Bootstrapping Algorithms¶

Bootstrapping is the standard method for constructing a discount curve from market quotes. The algorithm sequentially extracts discount factors from increasingly longer-maturity instruments, using previously determined values to solve for new ones.

Overview of the Bootstrapping Process¶

The Problem¶

Market data comes in the form of: - Deposit rates (short maturities) - Futures prices (medium maturities) - Swap rates (longer maturities)

Each instrument depends on multiple discount factors. Bootstrapping solves for discount factors sequentially from short to long maturities.

General Principle¶

Given: 1. A market quote for an instrument with maturity \(T_n\) 2. All discount factors \(P(0, T_1), \ldots, P(0, T_{n-1})\) for shorter maturities

Solve for \(P(0, T_n)\) such that the instrument is correctly priced.

Stage 1: Money Market Deposits¶

Instrument Description¶

A deposit with maturity \(T\) and rate \(R_d\) pays: - Invest 1 at time 0 - Receive \(1 + R_d \cdot \delta\) at time \(T\)

where \(\delta\) is the day count fraction (typically ACT/360).

Bootstrapping Formula¶

The deposit rate directly implies the discount factor:

\[ P(0, T) = \frac{1}{1 + R_d \cdot \delta} \]

Example¶

Maturity	Deposit Rate	Day Count	Discount Factor
1M	5.00%	31/360	0.995719
2M	5.10%	62/360	0.991311
3M	5.15%	92/360	0.986932

Stage 2: Futures (Eurodollar/SOFR Futures)¶

Instrument Description¶

A 3-month interest rate future with price \(F\) implies a forward rate:

\[ R_{\text{fut}} = \frac{100 - F}{100} \]

for the period from the futures expiry \(T_1\) to \(T_2 = T_1 + 0.25\).

Convexity Adjustment¶

Futures rates differ from forward rates due to daily margining. The convexity adjustment is:

\[ F(0; T_1, T_2) \approx R_{\text{fut}} - \text{CA}(T_1) \]

A common approximation:

\[ \text{CA}(T_1) \approx \frac{1}{2} \sigma^2 T_1 (T_2 - T_1) \]

where \(\sigma\) is the rate volatility (typically 0.5-1.5%).

Bootstrapping Formula¶

Given \(P(0, T_1)\) from previous stage and forward rate \(F(0; T_1, T_2)\):

\[ P(0, T_2) = \frac{P(0, T_1)}{1 + F(0; T_1, T_2) \cdot (T_2 - T_1)} \]

Sequential Application¶

Futures strip provides a chain:

\[ P(0, T_1) \to P(0, T_2) \to P(0, T_3) \to \cdots \]

Each step uses the forward rate from the corresponding future.

Stage 3: Interest Rate Swaps¶

Instrument Description¶

A payer swap with: - Fixed leg: pays fixed rate \(S\) on notional \(N\) - Floating leg: receives floating rate (e.g., 3M LIBOR/SOFR)

The swap rate \(S_n\) for an \(n\)-year swap is the rate that makes the swap have zero initial value.

Swap Valuation¶

The present value of a swap is:

\[ V_{\text{swap}} = \underbrace{\sum_{i=1}^{n} \delta_i P(0, T_i) \cdot (F_i - S)}_{\text{floating - fixed}} \]

where \(F_i\) is the forward rate for period \(i\).

At inception, \(V_{\text{swap}} = 0\) when \(S = S_n\) (the par swap rate).

Par Swap Rate Formula¶

The par swap rate satisfies:

\[ S_n = \frac{P(0, T_0) - P(0, T_n)}{\sum_{i=1}^n \delta_i P(0, T_i)} \]

For a spot-starting swap with \(T_0 = 0\) (so \(P(0, T_0) = 1\)):

\[ S_n = \frac{1 - P(0, T_n)}{\sum_{i=1}^n \delta_i P(0, T_i)} \]

Bootstrapping Formula¶

Rearranging to solve for the unknown \(P(0, T_n)\):

\[ P(0, T_n) = \frac{1 - S_n \sum_{i=1}^{n-1} \delta_i P(0, T_i)}{1 + S_n \delta_n} \]

This requires all previous discount factors \(P(0, T_1), \ldots, P(0, T_{n-1})\).

