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1
Introduction
In Chapter 2, 3, and 4, we usually takes the rate of return, yield rate, or yield to maturity
as given and compute some other quantities such as the present value of the cash–flow, the
size(s) of payments, and the number of payments. In the reality, it is common that we know
everything but the interest rate. Therefore, in Chapter 5, we are interested in solving the
unknown interest problem. The equation that we need to solve is based on the equation of
value, which states that the present value at time 0 of cash–inflow equals the present value
at time 0 of cash–outflow. In other words, we are interested in determining the rate that
will equate the present values of cash–inflow and cash–outflow. It is a simple idea, which
requires great knowledge in advance computation methods and numerical analysis theories
the equation that we need to solve is almost always non–linear.
Keep in mind that most of financial practitioners 50 years ago only have a desktop calculator,
which is not capable to perform many sophisticated calculations. The best they can to is
either to make some additional assumptions to make computation easier or to rely on some
analytical approximations.
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Internal Rate of Return
Supposed that we know in advance the size and the time of all cash–inflows and cash–outflows
that are generated by an investment.
P~in = R0 R1 R2 R3 · · · Rn and P~out = C0 C1 C2 C3 · · · Cn
Then by the equation of value (PV0 {P~in } = PV0 {P~out }) with the assumption of compound
interest accumulation function
R0 + R1 ν + R2 ν 2 + R3 ν 3 + · · · + Rn ν n = C0 + C1 ν + C2 ν 2 + C3 ν 3 + · · · + Cn ν n
Note that the reference time point can be picked arbitrarily since the above equation is based
on compound interest accumulation function.
By rearranging the terms, we have
(R0 − C0 ) + (R1 − C1 )ν + (R2 − C2 )ν 2 + (R3 − C3 )ν 3 + · · · + (Rn − Cn )ν n = 0, where ν =
1
1+j
The internal rate of return (IRR) is defined to be the rate j such that the above equation
holds. There are few things about internal rate of return you should be aware of. First of
all, the above equation is not linear in j; it is a polynomial of degree at most n. It is proved
in the algebra theory that no closed analytical formula exists in general. To obtains the root
of the above equation requires some iterative root–find algorithms, i.e. Bisection Method,
Secant Method, or Newton–Raphson Method. Secondly, the internal rate of return is not
unique. Please refer Q #5.1.6 e) for the non–uniqueness of IRR. Thirdly, since the internal
rate of return is not unique in general, there could be two or more rates satisfying the above
equation. It raises some difficulties on how to choose the “right” rate.
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Dollar–Weighted Rate of Return
OK. So we can not solve the following equation
(R0 − C0 ) + (R1 − C1 )ν + (R2 − C2 )ν 2 + (R3 − C3 )ν 3 + · · · + (Rn − Cn )ν n = 0
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with a desktop calculator. Can we makes some addition assumptions or apply some approximations so that the equation is easy enough by hand?
• Assumption: All cash–flows occur within one year
• Approximation: The compound–interest accumulation function is approximated by simple–interest accumulation function
Remark: Since the simple interest rate that solves the equation of value depends on the
reference time, the reference time is always set to be the end of year. Therefore, the equation
of value based on the simple interest rate accumulation function becomes
(R0 −C0 )[1+(1−t0 )j]+(R1 −C1 )[1+(1−t1 )j]+(R2 −C2 )[1+(1−t2 )j]+· · ·+(Rn −Cn )[1+(1−tn )j] = 0
, where t0 , t1 , t2 , . . . , tn are the times of transactions, and tk ∈ [0, 1] for k = 0, 1, 2, 3, . . . , n
The rate j solve above equation is called dollar–weighted rate of return.
Since we are interested in calculating the rate of return (yield rate) within one–year, we set
t0 = 0 and tn = 1. R0 − C0 and Rn − Cn are interpreted as the account balance at the
beginning of year and at the end of year, respectively.
The equation of value can be summarized as following:
accumulated amount of initial balance at the end of year
+ all deposits accumulated to the end of year with simple interest
= all withdrawals accumulated to the end of year with simple interest
+ balance at the end of year
Note that the above equation is the same one we would use to solve for the IRR, but for
IRR, we would use compound interest instead of simple interest.
Example 3.1 A pension fund receives contributions and pays benefits from time to time.
The fund value is reported after every transaction and at year end. The details during the
year 2005 are as follows:
Date
Jan 1, 2005
Ma 1, 2005
Sep 1, 2005
Nov 1, 2006
Jan 1, 2006
Amount
1,000,000
1,240,000
1,600,000
1,080,000
900,000
Contribution Received:
Date
Feb 28, 2005
Aug 31, 2005
Amount
200,000
200,000
Benefits Paid:
Date
Oct 31, 2005
Dec 31, 2005
Amount
500,000
200,000
Fund Values:
2
Find the dollar–weighted rate of return.
