03964.com

文档资料库 文档搜索专家

文档资料库 文档搜索专家

Case 7 - Solution / expected answers

Note: For extreme precision, all intermediate computations have been carried out with exact values. Figures of intermediate results in this solution may be rounded for simplicity of notation only. Final results are the only ones to be rounded according to the rules of the game. PART 1 Question 1: The correct answers are presented in table 1: Table 1 Maturity Date 15/02/2012 15/08/2012 01/01/2013 15/02/2013 15/08/2013 01/01/2014 15/02/2014 15/08/2014 01/01/2015 15/02/2015 15/08/2015 01/01/2016 15/02/2016 15/08/2016 01/01/2017 15/02/2017 Interest rate 0.0625% 0.0625% 0.0625% 0.0934% 0.2174% 0.3126% 0.3165% 0.3321% 0.3441% 0.3946% 0.5979% 0.7539% 0.8009% 0.9911% 1.1363% 1.1363%

Explanations : We obtain the zero coupon interest rates using the bootstrapping technique outlined in technical document 2. The zero coupon rate for 01/01/2013 is computed using the price quote on the zero coupon bond with a 1 year maturity. Deducing the interest rate from the bond pricing formula P=N*e-rT, we get: 1 = 1 1 100 ? ln ? ? = ? ln ? ? = 0.0625% 1 99.9375

To get the zero coupon rate for the 2 year maturity, it’s necessary to deduce the rate from the bond price equation: = 1 ? ?1 1 + (2 + ) ? ?2 2 , where c i is the coupon payment for period i, r i is the corresponding zero coupon rate and T i is the time from today in years. 2 = ? ln ??

2

finding the rates for maturities 3, 4 and 5 years from now (01/01/2012). 3 = 1 3 + ? ln ? ? 3 ? 1 ? ?1 1 ? 2 ? ?2 2

1

2 +

1 ?

?1 1

? = 2 ? ln ?99.875?0.25? ?0.0625%?1 ?The same method applies for

1

0.25+100

5 =

1 5 + ? ln ? ? ? ? 5 ? 1 ? 1 1 ? 2 ? 2 2 ? 3 ? ?3 3 ? 4 ? ?4 4

4 =

1 4 + ? ln ? ? ? 1 1 ? 2 ? ?2 2 ? 3 ? ?33 4 ? 1 ?

The rates between the January 1 data points are interpolated linearly. For example, the zero coupon rate for maturity on 15/02/2013 is found as follows: (0.0625%*320 + 0.3126%*45)/365 = 0.0934% , where 320 is the number of days from 15/02/2013 to 01/01/2014, 45 is the number of days from 01/01/2013 to 15/02/2013 and 365 is the number of days from 01/01/2013 to 01/01/2014. Similarly, the zero coupon rate for maturity on 15/08/2013 is (0.0625%*139 + 0.3126%*226)/365 = 0.2174%, where 139 is the number of days from 15/08/2013 to 01/01/2014, 226 is the number of days from 01/01/2013 to 15/08/2013 and 365 is the number of days from 01/01/2013 to 01/01/2014. The same method applies to all rates with maturities in between the full year marks.

Finally, the curve is flat outside the scope of the given data, thus, the zero coupon rate for maturities on 15/02/2012 and 15/08/2012 equals the rate at 01/01/2013 (0.0625%) and the rate for maturity on 15/02/2017 equals the one on 01/01/2017 (1.1363%). Question 2: The correct answers are presented in table 2: Table 2 Bond Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Explanation The value of a bond is equal to the present value of all cash-flows associated with the bond: all future coupon payments discounted at the appropriate discount rates and the discounted value of the principal payment at maturity. = ?

Maturity Date 15/02/2013 15/02/2013 15/08/2013 15/08/2013 15/02/2014 15/02/2014 15/08/2014 15/08/2014 15/02/2015 15/02/2015 15/08/2015 15/08/2015 15/02/2016 15/02/2016 15/08/2016 15/08/2016 15/02/2017 15/02/2017

Annual Coupon Rate Bond Value (%) 5.625% 1.625% 2.375% 7.000% 2.250% 3.375% 3.250% 6.125% 3.000% 3.375% 3.250% 4.125% 3.000% 2.625% 3.500% 6.000% 3.500% 6.500% 108.33 102.33 104.39 113.63 104.94 107.75 108.85 117.45 109.22 110.53 110.77 114.25 110.12 108.45 112.79 125.13 113.26 129.48

=0

( )

