Chapter 4

Chapter 4 - Time Value of Money

SEEM2440A/B 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 interestP : principalN : number of interest periodsi : interest rate per interest period.

Example 4.1: You borrowed $10,000 for 3 years at a simpleinterest rate of 10%. How much will you repay at the end of 3rdyear?I = (10, 000)(3)(0.1) = $3, 000. You need to pay $13,000.


Compound interest:Suppose the interest accumulates as follows:Period, p Amount owed Interest amount Amount owed

at the start of p for p at the end of p1 $10,000 $1,000 $11,0002 $11,000 $1,100 $12,1003 $12,100 $1,210 $13,310

You would repay $13,310. Total interest payment would be$13,310-$10,000=$3,310>$3,000 (interest in simple interestcalculation).Interest accumulates over the earned interests in the previousperiods. 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.


Economic equivalence

When comparing the alternatives, we should consider:

the interest ratethe amounts of money involvedthe timing of the monetary receipts or expensesthe 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 itis equivalent to $12,100 two years from now under the fixedcompound interest rate of 10%. One should be indifferentbetween the following three alternatives:

1 Pay/receive $10,000 now2 Pay/receive $11,000 one year from now3 Pay/receive $12,100 two years from now


Example 4.2 (3 plans for repayment of $10,000 in three yearsat 10% per year)You owe $10,000 to your credit card company. You have 3plans to pay the balance back:

Plan 1: Pay interest due at end of each year and principal atend of third year.Year Amount owed Interest Principal payment Total pmnt.1 10,000 1,000 0 1,0002 10,000 1,000 0 1,0003 10,000 1,000 10,000 11,000

Total: 3,000 13,000


Plan 2: Pay principal and interest in one payment at the end ofthird year.Year Amount owed Interest Principal payment Total pmnt.1 10,000 1,000 0 02 11,000 1,100 0 03 12,100 1,210 10,000 13,310

Total: 3,310 13,310

Plan 3: Pay off the debt in 3 equal end-of-year payments.Year Amount owed Interest Principal payment Total pmnt.1 10,000 1,000 3,021 4,0212 6,979 697.9 3,323 4,0213 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 rateof 10%! (Why? You can answer it at the end of this chapter.)


Cash-flow diagrams

i = effective interest rate per interest periodN = number of compounding periodsP = the present sum of moneyF = future sum of moneyA = end-of-period cash flows in a uniform series for a specifiednumber of periods, starting at the end of first period andcontinuing through the last period.

In a cash-flow diagram:horizontal line represents time scale,arrows represent cash flows. Downward arrows representexpenses (negative cash flows or cash outflows) andupward arrows represent receipts (positive cash flows orcash 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.


General setting

Beginning of period 1

End of period 1 F

01 32 N-1 N

i % per period

P


Example 4.3

Show the cash flow diagrams for Plans 2 and 3 from the creditcard company’s viewpoint. Identify i, P, F, A.


Solution

1 2 3 = N0

End of Year (EOY)

P = $10,000

F = $13,310

Plan 2

1 2 3 = N0

End of Year (EOY)

P = $10,000

A= $4,021

Plan 3

i = 10% per year


Example 4.4

You are analyzing a project with five-year life. The projectrequires a capital investment of $50,000 now, and it willgenerate uniform annual revenue of $6,000. Further, theproject will have a salvage value of $4,500 at the end of the fifthyear and it will require $3,000 each year for the operations.Develop the cash-flow diagram for this project from theinvestor’s viewpoint.


Solution

$50,000

$6,000

$3,000

$6,000 $6,000 $6,000

$3,000 $3,000 $3,000

$4,500

20 3 4 5

$6,000

$3,000

1

End of Year (EOY)


Rules for performing calculations with cash flows

Three Rules for performing arithmetic calculations with cashflows:

1 Cash flows cannot be added or subtracted unless theyoccur 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).


Interest formulas relating P and F

P = Present Equivalent (Given)

F = Future Equivalent (Find)

21 … N − 1 N0

i = Interest rate per period

End of Period



Finding F when P is given

F = P(1 + i)N .

Notation: F = P(F/P, i%, N) where the factor in theparentheses is read "find F given P at i% interest perperiod for N interest periods".



P = Present Equivalent (Find)

F = Future Equivalent (Given)

21 … N − 1 N0

i = Interest rate per period

End of Period



Finding P when F is given

P = F(

11+i

)N= F(1 + i)−N .

Notation: P = F(P/F, i%, N) where the factor in theparentheses is read "find P given F at i% interest perperiod for N interest periods.


