Optimal execution and the Almgren Chriss model, financial engineering for quants lesson

A trader must sell a block of shares before the close. Selling at a steady pace keeps the temporary price impact low, but every share still held is exposed to price moves, and selling early cuts that exposure only by trading faster. Which inventory path balances the two effects, and how strongly should price risk pull the sales forward? This lesson writes the tradeoff as the mean-variance objective of the continuous Almgren-Chriss model, derives the optimal schedule from the Euler-Lagrange equation, proves that it is the unique global minimum, and shows how a single urgency parameter sets its shape.

Key terms

Inventory trajectory

Definition A trader holds $Q > 0$ units at time $0$ and must hold none at the deadline $T > 0$ . An inventory trajectory is a deterministic function $x$ on $[0, T]$ , fixed at time $0$ , with

x(0) = Q, \qquad x(T) = 0,

that is absolutely continuous with $\int_0^T \dot x(t)^2\,dt < \infty$ . Write $\mathcal{A}$ for this admissible class. The selling speed is $v(t) = -\dot x(t)$ in units per unit of time and the cumulative sales are $y(t) = Q - x(t)$ , so every admissible path has $\int_0^T v(t)\,dt = Q$ . The class does not assume monotone selling, and it contains paths that buy back ( $v < 0$ on some interval) or go short ( $x < 0$ ). Because the path is fixed in advance, the schedule does not react to the prices it meets along the way.

In plain words A trajectory is the plan for how many units are still unsold at each moment. Its slope is the trading speed. The condition on the slope says that the plan has a finite trading cost in the model below, which rules out selling a block instantly.

Example Take $Q = 1$ normalized unit and $T = 1$ day. The linear trajectory $x(t) = 1 - t$ sells at the constant speed of $1$ unit per day and still holds $0.5$ units at midday. The trajectory $x(t) = \max(1 - 2t, 0)$ , which sells everything at speed $2$ during the morning, is also admissible. It holds nothing after midday, but it trades twice as fast while it trades.

Impact cost

Definition Temporary impact makes sales at speed $v(t)$ execute $\eta\, v(t)$ below the prevailing price, with a constant $\eta > 0$ in currency times time per squared unit. Over the schedule it costs

C_{\text{temp}}[x] = \eta \int_0^T \dot x(t)^2\,dt

currency units. Linear permanent impact lowers the prevailing price by $\gamma\, y(t)$ once $y(t)$ units have been sold, with a constant $\gamma \ge 0$ . The units sold at speed $v(t)$ then lose $\gamma\, y(t) v(t)\,dt$ , and since $y$ is absolutely continuous with $y(0) = 0$ and $y(T) = Q$ ,

\begin{aligned} C_{\text{perm}} &= \int_0^T \gamma\, y(t)\, \dot y(t)\,dt \\[0.4em] &= \tfrac12 \gamma Q^2 \end{aligned}

for every admissible path, monotone or not. A fixed half-spread $\beta \ge 0$ per unit traded costs $\beta \int_0^T \lvert v \rvert\,dt \ge \beta Q$ , with equality exactly for paths that never buy. For fixed $Q$ and $T$ , the Cauchy-Schwarz inequality gives

\begin{aligned} Q^2 &= \Bigl(\int_0^T v\,dt\Bigr)^2 \\[0.4em] &\le T \int_0^T v^2\,dt, \end{aligned}

so $C_{\text{temp}}[x] \ge \eta Q^2/T$ , with equality only when the speed is constant, equal to $Q/T$ , almost everywhere.

In plain words Trading faster moves the execution price further against each sale and also sells more units in each short interval, so the temporary cost grows with the square of the speed. Linear permanent impact costs the same whatever the timing, so it changes the total cost and not the schedule. Spreading the sales evenly is the cheapest way to pay temporary impact.

Example With the constructed coefficient $\eta = 0.04$ currency days per squared unit, $Q = 1$ and $T = 1$ day, the linear trajectory has $C_{\text{temp}} = 0.04$ . The morning trajectory of the previous card has $\int_0^T v^2\,dt = 2^2 \cdot \tfrac12 = 2$ , so its temporary cost is $0.08$ , twice as much. With a constructed permanent coefficient $\gamma = 0.01$ , both pay the same permanent cost $\tfrac12 \cdot 0.01 \cdot 1^2 = 0.005$ .