Note that today’s date is considered to be 01/01/2012. Since the bonds maturities are 15/02 and 15/08 every year and the coupon payments take place twice a year, the first cash flow date is 15/02/2012. The cash flows for the bonds, the corresponding discount rates and time periods (how much to discount) are in the table below:

Date of 15/02/12 15/08/12 15/02/13 15/08/13 15/02/14 15/08/14 15/02/15 15/08/15 15/02/16 15/08/16 15/02/17 cash flow Discount 0.0625% 0.0625% 0.0934% 0.2174% 0.3165% 0.3321% 0.3946% 0.5979% 0.8009% 0.9911% 1.1363% rate Time period 0.123288 0.621918 1.126027 1.621918 2.126027 2.621918 3.126027 3.621918 4.126027 4.624658 5.128767 (years) Bond Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 2.8125 0.8125 1.1875 3.5 1.125 1.6875 1.625 3.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 2.8125 0.8125 1.1875 3.5 1.125 1.6875 1.625 3.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 102.8125 100.8125 1.1875 3.5 1.125 1.6875 1.625 3.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 101.1875 103.5 1.125 1.6875 1.625 3.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 101.125 101.6875 1.625 3.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 101.625 103.0625 1.5 1.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 101.5 101.6875 1.625 2.0625 1.5 1.3125 1.75 3 1.75 3.25 101.625 102.0625 1.5 1.3125 1.75 3 1.75 3.25 101.5 101.3125 1.75 3 1.75 3.25 101.75 103 1.75 3.25 101.75 103.25

Pay attention when discounting the cash flows: there are no cash flows on 01/01 every year and you don’t need the zero coupon rates for maturities on 01/01 which were also computed before.

Question 3: The correct answers are presented in table 3: Table 3 Bond Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Explanation : The dirty price of a bond is the “all in price” of the bond as it includes the accrued interests on the bond. The dirty price is obtained by adding the accrued interest to the clean price. Dirty price = Clean price + accrued interest The accrued interest is computed as follows: Accrued interest = coupon amount ? Maturity Date 15/02/2013 15/02/2013 15/08/2013 15/08/2013 15/02/2014 15/02/2014 15/08/2014 15/08/2014 15/02/2015 15/02/2015 15/08/2015 15/08/2015 15/02/2016 15/02/2016 15/08/2016 15/08/2016 15/02/2017 15/02/2017 Annual Coupon Clean Price Dirty Price Rate (%) 5.625% 1.625% 2.375% 7.000% 2.250% 3.375% 3.250% 6.125% 3.000% 3.375% 3.250% 4.125% 3.000% 2.625% 3.500% 6.000% 3.500% 6.500% 100-22 99-4 98-20 102-8 97-18 99-0 98-16 103-21 97-18 98-12 97-21 100-1 96-15 95-9 97-23 106-19 97-13 109-9 102.81 99.74 99.52 104.89 98.41 100.27 99.73 105.97 98.70 99.65 98.88 101.59 97.60 96.27 99.04 108.86 98.73 111.74

Today’s date is considered to be 01/01/2012, thus, the last coupon payment on these bonds was 15/08/2011, that means that 139 days has passed since the last coupon date (number of

days from 15/08/2011 to 01/01/2012). The next coupon payment occurs on 15/02/2012, so there are 184 days in the current coupon period (number of days from 15/08/2011 to 15/02/2012). For example, the dirty price of bond 1 is: (100 + 22/32) + 100*5.625%/2*(139/184) = 102.81

Question 4: The correct answers are presented in table 4: Table 4 a0 Coefficients Explanation To find the coefficients, it’s necessary to set up a bond pricing model using the interest rates computed with the polynomial below as the corresponding discount rates for each cash flow period: () = 0 + 1 + 2 2 0.03800 a1 0.00 a2 0.0001

If the bond prices using this model (dependent on the three coefficients) are denoted with Q j and the dirty prices computed in question 3 with P j (j denotes the bond number), then the aim is to minimize the sum of squared errors between these two prices. Thus, use Excel solver to minimize:

2 ?( ? ) =1 18

by changing the values of the three coefficients (a 0 , a 1 and a 2 ), with the restriction that a 0 and a 1 are greater or equal to 0. a 1 needs to be positive also because it is required that the yield curve would not be inverted.

Question 5 The correct answers are presented in table 5 Table 5 a0 a1 a2 Long term rate Slope of the interest rate term structure curve Curvature of the interest rate term structure

Explanation Principal Component Analysis (PCA) has showed that there are three factors driving the interest rate term structure: long term rate, slope of the interest rate term structure and the curvature of the interest rate term structure. The first order parameter in a mathematical relationship determines the curve’s slope; the second order parameter determines its curvature.

