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33

1 + is

(6.9)

If R, =

1, then formula (6.7) can be obtained from formula (6.9). If / = j, then formula (6.9) becomes

L = E v% t=\

which is formula (6.8). Thus, the amortization method and the sinlcing fund method are equivalent if i = j.

It should be noted that we have implicitly assumed that the sinking fund deposit Rf - iL is positive. If it were negative, then it would mean that the payment in that year is not even sufficient to pay the interest on the loan. We would then have a negative sinking fund deposit, i.e. a withdrawal, from the sinking fund for that year. Example 6.14 is an illustration of this type.

Example 6.10 A borrower is repaying a loan at 5% effective with payments at the end of each year for 10 years, such that the payment the first year is \$200, the second year \$190, and so forth, until the 10th year it is \$110. (1) Find the amount of the loan. (2) Find the principal and interest in the fifth payment. 1. The amount of the loan is

L = 100a 2. We have:

+1( ) = 100(7.7217) + 10i5- = \$1227.83.

5 = 160.00

= lOOaj, + 10( )

/5 = iB = 100(1 -v*)+ 10(6-«5, )

= 100(1 - .74622) + 10(6 - 5.0757) = \$34.62 P5 = 5-/5 = 160.00 - 34.62 = \$125.38.

Example 6.11 Assuming the same payment pattern as in Example 6.10, find the amount of the loan if the borrower pays 6% effective on the loan and accumulates a sinking fund to replace the amount oftiie loan at 5% effective.

Using formula (6.9) the amount of the loan is

00»tt)1.05 + 10(P»)tT)1.os 1227.83 t,ng 89 1 + (.06 - .05) , 1 + (.01)(7.7217)

Balance 20,000 19,000 - 10,000 9000

5000

1000

Principal bterest

1000 600

10 1000 330 T

1000 300

1000 180

1000 60

20 1000 30

Figure 6.1 Time diagram for Example 6.12

Each year A pays \$1000 principal plus interest on die unpaid balance at 3%. The payments are shown in the lower portion of Figure 6.1. The price to at the end of the lOdi year is die present value of die remaining payments, i.e.

«51.05+ (l-05)«5).04l + C150a5,.05 + 30(Da)5,o5

1000

+ 30(1.05)-*(Da)5, 04] = 1000[4.3295 + (.78353)(4.4518)

150(4.3295) + 30

5 - 4.3295 .05

+ 30(.78353)

5-4.4518 .04

= \$9191

to die nearest dollar. The answer is less dian die outstanding loan balance of \$10,000,

since has a yield rate always in excess of 3 %.

Example 6.13 A borrows \$10,000 from and agrees to repay U wUh a series often installments at the end of each year such that each installment is 20% greater than the preceding installment. The rate of interest on the loan is 10% effective. Find the amount of principal repaid in the first three installments.

Using formula (4.39) we have

 [ -1 10,000 = Ri .1-.2

= 13.87182/?!

= 10.000 = \$720.89. 13.87182

This answer is less than the answer to Example 6.10(1), since these terms are less fevorable to the borrower.

Example 6.12 A borrows \$20,000 from and agrees to repay it with 20 equal annual installments of principal plus interest on the unpaid balance at 3% effective. After 10 years sells the right to futiire payments to C, at a price which yields 5% effective over the next 5 years and 4% effective over the final 5 years. Find the price which should pay to the nearest dollar.

We now compute the entries for the first three lines of the amortization schedule. For the first line

/, = / = (.1X10,000) = \$1000.00

p = R-[ = 720.89 - 1000.00 = -\$279.11.

What has happened is that the first payment is insufficient to pay interest on die loan, so the principal payment is negative. This results in an increase in the outstanding loan balance, i.e.

B, = - , = 10,000 + 279.11 = \$10,279.11.

The calculations for die second line of die amortization schedule are:

R2 = 1.2 , = (1.2)(720.89) = \$865.07 /2 = (Bj = (.1X10,279.11) = \$1027.91 P2 = R2-I2 = 865.07 - 1027.91 = - \$162.84 B2 = B, -P2 = 10,279.11 + 162.84 = \$10,441.95.

Continuing for the third line we have:

3 = 1.22 = (1.2)(865.07) = \$1038.08 /3 = /82 = (.1)(10,441.95) = \$1044.20

= 3 -/3 = 1038.08 - 1044.20 = - \$6.12 = = 10,441.95 + 6.12 = \$10,448.07.

