Part IV · 第 13 章反馈控制 · 13.4Part IV · Ch. 13 Feedback Control · 13.4

模型预测控制 MPCModel Predictive Control

模型预测控制在每个控制周期里用模型预测未来几秒，在力矩、速度、障碍等约束下解一个优化问题，只执行第一步，下个周期再重新求解。这一节讲清楚它为什么是反馈、时域之外该放什么（LQR 的终端代价）、它和"把 LQR 输出截断"有什么不同、它在什么时候会把自己逼进死胡同（递归可行性），以及怎样在几毫秒内把它解完：稠密与稀疏 QP、热启动，再到非线性 MPC 与 MPPI 在自行车模型上的路径跟踪。 Model predictive control uses a model to predict the next few seconds at every control cycle, solves an optimization problem under torque, speed and obstacle constraints, applies only the first step, and solves again at the next cycle. This lesson explains why that is feedback, what belongs beyond the horizon (the LQR terminal cost), how it differs from clipping an LQR output, when it drives itself into a dead end (recursive feasibility), and how to solve it in a few milliseconds: condensed versus sparse QPs, warm starts, and finally nonlinear MPC and MPPI tracking a path with a bicycle model.

约 45 分钟~45 min 前置：6.8 线性规划与二次规划 · 6.11 采样优化 · 6.12 动态规划 · 6.13 轨迹优化 · 9.1 移动机器人 · 13.3 LQRPrereqs: 6.8 LP & QP · 6.11 Sampling-based optimization · 6.12 Dynamic programming · 6.13 Trajectory optimization · 9.1 Mobile robots · 13.3 LQR 7 个交互7 interactives

1为什么机器人需要它Why robots need this

13.3 节的 LQR 给出一个漂亮的答案：$\mathbf u=-K\mathbf x$，一次矩阵乘法，在它的假设下是最优的。可是它的假设里没有"不能"：没有力矩上限，没有关节限位，没有速度上限，没有障碍物，也没有"足底力必须在摩擦锥里"。真实机器人处处是"不能"：

执行器饱和：电机力矩、电流、舵机转速都有上限。LQR 算出 50 N·m，电机只能给 20 N·m，截断之后 LQR 的最优性和稳定性证明都不再成立（第 3 节）。
状态约束：关节限位、速度上限、车道边界、与人保持的距离。这些是"永远不许违反"的，而不是代价里的一项。
预瞄：前方 20 米有个急弯，或者参考轨迹在 2 秒后会突然变化。一个只看当前误差的反馈律，等误差出现才反应，已经晚了。
接触：四足机器人的足底只能推、不能拉，摩擦力不能超过 $\mu f_z$。这些不等式决定了身体能怎样加速。

模型预测控制（Model Predictive Control，MPC）的想法很直接：既然有模型，就在线解一个有限时域的最优控制问题，把所有约束都写进去；只执行第一步，下个周期用新测到的状态再解一次。6.8 节已经看到，线性模型加二次代价加线性约束，这个问题就是一个 QP，并用一辆一维小车演示了"碰到约束就饱和"。这一节回答 6.8 节没回答的问题：为什么"只执行第一步再重解"是反馈；时域之外的未来该怎么算（终端代价，与 LQR 的精确关系）；它与"把 LQR 截断"差在哪里；它什么时候会把自己逼进无解的状态（递归可行性）；以及怎样在几毫秒内解完（稠密与稀疏的 QP、热启动、实时迭代）。最后把它用到非线性的自行车模型上，并和 6.11 节的采样式 MPC（MPPI）对比。

The LQR of Lesson 13.3 gives a beautiful answer: $\mathbf u=-K\mathbf x$, one matrix product, optimal under its assumptions. But its assumptions contain no "cannot": no torque limit, no joint limit, no speed limit, no obstacle, no "foot forces must stay in the friction cone". Real robots are full of "cannot":

Actuator saturation: motor torque, current and servo speed all have limits. LQR asks for 50 N·m, the motor can give 20 N·m, and once you clip, LQR's optimality and its stability proof no longer apply (Section 3).
State constraints: joint limits, speed limits, lane boundaries, a distance to keep from people. These must never be violated; they are not one more term in a cost.
Preview: there is a sharp bend 20 m ahead, or the reference will change abruptly in 2 s. A feedback law that only looks at the current error reacts once the error has appeared, which is already too late.
Contact: a quadruped's feet can push but not pull, and friction cannot exceed $\mu f_z$. These inequalities decide how the body can accelerate.

The idea of model predictive control (MPC) is direct: since you have a model, solve a finite-horizon optimal control problem online with every constraint written in; apply only the first step, and solve again at the next cycle from the newly measured state. Lesson 6.8 already showed that with a linear model, a quadratic cost and linear constraints this problem is a QP, and used a 1D cart to show inputs "saturating the moment a constraint binds". This lesson answers what 6.8 left open: why "apply the first step and re-solve" is feedback; how to account for the future beyond the horizon (the terminal cost, and its exact relation to LQR); how it differs from clipping LQR; when it drives itself into a state with no solution (recursive feasibility); and how to solve it in a few milliseconds (condensed and sparse QPs, warm starts, real-time iterations). Finally we apply it to a nonlinear bicycle model and compare it with the sampling-based MPC (MPPI) of Lesson 6.11.