The beginning account balance is $1,000,000, and the ending balance is $900,000.
306
The cash–inflow are $200,000 at time t = 1 − 365
and $200,000 at time t = 1 −
61
and $200,000 at time t = 1 − 365
cash–outflow are $500,000 at time t = 1 − 365
365
61
;
365
the
306
122
61
i) + 200000(1 +
i) = 500000(1 +
i) + 200000 + 900000
365
365
365
(500000)(61)
(200000)(306) (200000)(122)
+
]i = 1600000 +
i
1400000 + [1000000 +
365
365
365
⇒ i = 0.1737681
1000000(1 + i) + 200000(1 +
Note that you need the times, sizes, types of transactions as well as the beginning and the
ending account balance to find the dollar–weighted rate of return.
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Time–Weighted Rate of Return
The time–weighted rate of return for one year period is found by compounding the return over
successive parts of the year. We essentially partition the one year period into some fractional
interval with respect to the dates of transactions (deposits and/or withdrawals), and then
calculate the rate of return over each fractional time period. Then the dollar–weighted rate
of return is the compounding the rate in successive intervals.
Example 4.1 A pension fund receives contributions and pays benefits from time to time.
The fund value is reported after every transaction and at year end. The details during the
year 2005 are as follows:
Date
Jan 1, 2005
Ma 1, 2005
Sep 1, 2005
Nov 1, 2006
Jan 1, 2006
Amount
1,000,000
1,240,000
1,600,000
1,080,000
900,000
Contribution Received:
Date
Feb 28, 2005
Aug 31, 2005
Amount
200,000
200,000
Benefits Paid:
Date
Oct 31, 2005
Dec 31, 2005
Amount
500,000
200,000
Fund Values:
Q.1 Find the time–weighted rate of return.
Here, the year 2005 is partitioned into 4 smaller intervals: [01/01/2005, 03/01/2005),
[03/01/2005, 09/01/2005), [09/01/2005, 11/01/2005), and [11/01/2005, 01/01/2006).
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Partition
Beginning
Balance
Ending
Balance
Before Transaction
Rate of Return
[01/01/2005,
03/01/2005)
1000000
1040000
1040000
1000000
− 1 = .04
[03/01/2005,
09/01/2005)
1240000
1400000
1400000
1240000
− 1 = .129032
[09/01/2005,
11/01/2005)
1600000
1580000
1580000
1600000
− 1 = −.0125
[11/01/2005,
01/01/2006)
1080000
1100000
1100000
1080000
− 1 = .0185185
Therefore, the time–weighted return is
(1 + 04)(1 + 0.129032)(1 − 0.0125)(1 + 0.0185185) − 1 = 0.0.1809886
or it can also be solved by
1040000 1400000 1580000 1100000
− 1 = 0.18098865
1000000 1240000 1600000 1080000
Q.2 Find the internal rate of return.
122
61
306
1000000(1+i)+200000(1+i) 365 +200000(1+i) 365 = 500000(1+i) 365 +200000+900000
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One More Example:
Month
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Total
# of Days
31
28
31
30
31
30
31
31
30
31
30
31
365
# of Day till The End of Year
334
306
275
245
214
184
153
122
92
61
31
0
Example 5.1 An association had a fund balance of $75 on January 1 and $60 on December
31. At the end of every month association treasurer deposited $10 from the membership fees.
Withdrawals
Feb. 28
$5
Jun. 30 $25
Oct. 15 $80
Oct. 31 $35
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What is the dollar–weighted rate of return?
There are 12 deposits of $10 and 4 withdrawals on Feb. 28, Jun. 30, Oct 15, and Oct 31.
There are 16 transactions in total.
R0 − C0 R1 − C1 R2 − C2 R3 − C3 · · · R14 − C14
= 75 10 5 10 10 10 −15 10 10 10 −80 −25 10 −65
334
184
, where t0 = 0, t1 = 1 − 365
, t2 = 1 − 306
, t3 = 1 − 275
, t4 = 1 − 245
, t5 = 1 − 214
, t6 = 1 − 365
,
365
365
365
365
122
92
77
61
31
153
t7 = 1 − 365 , t8 = 1 − 365 , t9 = 1 − 365 , t10 = 1 − 365 , t11 = 1 − 365 , t12 = 1 − 365 , and
0
t13 = 1 − 365
Therefore, 75(1 + j) + 10(1 + 334
j) + 5(1 + 306
j) + · · · + (−65) = 0
365
365
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Basic Textbook Questions to Practice
• 5.2.1 ˜ 5.2.7
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