Sequential Application¶

Given swap rates for maturities 1Y, 2Y, 3Y, ..., 30Y:

Use deposits/futures to get \(P(0, T)\) for \(T \leq 1Y\)
Use 1Y swap rate to get \(P(0, 1Y)\) (or verify consistency)
Use 2Y swap rate + \(P(0, 1Y)\) to get \(P(0, 2Y)\)
Continue inductively...

Complete Bootstrapping Algorithm¶

``` Input: deposits {(T_d, R_d)}, futures {(T_1, T_2, F)}, swaps {(T_n, S_n)}

Stage 1: Deposits (short end)¶

For each deposit (T_d, R_d): δ = DayCount(0, T_d) P(0, T_d) = 1 / (1 + R_d × δ)

Stage 2: Futures (middle)¶

For each future (T_1, T_2, F) in chronological order: R_fut = (100 - F) / 100 F_fwd = R_fut - ConvexityAdjustment(T_1, T_2) P(0, T_2) = P(0, T_1) / (1 + F_fwd × (T_2 - T_1))

Stage 3: Swaps (long end)¶

For each swap (T_n, S_n) in chronological order: annuity = Σ_{i=1}^{n-1} δ_i × P(0, T_i) P(0, T_n) = (1 - S_n × annuity) / (1 + S_n × δ_n)

Output: Discount curve {P(0, T)} ```

Interpolation Between Nodes¶

The Need for Interpolation¶

Bootstrapping produces discount factors at discrete maturities. Pricing arbitrary instruments requires values at any maturity.

Common Interpolation Methods¶

Linear on Log-Discounts:

\[ \log P(0, T) = \frac{T_2 - T}{T_2 - T_1} \log P(0, T_1) + \frac{T - T_1}{T_2 - T_1} \log P(0, T_2) \]

Equivalent to linear interpolation on continuously compounded zero rates
Simple, but produces discontinuous forward rates

Cubic Spline on Zero Rates:

Fit a cubic spline \(z(T)\) through the bootstrapped zero rates.

Produces smooth zero curve
May produce oscillating forward rates

Monotone Convex:

Specialized method ensuring: - Positive forward rates - Monotone forwards in each segment - Local control (changes don't propagate)

Forward Rate Implications¶

The choice of interpolation strongly affects forward rates:

Method	Zero Curve	Forward Curve
Linear on log-discounts	Piecewise linear	Piecewise constant (jumps at nodes)
Cubic spline	Smooth	Can oscillate, may go negative
Monotone convex	Smooth	Monotone within segments

Multi-Curve Bootstrapping¶

Post-2008 Reality¶

After the 2008 financial crisis, the single-curve framework broke down: - Discounting curve: Based on OIS (overnight indexed swap) rates - Forwarding curves: Tenor-specific (1M, 3M, 6M LIBOR/SOFR)

Dual-Curve Bootstrap¶

Step 1: Bootstrap the OIS curve for discounting

\[ P^{\text{OIS}}(0, T_n) = \frac{1 - S_n^{\text{OIS}} \sum_{i=1}^{n-1} \delta_i P^{\text{OIS}}(0, T_i)}{1 + S_n^{\text{OIS}} \delta_n} \]

Step 2: Bootstrap each tenor curve using OIS for discounting

For a 3M tenor curve, use basis swaps or directly bootstrap from instruments referencing 3M rates, discounting cashflows with \(P^{\text{OIS}}\).

Tenor Basis Spreads¶

The spread between curves of different tenors reflects credit and liquidity risk:

\[ \text{Basis}_{3M/6M} = S_{3M} - S_{6M} \]

This spread was negligible pre-crisis but became significant afterward.

Numerical Considerations¶

Solver Techniques¶

For complex instruments (e.g., in-arrears swaps), analytic inversion may not be possible. Use:

Newton-Raphson: For single discount factor per instrument
Multidimensional optimization: For simultaneous fitting

Error Handling¶

Common issues and remedies:

Issue	Symptom	Remedy
Negative discount factor	\(P(0,T) \leq 0\)	Check input data, adjust interpolation
Non-monotone	\(P(0,T_1) < P(0,T_2)\)	Verify instrument ordering
Discontinuous forwards	Jumps at bootstrap nodes	Use smoother interpolation
Poor convergence	Solver fails	Check for data inconsistency

Verification¶

After bootstrapping, verify by repricing all input instruments:

\[ \max_i |V_i^{\text{model}} - V_i^{\text{market}}| < \epsilon \]

A well-constructed curve should reprice inputs to within bid-ask tolerance.

Example: USD Curve Bootstrap¶

Market Data (Illustrative)¶

Instrument	Maturity	Quote
Deposit	1M	5.25%
Deposit	3M	5.35%
Future (Jun)	6M	94.50
Future (Sep)	9M	94.30
Swap	1Y	5.00%
Swap	2Y	4.80%
Swap	5Y	4.50%
Swap	10Y	4.40%

Bootstrap Results¶

Maturity	Discount Factor	Zero Rate
1M	0.9957	5.25%
3M	0.9867	5.35%
6M	0.9738	5.43%
9M	0.9604	5.53%
1Y	0.9524	4.88%
2Y	0.9074	4.85%
5Y	0.8007	4.45%
10Y	0.6440	4.40%

Key Takeaways¶

Bootstrapping sequentially extracts discount factors from short to long maturities
Three stages: deposits → futures → swaps
Interpolation method critically affects forward rates
Post-2008: Multi-curve framework with OIS discounting is essential
Verification by repricing inputs ensures curve consistency

Exercises¶

Exercise 1. A 6-month deposit quotes a simply compounded rate of \(L = 3.0\%\) with ACT/360 day count and 182 days to maturity. Compute the discount factor \(P(0, 0.5)\) using \(P = 1/(1 + L \cdot \delta)\) where \(\delta = 182/360\). Then compute the continuously compounded zero rate \(R\) such that \(P = e^{-R \cdot 0.5}\).

Solution to Exercise 1

The day count fraction is:

\[ \delta = \frac{182}{360} \approx 0.50556 \]

The discount factor follows from the simple compounding formula:

\[ P(0, 0.5) = \frac{1}{1 + L \cdot \delta} = \frac{1}{1 + 0.03 \times 0.50556} = \frac{1}{1.01517} \approx 0.98506 \]

To find the continuously compounded zero rate \(R\), solve \(P = e^{-R \cdot 0.5}\):

\[ R = -\frac{\ln P(0, 0.5)}{0.5} = -\frac{\ln(0.98506)}{0.5} = -\frac{-0.01507}{0.5} \approx 0.03014 = 3.014\% \]

Verification: \(e^{-0.03014 \times 0.5} = e^{-0.01507} \approx 0.98506\), which matches.

Note that the continuously compounded rate (3.014%) is slightly higher than the simple rate (3.0%) because continuous compounding accumulates interest more frequently, so a higher stated rate is needed to produce the same discount factor.

Exercise 2. Given deposit-derived discount factors \(P(0, 0.25) = 0.9925\) and \(P(0, 0.5) = 0.9845\), and a 1-year par swap rate of \(S_1 = 3.2\%\) with semi-annual payments, bootstrap the 1-year discount factor \(P(0, 1.0)\) by solving \(S_1 = \frac{1 - P(0,1)}{0.5 \cdot P(0, 0.5) + 0.5 \cdot P(0, 1)}\).

Solution to Exercise 2

The par swap rate formula for a 1-year swap with semi-annual payments at \(T = 0.5\) and \(T = 1.0\) is:

\[ S_1 = \frac{1 - P(0, 1)}{0.5 \cdot P(0, 0.5) + 0.5 \cdot P(0, 1)} \]

Substituting \(S_1 = 0.032\), \(P(0, 0.5) = 0.9845\) (using the 0.5-year discount factor for the semi-annual payment schedule):

\[ 0.032 = \frac{1 - P(0, 1)}{0.5 \times 0.9845 + 0.5 \times P(0, 1)} \]

Let \(x = P(0, 1)\). Then:

\[ 0.032 \bigl(0.49225 + 0.5x\bigr) = 1 - x \]

\[ 0.015752 + 0.016x = 1 - x \]

\[ 1.016x = 0.984248 \]

\[ x = P(0, 1) = \frac{0.984248}{1.016} \approx 0.96876 \]

Verification: plugging back,

\[ S_1 = \frac{1 - 0.96876}{0.5 \times 0.9845 + 0.5 \times 0.96876} = \frac{0.03124}{0.49225 + 0.48438} = \frac{0.03124}{0.97663} \approx 0.03199 \approx 3.2\% \]

which confirms the result.

Exercise 3. Explain why bootstrapping must proceed sequentially from short to long maturities. What would happen if you tried to extract \(P(0, 5)\) from a 5-year swap rate without first determining all intermediate discount factors?

Solution to Exercise 3

Bootstrapping must proceed sequentially because each longer-maturity instrument depends on all shorter-maturity discount factors. The par swap rate formula is:

\[ S_n = \frac{1 - P(0, T_n)}{\sum_{i=1}^n \delta_i P(0, T_i)} \]

Rearranging for the unknown \(P(0, T_n)\):

\[ P(0, T_n) = \frac{1 - S_n \sum_{i=1}^{n-1} \delta_i P(0, T_i)}{1 + S_n \delta_n} \]

This formula contains \(P(0, T_1), \ldots, P(0, T_{n-1})\) on the right-hand side. If you attempted to extract \(P(0, 5)\) from a 5-year swap rate directly, you would need \(P(0, 1), P(0, 2), P(0, 3), P(0, 4)\) — all of which appear in the annuity sum. Without these intermediate values, the equation has \(n\) unknowns and only one equation, making it under-determined.

Concretely, a 5-year annual swap has five fixed-leg payments, each discounted at a different discount factor. The single swap rate equation cannot determine five unknowns simultaneously. By solving from shortest to longest maturity, each step introduces exactly one new unknown, keeping the system uniquely solvable.

Exercise 4. Eurodollar futures quote a rate that differs from the forward rate by a convexity adjustment. If the futures rate is \(F_{\text{fut}} = 3.5\%\) and the convexity adjustment is estimated at 5 basis points, what forward rate should be used for bootstrapping? Explain the economic reason for the convexity adjustment.

Solution to Exercise 4

The forward rate used for bootstrapping should be the futures rate minus the convexity adjustment:

\[ F(0; T_1, T_2) = F_{\text{fut}} - \text{CA} = 3.5\% - 0.05\% = 3.45\% \]

The economic reason for the convexity adjustment arises from the difference between futures and forward contracts. Futures are marked to market daily, meaning gains and losses are settled each day. When rates rise, the futures position loses money, and this loss must be financed at the now-higher prevailing rate. Conversely, when rates fall, the gain is reinvested at the lower prevailing rate. This creates a systematic disadvantage for long futures positions (those betting on rate declines) relative to equivalent forward positions.

Mathematically, this is a covariance effect. The futures rate equals the expected rate under the risk-neutral measure, while the forward rate equals the expected rate under the \(T_2\)-forward measure. The convexity adjustment captures the Radon-Nikodym derivative between these measures:

\[ \text{CA} \approx \frac{1}{2} \sigma^2 T_1 (T_2 - T_1) \]

where \(\sigma\) is the rate volatility. The adjustment increases with maturity (\(T_1\)) and with volatility, reflecting that the daily-settlement effect compounds over longer horizons.

Exercise 5. A bootstrapping algorithm uses deposit rates for maturities up to 6 months, futures for 6 months to 2 years, and swap rates from 2 to 30 years. Describe the potential discontinuity in the forward rate curve at the transition points between instrument types. How can this be mitigated?

Solution to Exercise 5

At each transition point between instrument types, different pricing methodologies and conventions meet, which can create discontinuities in the implied forward rate curve.

Deposit-to-futures transition (~6 months): Deposit rates are directly observed simple rates, while futures imply forward rates over 3-month intervals. The deposit-derived curve reflects spot lending rates, while the futures strip reflects market expectations of future rates (adjusted for convexity). These can embed different credit/liquidity premia, causing a jump in the forward curve at the switchover maturity.

Futures-to-swaps transition (~2 years): Futures provide forward rates for discrete 3-month intervals, while swaps imply discount factors through the par rate formula. The granularity differs: futures give quarterly forward rates, while annual swap rates constrain the average rate over multi-year horizons. Additionally, the convexity adjustment applied to futures introduces model dependence that does not affect swap quotes.

