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Money-Time Relationships
Part II – Tricks and Techniques
Overview
• Important assumptions and how to break them
• Deferred annuities– Slow (but simple) use only P/F
– Faster - Combining P/F and P/A
• Linear Gradients– P/G, A/G, etc.
– Sometimes P/F is easier
• Exponential growth (or decay)
• Compounding intervals more often than cash flows
• Variable interest rates require F/P or P/F
Important Assumptions
Factor Assumption Violations
All (P/F,F/P,P/A,etc.) Constant interest rate for N periods
Variable interest rates
(5% for 3 years, then 6% for 2 years, then 4% for 6 years)
(P/A,i%,N)
(F/A,i%,N)
Constant cash flows Cash Flows that grow or shrink
(P/A,i%,N) A/P
(F/A,i%,N) A/F
Cash flow interval matches interest compounding interval
Interest compounded more often (monthly) than cash flows (quarterly/yearly)
(P/A,i%,N) A/P
(F/A,i%,N) A/F
Cash flows start at end of year 1 and finish at end of year N
Cash flow starts at a future year (year 10)
Dealing with Violations IProblem Solution TechniquesChanges in i% Find breakdown of problem where i
% is piecewise-constant. Use P/F or F/P only. Avoid using P/A, etc.
Variable Cash Flows
($A grows or shrinks)
If $A is piecewise-constant, with changes in the level, use deferred annuity technique.
If $A changes every period, shortcuts exist only for special cases (linear, exponential).
For the general case, you can always use P/F or F/P and sum up the results.
Dealing with Violations II
Problem Solution Techniques
Interest compounded more often (monthly) than cash flows (quarterly/yearly)
Use time periods that match the cash flows.
Adjust the monthly interest rate I up to the new period.
I* = (1+I)4-1 I*=(1+I)12-1
First Payment of Annual Cash flow starts at a future year, not year 1 (example: need PV of cash flow that starts at year 10 with last payment in year 20)
Use deferred annuity formula.
Apply appropriate combinations of P/F and P/A.
example: (P/F,i%,9)*(P/A,i%,10)
Deferred annuitiesCash flow starts at beginning of year J+1Last payment at end of year N
0 J J+1 N
$A/year
Present Value $P = (P/F,i%,J)*(P/A,i%,N-J)*A
$P ?
Why not just P/A?
0 J J+1 N
$P ?$A/year
(P/A,i%,N-J)*$A gives the value of the cash flow in units of “year J dollars”. This is probably not what you wanted as your final result.
Remember: P/A gives a dollar value that is timed one year before the start of the cash flows. You need to use P/F or F/P to move this value to other years!
That’s why $P = (P/F,i%,J)(P/A,i%,N-J)*$A
Variable cash flows
• For general cases, use P/F or F/P• For linear cases (e.g. –3000,-1000,1000,3000,5000), there is a gradient
method involving gradient factors P/G, F/G, etc.• For exponential cases, (e.g. 1000,1100,1210,…)
there is a convenience interest rate method• Use of excel together with P/F or F/P is often the
best solution technique. If you are confused, then use P/F or F/P together with a table. This keeps the analysis simple and easy to follow.
Exponential Decay ExampleA watch manufacturer expects a revenue of
$100,000 for the first month. The revenue declines by 10% each month and ends after the 12th month.
Calculate the Present Value given i=1%/month.
0 1 2 3 4 5 6 7 8 9 10 11 12
$P?
$100000$59049
$31,381
Exponential Growth/Decay Slow but simple (Excel + P/F)
Time (months) Cash Flow P/F factor Contribution to PV1 $100,000 0.9901 $99,0102 $90,000 0.9803 $88,2273 $81,000 0.97059 $78,6184 $72,900 0.96098 $70,0555 $65,610 0.95147 $62,4266 $59,049 0.94205 $55,6277 $53,144 0.93272 $49,5688 $47,830 0.92348 $44,1709 $43,047 0.91434 $39,359
10 $38,742 0.90529 $35,07311 $34,868 0.89632 $31,25312 $31,381 0.88745 $27,849
PV= $681,235(sum)
Exponential Growth/Decay Convenience Interest Rate Method
Another way to calculate annuities with a growth (or decay) factor is to adjust the interest rate factor and use a special formula.
