Why Use More Than One Swap?
A single swap can be sized to offset one selected interest-rate sensitivity, but a portfolio may respond differently to rate changes at different maturities. For example, its value may rise when short-term rates rise and fall when long-term rates rise. Matching just one overall duration can leave those maturity-specific exposures intact. Multiple swaps provide additional sensitivity profiles that may be combined to offset more than one exposure.
The central task is to find notionals whose combined sensitivity vector is the negative of the portfolio’s sensitivity vector. This is a linear-system problem. An exact solution is available only if the candidate swaps can generate the required offset; having several swaps does not by itself guarantee that they can.
Set Up the Hedge System
Choose \(k\) maturity-rate factors to track. For example, with three factors, the entries might correspond to small changes in rates associated with the 2-year, 5-year, and 10-year points. The factors are treated as independent for the hedge calculation: a change in one factor does not imply a change in the others.
Let \(E_j\) be the portfolio’s approximate present-value change, in thousands of dollars per basis point, for a one-basis-point increase in factor \(j\). Collect these exposures in the vector \(E\). For each candidate swap \(r\), let \(h_{jr}\) be its corresponding sensitivity in thousands of dollars per basis point per $1 million of notional. Let \(N_r\) be its notional in millions of dollars. A positive notional means the stated swap direction; a negative notional means the same swap in the opposite direction.
Here \(m\) is the number of available swaps. Since a swap’s sensitivity scales with notional under the first-order model, the hedge’s total exposure is \(HN\). The combined exposure is \(E+HN\), so the hedge equation is:
The entries of \(H\) may be supplied by a sensitivity analysis or obtained by repricing the swaps under small, specified rate changes. The examples below take the profiles as given. They assume a fixed valuation date, a stated swap direction, and small rate changes for which the linear approximation is appropriate. They do not model nonlinear effects, changing basis relationships, or cash-flow mismatch.
Why the Matrix Equation Works
For a small change in the selected rate factors, let \(\Delta r_j\) denote the change in factor \(j\), measured in basis points. To first order, the portfolio’s present-value change is the sum of its sensitivities times the factor changes. A swap’s change is approximated in the same way, and its sensitivities are multiplied by its notional. Because present-value changes add across positions, the total change is the portfolio’s change plus the changes from all swaps.
To make this first-order change zero for every possible combination of small factor changes, each coefficient must be zero. If any coefficient were nonzero, changing just that one factor would leave a nonzero approximate value change. Thus the condition is \(E+HN=0\), not merely that the total sensitivity for one chosen rate movement is zero.
Solving for Notionals
Each column of \(H\) describes the exposure pattern contributed by one swap per $1 million of notional. Solving the hedge means combining those columns to produce \(-E\). With two independent factors and two swaps, this is usually a pair of simultaneous equations. With more factors, the same principle applies to the full system.
Worked Example: Offset Two Maturity Exposures
A hypothetical portfolio has exposures of \(-7\) and \(+3\) thousand dollars per basis point for its short- and long-maturity factors, respectively. A payer-fixed 2-year swap has profile \((4,1)\); a payer-fixed 5-year swap has profile \((1,5)\). Each entry is in thousands of dollars per basis point per $1 million of notional. Find notionals that exactly offset both exposures to first order.
Method and sign convention. Let \(x\) and \(y\) be the notionals in millions for the 2-year and 5-year payer-fixed swaps. The target hedge exposure is \(-E=(7,-3)\). A negative answer for either notional means use the opposite, receiver-fixed direction for that swap.
Set up the equations. The first factor receives \(4x+y\) thousand dollars per basis point from the swaps; the second receives \(x+5y\). Therefore:
From the first equation, \(y=7-4x\). Substitute this into the second:
Then \(y=7-4(2)=7-8=-1\). The hedge is $2 million payer-fixed in the 2-year swap and $1 million receiver-fixed in the 5-year swap.
Check. The swap exposures are \((4(2)+(-1),\,2+5(-1))=(7,-3)\). Adding the portfolio vector gives \((-7,+3)+(7,-3)=(0,0)\), as required.
Common mistake avoided. Treating both notionals as positive would use \((2,1)\), producing swap exposures \((9,7)\), not the required \((7,-3)\). The negative notional is meaningful: it reverses the specified swap direction.
Worked Example: Solve a Three-Factor Hedge
A hypothetical portfolio has three exposures, in thousands of dollars per basis point: \((-5,-6,+1)\). Candidate payer-fixed swaps have the following profiles per $1 million of notional: 2-year \((3,1,0)\), 5-year \((1,4,1)\), and 10-year \((0,1,3)\). Find notionals that exactly offset the portfolio to first order.
Let \(x,y,z\) be the notionals, in millions, of the 2-year, 5-year, and 10-year payer-fixed swaps. The required hedge vector is \((5,6,-1)\), giving:
From the first equation, \(y=5-3x\). Substitution into the third gives:
Substitute both expressions into the middle equation:
This result conflicts with the intended simple notionals; use the equations themselves to verify carefully. In fact, substituting \(x=1.2\) gives \(y=5-3.6=1.4\) and \(z=1.2-2=-0.8\). The middle equation then gives \(1.2+4(1.4)-0.8=6\), so these are the solution: $1.2 million payer-fixed in the 2-year swap, $1.4 million payer-fixed in the 5-year swap, and $0.8 million receiver-fixed in the 10-year swap.
Check. The combined swap profile is \((3(1.2)+1.4,\ 1.2+4(1.4)-0.8,\ 1.4+3(-0.8))=(5,6,-1)\). Adding \((-5,-6,+1)\) yields \((0,0,0)\).
Common mistake avoided. A solution should not be accepted because it looks plausible or because one equation balances. Substitution into every maturity-factor equation is the check; it catches sign and arithmetic errors.
Can the Swaps Span the Required Hedge?
A collection of swaps can produce only vectors formed by combinations of its sensitivity-profile columns. In linear algebra, these combinations form the column space, or span, of \(H\). An exact hedge exists precisely when \(-E\) is in that span. If the system is square and the columns are independent, a solution exists for every target vector and is unique. If the columns are dependent, different notionals may produce the same exposure, or the target may be unattainable. If there are more factors than independent swap profiles, some target vectors will generally be unattainable.
For a square matrix, a nonzero determinant is one test that the columns are independent. For a rectangular matrix, rank gives the corresponding general test: the target must lie in the column space. Counting swaps alone is not enough. Three swaps whose sensitivity profiles are dependent do not necessarily provide three independent directions of control.
Worked Example: Detect an Unspanned Exposure
A hypothetical portfolio has three maturity-factor exposures \((-4,-4,-2)\), in thousands of dollars per basis point. Only two payer-fixed swaps are available, with profiles \((2,1,0)\) and \((0,1,2)\) per $1 million of notional. Determine whether the portfolio can be exactly hedged with these swaps. If not, match the first and third factors and report the remaining exposure.
Let \(x\) and \(y\) be the two notionals in millions. The target vector is \((4,4,2)\), so an exact hedge would require:
The first equation requires \(x=2\), and the third requires \(y=1\). These values give \(x+y=3\), not \(4\). Therefore the equations are inconsistent, and no exact hedge exists with the two available profiles.
Matching the first and third factors gives \(x=2\), \(y=1\), and the swap exposure \((4,3,2)\). After adding the portfolio exposure:
The residual is a loss of one thousand dollars per basis point for an increase in the second factor, to first order. The target vector is not in the span of the two swap profiles: their combinations have the form \((2x,x+y,2y)\), and matching outer entries fixes the middle entry at \(3\), not \(4\).
Common mistake avoided. Choosing notionals to cancel two factors and then describing the result as an exact hedge overlooks the third equation. An exact hedge must satisfy every factor equation simultaneously.
Practical Checks and Limitations
A solved sensitivity system is a local, first-order hedge, not a guarantee that the portfolio value will remain unchanged under all market movements. The approximation treats the specified sensitivity coefficients as constant for small changes. Larger rate movements can make the coefficients change, and the portfolio and swaps may respond differently because of nonlinear price effects, curve-shape changes outside the chosen factors, or differences in cash flows and floating-rate benchmarks.
Before interpreting a solution, check the direction, units, and factor definitions:
- Use one consistent swap direction when building each column of \(H\). A payer-fixed profile and a receiver-fixed profile have opposite signs.
- Keep the notional units consistent. If \(h_{jr}\) is stated per $1 million, \(N_r\) must be in millions.
- Use the same rate-change unit for portfolio and swap sensitivities, such as dollars per basis point.
- Verify the final combined vector \(E+HN\), factor by factor.
- If the equations are inconsistent, report that an exact hedge is unavailable from those instruments rather than presenting a partial match as exact.
These calculations are aligned with the SOA Exam FM (Financial Mathematics) syllabus. The key exam skill is not just solving simultaneous equations: it is first setting up the exposures with consistent signs and units, then checking whether the available instruments can generate the required offset.
Check Your Understanding
For each problem, use the stated sensitivity profiles and first-order model. Notionals are measured in millions of dollars; exposure entries are thousands of dollars per basis point.
- A portfolio has exposures \((-7,-5)\). Two payer-fixed swaps have profiles \((3,1)\) and \((1,3)\). Find the exact hedge notionals. (Answer: $2 million payer-fixed in the first swap and $1 million payer-fixed in the second.)
- A portfolio has exposures \((-4,-6,-5)\). Two payer-fixed swaps have profiles \((2,0,1)\) and \((0,2,1)\). Find notionals that exactly hedge all three factors. (Answer: $2 million in the first swap and $3 million in the second.)
- A portfolio has exposures \((-5,-8,-5)\). Three payer-fixed swaps have profiles \((2,1,0)\), \((1,3,1)\), and \((0,1,2)\). Find the three notionals. (Answer: $1.75 million in the first swap, $1.50 million in the second, and $1.75 million in the third.)
- A portfolio has exposures \((-2,-3,-2)\). Two payer-fixed swaps have profiles \((1,1,0)\) and \((0,1,1)\). Is an exact hedge possible? If you match the first and third factors, what is the residual vector? (Answer: No exact hedge is possible; using $2 million in each swap leaves residual exposure \((0,+1,0)\).)