2直觉：看得见几秒，就只为这几秒做打算Intuition: plan for the seconds you can see

夜里开车，车灯照亮前方 50 米。你按这 50 米里看到的东西决定怎么踩油门、打方向，可你不会把这个计划"执行完"：一秒钟后车灯照到了新的路面，你就按新看到的东西重新打算。计划很长，承诺很短：每次只兑现计划的第一小步。

MPC 每个周期做三件事：

Driving at night, your headlights light up 50 m of road. You decide throttle and steering from what you see in those 50 m, but you never "finish executing" that plan: a second later the headlights reveal new road and you re-plan from what you now see. Long plans, short commitments: each time you honour only the first small step of the plan.

At every cycle MPC does three things:

$$ \begin{aligned}&\text{1. measure } \mathbf x(t)\\ &\text{2. solve } \min_{\mathbf u_0,\dots,\mathbf u_{N-1}} \sum_{k=0}^{N-1}\ell(\mathbf x_k,\mathbf u_k)+V_f(\mathbf x_N)\ \ \text{s.t. model, constraints},\ \mathbf x_0=\mathbf x(t)\\ &\text{3. apply } \hl{u0}{\mathbf u_0^\star}\text{, discard the rest}\end{aligned} $$

下图是一辆小车（质点，控制量是加速度，$|u|\le1$ m/s²，$|v|\le1$ m/s），每 0.1 s 解一次时域为 $N$ 步的 QP。上面的轨道上拖动琥珀色的目标；中间的图里，绿色是已经走过的轨迹，蓝色虚线是此刻的计划，青色的淡线是过去每 0.5 s 留下的旧计划；下面是输入，琥珀色的那一根就是此刻真正执行的 $\mathbf u_0^\star$。试试"踢一脚"（瞬间加 0.8 m/s），再把模式切到"开环"（只在开始时规划一次，然后照着计划执行到底）。

Below is a cart (a point mass whose input is acceleration, $|u|\le1$ m/s², $|v|\le1$ m/s) that solves an $N$-step QP every 0.1 s. Drag the amber target on the rail at the top. In the middle plot, green is the path already travelled, the blue dashed curve is the plan made right now, and the faint cyan curves are older plans left behind every 0.5 s; the bottom plot shows the inputs, and the amber bar is the $\mathbf u_0^\star$ actually applied now. Try "Kick" (an instant +0.8 m/s), then switch the mode to "open loop" (plan once at the start, then execute the plan to the end).

FIG 1滚动时域：每步重新规划，只执行第一步Receding horizon: re-plan every step, apply only the first

代价 $\sum_k (p_k-p_{\rm ref})^2+0.1\,v_k^2+0.1\,u_k^2$，终端代价是同一问题的 LQR 矩阵 $P$（第 3 节）。轨道上的蓝点是当前计划里每隔 0.2 s 的预测位置。"模型误差"让真实小车的质量是模型以为的倍数（加速度变成 $u/\text{倍数}$），控制器并不知道。开环模式下拖动目标会触发一次新的规划；计划用完后 $u=0$。 Cost $\sum_k (p_k-p_{\rm ref})^2+0.1\,v_k^2+0.1\,u_k^2$; the terminal cost is the LQR matrix $P$ of the same problem (Section 3). The blue dots on the rail are the current plan's predicted positions every 0.2 s. "Model error" makes the true cart's mass a multiple of what the model believes (its acceleration becomes $u/\text{multiple}$); the controller does not know. In open-loop mode, dragging the target triggers one new plan; once the plan is used up, $u=0$.

试一试Try this

在滚动时域模式下按"踢一脚"，等小车回到目标 2 cm 以内且速度小于 0.02 m/s。注意被踢之后的第一个新计划：它是从被踢之后的状态出发的。In receding-horizon mode press "Kick" and wait until the cart is back within 2 cm of the target with speed below 0.02 m/s. Look at the first new plan after the kick: it starts from the kicked state.
切到"开环"，把模型误差调到 ×0.8 或更小、×1.2 或更大，按"重置"。计划用完半秒后，小车离目标超过 20 cm。同样的模型误差下，切回滚动时域再重置，误差去哪了？Switch to "open loop", set the model error to ×0.8 or below, or ×1.2 or above, and press "Reset". Half a second after the plan runs out, the cart is more than 20 cm from the target. With the same model error, switch back to receding horizon and reset: where did the error go?

核心洞见Key insight

MPC 是一个反馈律 $\mathbf u=\kappa_N(\mathbf x)$，只不过这个函数不是写成公式，而是"以 $\mathbf x$ 为初值解一个优化问题，取第一个输入"。反馈来自重新测量、重新求解：模型不准、被踢一脚，下一个周期的计划就从真实的状态出发。开环执行同一个最优计划，模型误差会原封不动地留在结果里。

MPC is a feedback law $\mathbf u=\kappa_N(\mathbf x)$, except that the function is not written as a formula but as "solve an optimization problem starting from $\mathbf x$ and take the first input". The feedback comes from measuring again and solving again: whether the model is off or the cart gets kicked, the next cycle's plan starts from the true state. Executing the same optimal plan open loop leaves every model error in the result.