Common ratio =f = (Ak- Ak-1)/ Ak-1 = -0.10
“Convenience rate” icr=[(1+i)/(1+f)]-1
=[1.01/0.90]-1=0.1222
Special formula PV = A1(P/A, icr%,N)/(1+f) =
= ($100000)(6.132)/(0.90)=$681,333
Exponential Growth/Decay Convenience Interest Rate Method
Common ratio =f = (Ak- Ak-1)/ Ak-1 = -0.10
“Convenience rate” icr=[(1+i)/(1+f)]-1=[1.01/0.90]-1=0.1222
PV = A1(P/A, icr%,N)/(1+f) == ($100000)(6.132)/(0.90)=$681,333 Compare with more careful excel+P/F method:
$681,235 (difference is due to rounding 6.132)
Linear Gradients
Cash flow increases or decreases _linearly_
(by the same _amount_ each period)
1 2
3 4 5
-3000-1000
10003000 5000 7000
6
Investment loses 3000 in year 1, 1000 in year 2, but earns 1000,3000,5000,7000 in years 3-6. What is the PV at i=8%?
Gradient Factors
0 1 2 3 4 5 …………. N
$0 $0
$1$2
$3$4
$(N-1)
(P/G,i%,N) is the present value (units: year 0$) of this cash flow(F/G,i%,N) is the future value (units: year N$) of this cash flow(A/G,i%,N) is the annuity value (units: $/year) of this cash flow
Year
Linear Gradients
1 2
3 4 5
-3000-1000
10003000 5000 7000
6
=
+
-3000
2000 4000 6000 8000 10000
annuity
Simple gradient
Linear Gradients
1 2
3 4 5
-3000-1000
10003000 5000 7000
6
Cash flow = annuity (-3000) + gradient (2000/year)
Gradient always starts at year 2.
PV = -3000 (P/A,8%,6) + 2000(P/G,8%,6)
(p.632) = (-3000*4.6229) + 2000 (10.523)
=-$13869 + $21046 = $7177
60 Seconds Investment ChallengeLet i=2%/month. There are two investments, A and B.
An investment costs $300 in terms of today’s dollars.
0 1 2 3 4….N….. 24 months
$0 $1 $2 $3 ….$N… $36 (end)
INVESTMENT A
INVESTMENT B
0 1 2 3 ….N….. 36 months
$21 $21 $21 $21 $21$0
Do you want to swap $300 for the PV of A, or the PV of B?
$21 (end)
Investment Challenge: Analysis of A
Let i=2%/month. The investment costs $300 in terms of today’s dollars.
$0 $1 $2 $3 ….$N… $36 (end)
INVESTMENT A
0 1 2 3 ….N….. 36 months
PV = $1 * (P/G,2%,36) + $1 * (P/A,2%,36)
= $392.04 + $25.49 = $417.53
Did you forget that cash flow for P/G begins in year 2? In this example, this mistake cost you money!
Investment Challenge:Analysis of B
Let i=2%/month. The investment costs $300 in terms of today’s dollars.
0 1 2 3 4….N….. 24 months
INVESTMENT B
$21 $21 $21 $21 $21$0
Evaluating B is simple, because it has a constant cash flow of $21. It starts in year 1, so we can use the P/A formula.
$PV = $21 * (P/A,24,2%) = $21 * 18.9139 = $397.20
Compounding intervals more often than cash flows
Solution: Change interest rate period to match cash flow period
Example: Interest compounded every month at 1%/month, but cash flows are every six months. Use i*=[(1+i)^N]-1=1.01^6-1
=1.06152-1
=6.152%/6-month period
Savings account example: Problem
Every month your bank pays 0.25% interest on your savings account balance.
Every 3 months you deposit $5000. How much do you have after 3 years?
Note: this is a F/A problem, except that the compounding interval of 1 month does not match the deposit (cash flow) interval of 3 months.
Savings account example: Analysis
Step 1: Change interest rate to 3-month rate:i*=[(1+.0025)3-1]=.0075187Step 2: Determine N. If a period is 3-months, then we
have N=12 periods in 3 years.Step 3: Notice that the cash flow every period is
constant, so we can use the FV formula, $FV = $5000 * (F/A,i*,12)Final Step: Calculate the F/A formula.$FV =$5000 * [(1.0075187)12-1]/(0.0075187)=$5000*12.50888=$62544.42
Variable interest rates and F/P (or P/F)
Rule: Use a separate F/P (or P/F) for each group of years or periods where the interest rate is constant.
Example: You have $5000 today. For 3 years you invest it at 4% per year, then for 5 more years at 5% per year, then 2 more for 3% per year. The future value is
$FV = $5000 * (F/P,4%,3) * (F/P,5%,5) * (F/P,3%,2) = $5000 * 1.1249 * 1.2763 * 1.0609
Summary• We learned some tricks for finding present and
future values in special situations.• The “trick that always works” is to make a table of
all cash flows in excel, apply appropriate P/F or F/P factors, and add up the result.
• To apply shortcuts for linear or geometric cases, one must pay careful attention to detail.
• Next week – Chapter 4– more applications– Return on investment (finding the i%)– comparing machines, investment plans, etc.