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ponent minus 1, that is, by we have the sum required, = 20(}})" b—20b. Wherefore the principal sought is (}})"a + 20(}})" b—20b = (}})" × (a + 20b) — 20b.

To resolve this formula we must separately calculate its first term (21)” × (a + 20b), which is nL. + L. (a + 20b), for the number which answers to this logarithm in the tables will be the first term, and if from this we subtract 206 we have the principal sought.

Suppose a principal of 1000l. placed out at 5 per cent, compound interest, and to it there be annually added 100%. besides its compound interest, and it be required to know to what it will amount at the end of 25 years. Here a=1000, b=100, n=25⚫ and the operation is as follows:

L.=0-021189299

Multiply by 25 we have 25L.=0.5297324750

L.(a+20b)=3.4771213135

=40068537885

The first part or number which answers to this logarithm is 10159-11; from which if we subtract 206=2000 we find the principal in question to be after 25 years 8159.17.

If it be required to know in how many years a principal of 1000l. under the above conditions would amount to 1,000,000l.; let n be the number of years required, and since a=1000,b=100, the principal at the end of n years will be ()" (3000)—2000, which sum must make 1,000,000l. ; whence results this equation:

3000 ()"-2000=1000000 Adding to both sides 2000 we have 3000 (3) n 1002000 Dividing both sides by 3000 we have ()=334 Using logarithms we have nL.-L.334, and dividing by L., we obtain ==

0-0211893

L.334 Now

L.

L.334-2.5237465 and L. =0.0211893, wherefore n= 2-5237465 If, lastly, the two terms of this fraction be multiplied by 10000000, we shall have n = 2-5237465,

211893 equal to one hundred and nineteen years one month and seven days, which is the time wherein the principal of 1000l. will be increased to 1,000,000%. In the case of an annual decrease instead of increase of the capital by a certan sum, we shall have the following gradations as the values of a, year after year, the interest being at 5 per cent., and, representing by b the sum annually abstracted from the principal,

After 1 year it would be

After 2 years

After 3 years

After n years

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This principal evidently consists of two parts, one whereof is (a, and the other to be subtracted therefrom, taking the terms inversely, forms a geometrical progression, as follows: b + (?!)b + (?})2 b + (21)3b + . . . .

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The sum of this progression has already been found 20 (31)"b-20b; if, therefore, this be subtracted from (2)", we have the principal required after n years = (})" (a−20b) + 20b. For a less period than a year, the exponent n becomes a fraction; for example, 1 day =5, 2 days = 385, and so on. It often happens that we wish to know the present value of a sum of money payable at the end of a number of years. Thus, as 20 pounds in ready money amount in a twelvemonth to 21 pounds, so, reciprocally, 21 pounds payable at the end of a year can be worth only 20 pounds. Therefore, if a be a sum payable at the end of a year, the present value of it is a. Hence, to find the present value of a principal a at the end of a year, we must multiply by ; to find its present value at the end of two years, it must be multiplied by (2)2a; and, in general, its value n years before the time of payment will be expressed by (29)"a.

Thus, suppose a rent of 100% receivable for 5 years, reckoning interest at 5 per cent., if we would know its value in present money, we have

For £100 due after 1 year, the present value is £95.239

after 2 years

90.704

after 3 years

86-385

after 4 years

82-272

after 5 years

78.355

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But for a great number of years such a calculation would become labor ous. may be facilitated as follows: - Let the annual rent a, commencing directly and con

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tinuing n years, it will be worth a+ (}} )a + (??)3a + (??)3a + (}} )*a . . . . + (37)′′a, which is a geometrical progression whose sum is to be found. We have therefore only to multiply the last term by the exponent, the product whereof is (29)**la, then subtract the first term, and the remainder is (39)**1u-a. Lastly, dividing by the exponent minus 1, that is, 'p or, which is the same, multiplying by-21, we have the sum requir‹ d, = — 21(34)**1a+21a, or 21a-21"1a, the value of which second term is easily calculated by logarithms.

CHAP. IV.

COMPOUND INTEREST AND ANNUITY TABLES.

As the architect is often called on to value property, we here add some practical observations on the subject, and a set of Tables for the ready calculation of such matters, which we shall at once explain.

