El patrimoni arquitectònic i l’imaginari

Problems concerning the value of an annuity whose payments form a geometric pro-gression are best solved by noting that the value of the annuity is the sum of the values of the individual payments of the annuity, and these values also form a geo-metric progression. So Equation (3.2.2), which gives the sum of a geogeo-metric series, may be used to compute the desired value of the annuity.

EXAMPLE 3.8.1

Problem: On June 15, 1975, Roy purchased an annuity-immediate with annual payments for twenty-five years. The first payment was $800 and the payments increased by 3% each year. The purchase price was based on an annual effective interest rate of 7%. Find this price.

Solution The price is the June 15, 1975 value of the annuity payments. The k-th payment occurred k years after the purchase was made and had amount

$800.1:03/^{k 1}: So, the value of this payment on June 15, 1975 was

$800.1:03/^{k 1}.1:07/ ^k:Therefore, the June 15 value of the annuity is X25 In Example (3.8.1), the 3% rate meant that each payment was for an amount 1.03 times that of the previous payment. On the other hand, to compute the present value of a payment, we multiply by an extra_1:07¹ beyond what we needed to compute the present value of the previous payment; this was because we need to bring back the value one more year. Hence, when looking at the present values of the payments, we found they formed a sequence where each term was^1:03_1:07times the previous term.

Their sum therefore was a geometric series.

More generally, suppose that an annuity has payments where each payment is 1 C g times its predecessor. Let i denote the effective interest rate for the payment period. If the annuity is an annuity-immediate lasting for n payment periods with initial payment P and i ¤ g, then the value of the annuity one period before the first

payment is and the value of the annuity one period before the first payment is given by

P .1 C i/ ¹

Note that if the situation is as described in the previous paragraph but i D g, then the calculation of the value of the annuity one period before the first payment is fine until we use formula (3.2.2) to sum the geometric series. But if i D g,

"

and therefore the value of the annuity one period before the first payment is nP .1 C i/ ¹.

SUMMARY 3.8.2

Suppose that an annuity has payments with each payment equal to .1 C g/ times its predecessor. Let i denote the effective interest rate for the payment period. If the annuity is an annuity-immediate last-ing for n payment periods, the initial payment is P , and i ¤ g, then the value of the annuity one payment period before the first payment is P ¹

Our next example involves a perpetuity rather than an annuity, and it has the further complication of a change in interest rate.

EXAMPLE 3.8.3

Problem:On January 1, 2002, Andrea inherits a perpetuity-immediate with annual payments. The first payment is $2,000 and after that the payments increase by 2%

each year. Find the value of this perpetuity on January 1, 2002 if the annual effec-tive rate of interest is 3% from January 1, 2002 through January 1, 2009 and 4%

thereafter.

Solution Indicating the payment amounts in thousands of dollars, a time diagram for this problem is as follows:

PAYMENT 2 2(1.02)


2.1:02/⁶ 2.1:02/⁷ 2.1:02/⁸

The January 1, 2002 value of the payments that occur in 2003 through 2009 is

$2;000.1:03/ ¹

The January 1, 2002 value of the deferred perpetuity that begins with the 2010

payment is

.1:03/ ⁷$2;000.1:02/⁷.1:04/ ¹

"

1 C

1:02 1:04

 C

1:02 1:04

2

C

#

D .1:03/ ⁷$2;000.1:02/⁷.1:04/ ¹ 1 1 ^1:02_1:04

!

 $93;398:65656:

Thus Andrea’s inheritance has a January 1, 2002 value of about $13;202:68688

C$93;398:65656  $106;601:34: 

3.9 ANNUITIES WITH PAYMENTS IN ARITHMETIC PROGRESSION Consider an annuity lasting n interest periods with a payment of P C Q.j 1/ at the end of the j -th interest period. The first payment is P and the payments increase by the constant amount Q each interest period, hence form an “arithmetic progression.”

We introduce the annuity symbol .IP;Qa/_{n i} to denote the present value of this an-nuity, and the annuity symbol .IP;Qs/_{n i} denotes its accumulated value at the time of the last payment.

