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# RecommendedProblem - rT Fe P-= The first and second...

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Problem 3.10 Let P and D being the price and the Macaulay duration of the 10-year, 8% coupon bond. The yield is 10%. Assume a face value of \$100. + + + - = n n F c c P 2 2 ) 2 / 1 ( 2 / 2 / ) 2 / 1 ( 1 2 / 2 / λ =\$87.54 ( 29 ( 29 years PV k D k k 84 . 6 05 . 1 100 10 05 . 1 4 ) 2 / ( 20 20 1 = × + × = = Problem 3.11 Using r A PV = , first compute modified duration D M then convert to Macaulay duration D : r r A A r dr dPV PV D M 1 1 2 = - - = - = then r r D r D M + = + = 1 ) 1 ( . Problem 3.12 a) P(A)= \$885.84 P(B)= \$771.68 P(C)= \$657.52 P(D)= \$869.57 b) D(A)= 2.72 D(B)= 2.84 D(C)= 3.00 D(D)= 1.00 c) Since C has the highest duration, it is the most sensitive. d) To immunize we have to match the prices and durations. So we have to choose bond D and any other bond. If we choose bond C, we will have the following equations: V(C)+V(D)=PV

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D(C)V(C)+D(D)V(D)=2PV where PV= 2 15 . 1 000 , 2 =\$1,512.29 Solving V(C)=\$756.15 = V(D) e) This is solved in d). Problem 3.15 The present value is given by:
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Unformatted text preview: rT Fe P-= . The first and second derivatives are given by rT TFe dr dP--= and rT Fe T dr P d-= 2 2 2 . The convexity is defined as 2 2 2 1 T dr P d P C = = . Problem 4.12 First, we need to write down the equation for present value ∑ = + = n k k k m k m s e x PV 1 / )) / 1 ( ( ) ( λ Next we differentiate it: ∑ = +-= n k k k m k m s e x m k d dPV 1 / )) / 1 ( ( ) / ( ) ( Therefore ∑ = + =-n k k k k m s x m k PV d dPV PV 1 ) / 1 ( ) / ( ) ( 1 ) ( ) ( 1 Since we may interpret (k/m) as time, this is a weighted average of times where the weights are determined by the present value of cash flow k divided by the total present value of the cash flow stream....
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RecommendedProblem - rT Fe P-= The first and second...

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