Chapter 4 - Time Value of Money
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Chapter 4 - Time value of money
Money-time relationships Money has a time value because it can earn interest (or profit) over time. Simple interest: Total interest is linearly proportional to the amount of loan (principal), the interest rate, and the number of interest periods. I = (P)(N)(i) I : total interest P : principal N : number of interest periods i : interest rate per interest period. Example 4.1: You borrowed $10,000 for 3 years at a simple interest rate of 10%. How much will you repay at the end of 3rd year? I = (10, 000)(3)(0.1) = $3, 000. You need to pay $13,000. SEEM2440A/B
Chapter 4 - Time value of money
Compound interest: Suppose the interest accumulates as follows: Period, p Amount owed Interest amount at the start of p for p 1 $10,000 $1,000 2 $11,000 $1,100 3 $12,100 $1,210
Amount owed at the end of p $11,000 $12,100 $13,310
You would repay $13,310. Total interest payment would be $13,310-$10,000=$3,310>$3,000 (interest in simple interest calculation). Interest accumulates over the earned interests in the previous periods. More common than simple interest. Generalize: Future equivalent of $10,000 (P) at the end of 3 (N) periods is P(1 + i)N = $10, 000(1.1)3 .
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Chapter 4 - Time value of money
Economic equivalence When comparing the alternatives, we should consider: the interest rate the amounts of money involved the timing of the monetary receipts or expenses the manner in which the interest (or profit) is paid Back to the example with compound interest: $10,000 now is equivalent to $11,000 one year from now, and it is equivalent to $12,100 two years from now under the fixed compound interest rate of 10%. One should be indifferent between the following three alternatives: 1
Pay/receive $10,000 now
2
Pay/receive $11,000 one year from now
3
Pay/receive $12,100 two years from now SEEM2440A/B
Chapter 4 - Time value of money
Example 4.2 (3 plans for repayment of $10,000 in three years at 10% per year) You owe $10,000 to your credit card company. You have 3 plans to pay the balance back: Plan 1: Pay interest due at end of each year and principal at end of third year. Year Amount owed Interest Principal payment Total pmnt. 1 10,000 1,000 0 1,000 2 10,000 1,000 0 1,000 3 10,000 1,000 10,000 11,000 Total: 3,000 13,000
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Chapter 4 - Time value of money
Plan 2: Pay principal and interest in one payment at the end of third year. Year Amount owed Interest Principal payment Total pmnt. 1 10,000 1,000 0 0 2 11,000 1,100 0 0 3 12,100 1,210 10,000 13,310 Total: 3,310 13,310 Plan 3: Year 1 2 3
Pay off the debt in 3 equal end-of-year payments. Amount owed Interest Principal payment Total pmnt. 10,000 1,000 3,021 4,021 6,979 697.9 3,323 4,021 3,656 365.6 3,656 4,021 Total: 2,063.5 12,063
Plan 1,2 and 3 are all equivalent at the compound interest rate of 10%! (Why? You can answer it at the end of this chapter.) SEEM2440A/B
Chapter 4 - Time value of money
Cash-flow diagrams i = effective interest rate per interest period N = number of compounding periods P = the present sum of money F = future sum of money A = end-of-period cash flows in a uniform series for a specified number of periods, starting at the end of first period and continuing through the last period. In a cash-flow diagram: horizontal line represents time scale, arrows represent cash flows. Downward arrows represent expenses (negative cash flows or cash outflows) and upward arrows represent receipts (positive cash flows or cash inflows). The cash-flow diagram is dependent on the point of view. In the course, without explicitly mention, the company’s (investor’s) point of view will be taken. SEEM2440A/B
Chapter 4 - Time value of money
General setting
Beginning of period 1
0
1
End of period 1
2
3
F
N-1
N
i % per period P
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Chapter 4 - Time value of money
Example 4.3
Show the cash flow diagrams for Plans 2 and 3 from the credit card company’s viewpoint. Identify i, P, F, A.