PAYMENT: P P CQ P C2Q


P C.n 1/Q

TIME: 0 1 2 3


n

VALUE: .IP;Qa/_{n i} .IP;Qs/_{n i}

FIGURE (3.9.1)

Our next task is to derive a concise formula for the accumulated value .IP;Qs/_{n i}. The accumulated value .I_P;Qs/_{n i} is obtained by adding the accumulated values of the individual payments. Therefore,

.IP;Qs/_{n i} D P .1 C i/^{n 1}C .P C Q/.1 C i/^{n 2} C .P C 2Q/.1 C i/^{n 3}

C


C .P C .n 1/Q/:

(3.9.2)

Multiplying Equation (3.9.2) by .1 C i/; we have

.1 C i/.I^P;Qs/_{n i}D P .1 C i/ⁿC .P C Q/.1 C i/^{n 1} C .P C 2Q/.1 C i/^{n 2}

C


C .P C .n 1/Q/.1 C i/:

This may be combined with Equation (3.9.2) to obtain i .IP;Qs/_{n i} D .1 C i/.I^P;Qs/_{n i} .IP;Qs/_{n i}

D P .1 C i/ⁿC Q.1 C i/^{n 1}C Q.1 C i/^{n 2}C


C Q.1 C i/ .P C .n 1/Q/

D P

.1 C i/ⁿ 1 C Qh

.1 C i/^{n 1}C .1 C i/^{n 2}C


C .1 C i/ C 1 Qn

D P

.1 C i/ⁿ 1

C Q s^{n i} n
: So, dividing by i yields the equation

.IP;Qs/_{n i} D P

.1 C i/ⁿ 1 i

 CQ

i .sn i n/:

Recalling Equation (3.2.10), this may be rewritten as

.3:9:3/ .IP;Qs/_{n i} D P s^{n i}CQ

i sn i n
:

This completes our task of finding a simple formula for .IP;Qs/_{n i}. Moreover, if you multiply Equation (3.9.3) by vⁿ, you have the companion equation

.3:9:4/ .IP;Qa/_{n i}D P a^{n i}C Q

i .an i nvⁿ/:

When P D Q D 1, it is customary to write .Ia/n i for .IP;Qa/_{n i}and .I s/_{n i} for .IP;Qs/_{n i}. Then, according to (3.9.3),

.I s/_{n i} D .I^1;1s/_{n i}D s^{n i}C1

i.sn i n/:

On the other hand, sn iC 1

i.sn i n/ D i sn i C s^{n i} n

i D ..1 C i/ⁿ 1/ C s^{n i} n i

D s_{nC1 i} .n C 1/

i :

Recalling (3.3.10) s_{nC1 i} 1 D Rs^{n i}, we have

.3:9:5/ .I s/n i D s_{nC1 i} .n C 1/

i D Rs^{n i} n

i :

Multiplying (3.9.5) by vⁿyields

.3:9:6/ .Ia/n i D Ra^{n i} nvⁿ

i :

There is also a special notation for .IP;Qa/_{n i} and for .IP;Qs/_{n i}when P D n and Q D 1; that is, when the sequence of payments is n; n 1; n 2, : : : ; 2; 1: We let .Da/_{n i} D .I^{n; 1}/a_{n i} and .Ds/_{n i}D .I^{n; 1}s/_{n i}:The D reminds us that this is a decreasing annuity. Observe that (3.9.4) gives us

.Da/_{n i} D na^{n i} 1

i.an i nvⁿ/:

On the other hand, nan i

1

i.an i nvⁿ/ D ni an i an iC nvⁿ

i D n.1 vⁿ/ an iC nvⁿ

i D n an i

i :

Therefore,

.3:9:7/ .Da/_{n i}D n an i

i :

Of course, one can multiply this by .1 C i/ⁿto obtain a formula for .Ds/_{n i}. EXAMPLE 3.9.8

Problem:Susan receives an annuity for her eighteenth birthday. The annuity pays

$2,000 on her nineteenth birthday and then has an annual payment on each of her birthdays through her thirty-fifth birthday. Each year the payments increase by $500.

Find the value of the annuity on Susan’s eighteenth birthday, assuming the annual effective rate of interest is 4.2%.

Solution The desired value is .I$2;000;$500a/_{17 :042}. According to (3.9.4), this is equal to $2;000a_{17 :042}C ^$500:042.a_{17 :042} 17.1:042/ ¹⁷/. Calculating, we find that this is approximately $23;958:24444 C $42;050:18436  $66;008:43:  EXAMPLE 3.9.9

Problem: At the end of each year Mr. Dunn deposits $6,000 into an investment fund. The fund pays out interest each year at an annual effective interest rate of 6%.

