time value of money -math

7/17/2019 Time Value of Money -Math

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SM Nahidul Islam

Dept. of Finance & Banking (2nd batch)

Problem -1: hat is the! p"esent #alue (toda$) of the follo%ing cash flo%s at an inte"est "ate of 2

pe"cent pe" $ea"'a. "ecei#ed fi#e $ea"s f"om no%.b.

"ecei#ed each $ea" beginning one $ea" f"om no% and

ending ten $ea"s f"om no%.c.

"ecei#ed each $ea" beginning one $ea" f"om no% and continuing fo"e#e".

Solution:

a. Given that,

Futu"e #alue (F*) + Numbe" of $ea"s (n) + , $ea"s-nnual inte"est "ate (i) + 2 + .2/"esent #alue (/*) +'

e kno% that0

/* + F* (1+i)−n

+ (1+.12)−5

+ ,1.3

b. Given that

-mount "ecei#ed at the end of each $ea" (-) + Numbe" of $ea"s (n) + $ea"s-nnual inte"est "ate (i) + 2 + .2/"esent #alue (/*-) +'

We know that,

/*- + - [ 1−(1+i)−n

i ]

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Problems Solutions



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! [ 1−(1+.12)−10

.12 ]

+ 4 ,.1,2

+ ,1,.2

c. Given that,

-mount "ecei#ed at the end of each $ea" (-) +

-nnual inte"est "ate (i) + 2 + .2

/"esent #alue of a pe"petual annuit$ (/*) +'

e kno% that0

/"esent #alue of a pe"petual annuit$ (/*) + A

i

+$100

.12

+ 566.66

Problem-": hat is the #alue of the follo%ing in#estments fi#e $ea"s f"om no% at an inte"est "ate of

5 pe"cent pe" $ea"0 2 pe"cent pe" 7ua"te"0 o"2

3 pe"cent pe" month depending on the compounding

pe"iod'a. in#ested toda$ %ith inte"est compounded annuall$.b. l in#ested toda$ %ith inte"est compounded 7ua"te"l$.c. in#ested toda$ %ith inte"est compounded monthl$.

Solution:

a. Given that,

/"esent #alue (/*) + Numbe" of $ea"s (n) + , $ea"s-nnual inte"est "ate (i) + 5 + .5Futu"e #alue (F*) +'




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We know that,

F* + /* (1+i)n

+ (1+.08)5

+ 4 .3186 + 31.86

b. Given that,

/"esent #alue (/*) + Numbe" of $ea"s (n) + , $ea"s-nnual inte"est "ate (i) + 5 +.5 Numbe" of times pe" $ea" inte"est is compounded (m) + 3

Futu"e #alue (F*) +'We know that,

F* + /* (1+i /m)n×m

+ (1+.08/4 )5× 4

+ 4 (1+.02)20

+ 4 .35,8 + 35.,8

c. Given that,

/"esent #alue (/*) + Numbe" of $ea"s (n) + , $ea"s-nnual inte"est "ate (i) + 5 +.5 Numbe" of times pe" $ea" inte"est is compounded (m) + 2Futu"e #alue (F*) +'

We know that,

F* + /* (1+i /m)n×m

+ (1+.08/12)5× 12

+ 4 (1+.0066)60

+ 4 .3585 + 35.85

Problem-#: Find the futu"e #alue of the follo%ing in#estments9 the inte"est "ate is 5 pe"cent pe" $ea"0

compounded annuall$:a. is in#ested each $ea" beginning one $ea" f"om no% and continuing th"ough $ea" 0

%hen the p"oceeds a"e %ithd"a%n.




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b.

is in#ested each $ea" sta"ting toda$ and continuing th"ough $ea" 0 %hen the p"oceeds

a"e %ithd"a%n.c.

is in#ested each $ea" beginning one $ea" f"om no% and continuing th"ough $ea" 8. ;he

p"oceeds a"e to be %ithd"a%n in $ea" .

Solution:

a. Given that,

-mount in#ested at the end of each $ea" (-) + Numbe" of $ea"s (n) + $ea"s-nnual inte"est "ate (i) + 5 +.5Futu"e #alue of o"dina"$ annuit$ (F*-) +'

We know that,

F*- + - [(1+i)n−1

i ]

+ [(1+0.08)10−1

0.08 ]

+ 4 3.3511

+ 335.11

b. Given that,

-mount in#ested at the sta"ting of each $ea" (-) + Numbe" of $ea"s (n) + $ea"s-nnual inte"est "ate (i) + 5 +.5

Futu"e #alue of annuit$ due (F*-) +'

We know that,

F*- + - [(1+i)n−1

i ] (<i)




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+ [(1+0.08)10−1

0.08 ] (<.5)

+ 4 ,.13,,

+ ,13.,,

c. Given that,

-mount in#ested at the sta"ting of each $ea" (-) + Numbe" of $ea"s (n) + 8 $ea"s-nnual inte"est "ate (i) + 5 +.5Futu"e #alue (F*) +'

We know that,

Futu"e #alue of o"dina"$ annuit$ (F*-) + - [(1+i)n−1

i ]

+ [(1+0.08)9−1

0.08 ]

+ 4 2.35,

+ 235.,

;otal futu"e #alue + 235.,(1+.08)1

+ 635.1,

Problem-$:




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a. If $ou put into a sa#ings ce"tificate toda$ pa$ing inte"est of pe"cent pe" $ea"0 ho%

much mone$ %ill $ou ha#e in the account afte" si= $ea"s if no %ithd"a%als a"e made f"om it

until then' >o% much of this is inte"est'

b. Suppose $ou %ithd"a% the inte"est e#e"$ $ea". hat %ill be $ou" total ea"nings' h$ does thisdiffe" f"om the inte"est ea"ned in a'

Solution:

a. ?i#en that0/"esent #alue (/*) + Numbe" of $ea"s (n) + 1 $ea"s-nnual inte"est "ate (i) + + .Futu"e #alue (F*) +'

We know that,

F* + /* (1+i)n

+ (1+.10)6

+ 4 .1 + .1

;otal inte"est + .1 !