Example 4.5: You deposit $50,000 now in a savings account atan interest rate of 8% per year. How much will you have in thisaccount at the end of fifth year? Assume the interest rate isconstant at 8% for the next five years.Solution: P = 50, 000, i = 8% per year, N = 5F = P(F/P, 8%, 5) = 50, 000(1 + 0.08)5 = $73, 466

Example 4.6: How much should you deposit in a savingsaccount now so that you will have $100,000 in this account 6years from now? Assume that the interest rate is 5% per year.Solution: F = 100, 000, i = 5% per year, N = 6P = F(P/F, 5%, 6) = 100, 000(1 + 0.05)−6 = $74, 627


General setting with annuities (A)

A A A A A

N01 32 N-1

i % per period

FP


Interest formulas relating P,F, and A

Rule: Uniform (equal) receipts, A, at the end of each period forN periods at an interest rate of i% per period, where the firstreceipt occurs at the end of the first period and the last oneoccurs 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 cashflow A. Then sum them up.Future equivalent at the end of period N of the cash flow thathappens at the end of period 1: A(F/P, i%, N − 1)Future equivalent at the end of period N of the cash flow thathappens at the end of period 2: A(F/P, i%, N − 2)...........Future equivalent at the end of period N of the cash flow thathappens at the end of period N: A(F/P, i%, 0)


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] =N−1∑n=0

(1 + i)N−1[

1(1+i)

]n

The last term comprises a geometric series of the formar0 + ar1 + ar3 + ... + arN−2 + arN−1, where a = (1 + i)N−1,r = (1 + i)−1.


Note that the sum of the first N terms in a geometric series is:N−1∑n=0

arn = a(1−rN)1−r .

Therefore we have the following:N−1∑n=0

(1 + i)N−1[

1(1+i)

]n=

(1+i)N−1− 1(1+i)

1− 1(1+i)

= (1+i)N−1i

We find:

F = A[

(1+i)N−1i

]Notation: F = A(F/A, i%, N).


Finding A when F is given

From F = A[

(1+i)N−1i

],

A = F[

i(1+i)N−1

]Notation: A = F(A/F, i%, N).


Example 4.7: You plan to deposit $2,000 to your savingsaccount at the end of every month for the next 15 monthsstarting from the next month. If the interest rate you can earn is2% per month how much money will accumulate immediatelyafter your last deposit at the end of the 15th month?Solution: A = 2, 000, i = 2% per month, N = 15 months. F =?F = A

[(1+i)N−1

i

]= 2, 000

[(1.02)15−1

0.02

]= $34, 587

Example 4.8: What uniform monthly amount should you depositin your savings account at the end of each month for thefollowing 10 months in order to accumulate $75,000 at the timeof the 10th deposit? Assume that the interest rate you can earnis 4% per month and the first deposit will be made next month.Solution: F = 75, 000, i = 4% per month, N = 10 months. A =?A = F

[i

(1+i)N−1

]= 75, 000

[0.04

(1.04)10−1

]= $6, 247


Finding P when A is given

Given P = F(P/F, i%, N) and F = A(F/A, i%, N), we can deriveP = A(P/A, i%, N) as follows:

P = A(P/A, i%, N) = [A(F/A, i%, N)](P/F, i%, N)⇒A[

(1+i)N−1i

](1 + i)−N

P = A[

(1+i)N−1i(1+i)N

]Notation: P = A(P/A, i%, N).

Finding A when P is given

From P = A[

(1+i)N−1i(1+i)N

],

A = P[

i(1+i)N

(1+i)N−1

]Notation: A = P(A/P, i%, N).


Example 4.9: How much should you deposit to your savingsaccount now at an annual interest rate of 10% to provide for 5end-of-year withdrawals of $15,000 each?Solution: A = 15, 000, i = 10% per year, N = 5 years. P =?P = A

[(1+i)N−1

i(1+i)N

]= 15, 000

[(1.1)5−10.1(1.1)5

]= $56, 862

Example 4.10: You plan to borrow a loan of $100,000 whichyou will repay with equal annual payments for the next 5 years.Suppose the interest rate you are charged is 8% per year andyou will make the first payment one year after receiving theloan. How much is your annual payment?Solution: P = 100, 000, i = 8% per year, N = 5 years. A =?A = P

[i(1+i)N

(1+i)N−1

]= 100, 000

[0.08(1.08)5

(1.08)5−1

]= $25, 046



Deferred annuities

Table 1: Ordinary annuitiesYear 0 1 2 3 ... N

Cash flow - A A A ... A

Table 2: Deferred annuitiesYear 0 1 2 3 ... J J+1 J+2 ... N

Cash flow - - - - ... - A A ... A

What is the present equivalent of the deferred annuity shown inTable 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 inTable 2 as of time N (FN)?