3动手玩：时域之外，和约束之内Play: beyond the horizon, and inside the constraints

时域之外的代价：终端代价 $P_f$The cost beyond the horizon: the terminal cost $P_f$

有限时域有一个天生的盲点：第 $N$ 步以后发生什么，优化器完全不管。对一个不稳定的系统，这可能是致命的。考虑线性化的倒立摆 $\ddot\theta=10\,\theta+u$（$u$ 是角加速度输入，零阶保持离散化，$\Delta t=0.1$ s），代价 $\theta^2+0.1\dot\theta^2+0.1u^2$。如果时域很短又没有终端代价，优化器会想：反正 $N$ 步之内摆还倒不下去，何必花力气？于是它几乎不出力，摆就真倒了。

解决办法来自 6.12 节的贝尔曼原理：如果时域结束后剩下的代价能精确算出来，有限时域就等价于无限时域。约束不起作用时，从 $\mathbf x_N$ 出发的无限时域最优代价正是 LQR 的 $\mathbf x_N^\top P\mathbf x_N$（$P$ 是离散代数 Riccati 方程的解，13.3 节）。把它当终端代价 $V_f(\mathbf x_N)=\mathbf x_N^\top\hl{Pf}{P_f}\mathbf x_N$，$P_f=P$：

A finite horizon has a built-in blind spot: whatever happens after step $N$, the optimizer does not care. For an unstable system this can be fatal. Take the linearized inverted pendulum $\ddot\theta=10\,\theta+u$ ($u$ an angular-acceleration input, zero-order-hold discretization, $\Delta t=0.1$ s) with cost $\theta^2+0.1\dot\theta^2+0.1u^2$. With a short horizon and no terminal cost the optimizer reasons: the pendulum will not have fallen within $N$ steps anyway, so why spend effort? It barely pushes, and the pendulum really falls.

The fix comes from Bellman's principle in Lesson 6.12: if the cost left over after the horizon can be computed exactly, a finite horizon is equivalent to an infinite one. When no constraint is active, the infinite-horizon optimal cost from $\mathbf x_N$ is exactly LQR's $\mathbf x_N^\top P\mathbf x_N$ ($P$ solves the discrete algebraic Riccati equation, Lesson 13.3). Use it as the terminal cost $V_f(\mathbf x_N)=\mathbf x_N^\top\hl{Pf}{P_f}\mathbf x_N$ with $P_f=P$:

FIG 2终端代价：时域多短才会出事？Terminal cost: how short can the horizon get?

相平面 (θ, θ̇)：拖动琥珀色初值PHASE PLANE (θ, θ̇) · drag the amber start

θ(t)θ(t)

蓝色实线是 MPC 的闭环轨迹，紫色虚线是无限时域 LQR，青色点线是 $t=0$ 时 MPC 的预测（空心大点是 $\mathbf x_N$）。琥珀色椭圆是 LQR 价值函数过初值的等值线 $\mathbf x^\top P\mathbf x=\mathbf x_0^\top P\mathbf x_0$。这里不加输入约束，所以 MPC 律是线性的，$\mathbf u_0^\star=-K_N\mathbf x$，读数给出 $K_N$ 和闭环矩阵的谱半径 $\rho$（$\rho<1$ 稳定）。"闭环代价 / LQR" 是 8 s 内两者实际付出的代价之比。 The solid blue curve is MPC's closed-loop trajectory, the dashed purple one is infinite-horizon LQR, and the dotted cyan one is MPC's prediction at $t=0$ (the large hollow dot is $\mathbf x_N$). The amber ellipse is the level set of LQR's value function through the start, $\mathbf x^\top P\mathbf x=\mathbf x_0^\top P\mathbf x_0$. There are no input constraints here, so the MPC law is linear, $\mathbf u_0^\star=-K_N\mathbf x$; the readout gives $K_N$ and the spectral radius $\rho$ of the closed-loop matrix ($\rho<1$ is stable). "Closed-loop cost / LQR" is the ratio of the costs the two actually pay over 8 s.

试一试Try this

选"无终端代价"，从 $N=10$ 往下减，找出摆仍然会倒（$\rho>1$）的最长时域。这个时域对应几秒？再看 $P_f=Q$ 时界限移动了多少。Choose "no terminal cost" and lower $N$ from 10 to find the longest horizon at which the pendulum still falls ($\rho>1$). How many seconds is that? Then see how far the limit moves with $P_f=Q$.
选 $P_f=P$，把时域减到 $N=1$：$K_N$ 与 $K_\infty$ 的差小于 $10^{-6}$。只看一步的 MPC，为什么和无限时域的 LQR 一模一样？Choose $P_f=P$ and cut the horizon to $N=1$: $K_N$ differs from $K_\infty$ by less than $10^{-6}$. Why is a one-step MPC identical to infinite-horizon LQR?