Almgren-Chriss model

Definition The continuous-time Almgren-Chriss model of a liquidation makes the following assumptions. Under the physical measure $\mathbb{P}$ the unaffected price is $S_0 + \sigma W_t$ , where $W$ is a $\mathbb{P}$ -Brownian motion, so the price is an arithmetic Brownian motion with zero drift and constant volatility $\sigma \ge 0$ in currency per unit per square root of time. Impact is linear as in the previous card, and the schedule $x$ in $\mathcal{A}$ is deterministic. Sales at time $t$ execute at

\begin{aligned} \tilde S_t &= S_0 + \sigma W_t \\[0.4em] &\quad - \gamma\, y(t) - \eta\, v(t). \end{aligned}

The execution cost is the implementation shortfall against the arrival price, $C = Q S_0 - \int_0^T \tilde S_t\, v(t)\,dt$ . Since $x$ is deterministic and of finite variation, the product rule gives $d(x W) = x\,dW + W\,dx$ , and with $x(T) = 0$ and $W_0 = 0$ this yields $\int_0^T W_t\, v(t)\,dt = \int_0^T x(t)\,dW_t$ . Hence

\begin{aligned} C &= \tfrac12 \gamma Q^2 + \eta \int_0^T v^2\,dt \\[0.4em] &\quad - \sigma \int_0^T x(t)\,dW_t . \end{aligned}

The Itô integral of the bounded deterministic function $x$ has mean zero and, by the Itô isometry, variance $\int_0^T x^2\,dt$ . The risk penalty is $\lambda \operatorname{Var}(C) = \lambda \sigma^2 \int_0^T x^2\,dt$ , with a risk weight $\lambda \ge 0$ in inverse currency units, and the model chooses $x$ in $\mathcal{A}$ to minimize $\mathbb{E}^{\mathbb{P}}[C] + \lambda \operatorname{Var}(C)$ . Dropping the constant $\tfrac12 \gamma Q^2$ leaves the functional

\begin{aligned} J[x] &= \eta \int_0^T \dot x(t)^2\,dt \\[0.4em] &\quad + \lambda \sigma^2 \int_0^T x(t)^2\,dt . \end{aligned}

Adaptive schedules that react to prices, nonlinear or transient impact, time-varying liquidity and participation limits are outside this model, and each needs its own derivation.

In plain words The model scores a schedule by its expected cost plus a charge for uncertainty. The expected cost comes from impact. The uncertainty comes from price moves on the units not yet sold, so a unit held for a long time adds more variance, and $\lambda$ says how many currency units of expected cost the trader would give up to remove one squared currency unit of variance.

Example Take the constructed inputs $Q = 1$ , $T = 1$ day, $\eta = 0.04$ , $\sigma = 0.2$ currency per unit per square root day and $\lambda = 1$ per currency unit. The linear trajectory has $\int_0^1 (1 - t)^2\,dt = \tfrac13$ , so $\operatorname{Var}(C) = 0.04/3 \approx 0.013333$ squared currency units, the standard deviation of its cost is about $0.1155$ , and its risk penalty is about $0.013333$ . Its objective is $J = 0.04 + 0.013333 \approx 0.053333$ . The morning trajectory has $\int_0^1 x^2\,dt = \tfrac16$ , half the variance, but its objective $0.08 + 0.04/6 \approx 0.086667$ is much higher.

Euler-Lagrange equation

Definition Let $x$ be admissible and let $h$ be absolutely continuous with square-integrable derivative and $h(0) = h(T) = 0$ , so that $x + \varepsilon h$ is admissible for every real $\varepsilon$ . The first variation of $J$ at $x$ in the direction $h$ is

\begin{aligned} \delta J[x; h] &= \frac{d}{d\varepsilon} J[x + \varepsilon h]\Big|_{\varepsilon = 0} \\[0.4em] &= 2 \int_0^T \bigl(\eta\, \dot x \dot h + \lambda \sigma^2 x h\bigr)\,dt . \end{aligned}

If $x$ is twice continuously differentiable, integration by parts gives

\begin{aligned} \delta J[x; h] &= 2\bigl[\eta\, \dot x h\bigr]_0^T \\[0.4em] &\quad + 2 \int_0^T \bigl(\lambda \sigma^2 x - \eta\, \ddot x\bigr) h\,dt, \end{aligned}

and the boundary term vanishes because $h(0) = h(T) = 0$ . At a minimizer the first variation is zero for every such $h$ , and since the bracket is continuous, the fundamental lemma of the calculus of variations forces it to vanish identically. This is the Euler-Lagrange equation

\begin{gathered} \eta\, \ddot x(t) = \lambda \sigma^2 x(t), \\[0.4em] 0 < t < T, \end{gathered}

a linear second-order equation with the boundary conditions $x(0) = Q$ and $x(T) = 0$ . On its own it is a necessary condition, met by every twice differentiable minimizer, and the next card shows that its solution is in fact the minimum.