FN = A(F/A, i%, N − J)


Example 4.11

i=10 % per period

1 2 38

56 7

49 10

11 12 13 14 15

4,000..……..4,000

1,000……..………………….1,000

5,000

2,000 2,000……..

P0=?

0


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.


Arithmetic gradient series

Cash flow series that increases or decreases by a constantamount (G) from one period to the next. See the cash flowdiagram in the next slide.


Cash flows in arithmetic gradient series

G

2G

(N-2)G

3G

(N-1)G

0 N1 32 4 N-1

i% per period


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 = G

i [(1 + i)N−1 +(1 + i)N−2 +(1 + i)N−3 + ...+(1 + i)1 + 1]− GNi

F = Gi

[(1+i)N−1

i

]− GN

i ⇒ F = Gi (F/A, i%, N)− GN

i .


Finding P when G is given

We know the formula to find F given G, and the formula to findP given F. We can use these formulas to find (P/G, i%, N) asfollows:P =

(Gi

[(1+i)N−1

i

]− GN

i

)(P/F, i%, N)

Recall (P/F, i%, N) = (1 + i)−N

Hence,

P = G(

1i

[(1+i)N−1

i(1+i)N − N(1+i)N

]).


Finding A when G is given

We know the formula to find F given G, and the formula to findA given F. We can use these formulas to find (A/G, i%, N) asfollows:A =

(Gi (F/A, i%, N)− GN

i

)(A/F, i%, N)

⇒ A = Gi −

GNi (A/F, i%, N)

(A/F, i%, N) = i(1+i)N−1

Hence,

A = G[

1i −

N(1+i)N−1

].


Example 4.12

1 32 50

12,000

14,000

18,000

4

16,000

1 32 50

b) A=?

4

10,000

1 32 50

a) P=?

4

i=10% per period


Solution

$12,000

20 3 4 5$10,000

1

$14,000

$16,000

$18,000

0

$10,000

1 2 3 4 5 0 1 2 3 4 5

$2,000

$4,000$6,000

$8,000

+


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.


Geometric gradient series

Cash flow series that increases or decreases by a constantpercentage (f ) from one period to the next. See the cash flowdiagram in the next slide.Note that Ak = (1 + f )Ak−1 and f = Ak−Ak−1

Ak−1, where f is the

constant rate of change.


Cash flows in geometric gradient series

i% per period

0 1 3 4 N-12 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


Interest formulas

Derive the expression for the present equivalent (P) of thegeometric 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−1A1,2 ≤ k ≤ N.Hence,P = A1

[(1 + i)−1 + (1 + f )(1 + i)−2 + ... + (1 + f )N−1(1 + i)−N

]⇒ P = A1(1 + i)−1

[1 +

(1+f1+i

)+(

1+f1+i

)2+ ... +

(1+f1+i

)N−1]


Using the summation formulaN−1∑n=0

xn = (1−xN)1−x , when x 6= 1, we

can derive the following expression:(Note that x =

(1+f1+i

), and x 6= 1 reduces to i 6= f )

P =

{A1[1−(P/F,i%,N)(F/P,f%,N)]

i−f f 6= iA1N(P/F, i%, 1) f = i

Note that f can be negative. For example, when f = −0.1,(F/P, f %, N) = (1− 0.1)N .


Example 4.13

(Blank&Tarquin) Chemical Engineers at a Coleman Industriesplant in the Midwest have determined that a small amount of anewly available chemical additive will increase the waterrepellency of Coleman’s tent fabric by 20%. The plantsuperintendent has arranged to purchase the additive througha 5-year contract at $7,000 per year, starting 1 year from now.He expects the annual price to increase by 12% per yearthereafter for the next 8 years. Additionally, an initial investmentof $35,000 was made now to prepare a site suitable for thecontractor to deliver the additive. Use i=15% per year todetermine the equivalent total present worth of all these cashflows.


Solution

0 1 2 5 6 13

8)12.1(000,7$

…

…

… …

$7,000

i = 15% per year

$7,000(1.12)

$35,000

End of Year (EOY)


Solution

35, 000 + 7, 000(P/A, 15%, 4) +7,000[1−(P/F,15%,9)(F/P,12%,9)]

0.15−0.12 (P/F, 15%, 4) = $83, 232.


