Solving Two-Asset Duration Matches

Why Two Conditions Require Two Asset Holdings

A liability portfolio has both a value today and a sensitivity to interest-rate changes. A two-asset duration match chooses holdings to match both targets: the present value of the assets equals the liability’s present value, and the assets’ duration equals the liability’s duration. The first condition aligns current values; the second aligns their first-order interest-rate sensitivities under a common yield assumption.

The two conditions produce a system of two linear equations. Solving it gives the required market values invested in each asset. Those market values are not necessarily the number of bonds or contracts to hold: to find quantities, divide each required market value by the corresponding asset’s price per unit.

Definition: A two-asset present-value and duration match uses asset holdings whose combined present value equals the liability present value and whose combined duration equals the liability duration. In this tutorial, asset holdings are first expressed as market values at the valuation date.

Duration and the Matching Equations

Assume a flat effective annual interest rate $i$, fixed asset and liability cash flows, and the same valuation basis for all cash flows. For a stream of positive cash flows $C_t$ at times $t$, its present value is $\sum_t C_t v^t$, where $v=1/(1+i)$. Its Macaulay duration is the present-value-weighted average payment time:

D=\frac{\sum_t tC_t v^t}{\sum_t C_t v^t}.

For a liability, use the positive amounts of its payments in this calculation. If asset 1 and asset 2 have market values $A_1$ and $A_2$, and durations $D_1$ and $D_2$, their combined duration is the market-value-weighted average of their durations. Let $L$ and $D_L$ be the liability’s present value and duration. The matching equations are:

A_1+A_2=L,\qquad A_1D_1+A_2D_2=LD_L.

A duration quoted as modified duration can also be used if the liability and both assets use that same convention and valuation rate. Under a common flat annual yield, modified duration is Macaulay duration divided by $1+i$, so converting every duration by the same factor does not change the matched allocations.

Why the Weighted-Duration Equation Holds

For fixed cash flows valued at a common yield, modified duration measures the proportional first-order price response to a small yield change. A portfolio’s dollar price change is the sum of the dollar price changes of its components. Therefore its proportional response is weighted by each component’s share of the total market value.

Equivalently, write the portfolio value as $P=A_1+A_2$. If $D_j$ is Macaulay duration, the first derivative of asset $j$’s value with respect to the common yield is $-A_jD_j/(1+i)$. Adding those derivatives and dividing by the total value gives the portfolio’s modified duration:

-\frac{1}{P}\frac{dP}{di} =\frac{A_1D_1+A_2D_2}{(1+i)(A_1+A_2)}.

Thus, when $A_1+A_2=L$, matching the weighted duration numerator to $LD_L$ matches the liability’s first-order sensitivity as well. This is a local sensitivity match; it does not assert that asset and liability values will remain equal after every possible rate change.

Solving for the Two Market Values

When $D_1\ne D_2$, substitute $A_2=L-A_1$ into the duration equation:

$$ A_1D_1+(L-A_1)D_2=LD_L. $$

Rearranging gives $A_1(D_1-D_2)=L(D_L-D_2)$. The required allocations are therefore:

A_1=L\frac{D_2-D_L}{D_2-D_1},\qquad A_2=L\frac{D_L-D_1}{D_2-D_1}.

The fractions $A_1/L$ and $A_2/L$ are the portfolio weights. They add to 1, but either can be negative when the liability duration lies outside the range of the two asset durations. A negative allocation means a short position or borrowing exposure is needed; it is not an acceptable long-only holding.

Feasibility check: If only nonnegative holdings are allowed and $D_1\ne D_2$, both allocations are nonnegative exactly when $D_L$ lies between $D_1$ and $D_2$, including the endpoints. If $D_1=D_2$, the assets have the same duration: they can match the liability duration only if $D_L=D_1$. In that case, present value determines the total holding but not a unique split between assets.

