AS/AK/MATH2580 A Assignment AS/AK/MATH2580 A Assignment #3 Answers
Worked solutions using the value K = 25.

1. A good answer here will have five columns. Column A is payment number which is the numbers from 0 to 36. Suppose these are in cells A10 to A46. Column B will have the 36 payments in cells B11 to B46. Make sure the payment is rounded up not down to the cent. In this case the payment is 44500/a36|.00635646163 = 1386.839 =\$1386.84. In cell C11 put the formula E10*.00635646163 and copy this down to C46. (Thus C is the interest column.) In cell D11 put the formula B11-C11 and copy this down to D46. (Thus D is the repayment of principal column.) In cell E10 put the number \$44500. In E11 put the formula E10-D11 and copy this down to E46. Then the sheet is nearly complete. You should find the final balance in E46 to be slightly negative. You fix this by making the final payment in B46 a bit smaller than the others. In this case, you make it \$1386.81. In some blank cell type the formula sum(C23:C34). In this case the interest paid was \$1837.28.

2. (1209.78/1147.50) = (1+i)20-11 = (1+i)9. Thus (1+i) = 1.005889817. Hence 1209.78 = R(1.005889817)-(72-20+1) so R = 1651.4986. Finally, L = 1651.4986a72|.005889817 = 96683.061. The original loan was \$96683.06.

3. R = 170000/(a240|i) = \$1459.56 (where i = 1.0425(1/6)-1 = .0069610621) The Buyer's equity = 20000 + (170000-B125 = 190000 - (170000(1+i)125-1459.56s125|i) = \$74753.04.

4. 20000 = L(1+i)16-925s16|i where i = (1+0.075/2)(1/2)-1 = 0.01857744. Thus L = (20000+925s16|i)(1+i)-16 = 27599.98. The original amount of the loan was \$27599.98.

5. R1 = 172500/(a240|i) = \$1428.92 = old monthly payment (i = 1.04(1/6)-1 = .006558197) Balance after 24 months is 172500(1.04)4-1428.92s24|i = 164791.30. Adding the penalty, we get a balance of 164791.30+3(1428.92) = 169078.06. Then the new monthly payment is R2 = 169078.06/(a216|j) = \$1274.90 where j = (1.03)(1/6)-1 = .004938622. Thus the new payment is smaller than the old (by over \$150 per month) even after paying the penalty.

6. quarterly deposit of association = 202500/(s16|i) = 11627.78 where i = (1+.045/2)(1/2)-1 = 0.011187421. The quarterly deposit per cottager is 11627.78/80 = \$145.35.

7. Lender A: 12500/(a120|i) = \$147.64 (where i = (1+0.075/2)(1/6)-1 = .006154524.)

Lender B: 12500j+12500/(s120|r) = 66.81 + 89.54 = \$156.35. (where j = (1.0325)(1/6)-1 = .00534474 and r = (1.015)(1/6)-1 = .002484517.)

Thus she should choose Lender A even though Lender A is charging a higher interest rate.

8. P = (5000)(.04)a20|.03125+5000(1.03125)-20 = 5643.43. The price of the bond to yield 6.25% is \$5643.43.

9. P = C+(Fr-Ci)an|i = 10200+(250-357)a6|.035 = \$9629.84.

 time coupon interest on book value book value book value adjustment 0 \$9629.84 1 \$250.00 \$337.04 -\$87.04 \$9716.88 2 \$250.00 \$340.09 -\$90.09 \$9806.97 3 \$250.00 \$343.24 -\$93.24 \$9900.22 4 \$250.00 \$346.51 -\$96.51 \$9996.73 5 \$250.00 \$349.89 -\$99.89 \$10096.62 6 \$250.00 \$353.38 -\$103.38 \$10200.00

10. P20 = 1000+(42.50-31.25)a20|0.03125 = \$1165.45. For call date k = 10, 11, ..., 19 we have Pk = 1030+(42.50-32.1875)ak|0.03125 and the smallest of these is at k = 10. This early call value is \$1117.41. The smaller value, namely \$1117.41 is the price of the callable bond.

File translated from TEX by TTH, version 1.98.
On 26 Mar 2001, 16:35.