核心洞见Key insight

终端代价是"时域之外的未来"的替身。$P_f=P$ 时，只要约束不起作用，MPC 就等于 LQR，与 $N$ 无关：时域只在约束起作用时才需要长。没有终端代价时，时域必须长到足以"看见"不作为的后果；本例需要 $N\ge8$（0.8 s），而且 $N=8$ 时闭环代价仍是 LQR 的约 6 倍。工程上的结论：终端代价几乎总该放 LQR 的 $P$，它不花任何在线计算。

The terminal cost stands in for the future beyond the horizon. With $P_f=P$, as long as no constraint is active, MPC equals LQR whatever $N$ is: the horizon only needs to be long where constraints act. Without a terminal cost the horizon must be long enough to "see" the consequence of doing nothing; here that takes $N\ge8$ (0.8 s), and even at $N=8$ the closed-loop cost is about 6 times LQR's. The engineering conclusion: the terminal cost should almost always be LQR's $P$, which costs no online computation at all.

约束之内：MPC 与"截断的 LQR"Inside the constraints: MPC versus "clipped LQR"

最省事的处理约束的办法是：照算 LQR，再把输出截到 $[-u_{\max},u_{\max}]$。它常常够用，但它有两个盲点。第一，LQR 不知道将来会饱和：它以为刹车时能出任意大的减速度，于是放心地加速，等真到了刹车的时候才发现力气不够，只好冲过头（和 13.1 节的积分饱和是同一类病）。第二，它根本看不见状态约束：前面有堵墙，它也照样冲。MPC 把 $\hl{umax}{|u_k|\le u_{\max}}$ 和 $p_k\le p_{\rm wall}$ 都写进预测里，所以会提前刹车。

下图两辆小车从同一点出发去 $p=0$，前方有一堵可以拖动的墙。两者用同一个代价（$Q=\mathrm{diag}(q,0.2)$，$R=1$），MPC 的时域 $N=30$（3 s），终端代价 $P$。

The laziest way to handle constraints: compute LQR as usual and clip the output to $[-u_{\max},u_{\max}]$. It often works, but it has two blind spots. First, LQR does not know it will saturate later: it believes it can brake as hard as it likes, so it accelerates confidently and only discovers when braking time comes that it lacks the strength, and overshoots (the same family of illness as integrator windup in Lesson 13.1). Second, it cannot see state constraints at all: a wall ahead changes nothing for it. MPC writes both $\hl{umax}{|u_k|\le u_{\max}}$ and $p_k\le p_{\rm wall}$ into the prediction, so it brakes early.

In the figure two carts leave the same point for $p=0$, with a draggable wall ahead. Both use the same cost ($Q=\mathrm{diag}(q,0.2)$, $R=1$); MPC has horizon $N=30$ (3 s) and terminal cost $P$.

FIG 3截断的 LQR vs. 知道约束的 MPCClipped LQR vs. constraint-aware MPC

紫色 = 截断的 LQR（$u=\mathrm{clip}(-K\mathbf x)$），蓝色虚线 = MPC；红线是墙，绿色虚线是目标 $p=0$，琥珀色虚线是 $\pm u_{\max}$。每次改动都重新仿真 10 s（100 个控制周期，每个周期 MPC 解一个 30 变量的 QP），小车按两倍速重播。起点很远、电机又很弱时，MPC 也可能开到一个 QP 无解的状态（第 4 节的递归可行性问题），这时读数会显示不可行的步数，控制器退而满力刹车。 Purple = clipped LQR ($u=\mathrm{clip}(-K\mathbf x)$), dashed blue = MPC; the red line is the wall, the dashed green line the target $p=0$, the dashed amber lines $\pm u_{\max}$. Every change re-simulates 10 s (100 control cycles, each solving a 30-variable QP for MPC), and the carts replay at double speed. From a far start with a weak motor, MPC too can drive into a state where its QP has no solution (the recursive-feasibility problem of Section 4); the readout then counts the infeasible steps and the controller falls back to braking at full strength.

试一试Try this

让控制更激进（加大 $q$）或让电机更弱（减小 $u_{\max}$），直到截断的 LQR 越过墙 10 cm 以上，而 MPC 仍然停在墙前（越墙不到 0.5 cm）。Make the control more aggressive (raise $q$) or the motor weaker (lower $u_{\max}$) until clipped LQR crosses the wall by more than 10 cm while MPC still stops in front of it (less than 0.5 cm past).
反过来：找一组参数，让两条输入曲线在整个 10 s 内完全重合（$\max|u_{\rm LQR}-u_{\rm MPC}|<10^{-4}$），而输入确实碰到过 $u_{\max}$。截断并不总是错的：什么时候它恰好就是最优？The other way round: find parameters where the two input curves coincide for the whole 10 s ($\max|u_{\rm LQR}-u_{\rm MPC}|<10^{-4}$) while the input really does hit $u_{\max}$. Clipping is not always wrong: when does it happen to be optimal?