Worked Allocations and Constraint Checks

Worked Example: Matching a Four-Year Liability with Two Zero-Coupon Assets

At an effective annual rate of 5%, a hypothetical liability has present value $L=\$100{,}000$ and duration $D_L=4$ years. Two available assets are zero-coupon claims: asset 1 pays at the end of year 2, and asset 2 pays at the end of year 6. Find the market value allocated to each asset and the maturity amounts to purchase. Assume fractional holdings are permitted.

Method. A zero-coupon claim has duration equal to its payment time, so $D_1=2$ and $D_2=6$. Since $2<4<6$, a nonnegative match is possible.

Asset 1 market value.

A_1=100{,}000\frac{6-4}{6-2} =100{,}000\frac{2}{4} =\$50{,}000.

Asset 2 market value.

A_2=100{,}000\frac{4-2}{6-2} =100{,}000\frac{2}{4} =\$50{,}000.

To convert these market values to maturity amounts, accumulate each allocation to its payment date at 5%. Asset 1 requires a year-2 maturity amount of $\$50{,}000(1.05)^2=\$55{,}125.00$. Asset 2 requires a year-6 maturity amount of $\$50{,}000(1.05)^6=\$67{,}004.78$, rounded to cents.

Check. The asset present values total $\$50{,}000+\$50{,}000=\$100{,}000$. Their duration is $(50{,}000(2)+50{,}000(6))/100{,}000=4$ years, as required. This method avoids confusing the maturity amounts with the market values used in the duration weights.

Worked Example: Matching with a One-Year Asset and a Coupon Claim

At an effective annual rate of 5%, a hypothetical liability has two payments: $\$94{,}500$ at the end of year 1 and $\$33{,}075$ at the end of year 2. Asset A pays $\$105$ at the end of year 1. Asset B pays $\$105$ at the end of year 1 and $\$110.25$ at the end of year 2. Find the market value invested in each asset and the number of units to hold. Each quoted asset cash flow is per unit.

Method: calculate present values and durations before solving the allocation system. The year-1 liability payment has present value $\$94{,}500/1.05=\$90{,}000$; the year-2 payment has present value $\$33{,}075/(1.05)^2=\$30{,}000$. Thus $L=\$120{,}000$, and

D_L=\frac{90{,}000(1)+30{,}000(2)}{120{,}000} =\frac{150{,}000}{120{,}000} =1.25\text{ years}.

Asset A has a price per unit of $\$105/1.05=\$100$ and duration $D_A=1$. Asset B has a price per unit of $\$105/1.05+\$110.25/(1.05)^2=\$100+\$100=\$200$. Its duration is $(100(1)+100(2))/200=1.5$ years.

Market-value allocations. Applying the two-asset formulas with $D_1=1$ and $D_2=1.5$:

A_1=120{,}000\frac{1.5-1.25}{1.5-1} =120{,}000\frac{0.25}{0.5} =\$60{,}000,

A_2=120{,}000\frac{1.25-1}{1.5-1} =120{,}000\frac{0.25}{0.5} =\$60{,}000.

The required unit counts are $\$60{,}000/\$100=600$ units of asset A and $\$60{,}000/\$200=300$ units of asset B.

Check. The asset values total $\$120{,}000$. Their duration numerator is $\$60{,}000(1)+\$60{,}000(1.5)=\$150{,}000$, so their duration is $\$150{,}000/\$120{,}000=1.25$ years. Both positions are nonnegative. The common mistake avoided here is applying duration weights to unit counts rather than to market values.

Worked Example: Detecting a Short Position Required by the Match

At an effective annual rate of 4%, a hypothetical liability pays $\$43{,}264$ at the end of year 2 and $\$44{,}994.56$ at the end of year 3. The two available assets are zero-coupon claims paying at the end of years 3 and 7. Find the required market-value positions and decide whether the match is possible under a no-short-sales rule.

