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Math 618 Answer to Selected Problems in Section 4.2 Autumn 2005 Formula Review If the bond is redeemed at par (i.e. C = F), then one form of the price P is P = F + F(r−j)an|j For bond amortization, we have OBt = braceleftBigg F[1 + (r−j)a n−t|j] 0 ≤ t < n 0 t = n Kt = braceleftBigg Fr 0 < t < n F + Fr t = n PRt = braceleftBigg F(r−j)vn−t+1 0 < t < n F(r−j)v + F t = n It = F[j + (r−j)(1−vn−t+1)] 0 < t ≤ n PRt = PRt−1(1 + j) = PRt−2(1 + j)2 = ··· Problem 4.2.5 We have j = .035, F = 100, P −F = 36, and PR5 = 1. Also v = 11+j = .9661835749. PR5 = 1 ⇒ F(r−j)vn−4 = 1 ⇒ F(r−j)vn = 1·v4 ⇒ F(r−j)vn = .8714422278. Using the formula P = F + F(r−j)an|j and an|j = vn ·sn|j, we have 36 = P −F = F(r−j)an|j = F(r−j)vn ·sn|.035 Using the value (in red) obtained earlier, the last equation becomes 36 = .8714422278sn|.035 or sn|.035 = 41.31082800. Solving this last equation (using financial calculator, for example), we get n = 25.99905425. Answer 26 half-years, or 13 years Problem 4.2.7 Given: n = 30, PR2 = 977.19, and PR4 = 1046.79. Find: P −F. Solution 1. Using PR2 = F(r − j)v29 and PR4 = F(r − j)v27, we have F(r − j)v29 = 977.19, F(r − j)v27 = 1046.79. Dividing side by side, we get v2 = 977.19/1046.79 = .9335110194, and so v = .96618374. Substituting this value of v back to either PR2 or PR4 (in red above), we get F(r −j) = 2650.006905. Since v = 11+j, j = .03499982312. Subtracting F from both sides of the formula P = F + F(r−j)an|j, we get P −F = F(r−j)an|j = F(r−j)a30|j = 48739.15540. Solution 2. Since PR4 = PR2(1 + j)2, we have 1046.79 = 977.19(1 + j)2. Solving for j, we get j = .0349998231. PR2 = PR1(1 + j) ⇒ PR1 = PR2/(1 + j) = ··· = 944.1450889. We then have P −F = PR = PR1 + PR2 + PR3···+ PR30 = PR1 + PR1(1 + j) + PR1(1 + j)2 +···+ PR1(1 + j)29 = PR1[1 + (1 + j)2 +···+ (1 + j)29] = PR1s30|j = 48739.15510 Answer $48,739

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