Example 4.14

10,0009,500

9,000

8,0008,500

1 32 50 4 6 7 8 9

i=10% per period

7,5007,000

2,000 2,500X X X X

1 32 54 6 7

X/3 X/3 X/3

8 90

X=?


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


Example 4.15

Suppose that the parents of a young child decide to makeannual deposits into a savings account, with the first depositbeing made on the child’s 5th birthday and the last depositbeing made on the 15th birthday. Then, starting on the child’s18th birthday, four end-of-year withdrawals will be made at theamounts of $2,000, $2,400, $2,800, and $3,200, respectively. Ifthe effective annual interest rate is 8% during this period oftime, what are the annual deposits in years 5 through 15?


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


Nominal and effective interest rates

You are given two 1-year fixed deposit saving plans. Theinterest 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?


In Plan 1, the interest rate "12% per year" is known as theeffective interest rate per year, the same as the one we usebefore.

In plan 2, the interest rate "12% per year compounded monthly"is known as the nominal interest rate per year with thecompounding period 1 month (or compounding frequency 12).


In general, the nominal and effective interest rates are relatedby

i =

(1 +

r(M)

M

)M

− 1 (NOM-EFF)

wherei: effective interest rate per period;r(M): nominal interest rate per period with compoundingfrequency M.

It is obvious that i = r(1).


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 annualinterest rate? b) Suppose your balance as of today (year 0) is$10,000. How much will you owe to the credit card company atthe end of 3 years if you do not make any payments during 3years?Solution:a) i =

(1 + 0.15

12

)12 − 1 = 16.07%b) Use formula (F/P,i%,N). We can use effective annual interestrate of i% and N=3.P(F/P, i%, N) = 10, 000(F/P, 16.07%, 3) = 15, 637Alternatively, we can use interest rate per month 0.15

12 = 0.0125and N=3(12)=36.P(F/P, i%, N) = 10, 000(F/P, 1.25%, 36) = 15, 639


Continuous compounding

What if M (number of compounding periods)→∞. What is theeffective (continuous) interest rate?We know that i =

(1 + r

M

)M − 1.lim

M→∞i = lim

M→∞

(1 + r

M

)M − limM→∞

1

Recall: limp→∞

(1 + 1

p

)p= e (natural logarithm base)

Let Mr = p

⇒ i = limp→∞

[(1 + 1

p

)pr]− 1 = lim

p→∞

[(1 + 1

p

)p]r− 1

⇒ i = er − 1We can obtain the interest formulas relating P,F,A, by using therelation i = er − 1.



Payment Period and Compounding Period

Payment Period (PP) and Compounding Period (CP) of cashflow:

Type 1: PP = CP (Done!)Type 2: PP > CPType 3: PP < CP


Type 2: PP > CP

The strategy in this case is to find the effective interest rate perPP 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 ofa LAN. What is the equivalent amount after the last payment, ifthese funds are taken from a pool that has been returning 10%per year compounded quarterly?


Solution

PP = 6 months; CP = 3 months.

The effective interest rate per 6-month i is given by

i =(

1 +0.05

2

)2

− 1 = 0.05063.

The equivalent amount after the last payment= 500(F/A, 5.063%, 14) = $9, 842.


Type 3: PP < CP; Case 1

Assumption: Interperiod cash flows earn interest.

Under the above assumption, the strategy is similar to the caseof Type 2 by finding the effective interest rate per PP. Differentfrom Type 2, we cannot apply the relation of the nominal andeffective interest rates in (NOM-EFF) directly. The detailedtreatment can refer to Example 4.18 below.


Example 4.18

Suppose you make $500 monthly deposits to a tax-deferredretirement plan that pays interest at a rate of 10% per yearcompounded quarterly. Compute the balance at the end of 10years.


Solution

PP = 1 month; CP = 3 months.

The effective interest rate per 1 month i is given by

i =(

1 +10%

4

)1/3

− 1 = 0.826%

The balance at the end of 10 years =$500(F/A, 0.826%, 120) = $101, 907.89.


Type 3: PP < CP; Case 2

Assumption: Interperiod cash flows earn no interest.

In this case,Deposits (negative cash flows) are all regarded asdeposited at the end of the compounding period.Withdrawals (positive cash flows) are all regarded aswithdrawn at the beginning of the compounding period.


Example 4.19

Consider Example 4.18 again. Suppose that money depositedduring a quarter (the compounding period) will not earn anyinterest. Compute the balance at the end of 10 years.


Solution

3 6 120

…

…

Months

0

A = $500

F

1 2 40

…

…

Quarters

0

A = $1,500

F

39


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.


Chapter 4

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