Thus, die amount of principal repaid in die first diree installments is

p, + P2 + 3 = -279.11 - 162.84 - 6.12 = - \$448.07

which could also be computed as Bo - b3. The principal repaid in the fourth and succeeding installments is positive and the loan is ultimately amortized to zero at the end of ten years. This example illustrates negative amortization for the first three years of the loan.

Example 6.14 What is the maximum amount that A can borrow from if A is willing to make four successive cmnual payments of 100, 100,1000 and \ 1000, if is to receive 12% effective on the loan, and if A is to replace the \ amount of the loan in a sinking fund held by earning 8% effective?

Let L be the amount of the loan. Annual interest on the loan is then .12l. The four successive sinking fiind deposits are dien 100 - .12l, 100 - .12l, 1000 - .12l, and 10( - . 12l. Since the accumulated value of the sinking fiind must equal die original amount of the loan, we have

(100-.12z.)54, 08 + 900*21.08

ioo55, og + 9005j, 08

1 + A2s

41.08

(100)(4.5061) + (900)(2.08) .jjq 1 + (.12)(4.5061)

However, tius answer is subject to criticism. It is evident diat die payments by A for dw first two years are not sufficient to pay 12% on die amount of die loan. Thus, 1 sinking fund will be negative. It is not reasonable to assume diat diis negative sinking fiind earns 8% because diat would imply diat A could borrow money at 8%. Since holds the sinking fiind, it is much more reasonable to assume that A would have to borrow at 12%. Thus, die interest deficiencies during die first two years should be capitalized and added to die amount of die loan. Let L be die amount of die loan under these conditions. We dien have

B2 = Lil.llf - lOOij, ,2

B2 = 1.2544z, -212.

However, Bj is die increased amount of die loan which must be accumulated in die sinking ftind. Thus, we have

B2 = (1000- .1282)521.0

B2 =

iooo5

21.08

1 + .125

21.08

1 + (.12)(2.08)

The original loan is then calculated as b; + 212

1.2544

= \$1495.96.

Llhus, L is slightiy less dian L, as would be expected. .7 AMORTIZATION WITH CONTINUOUS PAYMENTS

It is possible to develop formulas for amortization of loans with continuous ayments. Such formulas have conceptual and theoretical value, but are not jpdely used in practical applications.

We first examme the case in which the rate of payment is constant, sider a loan of being repaid with continuous payments at the rate of 1 period over n periods. We can readily generalize formulas (6.1) and (6.2) find the outstanding loan balance at time t, 0 < t < n

Bf = aj (6.10)

; = 1 (l+O-Tl

(6.11)

Also, we know that the continuous payments are partially interest and illy repayment of principal. Let /, be die instantaneous rate at which

interest is being paid and P, be the instantaneous rate at which principal is being repaid at time t. By general reasoning, the generalizations of formulas (6.3) and (6.4) become

I, = b- B, (6.12)

= 1-5-S,. (6.13)

We will verify formulas (6.12) and (6.13) if we are able to show that

dt

since the instantaneous rate of decrease in the outstanding loan balance must equal the rate of principal repayment. If we differentiate formula (6.10), we have

d- d l-v"-

dt "-1 dt dO+O"

(1 +O"log,(l +0

(4.26)

dt 8 = -(1 +i)-" = -v"- = -1 = 65,-1 = -P,

from formula (6.13).

In Section 4.5 we developed the identity

d - ix-

and indicated that it was possible to give the formula a verbal & . The discussion immediately above provides that verbal inte etation.

We next consider a generalization in which the rate of payment varies. Assume that payments on the loan are being made continuously at the rate of Rf per period. This notation is consistent with the notation used in Chapter 5 from the lenders vantage point. The original amount of the loan is the present value of the payments, i.e.

L = = j\Rds. (6.14)

The two formulas for the outstanding loan balance then become

,s-t

R.ds

= BqH+D- R(l+i)-ds.

(6.15) (6.16)

It is interesting to note that formula (6.16) is identical to formula (5.18), which was developed in a different context, upon making the standard substitution R = -C.

The above formulas have assumed a constant force of interest. It is relatively straightforward to generalize the above results even further by assuming a varying force of interest. The resulting formulas would be analogous to formula (5.19) and are left as exercises.

Example 6.15 Show that

v"ajdt

where unprimed symbols are based on force of interest and primed symbols are based on force of interest b. We have

varrdt =

f vv"drdt. 0 Jo

We now reverse the order of integration as illustrated in Figure 6.2.

Figure 6.2 Area of btegration for Example 6.15 The double integral now becomes

.n-l , ,n

vv"drdt = I

0 JO

vv"dtdr