Question 6 The correct answers are presented in table 6: Table 6 Bond Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Explanation The duration of a bond is computed as follows: =

1 ? ? ( ) ? =0

Maturity Date 15/02/2013 15/02/2013 15/08/2013 15/08/2013 15/02/2014 15/02/2014 15/08/2014 15/08/2014 15/02/2015 15/02/2015 15/08/2015 15/08/2015 15/02/2016 15/02/2016 15/08/2016 15/08/2016 15/02/2017 15/02/2017

Annual Coupon Rate (%) 5.625% 1.625% 2.375% 7.000% 2.250% 3.375% 3.250% 6.125% 3.000% 3.375% 3.250% 4.125% 3.000% 2.625% 3.500% 6.000% 3.500% 6.500%

Duration 1.0869 1.1141 1.5879 1.5297 2.0723 2.0476 2.5102 2.4268 2.9819 2.9658 3.4173 3.3701 3.8814 3.9087 4.2771 4.0876 4.7058 4.4416

where P is the bond price, PV(CF t ) is the present value of the cash flow at time t and T the maturity. For example, duration for bond 1 is thus computed as follows:

=

Note that for P, we use the prices computed in question 2 because those prices are based on the interest rate term structure found in question 1.

1 108.3282 ? [2.8125 ? ?0.0625%?0.1233 ? 0.1233 + 2.8125 ? ?0.0625%?0.6219 ? 0.6219 + 102.8125 ? ?0.0934%?1.1260 ? 1.1260] = 1.0869

Question 7 The correct answers are presented in table 7: Table 7 Bond Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Explanation Convexity is computed as follows: Maturity Date 15/02/2013 15/02/2013 15/08/2013 15/08/2013 15/02/2014 15/02/2014 15/08/2014 15/08/2014 15/02/2015 15/02/2015 15/08/2015 15/08/2015 15/02/2016 15/02/2016 15/08/2016 15/08/2016 15/02/2017 15/02/2017 Annual Coupon Rate (%) 5.625% 1.625% 2.375% 7.000% 2.250% 3.375% 3.250% 6.125% 3.000% 3.375% 3.250% 4.125% 3.000% 2.625% 3.500% 6.000% 3.500% 6.500% Convexity 2.2952 2.3607 4.1299 3.9519 6.4044 6.3123 8.9448 8.5797 12.0710 11.9899 15.3836 15.1165 19.2774 19.4489 23.1039 21.7951 27.5435 25.5529

=

( ) ? ( 2 + ) 1 ?? (1 + )2 =0

Please note that the term 1/(1+r t )2 cannot be brought in front of the summation equation in this case because a different interest rate applies for each period. For example, convexity for bond 1 is thus computed as follows: = 1 108.3282 2.8125 ? ?0.0625%?0.1233 ? (0.12332 + 0.1233) ?? (1 + 0.0625%)2 2.8125 ? ?0.0625%?0.6219 ? (0.62192 + 0.6219) + (1 + 0.0625%)2 102.8125 ? ?0.0934%?1.1260 ? (1.12602 + 1.1260) + ? = 2.2952 (1 + 0.0934%)2

Question 8 The correct answers are presented in table 8 Table 8 Bonds Weights in % Bonds Weights in % Explanation 1 2 3 4 5 6 7 8 9 0.00% 3.446% 1.04% 0.00% 17.84% 0.00% 2.23% 0.00% 37.39% 10 11 12 13 14 15 16 17 18 0.00% 5.22% 0.00% 0.00% 23.01% 2.726% 0.00% 7.10% 0.00%

The aim of this task is to determine the number of each bond that needs to be bought so as to meet the company’s liabilities most efficiently (at the lowest cost). To solve for the weights, it’s necessary to set up a financial model. First, you need to compute the cash inflow from the bonds for each given date (t) when a specific liability has to be matched. The cash inflow (coupon payment and/or principal payment) from each bond has to be dependent on the number of units of this specific bond included in the portfolio. ? = ? ,

=1 18

, where x i is the number of bonds i that are bought.

For each liability date, the cash inflow from all bonds must equal or exceed the cash outflow necessary to meet the liability. Do not forget to take into account the excess cash inflow which can be transferred to the next period (the excess cash doesn’t earn interest, though), thus, the cash inflow from bonds required for a period may be less than the liability if the excess cash from the previous period covers the difference. Second, you need to compute the cost of the portfolio which must be dependent on the number of bonds. The total cost of the portfolio is: = ?