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Chapter 4 - Time value of money
Solution
Plan 3
Plan 2 F = $13,310
A= $4,021
0
0 1
2
1
3=N
2
End of Year (EOY)
End of Year (EOY) P = $10,000
P = $10,000
i = 10% per year
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Chapter 4 - Time value of money
3=N
Example 4.4
You are analyzing a project with five-year life. The project requires a capital investment of $50,000 now, and it will generate uniform annual revenue of $6,000. Further, the project will have a salvage value of $4,500 at the end of the fifth year and it will require $3,000 each year for the operations. Develop the cash-flow diagram for this project from the investor’s viewpoint.
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Chapter 4 - Time value of money
Solution
$4,500
$6,000 $6,000 1
0
$3,000
2
$3,000
$6,000
$6,000
3
4
$3,000
$3,000
$6,000 5
$3,000
End of Year (EOY)
$50,000
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Chapter 4 - Time value of money
Rules for performing calculations with cash flows
Three Rules for performing arithmetic calculations with cash flows: 1
Cash flows cannot be added or subtracted unless they occur at the same point in time.
2
To move a cash flow forward in time by one time unit, multiply the magnitude of the cash flow by (1 + i).
3
To move a cash flow backward in time by one time unit, divide the magnitude of the cash flow by (1 + i).
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Chapter 4 - Time value of money
Interest formulas relating P and F
F = Future Equivalent (Find)
i = Interest rate per period 0 1
N−1
…
2
N
End of Period
P = Present Equivalent (Given)
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Chapter 4 - Time value of money
Interest formulas relating P and F
Finding F when P is given F = P(1 + i)N . Notation: F = P(F/P, i%, N) where the factor in the parentheses is read "find F given P at i% interest per period for N interest periods".
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Chapter 4 - Time value of money
Interest formulas relating P and F
F = Future Equivalent (Given)
i = Interest rate per period 0 1
N−1
…
2
N
End of Period
P = Present Equivalent (Find)
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Chapter 4 - Time value of money
Interest formulas relating P and F
Finding P when F is given N 1 P = F 1+i = F(1 + i)−N . Notation: P = F(P/F, i%, N) where the factor in the parentheses is read "find P given F at i% interest per period for N interest periods.
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Chapter 4 - Time value of money
Example 4.5: You deposit $50,000 now in a savings account at an interest rate of 8% per year. How much will you have in this account at the end of fifth year? Assume the interest rate is constant at 8% for the next five years. Solution: P = 50, 000, i = 8% per year, N = 5 F = P(F/P, 8%, 5) = 50, 000(1 + 0.08)5 = $73, 466 Example 4.6: How much should you deposit in a savings account now so that you will have $100,000 in this account 6 years from now? Assume that the interest rate is 5% per year. Solution: F = 100, 000, i = 5% per year, N = 6 P = F(P/F, 5%, 6) = 100, 000(1 + 0.05)−6 = $74, 627
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Chapter 4 - Time value of money
General setting with annuities (A)
A
A
1
2
A
A
A
N
0
3
N-1 i % per period F
P
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Chapter 4 - Time value of money
Interest formulas relating P,F, and A Rule: Uniform (equal) receipts, A, at the end of each period for N periods at an interest rate of i% per period, where the first receipt occurs at the end of the first period and the last one occurs at the end of the Nth period. Finding F when A is given Find the future equivalent at the end of Nth period of each cash flow A. Then sum them up. Future equivalent at the end of period N of the cash flow that happens at the end of period 1: A(F/P, i%, N − 1) Future equivalent at the end of period N of the cash flow that happens at the end of period 2: A(F/P, i%, N − 2) ........... Future equivalent at the end of period N of the cash flow that happens at the end of period N: A(F/P, i%, 0) SEEM2440A/B
Chapter 4 - Time value of money
F = A(F/P, i%, N − 1) + A(F/P, i%, N − 2) + ... + A(F/P, i%, 1) + A(F/P, i%, 0) ⇒ F = A[(1 + i)N−1 + (1 + i)N−2 + ... + (1 + i)1 + (1 + i)0 ]. Note that [(1 + i)N−1 + (1 + i)N−2 + (1 + i)N−3 + ... + (1 + i)1 + (1 + i)0 ] = in h N−1 P 1 (1 + i)N−1 (1+i) n=0
The last term comprises a geometric series of the form ar0 + ar1 + ar3 + ... + arN−2 + arN−1 , where a = (1 + i)N−1 , r = (1 + i)−1 .
