HW4 Solution_2018.pdf - Solution to IMSE4110 Homework 4 Due Date Please write your own solution to these questions on A4 size papers and submit them at

# HW4 Solution_2018.pdf - Solution to IMSE4110 Homework 4 Due...

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Solution to IMSE4110 Homework 4 March 21, 2018 Due Date: Please write your own solution to these questions on A4 size papers and submit them at the beginning of the class on Thursday, March 29, 2018, noon (12:00pm). Please, DO NOT COPY the answers from known solution; otherwise, 6 points will be deducted from such answers even if they are correct. Questions (100 points in total, 12.5 points for each question): 1. Which of the following three bonds (similar except for yield and maturity) has the least duration? A bond with 5% yield and 10-year maturity 5% yield and 20-year maturity 6% yield and 10-year maturity. Explain why? (+12.5 for 0 mistakes, +6 for 1 mistake, +3 for 2 mistakes, 0.5 for 3 mistakes) Answer: The bond with 6% yield and 10-year maturity. Recall that the duration of a bond is given by D = n i = 1 t i c i e - yt i B . So other things equal, the duration is less when yield ( y ) and when maturity ( n ) is shorter. The bond with the highest yield and shortest maturity must have the lowest duration. 2. A 14% annul-pay coupon bond has six years to maturity. The bond is currently trading at par. What is the modified duration of this bond? (+12.5 for 0 mistakes, +6 for 1 mistake, +3 for 2 mistakes, 0.5 for 3 mistakes) Answer: (Both methods are acceptable.) [Method 1 - annual compounding] Because the bond is trading at par, the yield (expressed with annual compounding) is given by the annual-pay coupon rate 14%. Then, the duration is given by D = 6 i = 1 i 14 ( 1 + 14% ) i 100 + 6 100 ( 1 + 14% ) 6 100 = 4 . 4331 . Next, the modified duration is given by D * = D 1 + y m = D 1 + 14% = 3 . 8887 . [Method 2 - continuous compounding] Because the bond is trading at par, it must hold that 6 i = 1 14 × e - yi + 100 × e - 6 y = 100 . This gives the yield y 13 . 1%. Then, the duration is given by D = 6 i = 1 i 14 e - 13 . 1% × i 100 + 6 100 e - 13 . 1% × 6 100 = 4 . 4330 Because the yield y here is expressed with continuous compounding, the modified duration is equal to the duration due to D * = lim m D 1 + y m = D = 4 . 4330 .

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