=1 18

Note that you need to use the dirty bond prices computed in question 3.

Use Excel solver to minimize the cost of portfolio by changing the number of bonds variable for each bond (starting values are 100 for all bonds), with a constraint that for every period, the cash inflow (from bonds + excess cash from previous period) is equal or larger than the liability. You may have to tick the box “Make unconstrained variables non-negative” Each bond’s weight (value based) in the portfolio is then the number of bonds bought times the price of the bond divided by the total cost of the portfolio.

=

?

Question 9 The correct answers are presented in table 9 Table 9 Bonds Weights in % Bonds Weights in % 1 2 3 4 5 6 7 8 9 0.00% 15.36% 0.00% 0.00% 9.55% 0.00% 0.00% 0.00% 37.21% 10 11 12 13 14 15 16 17 18 0.00% 5.20% 0.00% 0.00% 22.90% 2.714% 0.00% 7.07% 0.00%

The same process as in the previous question applies here. You need to match the liabilities for the company at a minimum cost. The only difference is an additional restriction that the duration of the portfolio must equal 3.

The duration of the portfolio equals the weighted average of the bond durations.

= ?

=1

18

PART 2 Question 1 The probability that JH Brothers bond will be rated below IG in three years is 2.762%. Explanation First, it’s necessary to transform the one year rating transition matrix into a 3 year transition matrix. It’s convenient to use MMULT function to first multiply the one year matrix with itself to get the two year matrix and then to multiply the one and two year matrices to get the three year transition matrix below: From / To A ABBB+ BBB BBBBB+ BBA ABBB+ BBB BBBBB+ BB-

90.55 9.16 0.32 0.01 0.00 0.00 0.00

9.16 82.69 8.80 0.54 0.01 0.00 0.00

0.29 7.77 78.87 14.21 0.95 0.03 0.00

0.00 0.38 11.36 72.51 14.52 1.17 0.14

0.00 0.01 0.52 9.97 69.64 16.60 5.56

0.00 0.00 0.11 2.44 10.95 70.16 58.00

0.00 0.00 0.01 0.32 3.93 12.05 36.30

At a second step, look at the transition distribution for BBB (the red row). The probability that the rating will drop from BBB to either BB+ or BB- (below investment grade)is 2.44%+0.32% = 2.762%.

Question 2 The probability that JH Brothers defaults within 2 years is 1.98%. Explanation In our case, we have a Poisson process with parameter Lambda equal to 0.01t, the probability of no default (zero occurrences) in 2 years is P[X t =0] = exp(-λt) = exp(0.01*2)=98.02%. The default probability is (1 - probability of survival) = 1 – 98.02% = 1.98%

Question 3 The CDS spread is 60 bps. Explanation The expected value of the default leg is the sum of the present values of the payment in case of default times the probability of default for the given time periods. The expected value of the premium leg is the present value of the fees times the probability of payment (the company is still solvent). The fair price of a CDS (the spread) is such that the expected values of the two legs are equal. The CDS Spread is quoted per annum in basis points (1 basis point = 0.01%). The value of the default leg is: = ? ? ? ? ?

=0

, where LGD is the loss given default, N is the notional of the credit (what is to be insured), r is the interest rate, T is the maturity and PD t is the probability of default at period t. Note that you need to take steps of 0.5 with t because the default may occur twice a year at the time of premium payments. The value of the premium leg is = ?

=0

, where spread is the annual CDS spread rate, N is the notional of the credit (what is to be insured), r is the interest rate, T is the maturity and PS t is the probability that the company is

? ? ? ? 2

still solvent at period t. Note that you need to take steps of 0.5 with t and the annual spread rate is divided by 2 because the premium payments occur semi-annually. Setting the two legs equal, we can solve for the CDS spread: = 2 ? ?

? ∑ ? =0 ? ∑=0 ?

Note that the spread rate does not depend on the size of the credit. The default and survival probabilities are computed in the table below. PS (probability of survival) equals exp(-0.01*t) according to the Poisson process; PD (probability of default) at time t is PS one period before (t-1) minus PS at time t. T (years) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 PS 99.50% 99.00% 98.51% 98.02% 97.53% 97.04% 96.56% 96.08% 95.60% 95.12% PD 0.499% 0.496% 0.494% 0.491% 0.489% 0.486% 0.484% 0.482% 0.479% 0.477%

From the spread formula, we get a CDS spread of 60.15 bps (0.60%).

