ELS CONTINGUTS HISTÒRICS. QUINA HISTÒRIA ENSENYAR

An annuity lasting n interest periods with payments at the end of each interest period is called an annuity-immediate. Students are often confused by the term annuity-immediate as they view “immediate” as meaning “occurring without delay.”

However, when we speak of an annuity-immediate, the “immediate” signifies “of or near the future time.” Just as the immediate successor to the monarchy must wait until the present king or queen steps down or dies, the owner (the annuitant) of an annuity-immediate must wait until the end of one period to receive his or her first annuity payment. This is in contrast to the annuitant of an annuity-due who is due to receive a payment as soon as the annuity begins. We introduce annuities-due in Section (3.3).

By a basic annuity-immediate, we mean an annuity-immediate with level pay-ments, each of which is equal to 1. Thanks to the distributive law, the value at a given time of a level annuity-immediate with payments of Q is just Q times the value of the basic annuity-immediate.

Provided that the growth of money is governed by an increasing accumulation function, receiving a basic annuity-immediate lasting n periods is less advantageous than receiving $n immediately. The present value of the basic annuity immediate lasting n periods equals the sum of the present value of the n end-of-period payments, namely

v.1/ C v.2/ C v.3/ C


C v.n/:

When we have compound interest at an effective rate of i per interest period, v.t/ D .1 C i/ ^t D v^t;and this sum is the geometric series v C v²C v³C


C vⁿ. EXAMPLE 3.2.1

Problem:Marguerite receives an allowance of $20 at the end of each month. Her four-year old brother Stevie receives an allowance of $1 at month’s end. Find the value on January 1 of each of their allowances for the three-month period consisting of January, February, and March, using a monthly effective interest rate of .5%.

Solution Stevie receives $1 at the end of each of the three months. The January 1 value of his end of January payment is $1=1:005  $:99502, of his end of February payment is $1=.1:005/² $:99007, and of his end of March payment is $1=.1:005/³

 $:98515. The sum of these is $1=1:005 C $1=.1:005/²C $1=.1:005/³  $2:97.

Each month big sister Marguerite receives $20, so the January 1 value of her three allowance payments is $20=1:005C$20=.1:005/²C$20=.1:005/³D 20.$1=1:005C

$1=.1:005/²C $1=.1:005/³/  $59:40. Note that $59:40 D 20  $2:97. 

We next recall that there is a nice algebraic expression for the sum of a geometric series. Specifically,

(3.2.2) c C cr C cr²C


C cr^{n 1}D c.1 rⁿ/

1 r for r ¤ 1:

This is easy to derive. Let S D c C cr C cr²C


C cr^{n 1}, then rS D cr C cr²C c r³


C crⁿ;and .1 r /S D S rS D .c C cr C cr²C


C cr^{n 1}/ .c r C c r²C cr³


C crⁿ/ D c c rⁿD c.1 rⁿ/. Solving for S , the left-hand side of (3.2.2), we obtain^{c.1 r}_{1 r}^n/:

When c D r D v, since v.1 C i/ D 1, formula (3.2.2) gives

(3.2.3) v C v²C v³C


C vⁿD v.1 vⁿ/

1 v D .1 C i/v.1 vⁿ/

.1 C i/.1 v/ D 1 vⁿ i : The present value of the basic annuity-immediate lasting n interest periods is used frequently. If the accumulation function to be used is clear, the present value is denoted by the annuity symbol a_n. If a compound interest accumulation function is operative, and the effective interest rate for the payment period is i , then you often write a_{n i} instead of just a_n. This allows you to emphasize the effective interest rate used. The annuity symbol a_{n i}is read “a angle n at interest rate i .” Equation (3.2.3) may now be rewritten

(3.2.4) an i D v C v²C v³C


C vⁿD 1 vⁿ i :

The identity an iD ^{1 v}i ⁿ can be rewritten as the present value relation 1 D ia^{n i}C 1vⁿ; a unit invested for n years will produce end-of-year interest of i for each of the nyears, as well as the return of the unit investment at the end of the n years.