+ .1

b. ;otal ea"nings + 4 4 1

+ 1

;his diffe"s f"om the inte"est calculated in pa"t a b$ .1 (.1 @ 1) because the inte"est in pa"t a is

calculated on the compounded inte"est "ate but in pa"t b the inte"est is calculated on the simple inte"est

"ate.

Problem -%:

>o% much mone$ should $ou put in a sa#ings account at the end of each month in o"de" to ha#e0 in the account at the end of ten $ea"s' -ssume that the same amount %ill be sa#ed each

month and that the account pa$s1

2 pe"cent inte"est pe" month.

Solution:




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Given that,

Futu"e #alue of o"dina"$ annuit$ (F*-) + 0 Numbe" of $ea"s (n) + $ea"s

Numbe" of times pe" $ea" inte"est is compounded (m) + 2-nnual inte"est "ate (i) + 1

Monthl$ inte"est "ate (iAm) +1

2 + .,

-mount in#ested at the end of each month (-) +'

We know that,

- + F*-

[

i /m

(1

+i

/m

)

n×m

−1

] + 0 [ .005

(1+.005)10×12−1 ]

+ 1.2

Problem & ':

Solution:




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Problem -(:




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Solution:




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Problem- ):


*lternative +:

ash flo% st"eam:



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oan amortiation scheule: Coan Messineo bo""o%ed ,0 at a 3 annual "ate of inte"est to

be "epaid o#e" 6 $ea"s. ;he loan is amo"tied into th"ee e7ual0 annual0 end!of!$ea" pa$ments.

a. alculate the annual0 end!of!$ea" loan pa$ment.

b. /"epa"e a loan amo"tiation schedule sho%ing the inte"est and p"incipal b"eakdo%n of each of the th"ee loan pa$ments.

c. E=plain %h$ the inte"est po"tion of each pa$ment declines %ith the passage of time.

Solution:

a. - + ,0 ÷ (/*IF-306)

- + ,0 ÷ 2.622

- + 103,8.8,

alculato" solution: 1031.8

b. End of oan Beginning of /a$ments End of Gea"

Gea" /a$ment Gea" /"incipal Inte"est /"incipal /"incipal

103,8.8, ,0. 20. 306,8.8, 013.,

2 103,8.8, 013., 0358.1 308.63 ,0118.

6 103,8.8, ,0118. 86.1 ,0111.8

(;he diffe"ence in the last $ea"Hs beginning and ending p"incipal is due to "ounding.)

c. ;h"ough annual end!of!the!$ea" pa$ments0 the p"incipal balance of the loan is declining0 causing

less inte"est to be acc"ued on the balance.




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1. If $ou %ish to ha#e 0 ten $ea"s f"om no%0 ho% much mone$ must $ou in#est toda$ in a

sa#ings ce"tificate that pa$s 5 pe"cent pe" $ea"'

". -ssume that it is no% Canua"$ 0 2,. n Canua"$ 0 210 $ou %ill deposit 0 into a sa#ings

account that pa$s 5 pe"cent.a. If the bank compounds inte"est annuall$0 ho% much %ill $ou ha#e in $ou" account on Canua"$

0 28'b. hat %ould $ou" Canua"$ 0 280 balance be if the bank used 7ua"te"l$ compounding "athe"

than annual compounding'c. Suppose $ou deposited the 0 in 3 pa$ments of 2, each on Canua"$ of 210 20

250 and 28. >o% much %ould $ou ha#e in $ou" account on Canua"$ 0 280 based on 5

pe"cent annual compounding'. Suppose $ou deposited 3 e7ual pa$ments in $ou" account on Canua"$ of 210 20 250

and 28. -ssuming an 5 pe"cent inte"est "ate0 ho% la"ge %ould each of $ou" pa$ments ha#e to

be fo" $ou to obtain the same ending balance as $ou calculated in pa"t a'

#. Suppose M". J takes out a tk. ,0 fi#e $ea" loan at 8. ;he loan is pa$able at , annual

installments. /"epa"e the loan amo"tiation schedule.

$. Find the /* of the st"eam of cash flo%s sho%n in the follo%ing table. -mong the fi"mKs

oppo"tunit$ cost is 2.

Gea" ash

Flo%263,

2060301050

%. Lishi Singh has 0, to in#est. >is in#estment counselo" suggests an in#estment that pa$s no

stated inte"est but %ill "etu"n 20 at the end of 6 $ea"s.a. hat annual "ate of "etu"n %ill Lishi ea"n %ith this in#estment'b. Lishi is conside"ing anothe" in#estment0 of e7ual "isk0 that ea"ns an annual "etu"n of 5.

hich in#estment should he make0 and %h$'

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0ercise

time value of money -math

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