Swap Curve Calibration with Log-Linear Interpolation and Gauss-Newton

# swap curve calibration with interpolation using newton-like method


# swap curve calibration with interpolation using newton-like method












suppose 2 swap market quotes for 1Y and 2Y and that swap payments occur semi-annually.

calibrating / obtaining the discount factors means finding 4 unknowns / discount factors that reproduce the market quotes such that

$$ \textrm{min } f = \sum_{i=1}^2 \epsilon_i \\ \textrm{ with } \epsilon_i = (mktQuote_i - modelQuote_i)^2 \\ \textrm{ and modelQuote}_i := floatLeg_i / fixLeg_i = \frac{\sum_{j=1}^N (\delta_j * fwrd_j * DF_j)}{\sum_{j=1}^N (\delta_j * DF_j)} $$ for j = 1, 2, ..., N payments on swap "i".

considering log-linear interpolation on the discount factors (DF) is to be used, is it possible to set up a nice Newton-like scheme, that is, $(DF_{k+1} = DF_k - J^{-1}(DF_k) * f(DF_k))$, with the Jacobian a square matrix, to solve this problem ?

References / articles / books that discuss this topic ?

## Answer by Attack68 (score 3, accepted)

https://quant.stackexchange.com/a/79465

You cannot solve for 4 degrees of freedom with two fixed parameters. That is an underspecified curve and it will (even though solvable, with infinite solutions) lead to chaotic behaviour within your solver.

In your case, however, you are adding hyper parameters, in the form of interpolation, so that 2 of the discount factors are implied, via interpolation from the other 3 (the first discount factor at the start of your curve is fixed at 1.0).

Yes, it is possible to setup a newton like solver. The one below solves for two parameters with two instruments.

```
# PYTHON
from rateslib import *

curve = Curve(
    nodes={  # initialise 3 discount factors
        dt(2024, 5, 22): 1.0,
        dt(2025, 5, 22): 1.0,
        dt(2026, 5, 27): 1.0,
    },
    calendar="nyc",
    interpolation="log_linear"
)

solver = Solver(
    curves=[curve],
    instruments=[
        IRS(dt(2024, 5, 24), "1y", frequency="S", spec="usd_irs", curves=curve),
        IRS(dt(2024, 5, 24), "2y", frequency="S", spec="usd_irs", curves=curve),
    ],
    s=[5.16, 4.76],
    algorithm="gauss_newton"
)
SUCCESS: `func_tol` reached after 3 iterations (gauss_newton), `f_val`: 4.139106658800514e-14, `time`: 0.0023s
```

Relevant intermediate discount factors would then be known:

```
curve[dt(2024, 11, 24)]  # 0.974328
curve[dt(2025, 11, 24)]  # 0.927691
```

As well as those that have parametrised your curve:

```
curve.nodes
# {datetime.datetime(2024, 5, 22, 0, 0): 1.000000,
#  datetime.datetime(2025, 5, 22, 0, 0): 0.950244,
#  datetime.datetime(2026, 5, 27, 0, 0): 0.905908}
```

You can read more about this `Solver` at https://rateslib.readthedocs.io/en/latest/c_solver.html

The books I have personally written on this are "Coding Interest Rates: FX, Swaps and Bonds" which just documents algorithms used in rateslib, and "Pricing and Trading Interest Rate Derivatives: A Practical Guide to Swaps" which is broader and contains more generalist info related to swaps trading. (https://www.amazon.com/Pricing-Trading-Interest-Rate-Derivatives-dp-0995455546/dp/0995455546/ref=dp_ob_title_bk)

### Edit

Solver gradients are available. The API docs are at https://rateslib.readthedocs.io/en/latest/api/rateslib.solver.Gradients.html#rateslib.solver.Gradients although these don't make a huge amount of sense without the deriving documents.

The key gradients you are requesting are:

```
solver.grad_v_rT  # alias of solver.J
# array([[-103.8040534 ,   -2.38011349],
#       [  -1.45812049,  -54.08994059]])

solver.grad_s_vT
# array([[-0.00963949,  0.00042417],
#       [ 0.00025986, -0.01849916]])
```

The former is used in the calibration part where sensitivities to the rates determine by how much to adjust the discount factors.

The latter is used in risk sensitivities when you want to measure the economic sensitivity of a portfolio to the movement in the rates of instruments. The automatic differentiation returns sensitivities to discount factors and these have to be transformed to sensitivities to instruments rates. I.e

$$ \nabla_\mathbf{s} P = \nabla_\mathbf{s} \mathbf{v^T} \nabla_\mathbf{v} P $$