IMPORTANT FACT 3.2.5

With an annuity-immediate, you must wait one period to start receiv-ing payments. The present value of a basic annuity-immediate lastreceiv-ing n periods is denoted by an. It gives the value of the annuity one period before the first payment.

It is also useful to have a symbol giving the accumulated value of a basic annuity-immediate lasting n periods at the time of its final payment. The annuity sym-bol sn or sn i denotes this accumulated value. Of course the latter is only used when the accumulation is governed by the compound interest accumulation function a.t / D .1 C i/^t. Read the symbol sn ias “s angle n at interest rate i .”

IMPORTANT FACT 3.2.6

With an annuity-immediate lasting n periods, the final payment is made at time n. The accumulated value of a basic annuity-immediate lasting nperiods at the time of the last payment is denoted sn.

Note that sn and an each measure the value of the same sequence of payments, but sn measures the value n interest periods later.

PAYMENT: 1 1 1


1

TIME: 0 1 2 3 n

VALUE: an sn

FIGURE (3.2.7)

When a.n/ D .1 C i/ⁿ, we have the important equalities (3.2.8) sn iD .1 C i/ⁿan i and an iD vⁿsn i: More generally,

(3.2.9) sn D a.n/aⁿ and an D v.n/sⁿ:

It follows from (3.2.4) and (3.2.8), or directly from the definition of sn i, that

(3.2.10) sn i D .1 C i/^{n 1}C .1 C i/^{n 2}C


C 1 D .1 C i/ⁿ 1

i :

Here .1 C i/^{n k} is the accumulated value at time n of the k-th payment.

EXAMPLE 3.2.11 An unknown present value

Problem:Candace wishes to buy a new car. Her budget allows her to make monthly payments of $500, and she has saved $1,800 for a down payment. Candace quali-fies for a 36-month auto loan which has a nominal interest rate of 4.8% convertible monthly. How expensive a car can Candace buy?

Solution The monthly interest rate is ^i.12/₁₂ D 4:8%=12 D :4% D :004, and Can-dace is prepared to make 36 successive end-of-month payments of $500. The present value of her payments must equal the balance due on the car after the down pay-ment. Therefore, she can take out a loan for $500a_{36 :004} D $500

h1 .1:004/ ³⁶ :004

i



$16;732:94. Since she has $1,800 for a down payment, she can buy a car costing

$16;732:94 C $1;800 D $18;532:94. 

EXAMPLE 3.2.12 An unknown contribution

Problem:Seth wishes to borrow $4,400 so he can pay his tuition. He qualifies for a two-year loan with a monthly effective interest rate of .25% and level payments.

Find the amount of Seth’s monthly payments under this loan.

Solution The loan lasts twenty-four months. Therefore, if Q is the payment amount, the time 0 equation of value stating that his payments have present value $4,400 is

$4;400 D Qa24 :0025. So, Q D a^$_{24 :0025}^4;400 D .$4;400/.:0025/

1 .1:0025/ ²⁴  $189:1173327: Since Seth’s monthly payments must be an integral number of cents, they are $189.12.

These payments will in fact repay a loan of $4,400.06 since $189:12a_{24 :0025} 

$4;400:062057114: (The display will just show “PV = 4,400.062057," but if you subtract 4,400.06, you obtain .002057114.) Thus the last payment may be reduced by :062057114.1:0025/²⁴  :065889578  :07: The last payment should therefore

be $189.05. 

We note that in the solution to Example (3.2.12), we used the following basic fact.

IMPORTANT FACT 3.2.13

If a loan of amount L is to be repaid by n level end-of-period payments of Q and the effective interest rate for the payment period is i , then Q D a^Ln i:

We also note the following useful algorithm.

ALGORITHM 3.2.14

Suppose that a loan of amount L is to be repaid by n end-of-period payments, the effective interest rate for the payment period is i , and it is stipulated that the payments be level except that the last payment be slightly reduced if that is necessary so that the borrower does not repay more than the loan amount (rounded to the nearest penny).