In plain words Bending the path slightly around some moment changes the impact cost through the speed and the risk through the inventory held. On the best path these two marginal effects cancel at every instant. The term $\eta\, \ddot x$ comes from the cost of changing speed, and the term $\lambda \sigma^2 x$ from the risk of the inventory still held.

Example With the constructed inputs, $\lambda \sigma^2 = 0.04$ . The linear trajectory $x(t) = 1 - t$ has $\ddot x = 0$ , while $\lambda \sigma^2 x(t) = 0.04(1 - t) > 0$ before the deadline, so it does not solve the equation. In the direction $h(t) = t(1 - t)$ its first variation is

\begin{aligned} \delta J &= -0.08 \int_0^1 (1 - 2t)\,dt \\[0.4em] &\quad + 0.08 \int_0^1 t(1 - t)^2\,dt \\[0.4em] &= 0 + \tfrac{0.08}{12} \approx 0.006667 . \end{aligned}

Because $J$ is quadratic, $J[x + \varepsilon h] = J[x] + \varepsilon\, \delta J + \varepsilon^2 J[h]$ exactly, with $J[h] = 0.04 \cdot \tfrac13 + 0.04 \cdot \tfrac{1}{30} \approx 0.014667$ . The best step $\varepsilon = -\delta J/(2 J[h]) \approx -0.2273$ lowers the midday inventory and the objective, from $0.053333$ to about $0.052576$ .

Optimal execution schedule

Definition Assume $Q > 0$ , $T > 0$ , $\eta > 0$ , $\lambda \ge 0$ and $\sigma \ge 0$ , and let $\kappa = \sqrt{\lambda \sigma^2/\eta}$ . If $\kappa > 0$ , the unique minimizer of $J$ over $\mathcal{A}$ is

x^*(t) = Q\, \frac{\sinh\bigl(\kappa (T - t)\bigr)}{\sinh(\kappa T)},

with selling speed

v^*(t) = Q\kappa\, \frac{\cosh\bigl(\kappa (T - t)\bigr)}{\sinh(\kappa T)} .

If $\kappa = 0$ , it is the linear schedule $x^*(t) = Q(1 - t/T)$ . The function $x^*$ solves $\ddot x = \kappa^2 x$ , and among the solutions $A \sinh(\kappa(T - t)) + B \cosh(\kappa(T - t))$ the endpoint conditions force $B = 0$ and $A = Q/\sinh(\kappa T)$ . For any admissible $x$ , the difference $h = x - x^*$ vanishes at both ends, and the quadratic form of $J$ gives

\begin{aligned} J[x] - J[x^*] &= \delta J[x^*; h] \\[0.4em] &\quad + \eta \int_0^T \dot h^2\,dt \\[0.4em] &\quad + \lambda \sigma^2 \int_0^T h^2\,dt . \end{aligned}

The first variation is zero by the integration by parts of the previous card, which needs only that $x^*$ is smooth and $h$ is absolutely continuous with zero endpoints. The remaining terms are nonnegative, and they vanish only if $\dot h = 0$ almost everywhere, which makes $h$ constant and hence zero. Integrating $\eta\, \dot x^{*2}$ by parts and using the equation gives the minimum value

\begin{aligned} J[x^*] &= \eta\, Q\, v^*(0) \\[0.4em] &= \eta \kappa Q^2 \coth(\kappa T), \end{aligned}

which tends to $\eta Q^2/T$ as $\kappa \to 0$ . For $\kappa > 0$ the schedule is positive before $T$ and strictly decreasing, it never buys, and its speed falls from $Q\kappa \coth(\kappa T)$ at the start to $Q\kappa/\sinh(\kappa T) > 0$ at the deadline. Since $\sinh(u)/u$ increases for $u > 0$ , the schedule lies strictly below the linear one at every time strictly between $0$ and $T$ . The hypothesis $\eta > 0$ is needed. With $\eta = 0$ and $\lambda \sigma^2 > 0$ , paths that sell ever faster make $J$ as small as desired, but no admissible path reaches $J = 0$ .

