5.7 Optimal execution: Almgren–Chriss · M05 Intraday and execution

Motivation

Trading fast reduces exposure to price moves during execution but increases impact; trading slowly does the opposite. Almgren and Chriss (2000) formulate this as a mean–variance problem and solve it in closed form under linear impact. The solution, a $\sinh$ -shaped trajectory, is the backbone of institutional execution algorithms and a clean example of a constrained quadratic optimisation with a linear-recurrence solution.

5.7.1The execution problem#

Setup

Liquidate $X$ shares over $[0, T]$ in $N$ intervals of length $\tau = T/N$ , at times $t_k = k\tau$ . Holdings $x_0 = X$ , $x_N = 0$ ; the $k$ -th child order sells $n_k = x_{k-1} - x_k$ . The unaffected price evolves as

S_k = S_{k-1} + \sigma\sqrt{\tau}\,\xi_k - \gamma\, n_k, \tag{5.16}

with $\xi_k$ i.i.d. $N(0,1)$ , so $\sigma$ is price volatility in ₹ per $\sqrt{\text{unit time}}$ , and $\gamma$ is permanent impact (₹ per share sold). The $k$ -th child order executes at

\tilde S_k = S_{k-1} - \varepsilon\operatorname{sgn}(n_k) - \frac{\eta}{\tau}\,n_k, \tag{5.17}

where $\varepsilon$ is a fixed cost (half-spread plus fees) and $\eta$ the temporary impact coefficient: trading at rate $n_k/\tau$ shares per unit time costs $\eta n_k/\tau$ per share, and the effect does not persist. Implementation shortfall is $\text{IS} = XS_0 - \sum_k n_k\tilde S_k$ . The trader minimises

U(x) = E[\text{IS}] + \lambda\operatorname{Var}[\text{IS}], \qquad \lambda \ge 0 \text{ the risk aversion}.

5.7.2The optimal trajectory#

Proposition 5.5Almgren–Chriss, 2000

(a) For any deterministic schedule,

\text{IS} = -\sum_{k=1}^{N}\sigma\sqrt\tau\,\xi_k\,x_k + \tfrac12\gamma X^2 + \varepsilon\sum_k|n_k| + \Big(\frac{\eta}{\tau} - \frac{\gamma}{2}\Big)\sum_k n_k^2, \tag{5.18}

\operatorname{Var}[\text{IS}] = \sigma^2\tau\sum_{k=1}^{N} x_k^2, \tag{5.19}

E[\text{IS}] = \tfrac12\gamma X^2 + \varepsilon\sum_k|n_k| + \frac{\tilde\eta}{\tau}\sum_k n_k^2, \qquad \tilde\eta = \eta - \tfrac12\gamma\tau. \tag{5.20}

(b) If $\tilde\eta > 0$ and $\lambda > 0$ , the unique minimiser of $U$ over schedules is

\begin{gathered} x_j = X\,\frac{\sinh\big(\kappa(T - t_j)\big)}{\sinh(\kappa T)}, \quad j = 0,\dots,N, \\ \text{where}\quad \frac{2}{\tau^2}\big(\cosh(\kappa\tau) - 1\big) = \tilde\kappa^2 := \frac{\lambda\sigma^2}{\tilde\eta}. \end{gathered} \tag{5.21}

For $\lambda = 0$ the minimiser is the straight line $x_j = X(1 - t_j/T)$ (TWAP).

(c) As $\tau \to 0$ , $\kappa \to \sqrt{\lambda\sigma^2/\eta}$ and the trajectory converges to the solution of the continuous problem $\min \int_0^T \big(\eta\dot x^2 + \lambda\sigma^2x^2\big)\,dt$ , $x(0) = X$ , $x(T) = 0$ :

x(t) = X\,\frac{\sinh\big(\kappa(T - t)\big)}{\sinh(\kappa T)}, \qquad \kappa = \sqrt{\lambda\sigma^2/\eta}. \tag{5.23}

Proof

(a) Iterating (5.16), $S_{k-1} = S_0 + \sum_{j=1}^{k-1}(\sigma\sqrt\tau\xi_j - \gamma n_j)$ . Then

\begin{aligned} \sum_{k=1}^N n_k S_{k-1} &= XS_0 + \sum_{k=1}^N n_k\sum_{j<k}(\sigma\sqrt\tau\xi_j - \gamma n_j) \\ &= XS_0 + \sum_{j=1}^{N}(\sigma\sqrt\tau\xi_j - \gamma n_j)\sum_{k>j}n_k \\ &= XS_0 + \sum_{j=1}^{N}(\sigma\sqrt\tau\xi_j - \gamma n_j)\,x_j, \end{aligned}

using $\sum_k n_k = X$ and $\sum_{k>j} n_k = x_j$ . With (5.17),

\begin{aligned} \text{IS} &= XS_0 - \sum_k n_k S_{k-1} + \sum_k n_k\Big(\varepsilon\operatorname{sgn}(n_k) + \frac{\eta}{\tau}n_k\Big) \\ &= -\sum_j\sigma\sqrt\tau\xi_j x_j + \gamma\sum_j n_j x_j + \varepsilon\sum_k|n_k| + \frac{\eta}{\tau}\sum_k n_k^2. \end{aligned}

For the permanent-impact term, $x_{j-1}^2 - x_j^2 = (x_{j-1} - x_j)(x_{j-1} + x_j) = n_j(2x_j + n_j)$ , so $n_jx_j = \tfrac12(x_{j-1}^2 - x_j^2) - \tfrac12 n_j^2$ . Summing and telescoping, $\sum_j n_jx_j = \tfrac12X^2 - \tfrac12\sum_j n_j^2$ . Substituting gives (5.18). The $\xi_j$ are independent with mean 0 and variance 1 and the schedule is deterministic, so (5.19) and (5.20) follow.

(b) For a sell programme with $x$ non-increasing, $\sum_k|n_k| = X$ , and $\tfrac12\gamma X^2 + \varepsilon X$ does not depend on the schedule. So we minimise

F(x_1,\dots,x_{N-1}) = \frac{\tilde\eta}{\tau}\sum_{k=1}^N(x_{k-1} - x_k)^2 + \lambda\sigma^2\tau\sum_{k=1}^{N-1}x_k^2,

with $x_0 = X$ , $x_N = 0$ . $F$ is a strictly convex quadratic when $\tilde\eta > 0$ (the first sum is a positive definite quadratic form in the interior variables given fixed endpoints, and the second is positive semi-definite), so its unique minimiser is the solution of the first-order conditions. Differentiating with respect to $x_j$ , $1 \le j \le N-1$ :

\begin{gathered} \frac{\tilde\eta}{\tau}\big(2(x_j - x_{j-1}) - 2(x_{j+1} - x_j)\big) + 2\lambda\sigma^2\tau x_j = 0 \\ \Longleftrightarrow\quad x_{j-1} - 2x_j + x_{j+1} = \tilde\kappa^2\tau^2 x_j. \end{gathered} \tag{5.22}

This is a linear second-order recurrence with constant coefficients. Trying $x_j = r^j$ gives $r + r^{-1} - 2 = \tilde\kappa^2\tau^2$ . Writing $r = e^{\kappa\tau}$ , $r + r^{-1} = 2\cosh(\kappa\tau)$ , so $\kappa$ solves $2(\cosh\kappa\tau - 1) = \tilde\kappa^2\tau^2$ , which has a unique positive solution $\kappa = \tau^{-1}\operatorname{arccosh}(1 + \tilde\kappa^2\tau^2/2)$ . The general solution is $x_j = Ae^{\kappa t_j} + Be^{-\kappa t_j}$ , equivalently $x_j = C\sinh(\kappa(T - t_j)) + D\sinh(\kappa t_j)$ . The boundary condition $x_N = 0$ forces $D = 0$ (since $\sinh(\kappa T) \ne 0$ ), and $x_0 = X$ gives $C = X/\sinh(\kappa T)$ , which is (5.21). The trajectory is strictly decreasing because $\sinh$ is increasing, which confirms the assumption $n_k > 0$ used to treat $\sum|n_k|$ as constant. For $\lambda = 0$ , (5.22) becomes $x_{j-1} - 2x_j + x_{j+1} = 0$ , whose solution with the boundary conditions is linear.

(c) As $\tau \to 0$ , $\tilde\eta \to \eta$ , and from $\cosh(\kappa\tau) = 1 + \kappa^2\tau^2/2 + O(\tau^4)$ the defining equation gives $\kappa^2 + O(\tau^2) = \tilde\kappa^2 \to \lambda\sigma^2/\eta$ . Independently, the Euler–Lagrange equation for $L = \eta\dot x^2 + \lambda\sigma^2x^2$ is $\frac{d}{dt}(2\eta\dot x) = 2\lambda\sigma^2 x$ , i.e. $\ddot x = \kappa^2 x$ , with solution (5.23) under the boundary conditions. (The continuous objective is the limit of $F$ : $\frac{\eta}{\tau}\sum n_k^2 = \eta\sum(n_k/\tau)^2\tau \to \eta\int\dot x^2dt$ and $\sigma^2\tau\sum x_k^2 \to \sigma^2\int x^2 dt$ .)∎

