Continuous Compounding and Annualization

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Gilbert Reeves
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1 Continuous Compounding and Annualization Philip A. Viton Januay 11, 2006 Contents 1 Intoduction 1 2 Continuous Compounding 2 3 Pesent Value with Continuous Compounding 4 4 Annualization 5 5 A Special Poblem 6 1 Intoduction In ou discussion of highway maintenance we will discuss optimal maintenance policies, in which we choose the best inteval to esuface the oads. This involves maintenance costs spead out ove time and we know, in pinciple, how to handle them: we convet eveything to pesent value. Howeve, if we want to attack the optimality poblem using calculus, then the discete-time methods developed in (eg) CRP 763 won't quite do. The eason is that if we conside, say, a 30-yea policy and then think of extending it one yea, we get a jump discontinuity in the esult, and this makes calculus effectively unusable. The solution is to conside time as vaying continuously, and this note discusses the analytics of continuous-time discounting. A elated poblem is that we 1

maintenance we will discuss optimal maintenance policies, in which we choose the best inteval to esuface the oads.

2 want to conside a single epesentative yea of the maintenance policy, and this equies that we annualize the elevant pesent value that is, convet it to an equivalent once-a-yea quantity. A nal poblem is that the paticula sequence of maintenance costs we shall be concened with is special: it consists of costs which ae incued peiodically, once evey so-many yeas (that is, the yeas in which we actually pefom the maintenance). The nal section consides this special poblem. 2 Continuous Compounding Suppose you deposit P.0/ today (at time 0) in a bank. At the end of 1 yea the bank pays you inteest, calculating it at an annual ate of. Then at the end of one yea, you P.0/ has become P.0/.1 C / Now suppose that the bank pays you inteest twice a yea, a six-month intevals. If is the annual ate, then it will compute each payment based on a ate of =2; and thee will be two payments. So at the end of yea 1 P.0/ will become h S.1; 2/ D P.0/ 1 C i 1 C 2 2 D P.0/ 1 C 2 2 whee in S.1; 2/ the st agument (1) indicates that this is at the end of 1 yea, and the second (2) indicates that compounding is done twice a yea. What about thee times a yea? The effective inteest ate is =3; and thee ae thee compounding peiods, so at the end of the yea P.0/ becomes S.1; 3/ D P.0/ 1 C 1 C 1 C D P.0/ 1 C 3 : 3 We can see whee this is going. If payments ae compounded k times a yea, then at the end of 1 yea P.0/ becomes S.1; k/ D P.0/ 1 C k : k 2

which we actually pefom the maintenance). The nal section consides this special poblem. 2 Continuous Compounding Suppose you deposit P.0/ today (at time 0) in a bank.

3 If you leave you money on deposit fo t yeas with compounding k times a yea thee will be kt sepaate compoundings, so at the end of t yeas you will have Note that S.t; k/ is a peiod-t quantity. S.t; k/ D P.0/ 1 C kt k Now suppose that we allow the numbe of annual compoundings k to become lage and lage. In the limit we aive at continuous compounding in effect, the bank is making you inteest payments all the time. Of couse, the effective ate pe payment the analog of =k gets smalle and smalle; but the numbe of payments gets lage and lage. What will P.0/ become the end of t yeas? To study this we need to look at what happens as the numbe of annual compoundings k tends to innity, that is: S.t; 1/ D lim 1 P.0/ C k!1 k D P.0/ lim 1 C k!1 k If we wite x D =k then we can wite the limit expession as lim.1 C x/1=x t k!1 (since.1 C x/ 1=x t D.1 C x/ t=x D.1 C x/ t=.=k/ D.1 C x/ kt /: Since x D =k then k! 1 means that x! 0: Note that the oute exponent doesn't depend on k, so we can wite this as t lim.1 C x/1=x x!0 Fom the Binomial Theoem the quantity (1 C x/ 1=x tends to e.d 2:718 3 : : : / as x tends to zeo, and we have lim x!0.1 C x/1=x t D e t : In othe wods, with continuous compounding, P.0/ will gow in t yeas to S.t; 1/ D P.0/e t kt kt 3

In the limit we aive at continuous compounding in effect, the bank is making you inteest payments all the time.

