University of Connecticut, School of Business Financial Modeling with C# FNCE5353 Spring 2021 Instructor: Dr. Sateesh Mane Contact email:
[email protected] © Sateesh R. Mane 2021 _ Please upload your solution to HuskyCT. 1. Please submit your solution as a docx or pdf file. 2. Other _le types will not be accepted. 3. Write your name and PeopleSoft ID and "FNCE5353" and "HW3" to identify yourself and the course. (first name)_(last name)_(PeopleSoft ID)_FNCE5353_HW3.[docx/pdf] 4. Start the answer to each question at the top of a new page in your file. 5. Type (or print) all your answers to the questions. 6. Screenshots (or scanned images, etc.) of handwritten solutions will NOT be accepted. _ You may submit multiple updates until the deadline. Only your last submission is graded. 3.4 Calculations _ Use Face=100 in all the calculations below. _ Calculate the fair value, Macaulay duration and modified duration for various newly issued bonds. 1. Use different values for c, n and y. Set n >= 4 in all cases. 2. Use f = 2 (semiannual coupons) but also try f = 4 (quarterly coupons), etc. 3. Also use zero-coupon bonds. _ Example cases are: 1. c = 4, n = 4, f = 2, y = … . 2. c = 3.5, n = 5, f = 4, y = … . 3. zero-coupon bond: c = 0, n = 7, y = … . 1. Set y = 5% FV = [2 / (1 + 0.025)] + [2 / (1 + 0.025)2] + [2 / (1 + 0.025)3] + [(100 + 2) / (1 + 0.025)4] ≈ 98.1190129 DMac = (1/98.1190129) * {[2 / (1 + 0.025) * 1] + [2 / (1 + 0.025)2 * 2] + [2 / (1 + 0.025)3 * 3] + [102 / (1 + 0.025)4 * 4]} ≈ 3.882610774 Dmod = DMac / (1 + 0.025) ≈ 3.78791295 2. Set y = 2% FV = [0.875 / (1 + 0.005)] + [0.875 / (1 + 0.005)2] + [0.875 / (1 + 0.005)3] + [0.875 / (1 + 0.005)4] + [100.875 / (1 + 0.005)5] ≈ 101.8471999 DMac = (1/101.8471999) * [(0.875/1.005 * 1) + (0.875/1.0052 * 2) + (0.875/1.0053 * 3) + (0.875/1.0054 * 4) + (100.875/1.0055 * 5)] ≈ 4.914938656 Dmod = DMac / 1.005 ≈ 4.890486225 3. Set y = 3% FV = 100 / (1 + 0.03)7 ≈ 81.30915113 DMac = (1/81.30915113) * [100 / (1 + 0.03)7] * 7 ≈ 7 Dmod = DMac / (1 + 0.03) ≈ 6.796116505 _ Calculate the yield using bisection and Newton-Raphson. 1. Denote the target price by Btarget. Use values 80 <= Btarget <= 100. 2. Use a tolerance value of Btol = 10 -2 (one cent). 3. That is to say, iterate to calculate the yield y such that |B(y) - Btarget| <= Btol. _ Example cases are: 1. c = 4, n = 4, f = 2, Btarget = … . 2. c = 3.5, n = 5, f = 4, Btarget = … . 3. zero-coupon bond: c = 0, n = 7, Btarget = … . _ Count the number of iterations. In most cases Newton-Raphson should converge faster. 