核心洞见Key insight

截断的 LQR 只在"此刻"尊重约束，MPC 在"整个预测"里尊重约束。差别出现在现在的选择会让将来的约束起作用的时候：加速太猛，刹车时就会饱和。如果饱和只发生在一开始，之后线性律能接得住，截断恰好就是最优的（上面第二个任务）；可一旦"现在"决定了"将来够不够力"，只有能预测的控制器才算得对。

Clipped LQR respects the constraints "now"; MPC respects them over the whole prediction. The difference appears when today's choice makes tomorrow's constraint bind: accelerate too hard and braking will saturate. If saturation only happens at the very beginning and the linear law can take over afterwards, clipping happens to be optimal (the second task above); but as soon as "now" decides whether there will be enough strength "later", only a controller that predicts gets it right.

4第二个视角：MPC 会把自己逼进死胡同A second view: MPC can corner itself

约束带来一个 LQR 没有的新问题：优化问题可能无解。6.8 节讲过，初始状态本身违反硬约束时 QP 不可行。更隐蔽的是：此刻问题可行，MPC 自己选的动作却把系统带到了一个下一步不可行的状态。这叫失去递归可行性（recursive feasibility）。

还是那辆小车（$|u|\le1$），墙在 $p=0$，目标在 $p=-0.3$。在相平面 $(p,v)$ 里有两个重要的集合：

可挽救集 $\hl{Cinf}{\mathcal C_\infty}$：从这些状态出发，满力刹车能停在墙前。速度越大，需要的刹车距离越长（$v^2/2u_{\max}$），边界是一条抛物线。在这个集合外面，任何控制器都救不了。
$N$ 步可行集 $\hl{XN}{\mathcal X_N}$：MPC 的 $N$ 步问题此刻有解的状态。如果 $N$ 步之内的刹车距离不够把车停下，但 $N$ 步之内还没撞到墙，问题依然"可行"：撞墙发生在时域之外，优化器看不到。

$\mathcal X_N$ 比 $\mathcal C_\infty$ 大出来的那一条（下图红色）就是陷阱：此刻可行，却已注定撞墙。更糟的是，MPC 可以从安全的绿色区域自己开进红色区域：它为了尽快到达目标而加速，而墙在时域之外。

Constraints bring a problem LQR never had: the optimization can have no solution. Lesson 6.8 noted that the QP is infeasible when the initial state already violates a hard constraint. More insidious: the problem is feasible now, but the action MPC itself chooses carries the system into a state where the next problem is infeasible. This is losing recursive feasibility.

Same cart ($|u|\le1$), wall at $p=0$, target at $p=-0.3$. Two sets in the phase plane $(p,v)$ matter:

The safe set $\hl{Cinf}{\mathcal C_\infty}$: from these states, braking at full strength stops before the wall. The higher the speed, the longer the braking distance ($v^2/2u_{\max}$), so its boundary is a parabola. Outside it no controller can save the cart.
The $N$-step feasible set $\hl{XN}{\mathcal X_N}$: the states from which MPC's $N$-step problem has a solution now. If braking for $N$ steps is not enough to stop but the cart has not hit the wall within $N$ steps, the problem is still "feasible": the crash happens beyond the horizon, where the optimizer cannot see.

The strip by which $\mathcal X_N$ exceeds $\mathcal C_\infty$ (red in the figure) is the trap: feasible now, yet doomed. Worse, MPC can drive itself from the safe green region into the red one: it accelerates to reach the target quickly while the wall is beyond its horizon.

FIG 4递归可行性：此刻可行，不代表下一步可行Recursive feasibility: feasible now does not mean feasible next

相平面 (p, v)：拖动琥珀色起点PHASE PLANE (p, v) · drag the amber start

绿色：$N$ 步问题可行，且在可挽救集内。红色：$N$ 步问题可行，但已在可挽救集外（注定撞墙）。很淡的绿色（只在加终端约束时出现）：还救得了，但这个更严格的 $N$ 步问题不可行。绿色曲线是连续时间的刹车抛物线 $v=\sqrt{2u_{\max}(p_{\rm wall}-p)}$，与离散的可挽救集边界几乎重合。可行性用"满力刹车"精确判断：它让以后每一步的位置都最小。蓝色是闭环轨迹，青色点线是当前计划；问题不可行时画一个红色 ✕，控制器退而满力刹车。代价 $(p-p_{\rm ref})^2+0.1v^2+0.1u^2$，没有终端代价，$\Delta t=0.1$ s。 Green: the $N$-step problem is feasible and the state is in the safe set. Red: the $N$-step problem is feasible but the state is already outside the safe set (doomed). Very pale green (only with the terminal constraint): still savable, but this stricter $N$-step problem is infeasible. The green curve is the continuous-time braking parabola $v=\sqrt{2u_{\max}(p_{\rm wall}-p)}$, which almost coincides with the boundary of the discrete safe set. Feasibility is checked exactly with "full braking", which minimizes every future position. Blue is the closed-loop trajectory, dotted cyan the current plan; when the problem is infeasible a red ✕ is drawn and the controller falls back to full braking. Cost $(p-p_{\rm ref})^2+0.1v^2+0.1u^2$, no terminal cost, $\Delta t=0.1$ s.

