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16 Table 3.1 Fund Development  Withdrawals from Fund 2 Prior to Fund 1 for at 10%. Fund 1 Balance Fund 2 Balance Period  Before withdrawal  After withdrawal  Before withdrawal  After withdrawal    4.355     4.355  3.791  .436    3.791  3.170  .379    3.170  2.487  .317    2.487  1.736  .249    1.736  .910  .174    .910  .001*  .091   * zero,  except for .001  roundoff error   
The other extreme case would be to withdraw payments from Fund 1 whenever possible, and to withdraw from Fund 2 only when Fund 1 is exhausted. In view of the answer above, we will assume that Fund 1 becomes exhausted at the end of the fifth period. It will be necessary to confirm that the answer obtained is consistent with this assumption. At the end of the first period interest of is earned on Fund 1 and deposited into Fund 2. A withdrawal of 1 is then made with all of it coming from Fund 1. This leaves a new balance of  I in Fund 1 and . IK in Fund 2. At the end of the second period interest of {  I) is earned on Fund 1 and deposited into Fund 2. A withdrawal of 1 is then made with all of it coming from Fund 1. This leaves a new balance of 2 in Fund 1 and + (  I) = .2K in Fund 2. If this process is continued for two more periods, we have 4 in Fund 1, which we have assumed lies between 0 and 1. Also, we have [ +{ 1) + ( 2) + { 3)] = AK.6 in Fund 2. At the end of the fifth period interest of . 1(  4) is earned on Fund 1 and deposited into Fund 2. A withdrawal of 1 is then made with /sT  4 of it coming from Fund 1, which is now exhausted. The rest of the payment, i.e. l(K4) = 5 K, comes from Fund 2. The new balance in Fund 2 is ( .6) + ( 4)(5 ) = 1.5K6. At the end of the sixth period no interest is earned, since Fund 1 is exhausted. The fund balance in Fund 2 is unchanged, since Fund 2 does not earn interest. The final withdrawal of 1 comes entirely from Fund 2 exhausting it, so that we have Further analysis of annuity values at simple interest is instructive. Consider the investment of a sum of money for n periods at simple interest which is just sufficient to permit a withdrawal of 1 the end of each period for n periods. The concept of simple interest means that interest does not earn additional interest. Thus, the original amount of the investment is deposited in a fund which earns interest at rate but any interest earned is immediately transferred into a second fund which does not earn interest. The ambiguity arises whenever a withdrawal is made. How much should be withdrawn from the principal fund, which earns interest, and how much from the interest fund, which does not earn interest? Different answers will be obtained depending on how this allocation is made. In order to illustrate some of the possible answers which can be obtained, let n = 6 and i = 10%. Let the original investment be denoted by K. We will refer to the principal fund as "Fund 1," and the interest fund as "Fund 2." One extreme case would be to withdraw payments from Fund 2 whenever possible, and to withdraw from Fund 1 only when Fund 2 is exhausted. At the end of the first period interest of .1 : is earned on Fund 1 and deposited into Fund 2. A withdrawal of 1 is then made, with coming from Fund 2 and I  coming from Fund 1. This leaves a new balance in Fund 1 of (1  .1 = 1.1 :  1. At the end of the second period interest of . 1 1) is earned on Fund 1 and deposited into Fund 2. A withdrawal of 1 is then made, with [1  1] coming from Fund 2 and 1  .1(1.1 :  1) coming from Fund 1. This leaves a new balance in Fund 1 of (1  1)  [1  .1(1.1 : 1)] = (1 ) (1 + 1.1). If this process is continued for four more years, the balance in Fund 1 at the end of six years is (1.1) : [1 + 1.1 + (1.1)2 + (1 1)4 p i)5j and the balance in Fund 2 is zero. However, the balance in Fund 1 must also be zero after the final withdrawal. Thus (1.1)1 l(l.l) 4 (1.1)6(1.11) .1 which is the value of j using compound interest. We will discover in Chapter 6 that this is no accident. The results of this approach are summarized in Table 3.1.