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Chapter 4 - Time value of money
Note that the sum of the first N terms in a geometric series is: N−1 P n a(1−rN ) ar = 1−r . n=0
Therefore we have the following: h in (1+i)N−1 − 1 N−1 P (1+i) 1 (1 + i)N−1 (1+i) = = 1− 1 (1+i)
n=0
(1+i)N −1 i
We find: F=A
h
(1+i)N −1 i
i
Notation: F = A(F/A, i%, N).
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Chapter 4 - Time value of money
Finding A when F is given h i N From F = A (1+i)i −1 , A=F
h
i (1+i)N −1
i
Notation: A = F(A/F, i%, N).
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Chapter 4 - Time value of money
Example 4.7: You plan to deposit $2,000 to your savings account at the end of every month for the next 15 months starting from the next month. If the interest rate you can earn is 2% per month how much money will accumulate immediately after your last deposit at the end of the 15th month? Solution: N = 15 months. F =? h per15month, i h A N= 2,i000, i = 2% (1.02) −1 (1+i) −1 F=A = 2, 000 = $34, 587 i 0.02 Example 4.8: What uniform monthly amount should you deposit in your savings account at the end of each month for the following 10 months in order to accumulate $75,000 at the time of the 10th deposit? Assume that the interest rate you can earn is 4% per month and the first deposit will be made next month. Solution: N = 10 months. A =? h F = 75, i 000, i = 4% h per month, i i 0.04 A = F (1+i)N −1 = 75, 000 (1.04)10 −1 = $6, 247
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Chapter 4 - Time value of money
Finding P when A is given Given P = F(P/F, i%, N) and F = A(F/A, i%, N), we can derive P = A(P/A, i%, N) as follows: P h= A(P/A,ii%, N) = [A(F/A, i%, N)](P/F, i%, N) ⇒ N A (1+i)i −1 (1 + i)−N P=A
h
(1+i)N −1 i(1+i)N
i
i(1+i)N (1+i)N −1
i
Notation: P = A(P/A, i%, N). Finding A when P is given i h N −1 From P = A (1+i) , N i(1+i) A=P
h
Notation: A = P(A/P, i%, N). SEEM2440A/B
Chapter 4 - Time value of money
Example 4.9: How much should you deposit to your savings account now at an annual interest rate of 10% to provide for 5 end-of-year withdrawals of $15,000 each? Solution: h A N= 15, i 000, i = 10% h per iyear, N = 5 years. P =? (1+i) −1 (1.1)5 −1 P = A i(1+i)N = 15, 000 0.1(1.1)5 = $56, 862 Example 4.10: You plan to borrow a loan of $100,000 which you will repay with equal annual payments for the next 5 years. Suppose the interest rate you are charged is 8% per year and you will make the first payment one year after receiving the loan. How much is your annual payment? Solution: i 000, i = 8% h per year, i N = 5 years. A =? h P =N 100, 0.08(1.08)5 i(1+i) A = P (1+i)N −1 = 100, 000 (1.08)5 −1 = $25, 046
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Chapter 4 - Time value of money
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Chapter 4 - Time value of money
Deferred annuities Table 1: Ordinary annuities Year 0 1 2 3 Cash flow - A A A
... ...
Table 2: Deferred annuities Year 0 1 2 3 ... Cash flow - - - - ...
N A
J -
J+1 A
J+2 A
... ...
N A
What is the present equivalent of the deferred annuity shown in Table 2 as of time 0 (P0 )? P0 = A(P/A, i%, N − J)(P/F, i%, J) What is the future equivalent of the deferred annuity shown in Table 2 as of time N (FN )? FN = A(F/A, i%, N − J) SEEM2440A/B
Chapter 4 - Time value of money
Example 4.11
P0=?
5,000
4,000
..……..
4,000 2,000 …….. 2,000
6 0
1
2
3
4
1,000
7
8
9
5
10 11
……..………………….
12
13
14
15
1,000 i=10 % per period
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Chapter 4 - Time value of money
Solution
P0 = 4000(P/A, 10%, 5) − 1000(P/A, 10%, 8)(P/F, 10%, 2) + 5000(P/F, 10%, 11) + 2000(P/A, 10%, 4)(P/F, 10%, 11) = $14, 729.