试一试Try this

不加终端约束，从静止于 $p_0\approx-6$ m 出发（偏差 0.25 m 以内，可用"起点"按钮），找出能一路可行到达目标的最短时域 $N$（再短一步就会撞上不可行）。它和小车跑出的最高速度的刹车时间比，差多少？Without the terminal constraint, start at rest at $p_0\approx-6$ m (within 0.25 m; the "start" button helps) and find the shortest horizon $N$ that stays feasible all the way to the target (one step shorter runs into infeasibility). Compare it with the braking time from the top speed the cart reaches.
打开终端约束 $v_N\le0$（计划结束时必须已经停下或在后退），把时域缩到 $N\le8$（例如 $N=8$），从一个此刻可行的起点（例如静止于 −6 m）运行 MPC：它全程可行并到达目标。代价是什么？再试试 $N=3$：20 s 后它还在半路上。Turn on the terminal constraint $v_N\le0$ (the plan must end stopped or backing off), cut the horizon to $N\le8$ (say $N=8$), and run MPC from a start where the problem is feasible now (e.g. at rest at −6 m): it stays feasible all the way and reaches the target. What is the price? Then try $N=3$: after 20 s it is still only halfway.

核心洞见Key insight

递归可行性靠的是终端集：要求计划结束时落在一个"控制不变集" $\mathcal X_f$ 里，即一个进去以后有办法永远待在里面（且满足约束）的集合。这里 $\mathcal X_f=\{v\le0\}$：停下或后退的小车，令 $u=0$ 就永远不会撞墙。于是上一步的计划去掉第一步、末尾补一步 $u=0$，就是下一步的一个可行计划，可行性一步步传下去（第 6 节的证明）。代价是保守：$N$ 很短时，小车只敢以"$N$ 步内能停下"的速度前进，慢得像爬。

Recursive feasibility rests on a terminal set: require the plan to end inside a "control-invariant set" $\mathcal X_f$, a set that, once entered, can be stayed in forever (while satisfying the constraints). Here $\mathcal X_f=\{v\le0\}$: a cart that has stopped or is backing off never hits the wall with $u=0$. Then the previous plan with its first step dropped and one step $u=0$ appended is a feasible plan for the next problem, and feasibility is passed on step by step (the proof in Section 6). The price is conservatism: with a very short $N$, the cart only dares move as fast as it can stop within $N$ steps, and it crawls.

5第三个视角：几毫秒内解完A third view: solving it within milliseconds

同一个 QP，两种写法：稠密（缩并）与稀疏One QP, two ways to write it: condensed and sparse

MPC 的 QP 有两种标准写法，对应 6.13 节的打靶与配点：

稠密（缩并，condensed）：用动力学把所有状态消掉，$\mathbf X=S_x\mathbf x_0+S_u\mathbf U$，只剩 $Nm$ 个输入变量。Hessian $\hl{H}{H}$ 是 $Nm\times Nm$ 的稠密矩阵，因为每个状态都依赖之前所有的输入。
稀疏（sparse）：保留所有状态为变量，动力学作为等式约束。变量多了（$N(n+m)$ 个），但 KKT 矩阵 $\hl{KKT}{K}$ 是带状的：第 $k$ 步只和第 $k\pm1$ 步耦合。用 6.12 节的 Riccati 递推解它，代价随 $N$ 线性增长。

哪种更快，取决于状态维数 $n$、输入维数 $m$ 和时域 $N$。下图上方两幅是两种矩阵的非零结构，下方是一次分解所需的浮点运算数随 $N$ 的变化。

MPC's QP comes in two standard forms, the counterparts of shooting and collocation in Lesson 6.13:

Condensed (dense): use the dynamics to eliminate every state, $\mathbf X=S_x\mathbf x_0+S_u\mathbf U$, leaving only the $Nm$ inputs as variables. The Hessian $\hl{H}{H}$ is a dense $Nm\times Nm$ matrix, because every state depends on all earlier inputs.
Sparse: keep every state as a variable and the dynamics as equality constraints. There are more variables ($N(n+m)$), but the KKT matrix $\hl{KKT}{K}$ is banded: step $k$ couples only to steps $k\pm1$. Solved with the Riccati recursion of Lesson 6.12, its cost grows linearly in $N$.

Which is faster depends on the state dimension $n$, the input dimension $m$ and the horizon $N$. The figure shows the nonzero patterns of the two matrices in its top two panels and, below them, the floating-point operations one factorization needs as a function of $N$.

FIG 5稠密 vs. 稀疏：结构与运算量Condensed vs. sparse: structure and operation count

稠密 Hessian HCONDENSED HESSIAN H

稀疏 KKT 矩阵SPARSE KKT MATRIX

第一幅：紫色是稠密 $H$ 的非零元（全满）。第二幅：稀疏 KKT 矩阵，变量顺序 $(\mathbf u_0,\mathbf x_1,\mathbf u_1,\dots,\mathbf x_N)$ 之后是动力学的乘子；蓝色是代价的块（$R$、$Q$），绿色是动力学的块（$A$、$B$、$-I$）。下方：每次分解的运算量，稠密取 Cholesky 的 $(Nm)^3/3$，"每步缩并"再加上构造 $H$ 的约 $N^2n^2m$；稀疏取本页 Riccati 递推的朴素实现 $N(4n^3+6n^2m+4nm^2+m^3/3)$。这些是主导项的估计，真实求解器的常数不同（利用对称性约可减半），但增长的阶不变。 First panel: the nonzeros of the dense $H$ in purple (all of it). Second panel: the sparse KKT matrix, variables ordered $(\mathbf u_0,\mathbf x_1,\mathbf u_1,\dots,\mathbf x_N)$ followed by the dynamics multipliers; blue are cost blocks ($R$, $Q$), green are dynamics blocks ($A$, $B$, $-I$). Bottom: operations per factorization: condensed counts Cholesky's $(Nm)^3/3$, and "condense every step" adds about $N^2n^2m$ for building $H$; sparse counts a naive implementation of the Riccati recursion, $N(4n^3+6n^2m+4nm^2+m^3/3)$. These are leading-order estimates; real solvers have different constants (exploiting symmetry roughly halves them), but the orders of growth are the same.

试一试Try this

选"四旋翼"（$n=12,m=4$，线性定常）：从 $N=10$ 开始加长时域，停在稀疏刚刚变得更省的那个 $N$（交叉点或再多一步）。再切到"四足凸 MPC"（$m=12$），交叉点移到了哪里？为什么输入维数大时稠密吃亏？Choose "quadrotor" ($n=12,m=4$, LTI): lengthen the horizon from $N=10$ and stop at the $N$ where sparse has just become cheaper (the crossover or one step beyond). Then switch to "quadruped convex MPC" ($m=12$): where did the crossover go? Why does a large input dimension hurt the condensed form?

实时预算与热启动Real-time budgets and warm starts

控制器每 10 ms 要给出一个输入，可迭代求解器（ADMM、内点法、有效集法）需要的迭代数是不确定的。两条工程经验：

热启动（warm start）：这一周期的最优计划，和上一周期的计划"向前平移一步、末尾补一步"非常接近。从这里出发，而不是从零出发，求解器只需修一修。
迭代预算：给每个周期设一个迭代上限，到点就交出当前的近似解。热启动之后，几步迭代往往已经足够好；冷启动的几步迭代可能离最优解很远。

下图用 OSQP 式的 ADMM（6.8 节）解小车撞墙问题（$N=30$，墙在 0.05 m，$q=10$），两辆车同时跑：紫色每个周期从零开始，蓝色从平移后的上一个解开始。柱子是每个周期用掉的迭代数（对数坐标），半透明的柱子表示用完预算仍未收敛。

The controller must deliver an input every 10 ms, yet iterative solvers (ADMM, interior point, active set) need an uncertain number of iterations. Two engineering habits:

Warm starting: this cycle's optimal plan is very close to last cycle's plan "shifted one step forward with one step appended". Starting from there instead of from zero, the solver only has to touch it up.
An iteration budget: cap the iterations per cycle and hand over the current approximate solution when time is up. After a warm start a few iterations are often good enough; a few iterations from a cold start may be far from optimal.

The figure solves the cart-and-wall problem ($N=30$, wall at 0.05 m, $q=10$) with OSQP-style ADMM (Lesson 6.8), two carts running side by side: purple starts every cycle from zero, blue from the shifted previous solution. The bars are the iterations used per cycle (log scale); a translucent bar means the budget ran out before convergence.

FIG 6热启动：每周期只有几步迭代时谁还靠得住Warm starts: who is still reliable with a few iterations per cycle

ADMM 参数：$\rho=0.3$，$\sigma=10^{-6}$，松弛 $\alpha=1.6$，收敛判据为原始与对偶残差都低于 $10^{-3}$（绝对加相对）；约束行做了归一化，代价乘以 0.1（6.8 节说过的预处理）。矩阵只分解一次，每次迭代是一次回代。热启动把上一周期的原变量和对偶变量都平移一步。紫色 = 冷启动，蓝色 = 热启动，红色虚线是迭代预算。 ADMM settings: $\rho=0.3$, $\sigma=10^{-6}$, relaxation $\alpha=1.6$; converged when both primal and dual residuals are below $10^{-3}$ (absolute plus relative); constraint rows are normalized and the cost is scaled by 0.1 (the preconditioning mentioned in 6.8). The matrix is factored once, and each iteration is one back-substitution. The warm start shifts both the primal and the dual variables of the previous cycle by one step. Purple = cold start, blue = warm start, the dashed red line is the iteration budget.

试一试Try this

先在默认预算下看完一次运行，比较两者的平均迭代数。然后把预算降到 2 到 5 次之间，跑完 8 s：冷启动的小车越墙超过 5 cm，热启动的仍在 1 cm 以内。First watch one run at the default budget and compare the mean iterations. Then lower the budget to between 2 and 5 and let 8 s run: the cold-started cart crosses the wall by more than 5 cm while the warm-started one stays within 1 cm.