\.5Ke = I 7 = 4.67 = a This answer is valid, since the assumption was made that 4 < < 5. If an answer had been obtained outside this interval, the problem would have to be reworked until an answer consistent with the assumption about the exhaustion of Fund 1 is obtained. The results of this approach are summarized in Table 3.2. Table 3.2 Fund Development  Withdrawals from Fund I Prior to Fund 2 for at 10%  Fund 1 Balance  Fund 2 Balance   Before  After  Before  After  Period  withdrawal  withdrawal  withdrawal  withdrawal    4.667     4.667  3.667  .467  .467   3.667  2.667  .834  .834   2.667  1.667  1.101  1.101   1.667  .667  1.268  1.268   .667   1.268  1.002**     1.002  .002* 
* zero, except for .002 roundoff error *♦ calculated as 1.268 + .1(.667)  (1  .667) = 1.002 We next apply formula (3.38) directly and obtain 1.1 1.2 = 4.52 which lies between our two extreme answers already obtained. Apparently, formula (3.38) involves an implicit intermediate allocation of withdrawals from Fund 1 and Fund 2 between the two extremes. If we apply formula (3.41), we obtain 1 ( 1.1 + . • • + 1.5 =   = 4.69 which lies outside the range bounded by the two extremes. Thus, we must question whether formula (3.41) produces a result with any realworld significance at all, at least in this example. The results of this analysis are summarized in Table 3.3. Table 3.3 Comparison of flgi Computed at 10% Simple Interest on Four Different Bases Basis Value Low extreme (compound interest) Formula (3.38) High extreme Formula (3.41) 4.36 4.52 4.67 4.69 At this point the reader should be thoroughly convinced that finding annuity values at simple interest is treacherous indeed. Furthermore, simple discount would present just as many difficulties. The reader now should really appreciate the admonition given early in this section to avoid computing annuity values not involving compound interest if at all possible! Example 3.12 Compare the value of at 10% interest (1) assuming compound interest, (2) using formula (3.39), and (3) using formula (3.40). 1. Using compound interest at 10% = 7.72 , 2. From formula (3.39) = 1.6 1.1 1.2 = 7.23 . 3. From formula (3.40) = 1 + 1.1 + 1.2 + + 1.5 =7.50. Answer (3) involves simple interest being earned on each payment from the date of deposit to die end of die sixyear period of investment. Answer 1 is larger dian eidier of die other two, verifying the larger accumulation with compound interest. Example 3.13 Find , if &, = .021 for 0 < t B. If we accumulate each payment from the date of deposit to the end of the fiveyear period at the varying force of interest 5,, we have 51 = E « t=i = .24 + .21+ .16 .091 = 1.27125 + 1.23368 + 1.17351 + 1.09417 + 1 = 5.7726. Alternatively, formula (3.39) is applicable in diis case. We have
a{t) dr Olr so that = «(5) E = e* + e = 5.7726 . ,t1 a{t) 01 .21 = 25 (.01 .04 . .09 . .16 ,25) which agrees with the answer obtained from first principles above. EXERCISES 3.2 Annuityimmediate 1. A family wishes to accumulate $50,000 in a college education fund at die end of 20 years. If they deposit $1000 in die fund at die end of each of die fust 10 years and $1000 + X in die fund at die end of each of the second 10 years, find X to die nearest dollar if die fund earns 7% effective. 2. The cash price of a new automobile is $10,000. The purchaser is wilUng to finance die car at 18% convertible mondily and to make payments of $250 at the end of each month for four years. Find the down payment which will be necessary. 3. An annuity provides a payment of n at die end of each year for n years. The annual effective interest rate is I In. What is die present value of die annuity? 4. If = X and = y, express as a function ofArandy. 5. a) Show diat a = + v"a = va + . b) Show diat *5 1 = + (1 + Osi = (1 + OSFl + • c) Verbally interpret die results in {a) and ). 6. a) Show diat a =  v"s = (1 + 0"aml " where 0 < < m. b) Show diat ijTHi =  (1 + "Jn = "sn  where 0 < n < m. c) Verbally 1 1 1 die results in (a) and (b). 7. You are given the following annuity values: «71 =5.153 ayy = 7.036 =9.180 Find i. 8. Show diat 1 V *iol 3.3 Annuitydue 9. Find die present value of payments of $200 every six mondis starting immediately and continuing tiirough four years from die present, and $100 every six mondis diereafter dirough ten years from die present, if = .06. 10. A worker aged 40 wishes to accumulate a fiind for retirement by depositing $1000 at the beginning of each year for 25 years. Starting at age 65 die worker plans to make 15 annual withdrawals at the beginning of each year. Assuming that all payments are certain to be made, find the amount of each withdrawal starting at age 65 to die nearest dollar, if die effective rate of interest is 8% during die first 25 years but only 7% diereafter. 11. Find a if the effective rate of discount is 10%. 12. Derive formula (3.12). 13. a) b) c) Show diat = Show that s + 1  v  1 + (1 + 0". Verbally 1 1 1 the results in (a) and (b). 14. Show diat IpL + 4  1 = 1. 15. If «p and 5 = , show diat ap+q\ = 1 + iy 3.4 Annuity values on any date 16. Derive formulas (3.17), (3.18), and (3.19) algebraically. 17. Payments of $100 per quarter are made from June 7, Z through December 7, Z + 11, inclusive. If the nominal rate of interest convertible quarterly is 6%: a) Find the present value on September 7, Z  1. b) Find the current value on March 7, Z + 8. c) Find the accumulated value on June 7, Z + 12. 18. Show diat Y, (•n " ) = *T51 J=10 19. Annuities X and Y provide die following payments: End of Year Annuitv X Annuity Y 110 1 1120 2 0 2130 1 Annuities X and Y have equal present values at an annual effective interest rate i such diat v" = 1/2. Determine K.
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