Example Decomposition

# Example

Class Example: July 6th 2009
These examples also employ the method of Time Shifting

You're given a series of benefits that begins with $1 in period 1. The series doubles each period from periods 0 - 4, settles to a constant series in periods 5-7, then decreases by half each period for periods 8 - 11. The interest rate is 10% per period. What is the present value of these benefits?  Year Benefit 0 0 1$1 2 $2 3$4 4 $8 5$8 6 $8 7$8 8 $8 9$4 10 $2 11$1

This kind of problem lends itself to many equally valid approaches.
The key to solving these problems is to:
(1) Identify a decomposition that works.
(2) Apply what methods you know to solve your decompositions.

Note that some decompositions are more suitable to some methods of solving, and less suitable to others.
A common example is the geometric graded series. Tables exist for constant and linear series at standard interest rates,
but for geometric series you will need to calculate values in order to get a solution.

Here are two solutions that rely on two very different decompositions.

Method I: Decomposition into geometric/ constant series: Thinking along time

 Year Benefit Benefit1 Benefit2 Benefit3 0 $0$0 0 0 1 $1$1 0 0 2 $2$2 0 0 3 $4$4 0 0 4 $8$8 $0 0 5$8 0 $8 0 6$8 0 $8 0 7$8 0 $8$0 8 $8 0 0$8 9 $4 0 0$4 10 $2 0 0$2 11 $1 0 0$1

Benefit 1:
In periods 0-4, the value follows a geometric graded series. Since the value doubles, the value can be said to increase by
100% per period, so g=100%.
In factor notation, this is:

(1)
\begin{align} P_{1}=\1 \cdot \left(P|A_{1}, g=100\%, i=10\%, 4\right) \end{align} Benefit 2: In periods 5-7, the values are a constant series starting at period 4 (remember, a constant series has value 0 at period 0, and then a constant value for each following period) Since this series begins at period 4, we convert this to present value (value in period 0) by dividing out the factor of (1+i)^N In factor notation, this is: (2) \begin{align} P_{2}=\8 \cdot \frac{\left(P|A, i=10\%, 3\right)}{\left(1+0.1\right)^4} \end{align}

Benefit 3:
In periods 8-11, the values again follow a geometric graded series, this time decreasing by half each period.
So g=-50%. Since this series is offset by 7 periods, we must divide by a factor of (1+0.1)^7
In factor notation, this is:

(3)
\begin{align} P_{3}=\8 \cdot \frac{\left(P|A_{1}, g=-50\%, i=10\%, 4\right)}{\left(1+0.1\right)^7} \end{align} And so the overall present value of the benefit can be calculated: (P|A1, g=100%, i=10%, 4) = 11.03135 (P|A, i=10%, 3)/(1.1)^4 = 13.5887 (P|A1, g=-50%, i=10%, 4)/(1.1)^7 = 6.55003 (4) \begin{align} P_{total} = P_{1}+P_{2}+P_{3} = \31.17 \end{align}

Method II: The Scary Birthday Cake: Decomposition into 'layers' of constant series: Thinking in terms of values

In order to think through these problems, making a quick plot of the given series is a must.
Consider the values of the series of benefits. Working from the bottom to the top, the values go as 1, 2 ,4, 8.
We can break the series values into 'layers' that sum together:

(A) We start with a constant series of value 1.
(B) To get from a value of 1 to a value of 2, we add a constant series with value 1.
(C) To get from a value of 2 to a value of 4, we add a series of value 2.
(D) Finally, to reach a value of 8 from 4, we add a constant series of value 4.

Our decomposition is then just the four constant series:
(Remember: a constant series goes as (0,constant,constant,…)

(A) runs from 0 to 11, and has value 1:
It starts at period 0, and so we convert this to present value by dividing by (1+0.1)^0=1

(5)
\begin{align} P_{A}=\$1 \cdot \frac{\left(P|A, i=10\%, 11\right)}{\left(1+0.1\right)^0}={ \$ 6.495} \end{align}

(B) runs from 1 to 10, and has value 1:
It starts at period 1, and so we convert this to present value by dividing by (1+0.1)^1

(6)
\begin{align} P_{B}=\$1 \cdot \frac{\left(P|A, i=10\%, 9\right)}{\left(1+0.1\right)^1}={ \$ 5.235} \end{align}

(C) runs from 2 to 9, and has value 2:
To convert to present value: divide by (1+0.1)^2

(7)
\begin{align} P_{C}=\$2 \cdot \frac{\left(P|A, i=10\%, 7\right)}{\left(1+0.1\right)^2}={ \$ 8.047} \end{align}

(D) runs from 3 to 8, and has value 4:
To convert to present value: divide by (1+0.1)^3

(8)
\begin{align} P_{D}=\$4 \cdot \frac{\left(P|A, i=10\%, 5\right)}{\left(1+0.1\right)^3}={ \$ 11.39} \end{align}

The total present value is then:

(9)
\begin{align} P_{total}=P_{A}+P_{B}+P_{C}+P_{D}={ \31.17} \end{align} # Questions The following represent good sample questions that would help you prepare for an exam: Question: Given: Suppose you are lucky enough to have an asset that pays you10,000 the first month, but each following month the payout decreases by $100 for 20 months until it reaches a final payout of$8,000. The $8,000 payouts continue another 10 months (until it reaches month 30). The interest rate is 2% Find: Using factor notation, how could you express the present worth of the benefits earned over time from month 1, to month 30? Answer: First of all to avoid doing to much math in our head let's break down the benefits over time into the appropriate table spanning the months of interest (1-21).  Month Benefit 0 0 1$10,000 2 $9,900 3$9,800 4 $9,700 5$9,600 . . . . Notice the pattern! . . 20 $8,100 21$8,000 . . . . 30 $8,000 Remember after month 21 the benefits stays constant From the looks of this there seems to be two decompositions. From month 1 to month 21 there is a deduction of$100 every month but note that that deduction amount is always the same every month, this would appear to be a Linear Gradient series. However this is not purely a linear gradient series, remember that there is initially a big payout, but just decreasing in size every month. In order for things to add up correctly there must be a Constant series to accommodate the linear gradient series and make up for the positive payout occurring each month.

Now let's start decomposing the benefits over time into two separate series, a constant series, and a linear gradient series. As a check, add both values of the series to ensure that they equal the total cost

 Month Constant Linear Gradient Benefit 0 0 + 0 = 0 1 $10,000 + 0 =$10,000 2 $10,000 + -$100 = $9,900 3$10,000 + -$200 =$9,800 4 $10,000 + -$300 = $9,700 5$10,000 + -$400 =$9,600 . $10,000 . . . . .$10,000 . . . . . $10,000 . . . . .$10,000 . . . . 20 $10,000 + -$1,900 = $8,100 21$10,000 + -$2,000 =$8,000

Make note that the difference in costs between two consecutive months of the linear gradient series is the same ($100 difference). Now both series are ready to be placed into factor notation Constant series: A(P\A,i,n) First off remember that you are trying to find the present worth (or the principal if you are investing), given the monthly cost/benefit, the interest rate, and the number of times it occurs. In a constant series the monthly cost is a no-brainer, every month there is a$10,000 benefit, making this the "A" variable in factor notation. The interest rate was given as 2% and with no additional information for calculating effective interest rate it alone can be interpreted as the monthly interest rate (making this the "i" variable). To estimate the correct number of cycles check to see that the benefit/cost at time zero is zero and that the benefit/cost at month one is the initial constant value, start at zero and count the number of months that actually have benefits in them, in this case it's 21 months, making that our "n" variable.

Pconstant = A(P\A,i,n) = $10,000(P\A,i=2%,n=21) Linear Gradient Series: For this factor notation we must remember that we are trying to find present worth given that we know the gradient (or decrease in benefits per month), the interest rate, and the number of times this occurs. From subtraction of consecutive months from one another we know that the decrease in benefits each month is -$100, making this the "G" term. The interest rate, as before, is 2% (the "i" term). To find the correct number of times this happens, remember that linear gradients have no cost/benefit at either time zero or time n=1 (month one in this case). If there is zero cost at time zero and month one, then no time stepping is needed. The correct number of cycles is the number that there is actually a cost/benefit in that range. Particularly for this problem the first cost happens at month 2 and ends at month 21 which equals 20 (our "n" value).

Plinear gradient = G(P\G,i,n) = -$100(P\G,i=2%,n=21) And since each component is a single decomposition of a much greater one, they must be combined Hence; Ptotal =$10,000(P\A,i=2%,n=21) + -100(P\G,i=2%,n=21) The actual calculation (using the formulas from the text) is… (10) \begin{align} \begin{split} & 10000*\biggr[\frac{(1+0.02)^{21}-1}{0.02(1+0.02)^{21}}\biggl]+(-100)*\biggr[\frac{(1+0.02)^{21}-0.02(21)-1}{0.02^2(1+0.02)^{21}}\biggl]\\ & = 170,112.09 + -15,779.59\\ & = 154,332.51 \end{split} \end{align} Now we need to look at the rest of the problem, months 22-30. Each of these months have a constant positive payout of8000 . We will set up a simple constant series that will have to be time shifted back (see the next section for full details on time shifting).

In this series:
n = 9 (for months 22 through 30)
i = 2%
A = 8000
time shift n = 21

Pconstant = A(P\A,i,n) = \$8,000(P\A,i=2%,n=9)/(1+2%)^21

(11)
\begin{align} \begin{split} & \frac{8000*\biggr[\frac{(1+0.02)^{9}-1}{0.02(1+0.02)^9}\biggl]}{(1+.02)^21}\\ & = 43,081.97 \end{split} \end{align}

Therefore our total present worth is:

154,332.51 + 43,081.97 = 197,414.48

page revision: 10, last edited: 31 Mar 2016 18:48