Question 4 The estimated cost of the insurance is 23 041.66 Aces. Explanation The cost of the insurance is the expected value of the premium leg. 0.5% ? ? ? ? ? = ? 1 ? ? = 2 2

=0 =0 5

The assumptions are the same as in previous question, thus, we have the same probabilities of survival (table below) and the formula above gives the result 23 041.66 Aces. t (years) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 PS 0.9950 0.9900 0.9851 0.9802 0.9753 0.9704 0.9656 0.9608 0.9560 0.9512 Discounted PS 0.9851 0.9704 0.9560 0.9418 0.9277 0.9139 0.9003 0.8869 0.8737 0.8607

Question 5 The correct answers are presented in the table below Maturity 15/02/2012 15/08/2012 15/02/2013 15/08/2013 15/02/2014 15/08/2014 15/02/2015 15/08/2015 15/02/2016 15/08/2016 15/02/2017 Explanation The hazard rate in this case is computed by taking the difference between the risk bearing rate (part 1 question 4) and risk free rate (government bond curve, part 1 question 1) as follows: Maturity Risk Free rate Default rate hazard rate 15/02/2012 0.063% 3.83% 3.77% 15/08/2012 0.063% 3.83% 3.78% 15/02/2013 0.093% 3.85% 3.76% 15/08/2013 0.217% 3.87% 3.66% 15/02/2014 0.316% 3.90% 3.59% 15/08/2014 0.332% 3.94% 3.60% 15/02/2015 0.395% 3.98% 3.58% 15/08/2015 0.598% 4.02% 3.42% 15/02/2016 0.801% 4.07% 3.27% 15/08/2016 0.991% 4.13% 3.14% 15/02/2017 1.136% 4.20% 3.06% where hazard rate = default rate - risk free rate This is because hazard rate represents the risk premium per marturity, i.e. it is the premium that we add to the risk free rate to get the default rate. hazard rates 3.77% 3.78% 3.76% 3.66% 3.59% 3.60% 3.58% 3.42% 3.27% 3.14% 3.06%

Question 6 The CDS premium is equal to 188 bps. Explanation

Maturity 15/02/2012 15/08/2012 15/02/2013 15/08/2013 15/02/2014 15/08/2014 15/02/2015 15/08/2015 15/02/2016 15/08/2016 15/02/2017 Sum

t (years) 0 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.01

hazard rate 3.77% 3.78% 3.76% 3.66% 3.59% 3.60% 3.58% 3.42% 3.27% 3.14% 3.06%

PS 1 0.981 0.963 0.947 0.931 0.914 0.898 0.887 0.877 0.868 0.858

Disc PS 0.972 0.944 0.919 0.894 0.869 0.846 0.827 0.810 0.793 0.776 8.651

PD 0.019 0.018 0.016 0.016 0.017 0.016 0.011 0.010 0.009 0.010

Disc PD 0.019 0.018 0.016 0.015 0.016 0.015 0.010 0.009 0.008 0.009 0.135

Probability of survival (PS) equals exp(-hazard rate t *t) according to the Poisson process; probability of default (PD) at time t is PS one period before (t-1) minus PS at time t. The value of the default leg is: = ? ? ? ? = 60% ? 0.135 = 0.081

=0

The value of the premium leg is: = ?

=0

Note that the premium leg must be further discounted to today’s date (01/01/2012), so that the premium leg’s value is 4.325*spread*e-2%*(45/365) = 4.315*spread. The default leg does not have to be discounted further because if the default event occurs before the CDS begins, then the CDS is worthless as there is no possible credit default against which to purchase insurance. Setting the two legs equal, we can solve for the spread: 0.081 = 4.315 ?

? ? ? = ? 8.651 = 4.325 ? 2 2

=

0.081 = 0.0188 = 188 4.315

相关文章:

- 巴黎银行-2013-C7_solution.pdf
*巴黎银行-2013-C7*_solution - Case 7 - Solutio

- 浅基础作业-附答案.doc
- 浅基础作业-附答案 - 【习题 21】 (注岩 2012
*C7*) 多层建筑物、

- C7-3-5 系统检测记录.xls
*C7*-3-5 系统检测记录_建筑/土木_工程科技_专业资料。归档编号:*C7*-3-5 子...(GB50339-*2013*) 分包项目经理 系统检测 评定记录 检测结果 备注 该工程所检...

更多相关标签: