AS/AK/MATH2580 A AS/MATH2580 A Assignment #2 Answers
Worked solutions using the value K = 37.

  1. 1037500500...500-800-800...-800
    |||...|||...|
    012...606162...84
      i = 0.055/12 = 0.00458333333

    balance = 1037(1+i)84+500s60|i(1+i)24-800s24|i = $19711.22.

    Her balance just after her last withdrawal is $19711.22.

  2. ...-7370-7370-7370-7370
    RR...R
    |||...||||
    012...891011

    At time `8': Rs8|.062-7370a4|.062(1.062) = 0 so R = 2708.19

    Each of his deposits should be $2708.19.

  3. 700700...700747747...747
    |||...|||...|
    012...456...12
      i = 0.044/12 = 0.00366666666

    At time `1': 700a4|i(1+i)+747a8|i(1+i)-3 = 8599.09

    The amount needed on September 1, 1999 is $8599.09.

  4. 737737...737...737...737
    |||...|...|...|
    012...19...27...42
    |8%|7%|6%|

    At time `42': 737[s15|.015 + (1.015)15(s8|.0175 + (1.0175)8 s19|.02)] = 44312.86.

    The balance October 1, 1999 is $44312.86.

  5. For buying, NPV = (PV of revenue) - (PV of cost) = (9370a60|.0075+40000(1.0075)-60)-(400000) = $76932.50

    For leasing, NPV = (PV of revenue) - (PV of cost) = (9370a60|.0075)-(8000a60|.0075(1.0075)) = $63107.12.

    Thus buying is better and the savings is $13825.38. [Note: This is the same answer for all values of K because the machine monthly revenue cancels when you subtract.]

  6. The balance owing is $210000 and the interest owing for the first month is 210000(0.079/12) = $1382.50. If K < 38 the payment (1000+10K) is too small to cover interest and the mortgage can NEVER be paid off! In our case, K = 37 and the payment is $1370 which is too small. If K = 47, for example

    210000 = 1470an|.00658333 so (210000)(.00658333)/1470 = 1-(1.00658333)-n and (1.00658333)n = 16.8. Taking logs give n = (log 16.8)/(log 1.00658333) = 429.97

    Thus 430 monthly payments are needed. (That's 35 years and 10 months.)

  7. Let the list price be L. Then you can either pay 16 payments of L/16 or borrow money from a bank and pay a cash amount of 0.9L. If we regard the 16 payments as loan repayments then the implied interest rate i per month satisfies:

    0.9L = (L/16)a16|i or cancelling the L, a16|i = 0.9(16) = 14.4. The approximate formula for i gives i (1-(14.4/16)2)/14.4 = 0.013194. This gives j12 15.8%. Taking j12 = 16% gives a16|i = 14.32301. To get a value larger than 14.4 we must reduce the interest rate. Taking j12 = 15% gives a16|i = 14.42029. Hence

    j12-16
    17-16
    = 14.4-14.42029
    14.32301-14.42029
    = 0.2086

    Thus the equivalent rate charged by the store is j12 = 15.2086% [Note: All answers are the same since there was no K in this question.]

  8. i = (1.035)(1/6)-1 = .00575004 so 150000 = 1370an|.00575004. Solving for n using logs (see problem 6) gives n = 173.2. Take n = 173 months of full payments and let X be the 174th smaller payment. Then 150000 = 1370a173|.00575004 +X(1.00575004)-174. Thus X = 280.98.

    The mortgage requires 173 full payments of $1370 followed by a final 174th payment of $280.98.

  9. The value of the perpetuity on the date of the first payment is 10000+10000/j1 where 1+j1 = (1+(.05/4))4 = 1.05094337. Thus the value of the perpetual scholarship on the day of its first award is 10000+10000/.05094337 = $206288.82. The monthly interest rate is i = (1.0125)(1/3)-1 = .004149425. The accumulate value of 24 montly payments of R one month after the last payment is Rs24|.004149425(1.004149425) = 25.285348R. Setting this equal to 206288.82 gives R = 8158.43.

    Her monthly payment for 24 months is $8158.43. [Note: Answer independent of K.]

  10. The monthly rate is (1.06)1/12-1 = 0.004867551. The accumulation of the 12 payments of $2000 in the first year at the end of the first year is 2000s12|.004867551 = 24653.05654. This value is to rise by 3% each year. Our standard model for annuities with payments in geometric progression has first term R(1+d) so R(1.03) = 24653.05654 and R = 23935.01. The inflation-adjusted annual rate is (1.06/1.03)-1 = 0.02912621. The present value of the 10 annual payments is 23935.01a10|0.02912621 = 205083.97. The accumulated value is 205083.97(1.06)10 = 367274.16.

    The amount of the court award is $367274.16.


File translated from TEX by TTH, version 1.98.
On 3 Feb 2001, 16:26.