Mitigation strategies:

Overlapping instruments: Use instruments from both types in the overlap region and blend their contributions smoothly. For example, use both futures and the 2-year swap rate near the transition, assigning weights that gradually shift.
Smooth interpolation: Apply a global fitting method (e.g., smoothing splines with a penalty on the second derivative of the forward curve) rather than strict sequential bootstrapping. This distributes any misfit across the entire curve rather than concentrating it at transition points.
Simultaneous calibration: Instead of strict sequential bootstrapping, solve for all discount factors simultaneously by minimizing a penalized objective that prices all instruments correctly while penalizing roughness in the forward curve.
Stub rates: At transition points, use stub-period adjustments to ensure the last instrument of one type and the first instrument of the next type are consistent.

Exercise 6. Implement the bootstrapping step for a 3-year annual swap rate \(S_3 = 3.8\%\), given previously bootstrapped discount factors \(P(0,1) = 0.968\) and \(P(0,2) = 0.935\). Solve for \(P(0,3)\) and compute the implied 1-year forward rate \(f(0; 2, 3)\).

Solution to Exercise 6

The bootstrapping formula for the \(n\)-th swap discount factor is:

\[ P(0, T_n) = \frac{1 - S_n \sum_{i=1}^{n-1} \delta_i P(0, T_i)}{1 + S_n \delta_n} \]

With annual payments, \(\delta_i = 1\) for all \(i\), \(S_3 = 0.038\), \(P(0,1) = 0.968\), and \(P(0,2) = 0.935\):

\[ P(0, 3) = \frac{1 - 0.038 \times (1 \times 0.968 + 1 \times 0.935)}{1 + 0.038 \times 1} \]

Computing the numerator:

\[ 1 - 0.038 \times 1.903 = 1 - 0.072314 = 0.927686 \]

Computing the denominator:

\[ 1 + 0.038 = 1.038 \]

Therefore:

\[ P(0, 3) = \frac{0.927686}{1.038} \approx 0.89372 \]

The implied 1-year forward rate from year 2 to year 3 (simply compounded) is:

\[ f(0; 2, 3) = \frac{1}{T_3 - T_2}\left(\frac{P(0, 2)}{P(0, 3)} - 1\right) = \frac{1}{1}\left(\frac{0.935}{0.89372} - 1\right) = 1.04623 - 1 = 0.04623 \]

So \(f(0; 2, 3) \approx 4.62\%\).

Verification: the par swap rate implied by these three discount factors is:

\[ S_3 = \frac{1 - P(0,3)}{\sum_{i=1}^{3} P(0, T_i)} = \frac{1 - 0.89372}{0.968 + 0.935 + 0.89372} = \frac{0.10628}{2.79672} \approx 0.03800 = 3.80\% \]

which confirms consistency with the input swap rate.

Bootstrapping Algorithms¶

Overview of the Bootstrapping Process¶

The Problem¶

General Principle¶

Stage 1: Money Market Deposits¶

Instrument Description¶

Bootstrapping Formula¶

Example¶

Stage 2: Futures (Eurodollar/SOFR Futures)¶

Instrument Description¶

Convexity Adjustment¶

Bootstrapping Formula¶

Sequential Application¶

Stage 3: Interest Rate Swaps¶

Instrument Description¶

Swap Valuation¶

Par Swap Rate Formula¶

Bootstrapping Formula¶

Sequential Application¶

Complete Bootstrapping Algorithm¶

Stage 1: Deposits (short end)¶

Stage 2: Futures (middle)¶

Stage 3: Swaps (long end)¶

Interpolation Between Nodes¶

The Need for Interpolation¶

Common Interpolation Methods¶

Forward Rate Implications¶

Multi-Curve Bootstrapping¶

Post-2008 Reality¶

Dual-Curve Bootstrap¶

Tenor Basis Spreads¶

Numerical Considerations¶

Solver Techniques¶

Error Handling¶

Verification¶

Example: USD Curve Bootstrap¶

Market Data (Illustrative)¶

Bootstrap Results¶

Key Takeaways¶

Further Reading¶

Exercises¶