1. Set Btarget = 98.11 bisection: ylow = 0: FV = 2/1 + 2/1 2 + 2/13 + 102/14 = 108 yhigh = 1%: FV = 2/1.005 + 2/1.005 2 + 2/1.0053 + 102/1.0054 ≈ 105.9257435 yhigh = 2%: FV = 2/1.01 + 2/1.01 2 + 2/1.013 + 102/1.014 ≈ 103.9019656 yhigh = 3%: FV = 2/1.015 + 2/1.015 2 + 2/1.0153 + 102/1.0154 ≈ 101.9271923 yhigh = 4%: FV = 2/1.02 + 2/1.02 2 + 2/1.023 + 102/1.024 = 100 yhigh = 5%: FV = 2/1.025 + 2/1.025 2 + 2/1.0253 + 102/1.0254 ≈ 98.1190129 yhigh = 6%: FV = 2/1.03 + 2/1.03 2 + 2/1.033 + 102/1.034 ≈ 96.2829016 |B(y = 5%) – Btarget| = |98.1190129 – 98.11| = 0.0090129 < 10-2 = Btol The number of iterations under our condition Btarget = 98.11 could be 6. Newton-Raphson: y0 = 0: FV = 2/1 + 2/1 2 + 2/13 + 102/14 = 108 DMac = (1/108) * [(2/1 * 1) + (2/1 2 * 2) + (2/13 * 3) + (102/14 * 4)] ≈ 3.888888889 Dmod = DMac / 1 ≈ 3.888888889 f’(y0 = 0) = -3.888888889 * 108 ≈ -420 ΔFV / f’(y0 = 0) = Δy = (98.11 - 108) / (-420) ≈ 2.354761905% The y needs to increase at least 2.354761905%. 2. Set Btarget = 97.01 bisection: ylow = 0: FV = 0.875/1 + 0.875/1 2 + 0.875/13 + 0.875/14 + 100.875/15 = 104.375 yhigh = 1%: FV = 0.875/1.0025 + 0.875/1.0025 2 + 0.875/1.00253 + 0.875/1.00254 + 100.875/1.00255 ≈ 103.1016985 yhigh = 2%: FV = 0.875/1.005 + 0.875/1.005 2 + 0.875/1.0053 + 0.875/1.0054 + 100.875/1.0055 ≈ 101.8471999 yhigh = 3%: FV = 0.875/1.0075 + 0.875/1.0075 2 + 0.875/1.00753 + 0.875/1.00754 + 100.875/1.00755 ≈ 100.61118 yhigh = 4%: FV = 0.875/1.01 + 0.875/1.01 2 + 0.875/1.012 + 0.875/1.014 + 100.875/1.015 ≈ 99.3933211 yhigh = 5%: FV = 0.875/1.0125 + 0.875/1.0125 2 + 0.875/1.01253 + 0.875/1.01254 + 100.875/1.01255 ≈ 98.19331186 yhigh = 6%: FV = 0.875/1.015 + 0.875/1.015 2 + 0.875/1.0153 + 0.875/1.0154 + 100.875/1.0155 ≈ 97.01084689 yhigh = 7%: FV = 0.875/1.0175 + 0.875/1.0175 2 + 0.875/1.01753 + 0.875/1.01754 + 100.875/1.01755 ≈ 95.84562681 |B(y = 6%) – Btarget| = |97.01084689 – 97.01| = 0.00084689 < 10-2 = Btol The number of iterations under our condition Btarget = 97.01 could be 7. Newton-Raphson: y0 = 0: FV = 0.875/1 + 0.875/1 2 + 0.875/13 + 0.875/14 + 100.875/15 = 104.375 DMac = (1/104.375) * [(0.875/1 * 1) + (0.875/1 2 * 2) + (0.875/13 * 3) + (0.875/14 * 4) + (104.375/15 * 5)] ≈ 5.083832335 Dmod = DMac / 1 ≈ 5.083832335 f’(y0 = 0) = -5.083832335 * 104.375 ≈ -530.625 ΔFV / f’(y0 = 0) = Δy = (97.01 – 104.375) / (-530.625) ≈ 1.387985866% The y needs to increase at least 1.387985866%. 3. Set Btarget = 93.27 ylow = 0: FV = 100/1 7 = 100 yhigh = 1%: FV = 100 / [1 + (0.01/7)] 7 ≈ 99.00568988 yhigh = 2%: FV = 100 / [1 + (0.02/7)] 7 ≈ 98.02266262 yhigh = 3%: FV = 100 / [1 + (0.03/7)] 7 ≈ 97.05077437 yhigh = 4%: FV = 100 / [1 + (0.04/7)] 7 ≈ 96.08988334 yhigh = 5%: FV = 100 / [1 + (0.05/7)] 7 ≈ 95.13984974 yhigh = 6%: FV = 100 / [1 + (0.06/7)] 7 ≈ 94.20053574 yhigh = 7%: FV = 100 / (1 + 0.01) 7 ≈ 93.27180547 yhigh = 8%: FV = 100 / [1 + (0.08/7)] 7 ≈ 92.35352496 |B(y = 7%) – Btarget| = |93.27180547 – 93.27| = 0.00180547 < 10-2 = Btol The number of iterations under our condition Btarget = 93.27 could be 8. Newton-Raphson: y0 = 0: FV = 100/1 7 = 100 DMac = (1/100) * [100/1 7] * 7 = 7 Dmod = DMac / 1 = 7 f’(y0 = 0) = -7 * 100 = -700 ΔFV / f’(y0 = 0) = Δy = (93.27 – 100) / (-700) ≈ 0.9614285714% The y needs to increase at least 0.9614285714%.
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