(1) Compute _a^L

n i and round this to the nearest penny. Call the result Q₁.

(2) Compute Q1an i and round to the nearest penny. If this is L, then a slightly reduced final payment is not needed and the payments are all equal to Q1. (3) Otherwise, _a^L

n i is not an integral number of cents and we round-up to the next cent. Call this Q2. The first n 1payments are Q2.

(4) Compute E D Q²an i L. This is the amount beyond the loan amount that level payments of Q2would pay off.

(5) The final irregular payment is R D Q² E.1 C i/ⁿ.

Henceforth, use Algorithm (3.2.14) [or the equivalent BA II Plus calculator Algo-rithm (3.2.19)] to find the payments of an annuity set up to have level payments unless an alternate method is specified.

EXAMPLE 3.2.15 An unknown contribution

Problem:Mr. Liang wishes to accumulate $200,000 in a college fund for his new-born daughter. He wishes the money to be available to her on her eighteenth birth-day, and he is prepared to make level contributions on her first through eighteenth birthdays. The fund has an annual effective discount rate of 5%. How large will Mr. Liang’s contributions need to be?

Solution The annual interest rate earned by the fund is i D 1 :05^:05 D 19¹. Denote the amount of Mr. Liang’s contributions by Q. Then, we need

$200;000 D Qs18₁₉¹ D Q .1 C19¹/¹⁸ 1

1 19

!

 28:8331196Q:

Therefore Q  28:8331196^$200;000  $6; 936:47. In fact, these payments will produce

$200,000.07, so Mr. Liang could reduce his last payment by seven cents.  The basic principle used in Example (3.2.15) was

IMPORTANT FACT 3.2.16

If an amount B is to be accumulated by n level end-of-period payments of Qand the effective interest rate for the payment period is i , then Q D s^Bn i:

EXAMPLE 3.2.17 An unknown number of contributions

Problem:Hyun wishes to contribute $100 at the end of each month to his savings account until the account has a balance of at least $3,000, at which time he will buy his grandmother’s piano. Hyun’s savings account has an annual interest rate of 5.4%

convertible monthly. How many months it will take Hyun to accumulate the needed money, and how much money will he then have saved?

Solution The monthly interest rate is 5.4%/12 = .45% = .0045, and Hyun makes

$100 end-of-month payments. We wish to find the smallest integer n such that

$100s_{n :0045}  $3; 000. But $100s^{n :0045} D $100

.1:0045/ⁿ 1 :0045



, so what we need is the smallest integer n with ^.1:0045/_:0045^{n 1}  30: This inequality is equivalent to .1:0045/ⁿ  1 C 30.:0045/ D 1:135. Taking natural logarithms, we need n 

ln1:135

ln1:0045  28:2. So, n D 29 and Hyun’s balance is

$100s_{29 :0045}D $100

.1:0045/²⁹ 1 :0045



 $3;090:32:

Another method would have been to make a guess as to the number of payments and calculate the balance resulting from that guess. Had we found a balance less than

$3,000, we would have increased our guess and tried again. Had we guessed 30 or more, the balance would have been over $3,200 and we would have decreased our

guess. 

The fundamental relation between an i and sn i is given by (3.2.8). There is a second connection that we now derive using (3.2.10) and (3.2.4). First note that

i C 1

sn i D i C i

.1 C i/ⁿ 1 D i Œ.1 C i/ⁿ 1 C i

.1 C i/ⁿ 1 D i.1 C i/ⁿ .1 C i/ⁿ 1 : Multiplying this last expression by ^v_vⁿn D 1 and using v.1 C i/ D 1, we find

i C 1

s_{n i} D i

1 vⁿ D 1 a_{n i}: We have derived the relation

(3.2.18) i C 1

sn i D 1 an i

:

It is instructive to give a second demonstration of Equation (3.2.18) that involves interest theory interpretations of the terms involved. This latter argument will be important in Chapter 5 when we study sinking fund loans. Consider a loan of amount 1, to be repaid at the end of n periods, and let i denote the effective interest rate per period. The loan may be repaid by n level end-of-period payments. Recalling (3.2.12), these level payments will each be _a¹

n i. An alternate way to repay the loan would be as follows. At the end of each period, pay the lender the interest for that period, namely i . Also, make a deposit to a savings account (paying interest at an effective rate i per payment period) so that at time n the balance in the savings account will be 1. According to (3.2.16), the deposit amounts should be_s¹

n i. Finally, at time n, remove the balance 1 from the savings account, and give it to the lender.