4 3 Pesent Value with Continuous Compounding Suppose that someone offes you S.t/ to be eceived in yea t. To nd the pesent value of this we need to nd a quantity P.0/ such that you will be indiffeent between eceiving P.0/ now and S.t/ in yea t. If you maket oppotunities ae given by a banking system which compounds continuously at annual ate ; then at the end of t yeas you P.0/ will compound to P.0/e t In ode fo you to be indiffeent between S.t/ in peiod t note that since we ae now assuming continuous compounding, this is the same as what we wote as S.t; 1/ in the last section and eceiving P.0/ now and leaving it on deposit until peiod t we must have S.t/ D P.0/e t o, solving fo the yea-0 quantity, the pesent value: P.0/ D S.t/e t : In othe wods, with continuous compounding, the pesent value of S.t/ eceived in yea t is S.t/e t Fo efeence, hee is a table compaing the pesent value of $1 eceived at vaious times t and at vaious inteest (discount) ates, using continuous and discete time compounding; t e t 1.1C/ t :03 10 :7408 : :5488 : :4066 :4120 :05 10 :6065 : :3679 : :2231 :2314 :08 10 :4492 : :2019 : :0967 :0994 As we can see, the esults ae geneally quite close. 4

If you maket oppotunities ae given by a banking system which compounds continuously at annual ate ; then at the end of t yeas you P.0/ will compound to P.

5 4 Annualization Conside a poject which has a single up-font (time-0) cost and a benets extending though time. One way to evaluate the poject is to convet the steam of benets to its pesent value; then we can diectly compae the two. But we sometimes want to think about what happens at each yea of the poject, and in this case we need to compae the annual benet to some potion of the cost. To handle this, it is logical to think about evesing the idea of pesent value: that is, to constuct a steam of costs which is equivalent to the oiginal (time-0) cost. Since thee ae an innite numbe of ways to do this, we shall also equie that each of the costs in the constucted steam be the same: in othe wods we constuct an annuity which is equivalent to the oiginal time-0 cost. This pocess is known as annualization: it convets a single quantity to an equivalent annuity. Suppose you invest P.O/ now in some (public) poject that is pojected to last T yeas. We seek an (annual) annuity amount A that is equivalent in pesent value to P.0/ now. With continuous compounding, we will need to add up the pesent value at each possible time between 0 and T: With continuous time, this means that the pesent value of ou T -peiod annuity is an integal: A Z tdt td0 e t dt D A e T 1 D A 1 e T To nd the annuity amount A which is equivalent to a poject expenditue of P.0/ now, we must solve P.0/ D A 1 e T fo A; and the solution is A D P.0/ 1 e T D P.0/ 1 e T In othe wods, incuing a poject cost of P.0/ now is equivalent to incuing a cost of P.0/ 1 e T 5

But we sometimes want to think about what happens at each yea of the poject, and in this case we need to compae the annual benet to some potion of the cost.

6 in each of yeas 0 to T: This is the annualized cost ove a T yea poject life. If the poject lasts foeve, we need to see what happens as T! 1: By inspection, the tem e T tends to zeo and the annualized cost ove an innite poject lifetime is theefoe: P.0/: 5 A Special Poblem In ou discussion of oad maintenance we will conside a special situation: a maintenance policy incus a cost of C evey T yeas stating in yea T, and this patten continues indenitely. We want to nd the annualized cost of the policy. We do this in two steps: st, we nd the pesent value of the steam, and then we annualize that pesent value. Fist, what is the pesent value of the steam? The st cost comes in yea T I its pesent value is Ce T : The second comes in yea 2T I its pesent value is Ce 2T : So we'e looking at a seies of pesent value tems like Ce T C Ce 2T C Ce 3T C : : : D C.e T C e 2T C e 3T C : : : / whee the seies extends foeve. We handle this as follows: we st compute the value of the seies when it extends out fo J tems, and then we take the limit as J tends to innity. Ignoing the constant C fo the moment, ou J-tem seies is U D e T C e 2T C e 3T C C e JT We now use a tick vey much like the one we use when summing a geometic seies, except that this time we multiply each tem by e T : The esult is Subtacting, we obtain Ue T D e 2T C e 3T C C e JT C e.jc1/t : U Ue T D U.1 e T / D e T e.jc1/t (since eveything in between cancels out), o U D e T e.jc1/t 1 e T : 6

0/: 5 A Special Poblem In ou discussion of oad maintenance we will conside a special situation: a maintenance policy incus a cost of C evey T yeas stating in yea T, and this patten continues

7 Now, what happens as J tends to innity? The second tem in the numeato vanishes (tends to zeo), and we have U D e T 1 e T : We can make this look a bit neate by multiplying top and bottom by e T esult is 1 U D e T 1 : the and ou conclusion is that the pesent value of in innite steam of costs C incued evey T yeas is: C e T 1 Finally we annualize this pesent value. Since the steam of costs continues indenitely, the annualization is the one shown at the end of the last section, and we see that the annualized pesent value is C e T 1 : 7

top and bottom by e T esult is 1 U D e T 1 : the and ou conclusion is that the pesent value of in innite steam of costs C incued evey

AMB111F Financial Maths Notes

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