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Chapter 4 - Time value of money
Arithmetic gradient series
Cash flow series that increases or decreases by a constant amount (G) from one period to the next. See the cash flow diagram in the next slide.
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Chapter 4 - Time value of money
Cash flows in arithmetic gradient series
(N-1)G (N-2)G 3G 2G G
0
1
2
3
4
N-1
N
i% per period
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Chapter 4 - Time value of money
Interest formulas Finding F when G is given F = G(F/P, i%, N − 2) + 2G(F/P, i%, N − 3) + 3G(F/P, i%, N − 4) + ... + (N − 1)G(F/P, i%, 0) ⇒ F = G[(1 + i)N−2 + 2(1 + i)N−3 + 3(1 + i)N−4 + ... + (N − 1)(1 + i)0 ] (1) Multiply both sides of the expression in (1) with (1 + i): F(1+i) = G[(1+i)N−1 +2(1+i)N−2 +3(1+i)N−3 +...+(N−1)(1+i)1 ] (2) Subtract (1) from (2): Fi = G[(1 + i)N−1 + (1 + i)N−2 + (1 + i)N−3 + ... + (1 + i)1 − (N − 1)] F = Gi [(1 + i)N−1 + (1 + i)N−2 + (1 + i)N−3 + ... + (1 + i)1 + 1] − GN i F=
G i
h
(1+i)N −1 i
i
−
GN i
⇒F=
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G i (F/A, i%, N)
−
Chapter 4 - Time value of money
GN i .
Finding P when G is given We know the formula to find F given G, and the formula to find P given F. We can use these formulas to find (P/G, i%, N) as follows: i h N (P/F, i%, N) P = Gi (1+i)i −1 − GN i Recall (P/F, i%, N) = (1 + i)−N Hence, P=G
h 1 i
(1+i)N −1 i(1+i)N
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−
N (1+i)N
i .
Chapter 4 - Time value of money
Finding A when G is given We know the formula to find F given G, and the formula to find A given F. We can use these formulas to find (A/G, i%, N) as follows: (A/F, i%, N) A = Gi (F/A, i%, N) − GN i ⇒ A = Gi − GN (A/F, i%, N) i (A/F, i%, N) = (1+i)i N −1 Hence, A=G
h
1 i
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−
N (1+i)N −1
i
.
Chapter 4 - Time value of money
Example 4.12
18,000 16,000 14,000 12,000 10,000 1
0
2
3
4
5
i=10% per period a) P=?
0
b) A=?
1
2
3
4
5
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0
1
2
3
Chapter 4 - Time value of money
4
5
Solution $18,000 $16,000 $14,000 $12,000 $10,000 5
4
3
2
1
0
$8,000 $6,000
$10,000 $4,000
+ 0
1
2
3
4
5
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$2,000 0
1
2
3
4
Chapter 4 - Time value of money
5
Solution
P = 10000(P/A, 10%, 5) + 2000(P/G, 10%, 5) = $51, 631. A = P(A/P, 10%, 5) = 51631.47(A/P, 10%, 5) = $13, 620.
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Chapter 4 - Time value of money
Geometric gradient series
Cash flow series that increases or decreases by a constant percentage (f ) from one period to the next. See the cash flow diagram in the next slide. A −A Note that Ak = (1 + f )Ak−1 and f = kAk−1k−1 , where f is the constant rate of change.