TABLE FIRST Contains the amount of 17. put out to accumulate at compound interest for any number of years up to 100, at the several rates of 3, 4, 5, 6, 7, and 8 per cent. The amount of any other sum is found by multiplying the amount of 17. found in the table at the given rate per cent., and for the given time, by the pro osed sum. Example:-Required the amount of 755l. in 51 years, at 5 per cent. Amount of 17. for 51 years. at 5 per cent. is Given sum

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or 90901. 15s. 73%d.

12-040769
755

£9080 780595

TABLE SECOND contains the present value of 11. payable at the end of any number of years up to 100. The present value of any given sum payable at the expiration of any number of years is found by multiplying the present value of 17. for the given number of years, at the proposed rate per cent, by the given sum or principal.

Example: - Required the present value of 90907 payable 51 years hence, compound interest being allowed at 5 per cent.

By the table, the present value of 14 payable at the expiration of 51 years at 5 per cent. is

Given principal

or 7541. 188. 7d.

*083051
9090

£754.933590

TABLE THIRD contains the amount of an annuity of 11. for any number of years, and is thus used. Take out the amount of 17. answering to the given time and rate of interest : this multiplied by the given annuity will be the required amount.

Example:- Required the amount of an annuity of 271. in 21 years, at 5 per cent. compound interest.

Annuity of 17. in 21 years at 5 per cent.
Annuity given

or 9641. 88. 47d.

$5.719251

27

£964-419777

TABLE FOURTH shows the present value of an annuity of 17. for any number of years, &t 3, 4, 5, 6, 7, and 8 per cent., and is used as follows:

First, when the annuity commences immediately. Multiply the tabular number answering to the given years and rate of interest by the given annuity, and the product will be the value required. (This table provides for the percentage and to get back the principal. ) Example:-Required the present value of an annuity of 451., which is to continue 18 years, at the rate of 5 per cent.

Under 5 and opposite to 48 years is (years' purchase)
Annuity given

or 8137. 98. 5.

18.077157

45

£813-472065

Second, when the annuity does not commence till after a certain number of years. Multiply the difference between the tabular numbers answering to the time of commencement and end, at the proposed rate of interest, by the given annuity, the product will be the present value required.

Example.

An annuity of 401. is to commence 20 years hence, and is to continue 30 years; required its present value, the rate of interest being 4 per cent.

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TABLE FIFTH Contains the annuity which 14 will purchase, compound interest being allowed. The manner of using this table is obvious, from what has been said relative to the preceding tables.

Example.

What annuity for 10 years will 500% purchase, the rate of interest being 5 per cent.?

Under 5 and opposite to 10 is

Principal given

or 641. 15s. Ofd.

•1 29504 500

£64.752000

TABLES SIXTH, SEVENTH, and EIGHTH are for finding the value of annuities on single and joint lives, and were constructed by Simpson, on the London bills of mortality.

To find the value of an annuity for a single life, at a proposed rate of interest, within the limits of the table, take from Table VI. the number answering to the given age and proposed rate of interest, which multiplied by the given annuity, the product will be the value required.

Example.

What is the value of an annuity of 50l. upon a single life aged 40 years, according to the London bills of mortality, the rate of interest being 4 per cent.?

The value of an annuity of 17. for 40 years at 4 per cent. is -
Annuity

Value

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11.5

50

£575

To find the value of an annuity of two joint lives, multiply the number in Table VII. answering to the given ages, and at the proposed rate of interest, by the given annuity, and the product will be the required value.

Example.

What is the value of an annuity of 60%. for two joint lives, the one being 30 and the other 40 years, interest at 4 per cent.?

The number answering to 30 and 40 years at 4 per cent. is
Annuity

8.8

60

Value

£528-0

To find the value of an annuity for the longest of two given lives, proceed as directed in the case immediately preceding, but using Table VIII., and the product will be the value. Example.

What is the value of an annuity of 60%. for the longest of two lives, the one being 30 and the other 40 years, interest at 4 per cent.

The tabular number answering at 4 per cent. is
Annuity

Present value

15.9

60

£954.0

The first five tables which follow are printed from those of Smart; the remainder are from Simpson.

The calculations involving the valuation of annuities on lives are not very frequently imposed on the architect, but it is absolutely necessary he should be capable of performing them, as in the case of valuations of leases upon lives, which sometimes occur to him.

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