comfort, he is not only not bound, but he ought not, to transmit it to his successor." The principle thus stated is directly opposite to that which, as we have above observed, regulates civil cases, namely, that the occupier must keep in repair whatever may have been erected during his occupation. The judgment just cited continues in these words : “ If the successor may recover damages from the executor after such things have been remored by the testator, there can be no doubt he in his turn must maintain it; and if he maintain it he must also restore, and even rebuild when decayed; so that the benefice might become permanently saddled with a useless burden.” The duty to remove such erections does not, however, appear quite to have been thrown upon the estate of the erector : the same judgment says, “ the case now supposed is that of an erection, which, if the deceased had left out of repair, his successor could not have maintained any action for dilapidations, which he himself therefore would not be bound to keep in repair, which imposes no burden on him, and which he may remove; for it would be unreasonable to hold that he might not remove, however useless or unsuitable to the living, or even inconvenient to the occupation of the arsonage or glebe, that which for one of these reasons he was not bound to keep in repair.” Finally we would quote from the same judgment: " with regard to an ecclesiastical benefice, the character and object of the building to which the chattel is attached, and the manner in which it has been so attached, seem of very great consequence in determining whether there was any intention to separate it permanently and irrevocably from the personal estate.” In this case the plaintiffs (executors) were held justified in removing the framework and sashes, valued at 300l., of two hot houses, and might apparently have removed also the brickwork, repairing any waste or dainage to the freehold. With respect to that damage we have already referred to this The real difference between civil and ecclesiastical dilapidations may be thus stated :One man takes certain premises, engaging to pay a rent in order to derive advantages out of them, but having no interest in the freehold. The other man receives a salary to do certain services, the use of the house being a portion of that salary. In the latter case, if for a man's own private convenience he lays out a large sum on the freehold, that expendi. ture will seriously affect his successor. if he have to be burdened with large and expensive erections or decorations suitable perhaps for one of an aristocratic family, but quite foreign to the babits of a future rector of the village coming in as an ordinary occupant. Such are the general principles of the law of dilapidations; these, in their application, generally impose upon the out-going occupant or his representatives the payment of a sum for which special provision is rarely made during the occupancy: the misery thus entailed is sometiines evaded in civil cases by the lessee, who parts with the remainder of a lease to any one who will give something for it; or who (if the lessor be not caretul) assigns it i a man of straw. case. CHAP. III. CALCULATION OF INTEREST. Interest, or the value of the use of money, is usually expressed per cent, or after the rate per hundred on the principal lent. Thus, if we put out 500 pounds sterling at 5 per cent., it signifies that for every hundred pounds the lender is to receive five pounds per annum during the continuance of the loan. The solution of this question, which is one merely of simple interest, is so obvious, that it is unnecessary further to detain the reader upon it; and we therefore pass on to compound interest, or interest upon interest, which arises from the principal and interest taken together, as it becomes due at the end of each stated time of payment. In the resolution of this question, we are to consider that 1001. at the end of a year becomes 1051. Let a = principal. Its amount at the end of the year is found by saying, if 100 gives 105, what will a give? and we answer 100 which may be also expressed ***a, or a + xbxan Thus, by adding its twentieth part to the original principal, we have the principal at the end of the first year; adding to this last its twentiethi, we know the amount of the given principal in two years, and so on. Hence the annual increases to the principal may be easily computed. Suppose, for instance, the principal of 10001, Expressing the values in decimal fractions, it will be worth 21a After 4 years 1215-506 1215.506 1276.281 &c. The method above exhibited would, however, in calculations for a number of years, become very laborious, and it may be abridged in the following manner. Let the present principal =a; now, since a principal of 201. will amount to 211. at the end of a year, the principal a will amount to ffxa at the end of that time. At the end of the following year the same principal will amount to 203 xa = (b) xa. This principal of two years will, the year after, amount to (3})$ xa, which will therefore be the principal of three years ; increasing in this manner, at the end of four years the principal becomes (30)* x a. After a century it will amount to (34) 100 x , and in general (46)" xa is the amount of the principal after n years; a formula serving to determine the amount of principal after any number of years. The interest of 5 per cent., which has been taken in the above calculation, determined the fraction }}. Had the interest been reckoned at 6 per cent. the principal a would at the end of a year be (185) * a; at the end of two years to (183)2 x a; and at the end of n years to (188) " x a. Again, if the interest be at 4 per cent. the principal a will, after » years, be (188) " x a. Now all these formulæ are easily resolved by logarithms; for if, according to the first supposition, the question be (3!)" x a, this will be L.(*)" + L.Q, and as ({})" is a power, we have L. (3b)"=nL. it: so that the logarithm of the principal required is = n < L.3}+ L.a, and the logarithm of the fraction ist=L. 21 - L.20. We shall now consider what the principal of 10001. will amount to at compound interest of 5 per cent. at the end of 100 years. Here n=100. Hence the logarithm of the principal required will be =100L.Ft + L.1000, calculated as under : L.21 =1.3222193 =1.3010300 L.j=0.0211899 100 L. = 2:1189300 5.1189300 = Logarithm of the principal required, from the characteristic whereof the principal must be a number of six figures, and by the tables it will appear to be 131,5011. In the case of a principal of 3452. at 6 per cent. for sixty-four years, we have a=3452 and n=64. Principal at the end of the first year therefore = 189 = Hence the logarithm of the principal sought=64 L. + L.3452, which will be found to amount to 143,7631. When the number of years is very great, errors of considerable magnitude may arise from the logarithms not being sufficiently extended in the decimal places; but as our object here is only to show the principle on which these calculations are founded, we do not think it necessary further to pursue that subject. There is another case which now requires our consideration ; it is that of not only adding the interest annually to the principal, but increasing it every year by a new sum =b. The original principal a would then increase in the following manner : After 1 year, zda + b After n years, (3!)"