Nov 17, 2012 ... Class Notes – 11/17/12 ... Example 1: Jason deposits $300 into an account
paying 4%(2) ... in the account 6 months after the final deposit?
Class Notes ʹ 11/17/12 Chapter 5: Other Annuities Annuity due ʹ payments at the beginning of each period NOW 1 2 ͙. n-‐2 n-‐1 n An R R R R Sn (due)
(due)
Difference from ordinary annuities discussed in previous classes Æ Now they match up at the beginning, and are off at the end (opposite of ordinary with regards to the timeline) Sn (due) = Sn (1+i) Æ Sn (due) = R((1+i)n -‐1)(1+i) / i An (due) = An (1+i) Æ An (due) = R(1-‐(1+i)-‐n)(1+i) / i Example 1: Jason deposits $300 into an account paying 4%(2) every 6 months for 4 years. How much is in the account 6 months after the final deposit? *This is asking for a period AFTER the final deposit Sn (due) = ? R = 300
S8(due) = 300(1.028 ʹ 1)(1.02) / .02 = $2626.39
i = .04/2 = .02 n = 8 Example 2: On April 4, 2002, Ben began depositing $125 per month into an account earning 4.2%(12) with the last deposit on May 4, 2004. How much was in the account on June 4,2004? * This is asking for a period AFTER the final deposit Sn (due) = ? R = 125 i = .042/12 = .0035 n = 26 (2 years + 1 month+1 rule)
S26(due) = 125(1.003526-‐1)(1.0035) / .0035 = $3408.14
Example 3: Mara is paying $300 at the beginning of each month for the next 3.5 years to buy a car. If she is paying 4.8%(12), how much was borrowed to buy the car? An (due)= ? R = 300
A42(due) = 300(1-‐1.004-‐42)(1.004) / .004 = $11,623.52
i = .048/12 = .004 n = 42 Example 4: Luke deposits $150 per month into a savings account earning 3.6%(12) starting on 3/19/2002 with the last deposit on 12/19/2004. What is the cash equivalent of this on 3/19/2002? An (due)= ?
A34(due) = 150(1-‐1.003-‐34)(1.003) / .003
R = 150 = $4856.15 i = .036/12 = .003 n = 34 Example 5: Ani has an inheritance of $250000 from which he wants to withdraw money quarterly, starting immediately, for the next 8 years. If the money will earn 2%(4), how much are the payments? R = ?
R(1-‐1.005-‐32)(1.005)/.005 = 250000
An (due) = 250000
29.65080R = 250000
i = .02/4 = .005
R=250000/29.65080
n = 32
R = $8431.48
Example 6: If monthly deposits for 1.5 years into an account earning 3%(12) amount to $1500 one month after the final deposit, how much is each deposit? R = ?
R(1.002518 ʹ 1)(1.0025) / .0025
Sn (due) = 15000
18.43362R = 15000
i = .03/12 = .0025
R = 15000/18.43362
n=18
R = $813.73
Deferred Annuities ʹ first payment is delayed some number of periods (m). Î So the present value, An(def), is two or more periods before the 1st payment. Î Difference from annuities due Æ Now we are looking to a period before the first payment (we will either be finding the payment size or the present value) An (def) = An (1+i)-‐m Î An (def) = R(1-‐(1+i)-‐n)(1+i)-‐m / i Example 7: Find the value at 6%(4) on 3/4/2002 of quarterly rents of $2000 with the first made on 3/4/2004 and the last made on 3/4/2008. *Given date of An(def) and first R, you must subtract 1 from m An (def) = ? R = 2000 i = .06/4 = .015
An (def) = 2000(1 ʹ 1.015-‐17)(1.015-‐7) / .015 = $26864.38
n = 17 m = 7 Example 8: Upon graduating from the academy on May 15,2005, Jacen had loans in the amount of $80000. He repays these loans by making monthly payments for 15 years with the 1st due in 2 years. If he is charged 3%(12), how much are his payments? R = ?
R(1-‐1.0025-‐180)(1.0025-‐23) / .0025 = 80000
An (def) = 80000 136.72383R = 80000 i = .03/12 = .0025 R = 80000/136.72383 n = 180 (you are not given a date this time)
R = $585.12 m = 23
Forborne Annuity ʹ The future value is 2 or more periods after the final deposit. ÆThe final value is 2 or more periods after the final deposit (you will be moving Sn forward by p periods into the future) Sn (for) = Sn(1+i)p Æ Sn (for) = R(1+i)n ʹ 1)(1+i)p / i Example 9: Lando makes monthly deposits of $100 for 20 years into an account paying 3%(12). How much is in the account 5 years after the final deposit? Sn (for) = ? R = 100
S240 (for) = 100(1.0025240-‐1)(1.0025)60 / .0025
i = .03/12 = .0025 = $38136.11 n = 240 p = 60 Example 10: Han began depositing $1200 each quarter into an account paying 5%(4) on January 10, 2000. He made his last such deposit on July 10, 2004. How much is in the account on January 10, 2010? Sn(for) = R = 1200 i = .05/4 = .0125 n = 19 p = 22
S19 (for) = 1200(1.012519-‐1)(1.012522) / .0125 = $33588.12
Perpetuity ʹ payments go on forever Î Payments = interest earned during one payment period Present value (A) = R / i
if interest periods = payment periods
OR (A) = R / ((1+i)k ʹ 1) where there are k interest periods in 1 payment period Example 11: A fund is to be set up from which $200 each month can be withdrawn indefinitely. How much must be invested at 1.8%(12) to accomplish this? simple A = ?
A = 200 / .0015
R = 200 A= $133333.34 (*have to round cents up) i = .018/12 = .0015 Example 12: An alumnus donates $150000 to his alma mater which is then invested at 4.6%(2) on February 15, 2010. How large a scholarship can be given every six months and when is the first scholarship available? A = 150000 R = ? i = .046/2 = .023
150000 = R / .023 R = .023 (150000) Æ = $3450 on August 15, 2010 (6 months after date the $ is first invested)
Example 13: Another alumnus donates $200000 to his alma mater. This money is invested at 2.4%(12) and is to be used to give away a quarterly scholarship. How much is each scholarship? When is the first available? general A = 200000 R = ? i = .024/12 = .002 k = 3
R / (1.0023-‐1) = 200000 R = 200000(1.0023-‐1) R = $1202.40 each quarter starting in 3 months (or 1 quarter)
Example 14: Suppose a scholarship fund that will award a semiannual scholarship starting on August 20, 2010, of $3500 is to be set up. If money can be invested at 2.6%(4), how much must be invested and on what date? general A = ? R = 3500 i = .026/4 = .0065 k = 2 (2 quarters in ½ year)
A = 3500 / (1.00652-‐1) = $268358.61 on February 20, 2010
