Math 618 Answer to Problem 2.1.7 Autumn 2005 Smith: At the end of the 10th year, Smith had collected 10 dividend payments of $.80 × 100 = $80 each. The accumulated value of these payments at this moment is 80· s10|.06 = 1054.46. Smith also sold his 100 shares at $2.00 each, so the total proceed was $200. Then his total wealth at the end of 10th year was 1054.46 + 200 = 1254.46. At the end of the nth year (that is n − 10 years later from the end of the 10th year), his total wealth grows from $1,254.46 to 1,254.46(1 + .06)n−10. (1) Brown: Among the n years, Brown did not collect any dividend payment in the first 10 years. At the end of the nth year, he has collected n−10 dividend payments of $.40×100 = $40 each. The accumulated value of these payments at the end of the nth year is 40· sn−10|.06 = 40· (1 + .06) n−10 −1 .06 . Suppose he sold his stocks at $p each, and so his proceeds is $100p. Therefore, Brown’s total wealth at the end of the nth year is 40(1 + .06) n−10 −1 .06 + 100p . (2) In order for Brown’s investment matches Smith’s, the accumulated values in (1) and (2) are equal: 40(1 + .06) n−10 −1 .06 + 100p = 1,254.46(1 + .06) n−10. (3) If n = 15, (3) becomes 40(1 + .06) 15−10 −1 .06 + 100p = 1,254.46(1 + .06) 15−10, or 225.4837 + 100p = 1678.75. Solving, we have p = 14.53. If n = 20, (3) becomes 40(1 + .06) 20−10 −1 .06 + 100p = 1,254.46(1 + .06) 20−10, or 527.2318 + 100p = 2246.5468. Solving, we have p = 17.19. If n = 25, (3) becomes 40(1 + .06) 25−10 −1 .06 + 100p = 1,254.46(1 + .06) 25−10, or 931.0388 + 100p = 3006.3864. Solving, we have p = 20.75.