Mr. Dunn is only able to reinvest this interest at an annual effective interest rate of 4%. Assuming Mr. Dunn continues to make his $6,000 deposits for twenty years, that he always immediately deposits all the interest paid out into the 4% account, and that

he makes no withdrawals, what is the accumulated value of Mr. Dunn’s investments at the time of the last deposit?

Solution Mr. Dunn makes twenty $6,000 annual payments into the investment fund. Refer to the times he made these as times 1, 2, . . . , 20. If k 2 f1; 2; : : : ; 20g, then during the year that begins at time k, Mr. Dunn earns 6% interest on k $6,000 deposits. Therefore, at the end of the .k C 1/-st year (which begins at time k), he earns k.:06/.$6;000/ D $360k. These earnings are deposited into his 4% savings account. Thus, during the nineteen-year period from time 1 to time 20, the deposits into the 4% account form an annuity-immediate which is $360 times the one whose accumulated value is denoted .I s/_{19 :04}. So, the balance in the 4% account at time 20 is $360.I s/_{19 :04} D $360_Rs

19 :04 19 :04



 $88;002:71. There is also 20.$6;000/ D

$120;000 in the 6% fund, so at the time of Mr. Dunn’s last deposit (time 20), the accumulated value of his investments is $88;002:71 C $120;000 D $208;002:71:  Heretofore, the studied annuities with payments in arithmetic progression have all been annuities-immediate. However, we can also define annuity symbols for annuities-due. Let .IP;QRa/n i denote the present value of an annuity lasting n in-terest periods of an annuity with a payment of P C Q.k 1/at the beginning of the k-th interest period, and let .IP;QRs/n idenote its value at the end of the n-th period.

PAYMENT: P P CQ P C2Q


P C.n 1/Q

TIME: 0 1 2


n 1 n

VALUE: .IP;QRa/n i .IP;QRs/n i

FIGURE (3.9.10)

Comparing Figures (3.9.1) and (3.9.10) and recalling (3.9.3), we see that

.IP;QRs/n iD .1 C i/.I^P;Qs/_{n i}D .1 C i/



P sn iC Q

i .sn i n/

 :

Since .1 C i/s^{n i} D Rs^{n i}(3.3.8) and d D _1Ciⁱ (1.9.4),

.3:9:11/ .IP;QRs/n i D P Rs^{n i} CQ

d.sn i n/:

Multiplying (3.9.11) by vⁿyields

.3:9:12/ .IP;QRa/n i D P Ra^{n i}CQ

d .an i nvⁿ/:

Introduce

.I Ra/n iD .I^1;1Ra/^{n i}; .I Rs/n i D .I^1;1Rs/^{n i}; .D Ra/n iD .I^{n; 1}Ra/^{n i}; .D Rs/n iD .I^{n; 1}Rs/^{n i}: If one multiplies equation (3.9.5) by .1 C i/, one finds

.3:9:13/ .I Rs/n iD s_{nC1 i} .n C 1/

d D Rs^{n i} n

d :

Similarly, from (3.9.7) one obtains

.3:9:14/ .D Ra/n iD n an i

d :

Note that .Ia/_{n i}C .Da/n i D .n C 1/a^{n i}, and .I Ra/n iC .D Ra/n i D .n C 1/Ra^{n i}. These equalities give us alternate derivations of equations (3.9.7) and (3.9.14) [see Problem (3.9.8)].

EXAMPLE 3.9.15

Problem:Alfonso has an annuity that will pay $4,000 on October 30, 2010 and has annual payments through October 30, 2024. The payments increase by $800 each year. As soon as Alfonso receives a payment, he will deposit it in a savings account with a 4% annual effective interest rate, and he will not make any withdrawals. What will be the balance in Alfonso’s 4% account on October 30, 2025?