Each of these repayment scenarios have level costs to the borrower, utilize the interest rate i , and result in the loan being exactly repaid at time n. The level payments under the two schemes must be equal. This equality is (3.2.18).

The BA II Plus calculator has special buttons

N I/Y PV PMT FV

that, along with CPT, are useful for solving level annuity problems that have one unknown. These special keys appear on the third row of the BA II Plus calculator, and together they govern the so-called Time - Value - Money (or TVM)

worksheet. All five of the registers governed by the TVM buttons may simulta-neously be set to zero by pushing 2ND CLR TVM . This also happens when you press 2ND RESET but, as discussed in Chapter 0, the latter has additional consequences.

Recall that an annuity-immediate has end-of-period payments. In order for the TVM worksheet of the BA II Plus calculator to handle a problem involving a level annuity-immediate, it must be set to interpret payments as coming at the end of the period. This is the factory default setting, but if you have not just reset your calculator, you should confirm that your calculator is so designated by glancing at the display window. If the display contains “BGN" at the top of the right-hand side, the calculator is not as desired, and you should remedy the situation by pushing 2ND BGN 2ND SET 2ND QUIT . The calculator will be said to be in “END mode" when “BGN " is not displayed.

The registers PMT, PV, and FV accept dollar amounts, and the labels are abbreviations for payment, present value, and future value. One uses PMT to enter the dollar amount of each of the level annuity payments. This amount (and a cashflow amount in general) should be entered as positive if it is received (an inflow) and negative if it is paid out (an outflow). Of course whether a pay-ment is an inflow or an outflow depends on which party is being considered, but the important thing here is to take a consistent viewpoint, thereby minimizing the number of error messages and incorrect answers.

Choose a perspective, and then enter inflows as positive and outflows as negative.

If the annuity is to repay a loan , from the borrower’s perspective , the annuity payments will be out payments, while if the annuity is to accumulate money in a savings account, from the investor’s perspective, the annuity payments will be out payments. Sticking with a consistent viewpoint, in the first case there will be an inflow (the loan) at the beginning of the annuity and this should be entered in the register PV while zero should be entered in the FV register. In the latter case, the accumulated amount at the end of the annuity period (for an annuity-immediate, the time of the last payment) needs to go in FV as an inflow (one imagines the investor closing the account and withdrawing his money), while zero should be put in the PV . In these simple situations the entry in either PV or FV will be zero, but later there will be many examples where one enters nonzero numbers in both.

Registers I/Y, P/Y, and C/Y are all filled using the I/Y key, but P/Y and C/Yfirst require 2ND to be pushed to access their worksheet. Note that I/Y

is a shorthand for “interest rate per year" and is designed to store a nominal interest rate, P/Y stands for “payments per year," and C/Y indicates the number of interest conversion periods per year. The “per year" designation is somewhat confusing in that the user may choose a basic time period other than a year when choosing entries for registers I/Y, P/Y, and C/Y, for instance, a month if you have an effective monthly interest rate and monthly payments. It is essential that you consider the same basic period when filling all three of these registers!

Use I/Y to enter a nominal interest rate convertible m times per period as a percentage. A convenient choice form depends on what sort of an interest rate you know. When an effective interest rate per payment period is handy, enter that rate as a percentage. In this case, you will wish the P/Y and C/Y registers to each hold the value 1. (You may prefer to keep the number 1 in the P/Y and C/Y registers, and convert to an effective interest rate per payment period if one is not provided. This choice makes your calculator seem more similar to the Texas Instruments BA-35 calculator and is the factory default setting on the BA II Plus Professional calculator. The factory default setting on some BA II Plus calculators for the P/Y register and also for the C/Y register is 12, which is perfect if you have monthly payments and a nominal interest rate convertible monthly.)