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Chapter 4 - Time value of money
Cash flows in geometric gradient series
i% per period 0
1
3
2
4
N-1
N
A1 A2=(1+f)A1 A3=(1+f)2A1 A4=(1+f)3A1 AN-1=(1+f)N-2A1
AN=(1+f)N-1A1
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Chapter 4 - Time value of money
Interest formulas
Derive the expression for the present equivalent (P) of the geometric series. Use the previously derived formulas (A/P,i%,N) and (F/P,i%,N) to find A and F, respectively. P = A1 (P/F, i%, 1) + A2 (P/F, i%, 2) + ... + AN (P/F, i%, N) Note that (P/F, i%, N) = (1 + i)−N and Ak = (1 + f )k−1 A1 , 2 ≤ k ≤ N. Hence, P = A1 (1 + i)−1 +(1 + f )(1 + i)−2 + ... + (1 + f )N−1 (1 + i)−N N−1 2 1+f 1+f + + ... + ⇒ P = A1 (1 + i)−1 1 + 1+f 1+i 1+i 1+i
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Chapter 4 - Time value of money
Using the summation formula
N−1 P
xn =
n=0
(1−xN ) 1−x ,
when x 6= 1, we
can derive the following expression: 1+f (Note that x = 1+i , and x 6= 1 reduces to i 6= f ) ( P=
A1 [1−(P/F,i%,N)(F/P,f %,N)] i−f
A1 N(P/F, i%, 1)
f = 6 i f =i
Note that f can be negative. For example, when f = −0.1, (F/P, f %, N) = (1 − 0.1)N .
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Chapter 4 - Time value of money
Example 4.13
(Blank&Tarquin) Chemical Engineers at a Coleman Industries plant in the Midwest have determined that a small amount of a newly available chemical additive will increase the water repellency of Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive through a 5-year contract at $7,000 per year, starting 1 year from now. He expects the annual price to increase by 12% per year thereafter for the next 8 years. Additionally, an initial investment of $35,000 was made now to prepare a site suitable for the contractor to deliver the additive. Use i=15% per year to determine the equivalent total present worth of all these cash flows.
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Chapter 4 - Time value of money
Solution
End of Year (EOY) 0
1
2
…
5
…
6
13
… …
$7,000
$7,000(1.12)
$35,000
$7,000(1.12)8
i = 15% per year
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Chapter 4 - Time value of money
Solution
35, 000 + 7, 000(P/A, 15%, 4) + 7,000[1−(P/F,15%,9)(F/P,12%,9)] (P/F, 15%, 4) = $83, 232. 0.15−0.12
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Chapter 4 - Time value of money
Example 4.14 10,000 9,500 9,000 8,500 8,000
7,500 7,000
0
1
2
4
3
5
6
7
8
9
i=10% per period 2,000
2,500 X
0
X/3
X/3
X/3
1
2
3
4
X
X
X
5
6
7
8
X=? SEEM2440A/B
Chapter 4 - Time value of money
9
Solution
[10, 000(P/A, 10%, 7) − 500(P/G, 10%, 7)](P/F, 10%, 2) − 2, 000 = X(P/A, 10%, 7) − 2X 3 (P/A, 10%, 3) + 2, 500(P/F, 10%, 9). [10, 000(4.8684) − 500(12.763)](0.8264) − 2, 000 = X(4.8684) − 2X 3 (2.4869) + 2, 500(0.4241) ⇒ X = 9, 935.8
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Chapter 4 - Time value of money
Example 4.15
Suppose that the parents of a young child decide to make annual deposits into a savings account, with the first deposit being made on the child’s 5th birthday and the last deposit being made on the 15th birthday. Then, starting on the child’s 18th birthday, four end-of-year withdrawals will be made at the amounts of $2,000, $2,400, $2,800, and $3,200, respectively. If the effective annual interest rate is 8% during this period of time, what are the annual deposits in years 5 through 15?
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Chapter 4 - Time value of money
Solution
A= [2,000 (P/A,8%,4) + 400 (P/G,8%,4)] (P/F,8%,2) (A/F,8%,11) = [2, 000(3.3121)+400 (4.650)] (0.8573) (0.0601) = $437.14
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Chapter 4 - Time value of money
Nominal and effective interest rates
You are given two 1-year fixed deposit saving plans. The interest rate of the two plans are: Plan 1: 12% per year (compounded annually). Plan 2: 12% per year compounded monthly. Which plan will you choose?
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Chapter 4 - Time value of money
In Plan 1, the interest rate "12% per year" is known as the effective interest rate per year, the same as the one we use before. In plan 2, the interest rate "12% per year compounded monthly" is known as the nominal interest rate per year with the compounding period 1 month (or compounding frequency 12).
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Chapter 4 - Time value of money
In general, the nominal and effective interest rates are related by !M r(M) −1 (NOM-EFF) i= 1+ M where i: effective interest rate per period; r(M) : nominal interest rate per period with compounding frequency M. It is obvious that i = r(1) .