Solution The annuity has fifteen payments and we are looking for an accumu-lated value one period after the final payment. When i D :04, d D 1:04^:04. The desired value is therefore .I$4;000;$800Rs/15 :04 D $4;000Rs15 :04C:04=1:04^$800 .s_{15 :04} 15/

 $83;298:12457 C $20;800.20:02358764 15/  $187;788:75:  Perpetuities with payments in arithmetic progression also deserve mention. In this case, we replace the n in our present value annuity symbols by 1. For example, let .IP;Qa/_{1 i} denote the present value of a perpetuity-immediate that has an ini-tial payment of P and payments increasing by Q each interest period, and .I Ra/_{1 i} signify the present value of an annuity-due with a payment of k in the k-th interest period.

We may take limits (calculus!) in (3.9.4) and (3.9.12). In each case, we will need lim_n!1nvⁿ. Using l’Hospital’s rule, we find

lim

n!1nvⁿD lim

n!1

n

.1 C i/ⁿ D lim

n!1

1

.1 C i/ⁿln n D 0:

Recalling (3.4.2), from (3.9.4) we obtain, .3:9:16/ .IP;Qa/_{1 i} D P a1 iC Q

i a_{1 i}D P i C Q

i²:

Likewise, referencing (3.4.2) and (3.4.3), (3.9.12) gives us

.3:9:17/ .IP;QRa/_{1 i}D P Ra1 iCQ

d a_{1 i} D P d C Q

id:

Equation (3.9.17) may also be derived by multiplying (3.9.16) by .1 C i/: You may choose just to memorize (3.9.16), but be forewarned that the expression for .IP;QRa/1 i is not just obtained by replacing the i in the expression for .IP;Qa/_{1 i} by a d .

EXAMPLE 3.9.18

Problem:Rafael purchases a perpetuity-immediate. The perpetuity pays $1,000 at the end of each of the first eleven years and then has payments that increase by $180 each year. So, the payment at time 11 is $1,000, and the payment at time 12 is $1,180.

The purchase price was determined using a 5% annual effective interest rate. Find Rafael’s purchase price.

SolutionThink of the perpetuity as a level annuity-immediate paying $1,000 for ten years, followed by a deferred perpetuity-immediate. The perpetuity is deferred for ten years, has an initial payment of $1,000, and has payments that increase by $180 each year. The ten-year annuity has a present value of 1;000a_{10 :05} $7;721:73 and the deferred perpetuity has a present value

.1:05/ ¹⁰.I$1;000;$180a/_{1 :05}D .1:05/ ¹⁰

$1;000

:05 C $180 .:05/²



D .1:05/ ¹⁰.$20;000 C $72;000/

 $56;480:02 :

So, the purchase price is $7;721:73 C $56;480:02 D $64;201:75.  This section contains formulas for computing present value of annuities with payments in arithmetic progression. In particular, we draw your attention to Equa-tion (3.9.4), along with the more specialized EquaEqua-tion (3.9.6). We now wish to explain how, given these equations, you may use your BA II Plus calculator to efficiently calculate the quantitites.Ia/n i, and.IP;Qa/_{n i}:

According to Equation (3.9.6), to calculate.Ia/n i; you just need to calculate Ra^{n i} nvⁿand then divide your answer by the interest ratei: This can be done in BGN mode by first enteringn in the N register and also in the FV register, i as a percent in the I/Y register, 1 in the PMT register, and then keying

CPT PV ; next divide byi as a decimal rather than as a percent.

EXAMPLE 3.9.19

Problem: Use the BA II Plus calculator to compute.Ia/_{180 :5%}: SolutionPut your calculator in BGN mode and key

1 8 0 N FV  5 I/Y 1 C= PMT CPT PV .

At this point you will have computed _{Ra180 :5%} 180.1:005/ ¹⁸⁰, and the display should read “PV = 45.74919544". To complete the calculation, divide by_{i D :005} by pressing   0 0 5 . This gives the result$9,149.839088.  Let’s now move on to the more general.IP;Qa/_{n i}: You may rewrite equation (3.9.4) as

in the PMT register,^Qi_n in the FV register,i in the I/Y register,n in the N register, and then pressing

CPT PV .

Here we have calculated the present value of ann-period annuity-immediate that paysP in the first period and whose payments increase by Q each subse-quent period. Should you wish to find the accumulated value at the end of the period, since Equation (3.9.3) is equivalent to

.IP;Qa/_{n i}D .IP;QRs/n i are perhaps most easily obtained by first obtaining .IP;Qa/_{n i} and .IP;Qs/_{n i} respectively, and then multiplying by.1 C i/:

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