If you have a nominal interest rate convertiblem times per period, k pay-ments for that same period, and do not wish to convert to an effective interest rate per payment period, set the P/Y and C/Y registers to m and k respec-tively by first pushing 2ND P/Y , then enteringm, pushing # , enteringk, and finally pushing 2ND QUIT . For example, if you wish to have P/Y = 6 and C/Y=

4, depress

2ND P/Y 6 ENTER # 4 ENTER 2ND QUIT .

Ifm D k, you can skip pushing ^# and enteringk, since when a value is entered in P/Y , it is also automatically entered in C/Y . So, if you desire to have P/Y = 6and C/Y= 6, just push 2ND P/Y 6 2ND QUIT . Again note that the factory default entries for the registers P/Y and C/Y on the BA II Plus calculator (but not on the BA Plus Professional) are both 12. So, if you wish to store P/Y = 12 and C/Y = 12 on the BA II Plus calculator, you may accomplish this by simply keying 2ND P/Y 2ND CLR WORK . (A similar shortcut is available on the BA II Plus Professional if you want to enter 1 into the P/Y and C/Y registers.)

The register N is designed to hold the numberq of payments that the annuity has. This may be done directly by entering the numberq and then depressing N . (Alternatively, if you have an annuity lastingn periods with m payments per

period andm has been entered in the P/Y register, you may enter n and then push 2ND P/Y N to correctly fill the N register.)

The TVM worksheet is designed so that if values are entered in any order into the four of the registers and the key CPT is depressed, followed by the fifth TVM key, the value of the fifth TVM variable is displayed and simultaneously entered into the fifth TVM register. Of course this is only possible when consistent values are stored in the first four registers. If incompatible values are found, “ERROR 5"

is displayed. (A common cause of this error message appearing is for an inflow or outflow to be entered with the incorrect sign.) If the unknown is the interest rate, there may be a delay of a number of seconds before a display appears.

This is because the calculation requires an iterative method. It is possible that the calculator will fail to find a solution, even if one exists, if its limit of time for performing iterations has expired. In this case, “ERROR 7" will be displayed. Note that the TVM worksheet covers a situation where there is one borrower until the end of the annuity term, hence a unique yield rate.

Utilizing the Cash Flow worksheet is an alternate way of letting the BA II Plus calculator help solve annuity-immediate problems if the unknown is the present value of the sequence of annuity payments or the interest rate. (On the BA II Plus Professional calculator, the NPV subworksheet is more exten-sive, and the Cash Flow worksheet can also be used to compute unknown future values.) In the case of an unknown present value, the amount of the level payments needs to be entered as C01 and the number of payments is entered as F01. Then push NPV . The display will show “I =" and you need to enter the effective interest rate per payment period. Do this by entering the numerical value of the rate as a percentage and then depressing ENTER . Now push # , or "

so that “NPV =" appears. Pushing CPT will give you the previously unknown present value. If the interest rate per payment period is the unknown, enter the present value of the sequence of annuity payments as CFo, the negative of the amount of the individual level payments as C01, and the number of payments as F01. Now hit IRR CPT to display a computed value of the effective interest rate per payment period as a percentage. Again, it is possible to receive ERROR 5 if you have entered inconsistent values or ERROR 7 if the iteration fails to

so that “NPV =" appears. Pushing CPT will give you the previously unknown present value. If the interest rate per payment period is the unknown, enter the present value of the sequence of annuity payments as CFo, the negative of the amount of the individual level payments as C01, and the number of payments as F01. Now hit IRR CPT to display a computed value of the effective interest rate per payment period as a percentage. Again, it is possible to receive ERROR 5 if you have entered inconsistent values or ERROR 7 if the iteration fails to

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