 in two years.
– 47 –
3. Cash Flows
(a) For which nominal interest rate compounded annually are these investments comparable?
(b) For which nominal interest rate compounded semi–annually are these
investments comparable?
(c) At what price that equipment has to be sold now to be worth to buy it
today if prevailing nominal interest rate compounded annually is 6%
(d) At what price that equipment has to be sold now to be worth to buy it
today if prevailing nominal interest rate compounded semi–annually is
6%
[4.413%, 4.365 %, less than $78,335.71, less than $78,243.32]
Example 3.25. A parcel of land costs $500,000. For an additional $800,000
you can build a motel on the property. The land and motel should be worth
$1,500,000 next year. Suppose that common stocks with the same risk as this
investment offer a 10 percent expected return. Would you construct the motel? Why or why not?
Example 3.26. Would you accept these projects if prevailing interest rate is
5%, or 15% respectively?
Years
Cash Flow 1
Cash Flow 2
CF2: no, yes]
0
-$10,000
$10,000
1
$400
-$10,800
2
$100
-$9,990
– 48 –
3
$15,300
$10,700
[CF1: yes, yes,