In plain words The best schedule sells fastest at the start, when the most inventory is exposed, and slows down as the exposure shrinks, without ever stopping before the deadline. No search over paths is needed to see that it is best, because any change to it adds a positive quadratic amount to the objective.

Example With the constructed inputs, $\kappa = \sqrt{0.04/0.04} = 1$ per day. The schedule holds about $0.6997$ , $0.4434$ and $0.2150$ units at $t = 0.25$ , $0.5$ and $0.75$ days, against $0.75$ , $0.5$ and $0.25$ on the linear path, and its speed falls from about $1.3130$ to about $0.8509$ units per day. The identities $\sinh^2 u = \tfrac12(\cosh 2u - 1)$ and $\cosh^2 u = \tfrac12(\cosh 2u + 1)$ give

\begin{aligned} \int_0^T x^{*2}\,dt &= \frac{Q^2}{\sinh^2(\kappa T)} \\[0.4em] &\quad \times \Bigl(\frac{\sinh(2\kappa T)}{4\kappa} - \frac{T}{2}\Bigr) \\[0.4em] &\approx 0.294487 \end{aligned}

and

\begin{aligned} \int_0^T v^{*2}\,dt &= \frac{Q^2 \kappa^2}{\sinh^2(\kappa T)} \\[0.4em] &\quad \times \Bigl(\frac{\sinh(2\kappa T)}{4\kappa} + \frac{T}{2}\Bigr) \\[0.4em] &\approx 1.018548 . \end{aligned}

The temporary cost is about $0.040742$ and the risk penalty about $0.011779$ , so $J[x^*] \approx 0.052521$ , which equals $0.04 \coth(1)$ . The linear schedule scores $0.053333$ , so the optimum is about $1.5$ percent lower. Compared with constant speed it pays about $0.000742$ more impact cost and saves about $0.001554$ of risk penalty. The one-direction improvement of the previous card reached $0.052576$ , close to this minimum and above it.

Execution urgency

Definition The parameter

\kappa = \sqrt{\frac{\lambda \sigma^2}{\eta}}

measures execution urgency. It has units of inverse time, $1/\kappa$ is the time scale over which the optimal inventory decays, and the dimensionless number $\kappa T$ fixes the shape of $x^*(t)/Q$ as a function of $t/T$ . Urgency increases with the risk weight $\lambda$ and the volatility $\sigma$ , decreases with the temporary coefficient $\eta$ , and does not depend on $Q$ , $S_0$ or the permanent coefficient $\gamma$ . At each time $0 < t < T$ the inventory $x^*(t)$ is strictly decreasing in $\kappa$ , because

\begin{aligned} &\frac{\partial \ln x^*(t)}{\partial \kappa} \\[0.4em] &\quad = (T - t)\coth\bigl(\kappa(T - t)\bigr) \\[0.4em] &\qquad - T \coth(\kappa T) < 0, \end{aligned}

as $s \coth(\kappa s)$ increases in $s > 0$ . At midday $x^*(T/2) = Q/\bigl(2\cosh(\kappa T/2)\bigr)$ , and when $\kappa (T - t)$ is large, $x^*(t) \approx Q e^{-\kappa t}$ . When $\sigma > 0$ the minimized objective $\eta \kappa Q^2 \coth(\kappa T)$ increases with $\lambda$ , because a larger $\lambda$ raises $J$ of every path, so objective values computed with different $\lambda$ measure different criteria and cannot rank schedules against each other. The original Almgren-Chriss model trades at $N$ dates spaced $\tau = T/N$ apart. There the permanent cost has a schedule-dependent part $-\tfrac12 \gamma \sum_k n_k^2$ for the sales $n_k$ , the temporary coefficient becomes $\eta - \tfrac12 \gamma \tau$ , assumed positive, and the optimal inventory at the trading dates is again a hyperbolic sine, with a decay rate $\kappa_\tau$ that solves

\begin{aligned} &\frac{2\bigl(\cosh(\kappa_\tau \tau) - 1\bigr)}{\tau^2} \\[0.4em] &\quad = \frac{\lambda \sigma^2}{\eta - \gamma \tau/2} . \end{aligned}

As $\tau \to 0$ the rate $\kappa_\tau$ tends to $\kappa$ , so the continuous formulas are the limit of the discrete ones, not the same model.