Liability targets. The present values of the two payments are $\$43{,}264/(1.04)^2=\$40{,}000$ and $\$44{,}994.56/(1.04)^3=\$40{,}000$. Therefore $L=\$80{,}000$, and $D_L=(40{,}000(2)+40{,}000(3))/80{,}000=2.5$ years.

Asset durations and allocations. The zero-coupon asset durations are $D_1=3$ and $D_2=7$. The target duration is below both, so check the signs in the allocation formulas:

A_1=80{,}000\frac{7-2.5}{7-3} =80{,}000\frac{4.5}{4} =\$90{,}000,

A_2=80{,}000\frac{2.5-3}{7-3} =80{,}000\frac{-0.5}{4} =-\$10{,}000.

Constraint decision. The equations require a $\$90{,}000$ long position in the year-3 asset and a $\$10{,}000$ short position in the year-7 asset. The signed solution satisfies the equations, but it is not feasible under a no-short-sales rule.

Check. The signed market values sum to $\$90{,}000-\$10{,}000=\$80{,}000$. Their duration numerator is $\$90{,}000(3)-\$10{,}000(7)=\$200{,}000$, giving duration $\$200{,}000/\$80{,}000=2.5$ years. Treating the negative allocation as a positive holding would incorrectly give both the wrong present value and the wrong duration.

Common Exam Traps and Practical Checks

A solved pair of equations is only the beginning of the portfolio decision. Confirm that the solution fits the actual restrictions, such as no short sales, a minimum holding, or a maximum position. A mathematically valid signed solution may be unusable when the portfolio must be long-only. If a target is exactly one asset’s duration, the formula assigns all the market value to that asset and zero to the other.

Using quantities instead of values. The duration-weighted equation uses market values. Convert to units only after solving.
Mixing duration conventions. Do not combine Macaulay duration and modified duration in the same equation. Use a consistent convention and valuation basis.
Ignoring signs. A negative allocation indicates a short position, not a small positive holding.
Assuming a match is unique when durations coincide. If $D_1=D_2$, the duration equation cannot distinguish between the two assets.
Overstating what duration matching achieves. It matches present value and first-order sensitivity under the stated assumptions. It does not guarantee exact cash-flow matching or equal values after every rate movement.

A reliable procedure is to calculate $L$ and $D_L$, establish each asset’s duration on the same basis, solve for market values, translate those values into security units if needed, and finally check every holding restriction. This makes the arithmetic and the feasibility decision visible rather than leaving a negative or mis-scaled position hidden in the final answer. This topic is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, first identify the market-value allocations, then check any stated portfolio restriction.

A liability has present value $\$75{,}000$ and duration 5 years. Two assets have durations 2 and 8 years. Find the market value in each asset and state whether both holdings are positive. (Answer: $\$37{,}500$ in each asset; both positive.)
A liability has present value $\$120{,}000$ and duration 3 years. Asset 1 has duration 1 year and asset 2 has duration 5 years. Find the allocations. (Answer: $\$60{,}000$ in each asset.)
A liability has present value $\$90{,}000$ and duration 2 years. The two asset durations are 3 and 6 years. Find the signed solution and determine whether a long-only match exists. (Answer: $\$120{,}000$ in the duration-3 asset and $-\$30{,}000$ in the duration-6 asset; no long-only match.)
A liability has present value $\$50{,}000$ and duration 4 years. Both assets have duration 4 years, but their prices per unit are $\$250$ and $\$400$. What can be concluded about the matching holdings? (Answer: Any nonnegative market-value split totaling $\$50{,}000$ matches; the duration condition does not determine a unique split. If all value is placed in asset 1, hold 200 units; if all is placed in asset 2, hold 125 units.)
A liability has present value $\$200{,}000$ and duration 4 years. Asset A has duration 2 years and asset B has duration 6 years. A rule permits at most $\$125{,}000$ in asset A. Find the match and check the rule. (Answer: $\$100{,}000$ in each asset; the asset A limit is satisfied.)

What You'll Learn