Chapter006

Chapter006 - The Theory of Interest - Solutions Manual...

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The Theory of Interest - Solutions Manual 59 Chapter 6 1. ( a ) () 10 1000 1.10 $385.54 P == . ( b ) 10 1000 1.09 $422.41 P . ( c ) The price increase percentage is 422.41 385.54 .0956, or 9.56%. 385.54 = 2. The price is the present value of the accumulated value, so we have 20 10 .08 1000 1 1.1 $844.77. 2 P ⎛⎞ =+ = ⎜⎟ ⎝⎠ 3. ( a ) The day counting method is actual/360. In 26 weeks there are 26 7 182 ×= days. Using the simple discount method, we have 182 9600 10,000 1 and .0791, or 7.91%. 360 dd =− = ( b ) An equation of value with compound interest is 1 2 9600 10,000 1 and .0851, or 8.51% ii = . 4. We have 100, 105, .05, 5/105, .04, 5/.04 125, F Cr g i G = = = = = 20 105 1.04 47.921, K and 20 n = . Basic: ( ) ( ) 20 20 5 105 5 13.59031 105 .45639 $115.87 Pa v = + = . Premium/discount: ( ) 20 105 5 4.2 $115.87 = . Base amount: ( ) 20 125 105 125 1.04 $115.87 P = . Makeham: 5 47.921 105 47.921 $115.87 .04 105 P = . 5. We apply the premium/discount formula to the first bond to obtain ( ) 1136.78 1000 1000 .025 .02 n a
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The Theory of Interest - Solutions Manual Chapter 6 60 which can be solved to obtain 27.356 n a = . Now apply the premium/discount formula to the second bond to obtain ( )( ) 1000 1000 .0125 .02 27.356 $794.83. P =+ = 6. Since the present value of the redemption value is given, we will use Makeham’s formula. First, we find 45 .04. 1125 Fr g C == = Now () .04 225 1125 225 $945. .05 g PK CK i = + = 7. Since , n KC v = we have 450 1000 n v = and .45 n v = . Now we will apply the base amount formula ( ) 1 nn n P GC G vG v C v =+ − = − + and substituting values ( ) 1110 1 .45 450 and $1200. GG =−+ = 8. The price of the 10-year bond is 20 20 .035 1000 1.035 50 1213.19 Pa = . The price of the 8-year bond is 16 16 .035 1.035 .03 1213.19 PF F a = and solving ( ) 1213.19 $1291 to the nearest dollar. .576706 .03 12.09412 F + 9. Since n is unknown, we should use an approach in which n only appears once. We will use the base amount formula. First, we have ( ) 1000 .06 1200 .05 Fr G i = and ( ) 1200 1000 1200 1200 200 . P vv =−
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The Theory of Interest - Solutions Manual Chapter 6 61 If we double the term of the bond we have ( ) 2 50 1200 1000 1200 1250 200 . nn P vv += + = Thus we have a quadratic which reduces to 2 200 200 50 0 or 2 44 1 0 and factoring () 2 210 . n v = Thus, .5 n v = and ( ) 1200 200 .5 $1100 P =− = . 10. ( a ) The nominal yield is the annualized coupon rate of 8.40%. ( b ) Here we want the annualized modified coupon rate, so 42 22 2 8 . 0 0 % . 1050 Fr g C ⎛⎞⎛ == = ⎜⎟⎜ ⎝⎠⎝ ( c ) Current yield is the ratio of annualized coupon to price or 84 9.14% 919.15 = . ( d ) Yield to maturity is given as 10.00%. 11. Using the premium/discount formula, we have ( ) 1 11 1 . 5 1 . 5 P pi i a i a =+ =+ =+ and ( ) 2 1 . 7 5 1. 2 5 5. P iia i a p 12. Let X be the coupon amount and we have 7 5 X X = + so 20 X = . 13. We have 20 n = and are given that ( ) 2 19 8 PC i g v = −= . We know that the principal adjustment column is a geometric progression. Therefore, we have ( ) 8 11 12 18 1 10 10 8 8 at 4.5% 8 8 1.045 6.59589 $33.98. t t Pv v v va = + + = "
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The Theory of Interest - Solutions Manual Chapter 6 62 14. Since , ig > the bond is bought at a discount. Therefore, the total interest exceeds total coupons by the amount of the discount. We have ( )( ) ( ) () 10 .06 10 50 1000 .06 .05 500 10 7.36009 $573.60.
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This note was uploaded on 11/18/2010 for the course MATH 320 taught by Professor Dr.k during the Spring '10 term at Nevada.

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Chapter006 - The Theory of Interest - Solutions Manual...

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