Let's go step by step through your **Hard Stone Investment (HSI) CMBS Securitization** problem. --- ## **Given Data Recap** - **Mortgage Pool**: R100 million, 10 equal loans (R10m each). - 4 mature in 1 year (R40m), 6 mature in 2 years (R60m). - **Tranches**: - **A (Senior IG)**: R65m, coupon 7%, sold at par (R65m), credit support 35%. - **B (Junior Non-IG)**: R35m, coupon 10%, sold at discount (face R35m). - **X (IO Strip)**: R100m notional (interest only strip from Tranche A), sold at R100m, YTM 7%. - **Weighted Average Coupon of Pool**: 10% - **Tranche Sale Prices**: - A: Par (R65m) - B: Discount (R35m) - X: Par (R100m notional) - **Credit Losses**: R8m in year 2. - **YTM**: - A: 7% - B: 11% - X: 7% --- # **1. Coupon of Tranche X (IO)** **Tranche X (IO)** is the *Interest-Only* strip, which means it receives all the interest payments (the coupon cash flows) from the underlying mortgage pool, but none of the principal. - **Total pool coupon:** 10% × R100m = **R10m/year** - **Tranche X Coupon:** Receives the *excess* interest not paid to Tranches A and B, or, more commonly, receives all the interest from the underlying loans (if A and B are principal-only). But here, the description states IO is created from the *senior investment grade CMBS Tranche A valued at R100m*. Usually, IO means it gets all the interest from the underlying pool: - **Therefore, Coupon of Tranche X (IO) = 10% × R100m = R10m/year** **Answer 1:** > **Coupon for Tranche X (IO): R10 million per year** --- # **2. Cash Flows of Each Tranche in Year 1 and Year 2** Let's break this down: ## **a. Mortgage Pool Cash Flows** - **Year 1:** - Principal: 4 loans mature = R40m - Interest: 10% × R100m = R10m - **Year 2:** - Principal: 6 loans mature = R60m - Interest: 10% × R60m = R6m **(since only outstanding balance earns interest)** - **Credit loss:** R8m (assumed to occur in Year 2, reduces principal repaid) - **Total principal repaid in Year 2:** R60m - R8m = R52m **Summary Table:** | Year | Principal | Interest | Credit Loss | Total Cash Flow | |------|-----------|----------|-------------|-----------------| | 1 | R40m | R10m | 0 | R50m | | 2 | R52m | R6m | R8m | R58m | --- ## **b. Tranche Payment Waterfall** - **Tranche A is senior, B is subordinated** - **Tranche X gets all the interest (IO)** - **Principal is paid to A first, then B** - **Credit loss is absorbed by B (since A has 35% credit support = R35m, matching B's size)** ### **Year 1:** - **Interest:** R10m (all goes to Tranche X) - **Principal:** R40m (all goes to pay down Tranche A balance) #### **Tranche A:** - Receives R40m principal (reducing balance from R65m to R25m) - Receives nothing in interest (since IO strips it?) #### **Tranche B:** - Receives nothing in Year 1 (subordinated, no principal left after A is paid) #### **Tranche X:** - Receives all R10m interest ### **Year 2:** - **Interest:** Outstanding principal is now R60m (6 loans), so interest is 10% × R60m = R6m (goes to IO) - **Principal:** Scheduled R60m, but R8m loss → **R52m** actual - First, pay off remaining Tranche A (R25m) - Remaining principal available: R52m - R25m = R27m - Tranche B receives up to R27m principal (but B's face is R35m, so B suffers a loss of R8m) #### **Tranche A:** - Receives R25m principal (fully paid off) #### **Tranche B:** - Receives R27m principal (R8m shortfall, so only 77% of face value repaid) - Suffers all R8m credit loss #### **Tranche X:** - Receives R6m interest --- ## **Summary Table** | Year | Tranche A | Tranche B | Tranche X (IO) | |------|-----------|-----------|---------------| | 1 | R40m principal | 0 | R10m interest | | 2 | R25m principal | R27m principal | R6m interest | --- # **3. Weighted Average Maturity (WAM)** **a. Mortgage Pool:** - 4 loans: mature in 1 year (R40m) - 6 loans: mature in 2 years (R60m) \[ \text{WAM} = \frac{(R40m \times 1) + (R60m \times 2)}{R100m} = \frac{40 + 120}{100} = 1.6 \text{ years} \] **b. Tranche