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MA270 - Financial Math MIDTERM SOLUTIONS F21
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MA270 - Financial Math MIDTERM SOLUTIONS F21 (WRITTEN QUESTIONS 1 - 10). Answers shown with work.
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MA 270 – Midterm Test Solutions Page 1 of 2[10 marks] 1. Consider a 10-year bond that pays a semi-annual coupon of 3% on a face value of $1,000.
Note that the price function for this bond is P (y) = 1000[1 + (.015 y)a20,y ].
(a) Sketch the bond’s price as a function of its yield. That is, sketch the graph of P (y). Be
sure to label any important points and/or asymptotes.
Graph must be (i) decreasing, (ii) convex, (iii) pass through the point (.015, 1000), (iv)
have vertical intercept at P = 1300 and (v) have horizontal asymptote at P = 0 as
y ! 1.
(b) The bond is currently trading at $844.11. Is its yield higher or lower than 7% compounded
semi-annually? Explain your answer.
Lower. Price at 7% c.s.a. is P (.035) = 715.15, which is less than the observed price.
Because of the inverse relationship between price and yield, a higher price requires a lower
yield.
(c) Suppose I purchase the bond now, at a yield of 3% compounded semi-annually, and sell
it immediately after I receive the next coupon in six months. If the bond’s yield at that
time is 3.4% compounded semi-annually, then what is my total return?
Total return is
P1 + C P0
,
P0
where P0 = price at which bond was purchased, C = the dollar value of the coupon
received and P1 = price at which bond was sold. Bond was purchased for P0 = 1000[1 +
(.015 .015)a20,.015 ] = 1000 and sold for P1 = 1000[1 + (.015 .017)a19,.017 ] = 967.76.
Coupon was C = (.03/2)1000 = 15. Total return is therefore 967.76+15
1000
1000
= 0.01724,
or 3.448% on an annualized basis.
[6 marks] 2. The Government of Ontario has an outstanding bond that matures in exactly 3 years and
pays a semi-annual coupon of 4% on a face value of $1,000. The current market price of the
bond is $945.83.
(a) Using an initial guess of 9% compounded semi-annually, use one iteration of Newton’s
method to approximate the bond’s yield. Please express your answer as an annual
rate with semi-annual compounding. You may use the fact that P (.045) = 871.05 and
P 0 (.045) = 4, 741.82 for this bond.
Update formula is
B P (y0 ) 945.82 871.05
y1 = y0 + = .045 + = .045 .01576 . . . = 0.029230 . . . .
P 0 (y
0) 4741.82
So the bond’s yield is approximately 5.8% compounded semi-annually.
(b) What happens to the interval (4%, 9%) after one iteration of the bisection method? You
may use the fact that P (.0325) = 932.84.
Interval becomes (4%, 6.5%). Prices at endpoints are P (.02) = 1000 and P (.045) =
871.05. Midpoint is 6.5% c.s.a. and price at midpoint is P (.0325) = 932.84. So yields
in (4%, 6.5%) go with prices in (932.84, 1000), while yields in (6.5%, 9%) go with prices
in (871.05, 932.84). Since the observed price lies in (932.84, 1000) its yield must lie in
(4%, 6.5%).
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