+(35)– 16+(39)"-26+. ... 16+B This principal evidently consists of two parts, whereof the first =() "a, and the other, taken inversely, forms the series b + 3}b +(35) 6 +(34)96+ (38)”-'. This last series is evidently a geometrical progression, whose exponent = 34. Its sum, therefore, will be found by first multiplying the last term (3})^= 16 by the exponent it, which gives (35)"6. Subtract the first term b, and we have the remainder (38)"6-b; and lastly, dividing by the ex 2-5237465 2-5237465, ponent minus 1, that is, by so we have the sum required, = 20(34)*6–206. Wherefore the principal sought is (3})"a + 20(3})"b— 206 = (3})" *(a + 206)– 20b. To resolve this formula we must separately calculate its first term (3)" x(a + 206), which is nL. f} + L.(a +206), for the number which answers to this logarithm in the tables will be the first term, and if from this we subtract 200 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 100besides 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.3}=0021189299 L.(a + 206)=3•4771213135 =4-0068537885 The first part or number which answers to this logarithm is 10159·11.; from which if we subtract 200=2000 we find the principal in question to be after 25 years 8159.11. 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 d=1000,b=100, the principal at the end of n years will be (3')"(3000) – 2000, which sum must make 1,000,0001. ; whence results this equation: 3000 (3)" - 2000=1000000 Adding to both sides 2000 we have 3000 (38)"=1002000 Dividing both sides by 3000 we have (3})=334 L.334 Using logarithms we have nLf=L.934, and dividing by L. we obtain n= Now L. 35 L.994=2.5237465 and L. ff=0·0211893, wherefore n= 0-0211893. If, lastly, the two terms of this fraction be multiplied by 10000000, we shall have n= 211893' equal to one hun. | dred and nineteen years one month and seven days, which is the time wherein the prin cipal of 10001. will be increased to 1,000,000L In the case of an annual decrease in. stead 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 a – 6 (3:)*a - 3-6 (?!)a-(35) h-331-7 After n years (34)*a-(31)–16—(34)"–....-(*)6-b. This principal evidently consists of two parts, one whereof is (3)"a, and the other to be subtracted therefrom, taking the terms inversely, forms a geometrical progression, as follows: 6+(?!)6+ (3})?6+ ({}) 36+ .... (3)"16 The sum of this progression has already been found = 20 (3)"6–206; if, therefore, this be subtracted from ("a, we have the principal required after n years =()"(a – 20b) + 206. For a less period than a year, the exponent n becomes a fraction; for example, 1 day =sts, 2 days = sås, 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 fa. Hence, to find the present value of a principal a at the end of a year, we must multiply by my; to find its present value at the end of two years, it must be multiplied by ()a; and, in general, its value n years before the tiine of payment will be expressed by (3)"a. Thus, suppose a rent of 1002 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 86-385 82.272 after 5 years 78.355 Sum of the five terms £432.955 So that in present money, the value is 4321. 19s. Id. But for a great number of years such a calculation would become labor ous. It may be facilitated as follows: — Let the annual rent = 2, commencing directly and con 90.704 tinuing » years, it will be worth a +(IP)a +(79)*a +(39)*a+(79'a ....+(!?)"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 (39)** 'a, then subtract the first term, and the remainder is (39)"+-a. Lastly, dividing by the exponent minus 1, that is, g'n or, which is the same, multiplying by-21, we have the sum required, =-21(39)*+la+21a, or 21a-21*+'a, the value of which second term is easily calculated by logarithms. CHAP. IV. 9090 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 11. 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 11, 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 7551. in 51 years, at 5 per cent. Amount of 11. for 51 years. at 5 per cent. is 12.040769 Given sum 755 or 90901, 158, 7%d. £9080-780595 Table Second contains the present value of ll. payable at t'e 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 ll. for the given number of years, at the proposed rate per cent., by the given sum or principal. Example:— Required the present value of 90902. payable 51 years hence, compound interest being allowed at 5 per cent. By the table, the present value of 12 payable at the expiration of 51 years at 5 per cent. is *089051 Given principal or 7541. 188. 716d. £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 1!. 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 il. in 21 years at 5 per cent. 85.719251 Annuity given 27 or 9641. 88. 47d. £964.419777 Table Fourth shows the present value of an annuity of 11. for any number of years, at 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 che 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 48 years, at the rate of 5 per cent. Under 5 and opposite to 48 years is (years' purchase) 18:077157 Annuity given 45 or 8131, 9s. 5101. £813.472065 Serond, 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 ad 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. Under 4 per cent, and opposite to 20 is 13 590326 Under 4 per cent, and opposite to 50 (20+30) is 21.482184 Difference 7.891858 Annuity given 40 £315.67 4320 or 3151. 138, 5 fotka Table Fifth contains the annuity which 11. 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 5001. purchase, the rate of interest being 5 per cent. ? Under 5 and opposite to 10 is •1 29504 Principal given 500 £64.752000 or 641. 15s. O fod TABLES Sixty, 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 501. 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 11. for 40 years at 4 per cent. is 11.5 Annuity 50 Value £575 To and 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 601, 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 · £528.0 88 60 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 60l. 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 15.9 Annuity 60 Present value £9540 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, is in the case of valuations of leases upon lives, which sometimes occur to him. |