A:** - Receives R40m principal in Year 1, R25m in Year 2 - WAM = (R40m × 1 + R25m × 2) / R65m = (40 + 50) / 65 ≈ 1.385 years **c. Tranche B:** - Receives R27m principal in Year 2 (only, nothing in Year 1) - WAM = (R27m × 2) / R35m = 54 / 35 ≈ 1.543 years --- # **4. IRR of the CMBS and Mortgage Pool** **a. Mortgage Pool IRR** - Outflows: -R100m at t=0 - Inflows: R50m at t=1, R58m at t=2 \[ \text{IRR:} -100 + \frac{50}{(1+IRR)} + \frac{58}{(1+IRR)^2} = 0 \] Let’s solve: Set IRR = x \[ -100 + \frac{50}{1+x} + \frac{58}{(1+x)^2} = 0 \] Try x = 10% (0.1): \[ -100 + \frac{50}{1.1} + \frac{58}{1.21} = -100 + 45.45 + 47.93 = -6.62 \] Try x = 13% (0.13): \[ -100 + \frac{50}{1.13} + \frac{58}{1.2769} = -100 + 44.25 + 45.42 = -10.33 \] Try x = 7% (0.07): \[ -100 + \frac{50}{1.07} + \frac{58}{1.1449} = -100 + 46.73 + 50.67 = -2.6 \] Try x = 5% (0.05): \[ -100 + \frac{50}{1.05} + \frac{58}{1.1025} = -100 + 47.62 + 52.60 = +0.22 \] So, IRR is about 5%. **Answer: IRR of the Mortgage Pool is approximately 5%.** --- # **5. IRR of Each Tranche Based on Realized and Scheduled Cash Flows** ## **A. Tranche A:** - Outflow: -R65m (purchase price at t=0) - Inflow: R40m at t=1, R25m at t=2 - **No interest received** (since all goes to IO) \[ -65 + \frac{40}{(1+IRR)} + \frac{25}{(1+IRR)^2} = 0 \] Try IRR = 7% (0.07): \[ -65 + \frac{40}{1.07} + \frac{25}{1.1449} = -65 + 37.38 + 21.83 = -5.79 \] Try IRR = 5%: \[ -65 + \frac{40}{1.05} + \frac{25}{1.1025} = -65 + 38.10 + 22.68 = -4.22 \] Try IRR = 10%: \[ -65 + \frac{40}{1.1} + \frac{25}{1.21} = -65 + 36.36 + 20.66 = -7.98 \] So, IRR is less than 7%. Try IRR = 0%: \[ -65 + 40 + 25 = 0 \] So, IRR is 0%. This suggests that **Tranche A only receives its principal back, and no interest**. In reality, Tranche A holders expect to get a coupon, but in this structure, IO strips all the interest from the underlying pool, so Tranche A is a principal-only security. ## **B. Tranche B:** - Outflow: -R35m at t=0 - Inflow: R27m at t=2 (all principal repaid in year 2, none in year 1, and no interest) - **IRR:** \[ -35 + \frac{27}{(1+IRR)^2} = 0 \implies (1+IRR)^2 = \frac{27}{35} \implies (1+IRR) = \sqrt{\frac{27}{35}} = 0.878 \implies IRR = -12.2\% \] Negative IRR, as expected, since B suffers the full credit loss. ## **C. Tranche X (IO):** - Outflow: -R100m at t=0 - Inflows: R10m at t=1, R6m at t=2 \[ -100 + \frac{10}{(1+IRR)} + \frac{6}{(1+IRR)^2} = 0 \] Try IRR = 7% (0.07): \[ -100 + \frac{10}{1.07} + \frac{6}{1.1449} = -100 + 9.35 + 5.24 = -85.41 \] IRR is much lower; in fact, the IO strip is a premium bond, and these cash flows are much less than the initial outlay. So, this cannot be correct unless the IO is sold at a much lower price (probably should be sold at the NPV of the cash flows, not at par). Let's check the IRR: \[ -100 + 10/(1+IRR) + 6/(1+IRR)^2 = 0 \] Try IRR = -80% (very negative): \[ (1-0.8) = 0.2 -100 + 10/0.2 + 6/0.04 = -100 + 50 + 150 = 100 \] Try IRR = -50%: \[ 1-0.5 = 0.5 -100 + 10/0.5 + 6/0.25 = -100 + 20 + 24 = -56 \] Try IRR = -60%: \[ 1-0.6=0.4 -100 + 10/0.4 + 6/0.16 = -100 + 25 + 37.5 = -37.5 \] So, even with very negative IRR, the sum is still negative. This implies the sale price is much higher than the NPV of future cash flows, i.e., the IO is overpriced. **Alternatively**, IRR for IO is negative and cannot be positive at that price. --- # **6. Percentage Value Increase of the Pool** - **Total sale proceeds:** Tranche A (R65m) + Tranche B (R35m) + Tranche X (R100m) = **R200m** - **Original Pool Value:** R100m \[ \text{Percentage value increase} = \frac{200 - 100}{100} \times 100\% = 100\% \] **Note:** In reality, you cannot sell both principal tranches and the IO strip at face value; otherwise, the total proceeds double-count the interest. One would expect a much lower sale price for the IO strip. --- ## **Final Answers Table** | Question | Answer | |----------|--------| | 1. Coupon of Tranche X (IO) | **R10 million per year (Year 1); R6