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Chapter 4 - Time value of money
Example 4.16 You own a credit card which has a nominal interest rate of 15% per year, compounded monthly. a) What is the effective annual interest rate? b) Suppose your balance as of today (year 0) is $10,000. How much will you owe to the credit card company at the end of 3 years if you do not make any payments during 3 years? Solution: 12 − 1 = 16.07% a) i = 1 + 0.15 12 b) Use formula (F/P,i%,N). We can use effective annual interest rate of i% and N=3. P(F/P, i%, N) = 10, 000(F/P, 16.07%, 3) = 15, 637 Alternatively, we can use interest rate per month 0.15 12 = 0.0125 and N=3(12)=36. P(F/P, i%, N) = 10, 000(F/P, 1.25%, 36) = 15, 639
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Chapter 4 - Time value of money
Continuous compounding
What if M (number of compounding periods) → ∞. What is the effective (continuous) interest rate? r M − 1. We know that i = 1 + M r M lim i = lim 1 + M − lim 1 M→∞ M→∞ M→∞ p Recall: lim 1 + p1 = e (natural logarithm base) p→∞
Let
M r
=p
⇒ i = lim
p→∞
pr i h p ir h 1 + 1p − 1 = lim 1 + 1p −1 p→∞
⇒ i = er − 1 We can obtain the interest formulas relating P,F,A, by using the relation i = er − 1.
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Chapter 4 - Time value of money
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Chapter 4 - Time value of money
Payment Period and Compounding Period
Payment Period (PP) and Compounding Period (CP) of cash flow: Type 1: PP = CP (Done!) Type 2: PP > CP Type 3: PP < CP
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Chapter 4 - Time value of money
Type 2: PP > CP
The strategy in this case is to find the effective interest rate per PP by using the relation (NOM-EFF). Example 4.17: For the past 7 years, a quality manager has paid $500 every 6 months for the software maintenance contract of a LAN. What is the equivalent amount after the last payment, if these funds are taken from a pool that has been returning 10% per year compounded quarterly?
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Chapter 4 - Time value of money
Solution
PP = 6 months; CP = 3 months. The effective interest rate per 6-month i is given by 0.05 2 − 1 = 0.05063. i= 1+ 2 The equivalent amount after the last payment = 500(F/A, 5.063%, 14) = $9, 842.
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Chapter 4 - Time value of money
Type 3: PP < CP; Case 1
Assumption: Interperiod cash flows earn interest. Under the above assumption, the strategy is similar to the case of Type 2 by finding the effective interest rate per PP. Different from Type 2, we cannot apply the relation of the nominal and effective interest rates in (NOM-EFF) directly. The detailed treatment can refer to Example 4.18 below.
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Chapter 4 - Time value of money
Example 4.18
Suppose you make $500 monthly deposits to a tax-deferred retirement plan that pays interest at a rate of 10% per year compounded quarterly. Compute the balance at the end of 10 years.
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Chapter 4 - Time value of money
Solution
PP = 1 month; CP = 3 months. The effective interest rate per 1 month i is given by 10% 1/3 i= 1+ − 1 = 0.826% 4 The balance at the end of 10 years = $500(F/A, 0.826%, 120) = $101, 907.89.
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Chapter 4 - Time value of money
Type 3: PP < CP; Case 2
Assumption: Interperiod cash flows earn no interest. In this case, Deposits (negative cash flows) are all regarded as deposited at the end of the compounding period. Withdrawals (positive cash flows) are all regarded as withdrawn at the beginning of the compounding period.
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Chapter 4 - Time value of money
Example 4.19
Consider Example 4.18 again. Suppose that money deposited during a quarter (the compounding period) will not earn any interest. Compute the balance at the end of 10 years.
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Chapter 4 - Time value of money
Solution
F
Months 0
3
…
6
120
… A = $500 F
Quarters 0
1
…
2
39
… A = $1,500
SEEM2440A/B
Chapter 4 - Time value of money
40
Solution
The effective interest rate per quarter = 10% 4 = 2.5%. A = 3($500) = $1, 500 per quarter. So, F = $1, 500(F/A, 2.5%, 40) = $101, 103.83.
SEEM2440A/B
Chapter 4 - Time value of money