In plain words Urgency says how hard price risk pushes sales forward compared with the impact cost of trading fast. A trader with high urgency accepts larger impact early to cut exposure, and a trader with no urgency spreads the sales evenly.

Example Raising $\lambda$ from $1$ to $4$ in the constructed example doubles $\kappa$ to $2$ per day. Midday inventory falls from about $0.4434$ to about $0.3240$ units, the initial speed rises to about $2.0746$ and the final speed falls to about $0.5514$ units per day. The inventory integral drops from $0.294487$ to about $0.221318$ , while the temporary cost rises from $0.040742$ to about $0.047574$ . The new minimum, about $0.082985$ , is larger than $0.052521$ only because the penalty is larger. Under the $\lambda = 1$ criterion the $\kappa = 2$ schedule scores about $0.056427$ , above that criterion’s minimum, and under the $\lambda = 4$ criterion the $\kappa = 1$ schedule scores about $0.087860$ , above $0.082985$ . In the discrete model with $\lambda = 1$ , $\tau = 0.25$ days and the permanent coefficient $\gamma = 0.01$ of the impact cost card, the decay rate is about $1.0133$ per day, against the continuous value $1$ .

Check yourself

In the constructed example the order doubles to $Q = 2$ units, with $T = 1$ day, $\eta = 0.04$ , $\sigma = 0.2$ and $\lambda = 1$ unchanged. How do the shape of the optimal schedule, the midday inventory and the minimum objective change?

Show the answer

The urgency $\kappa = \sqrt{\lambda \sigma^2/\eta} = 1$ per day does not involve $Q$ , so $x^*(t)/Q$ is the same function as before. The schedule still holds about $44.34$ percent of the order at midday, which is now about $0.8868$ units. Both integrals in $J$ scale with $Q^2$ , so

\begin{aligned} J[x^*] &= 0.04 \cdot 1 \cdot 2^2 \coth(1) \\[0.4em] &\approx 0.210086, \end{aligned}

four times $0.052521$ . With linear impact and a variance penalty the tradeoff does not depend on the size of the order, which would no longer hold with nonlinear impact.

A colleague adds the constructed permanent impact $\gamma = 0.01$ to the example and argues that the schedule should now sell faster, to get ahead of its own price pressure. Is that right in the continuous model?

Show the answer

No. In the continuous model the permanent cost is $\int_0^T \gamma\, y\, \dot y\,dt = \tfrac12 \gamma Q^2 = 0.005$ for every admissible path, so the objective becomes $0.005 + J[x]$ and has the same minimizer. The urgency $\kappa$ involves only $\lambda$ , $\sigma$ and $\eta$ , and the minimum of $\mathbb{E}^{\mathbb{P}}[C] + \lambda \operatorname{Var}(C)$ is about $0.005 + 0.052521 = 0.057521$ . In the discrete model the permanent impact enters the schedule only through the coefficient $\eta - \tfrac12 \gamma \tau$ , and that effect disappears as $\tau \to 0$ . With $\tau = 0.25$ days, raising $\gamma$ from $0$ to $0.01$ moves the discrete decay rate from about $0.9974$ to about $1.0133$ per day, a slightly faster schedule, while the continuous rate stays at $1$ .

In the constructed example with $\lambda = 1$ , a participation limit caps the selling speed at $1.2$ units per day. Is $x^*$ still optimal, and what can be said about the best capped schedule?

Show the answer

The schedule $x^*$ starts at speed $\coth(1) \approx 1.3130$ and stays above $1.2$ until about $t = 0.1226$ days, so it violates the cap and is not a candidate. The proof of optimality compares $x^*$ with every path in $\mathcal{A}$ , and $x^*$ is the only path with $J = 0.052521$ , so every capped schedule has an objective strictly above $0.052521$ . Constant speed $1$ respects the cap and scores $0.053333$ , so the best capped schedule scores between these two values. Clipping $x^*$ at the cap is not a solution, because it sells about $0.0067$ units less by $t = 0.1226$ and, if it then follows the speed of $x^*$ , still holds that amount at the deadline. The cap has to enter the feasible set of the optimization, which is outside the model of this lesson.

Optimal execution and the Almgren Chriss model

Key terms

Inventory trajectory

Impact cost

Almgren-Chriss model

Euler-Lagrange equation

Optimal execution schedule

Execution urgency

Check yourself

How this lesson connects

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