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Homework answers / question archive / (1 pointy Two bonds, each with a face value of $16000, are redeemable at par in t
(1 pointy Two bonds, each with a face value of $16000, are redeemable at par in t.years and priced to yield ya = 6%. Bond 1 has a coupon rate og = 114% and sells for $22485.29. Bond 2 has coupon rate 12 = 3.9% and sells for $P. What is the value of P? Answer: $ 13527.01 Preview My Answers Submit Answers You have attempted this problem 1 time. Your overall recorded score is 0%.
Bond 1
Let the months be t
and the yield , we have for annual, so we will first convert it to monthly = 0.06/12 = 0.5%
Face value = 16000
Sales Value = 22485.29
Coupon value = 11.4%
So the equation will be
⇒ 22485.29 = (16000*0.114/12)*(v + v^2 +v^3 +…..+v^t) + 16000*v^t
where v = 1/(1+i) and i = 0.5%
⇒ 22485.29 = 152*v*(1 - v^t)/(v) + 16000*v^t
⇒ 22485.29 = 152*((1 - (1+0.005)^(-t))/(0.005)) + 16000*(1+0.005)^(-t)
So by putting the values in this ques , 0
at t = 119 , RHS = 22445.71508
at t = 120 , RHS = 22485.28864
at t = 121 , RHS = 22524.66531
So we get the value of t = 120 months
Bond 2
Let the months be t
and the yield , we have for annual, so we will first convert it to monthly = 0.06/12 = 0.5%
Face value = 16000
Sales Value = P
Coupon value = 3.9%
So the equation will be
⇒ P = (16000*0.039/12)*(v + v^2 +v^3 +…..+v^t) + 16000*v^t
where v = 1/(1+i) and i = 0.5%
⇒ P = 52*v*(1 - v^t)/(v) + 16000*v^t
⇒ P = 52*((1 - (1.005)^(-120))/0.005) + 16000*((1.005)^(-120))
⇒ P = $ 13477.94