5.7.3Half-life and the efficient frontier#

Reading the solution

$\kappa^{-1}$ is the trade's half-life scale: with $\kappa T \gg 1$ the holdings decay roughly like $e^{-\kappa t}$ , so most of the order is done within a few multiples of $1/\kappa$ .
$\kappa^2 = \lambda\sigma^2/\eta$ : more risk aversion or more volatility means faster trading; more temporary impact means slower.
The efficient frontier: as $\lambda$ rises from 0, $E[\text{IS}]$ rises and $\operatorname{Var}[\text{IS}]$ falls. At $\lambda = 0$ the frontier has zero slope, so a small move away from TWAP reduces variance at second-order cost in expectation. Every risk-averse trader should front-load at least a little.
In continuous time, for the solution (5.23), $E[\text{IS}] - \tfrac12\gamma X^2 - \varepsilon X = \eta\int_0^T\dot x^2dt = \frac{\eta X^2\kappa^2}{\sinh^2(\kappa T)}\Big(\frac{T}{2} + \frac{\sinh(2\kappa T)}{4\kappa}\Big)$ and $\operatorname{Var} = \frac{\sigma^2X^2}{\sinh^2(\kappa T)}\Big(\frac{\sinh(2\kappa T)}{4\kappa} - \frac T2\Big)$ (Problem 5.10).

5.7.4Liquidating 5 lakh shares in a day#

Worked example

Sell $X = 5$ lakh shares of a ₹1,500 stock over one trading day ( $T = 1$ ), in $N = 13$ slices of about 29 minutes. Daily volatility 1.8%, so $\sigma = \text{₹}27$ per share per $\sqrt{\text{day}}$ . Take $\eta = 2.5\times10^{-6}$ ₹ per share per (share/day), so that selling at a rate of 5 lakh shares a day costs $\eta \times 5\times10^5 = \text{₹}1.25$ per share (8.3 bp), and $\gamma = 2.5\times10^{-7}$ (the whole order moves the price permanently by ₹0.125). Results from ac_trajectory and ac_cost_mean_var:

Risk aversion $\lambda$ (per ₹)	$\kappa T$	First slice	Holdings after 6 of 13 slices ( $t \approx 0.46$ )	$E[\text{IS}]$	sd(IS)
0 (TWAP)	0	7.7%	53.8%	₹6.5 lakh (8.7 bp)	₹73.4 lakh (98 bp)
$10^{-8}$	1.71	13.2%	39.5%	₹7.3 lakh (9.7 bp)	₹61.2 lakh (82 bp)
$10^{-7}$	5.37	33.9%	8.4%	₹16.8 lakh (22.4 bp)	₹33.0 lakh (44 bp)
$10^{-6}$	16.07	70.9%	0.1%	₹44.8 lakh (59.7 bp)	₹11.4 lakh (15 bp)

Basis points are of the initial value ₹75 crore. The trade-off is stark: halving the risk from TWAP (98 bp sd to 44 bp) multiplies the expected cost by about 2.6 (8.7 to 22.4 bp). With this volatility, the risk of a slow execution is far larger than its expected cost; which point to choose depends on $\lambda$ , i.e. on how much the desk dislikes a bad day relative to an expected cost.

Code Exercise m05_almgren_chriss (in the app: Exercises → m05_almgren_chriss) implements (5.19)–(5.22), and ac_numerical minimises $U$ directly with scipy.optimize.minimize (BFGS with an analytic gradient). The tests require the closed form and the numerical optimum to agree to $10^{-6}$ of $X$ for several $(\lambda, N)$ , check that $\lambda = 0$ gives TWAP, that random perturbations of the closed form always increase $U$ , that the frontier is monotone, and that with $N = 4{,}000$ the discrete solution matches the continuous one (5.23) to $10^{-4}$ .

Practitioner notes

The parameters $\eta$ , $\gamma$ and $\sigma$ must be estimated, and impact is concave rather than linear (5.5). Practitioners use the AC structure with a power-law temporary impact and solve numerically, and they re-optimise during the day as prices and volumes arrive.
Volume is not uniform. A practical schedule blends the AC urgency with the VWAP volume profile.
The model assumes no alpha. If you expect the price to drift against you (you are buying because you expect a rise), add a linear drift term; the optimal schedule becomes more front-loaded (Almgren and Chriss 2000 treat drift as an extension; Problem 5.11 asks you to derive it).