次优也稳定Suboptimal is still stable

热启动在理论上也站得住：第 6 节的稳定性证明里，下一周期的候选计划正是"平移后的上一计划"，它已经让代价下降。所以只要求解器从这个候选出发、并且不让代价变大、不违反约束，即使远没有收敛，闭环依然稳定（Scokaert、Mayne 与 Rawlings，1999："可行即稳定"）。FIG 6 的 ADMM 并不严格满足这些条件（它的中间迭代可能略微违反约束，代价也不保证单调下降），但道理相同：热启动的起点就是平移后的上一计划，离最优解已经很近，所以几步迭代就够用；冷启动的几步迭代离最优解很远，也没有这种保证。

Warm starting also stands on theory: in the stability proof of Section 6, the candidate plan for the next cycle is exactly "the previous plan, shifted", and it already decreases the cost. So as long as the solver starts from that candidate, never increases the cost and never violates a constraint, the closed loop stays stable even far from convergence (Scokaert, Mayne & Rawlings, 1999: "feasibility implies stability"). The ADMM of FIG 6 does not strictly meet these conditions (its intermediate iterates can slightly violate constraints, and its cost need not decrease monotonically), but the same reason is at work: the warm start begins at the shifted previous plan, already close to optimal, so a few iterations suffice; a few iterations from a cold start are far from optimal and come with no such guarantee.

6再看数学The math

提示：把鼠标悬停在公式中带底色的符号上，对应的图元素会被高亮（FIG 1 的第一个输入柱、FIG 2 的琥珀色价值椭圆、FIG 3 的 $\pm u_{\max}$ 线、FIG 4 的红色区域或绿色边界、FIG 5 的两个矩阵、FIG 6 的热启动柱）。Tip: hover the highlighted symbols in the formulas; the matching element lights up (FIG 1's first input bar, FIG 2's amber value ellipse, FIG 3's $\pm u_{\max}$ lines, FIG 4's red region or green boundary, FIG 5's two matrices, FIG 6's warm-start bars).

问题The problem

线性 MPC 在每个采样时刻，以测得的 $\mathbf x(t)$ 为初值解

At every sampling instant, linear MPC solves, from the measured $\mathbf x(t)$,

$$ \begin{aligned} V_N(\mathbf x)=\min_{\mathbf u_0,\dots,\mathbf u_{N-1}}\ &\sum_{k=0}^{N-1}\big(\mathbf x_k^\top Q\mathbf x_k+\mathbf u_k^\top R\mathbf u_k\big)+\mathbf x_N^\top\hl{Pf}{P_f}\mathbf x_N\\ \text{s.t.}\quad &\mathbf x_{k+1}=A\mathbf x_k+B\mathbf u_k,\ \ \mathbf x_0=\mathbf x,\ \ \hl{umax}{\mathbf u_k\in\mathcal U},\ \ \mathbf x_k\in\mathcal X,\ \ \mathbf x_N\in\mathcal X_f, \end{aligned} $$

并执行 $\kappa_N(\mathbf x)=\hl{u0}{\mathbf u_0^\star(\mathbf x)}$。$\mathcal U,\mathcal X$ 是多面体（线性不等式），$\mathcal X_f$ 是终端集。跟踪一个参考 $\mathbf x_{\rm ref}$ 时，把 $\mathbf x$ 换成 $\mathbf x-\mathbf x_{\rm ref}$（参考须是平衡点，或者另加前馈）。

and applies $\kappa_N(\mathbf x)=\hl{u0}{\mathbf u_0^\star(\mathbf x)}$. $\mathcal U,\mathcal X$ are polyhedra (linear inequalities) and $\mathcal X_f$ is the terminal set. To track a reference $\mathbf x_{\rm ref}$, replace $\mathbf x$ by $\mathbf x-\mathbf x_{\rm ref}$ (the reference must be an equilibrium, or a feedforward is added).

缩并成稠密 QPCondensing into a dense QP

把动力学展开：$\mathbf x_k=A^k\mathbf x_0+\sum_{j\lt k}A^{k-1-j}B\mathbf u_j$。叠成向量 $\mathbf X=(\mathbf x_1,\dots,\mathbf x_N)$、$\mathbf U=(\mathbf u_0,\dots,\mathbf u_{N-1})$：

Unroll the dynamics: $\mathbf x_k=A^k\mathbf x_0+\sum_{j\lt k}A^{k-1-j}B\mathbf u_j$. Stacking $\mathbf X=(\mathbf x_1,\dots,\mathbf x_N)$ and $\mathbf U=(\mathbf u_0,\dots,\mathbf u_{N-1})$:

$$ \mathbf X=S_x\mathbf x_0+S_u\mathbf U,\quad S_x=\begin{bmatrix}A\\A^2\\ \vdots\\ A^N\end{bmatrix},\quad S_u=\begin{bmatrix}B&0&\cdots&0\\ AB&B&\cdots&0\\ \vdots&&\ddots&\\ A^{N-1}B&A^{N-2}B&\cdots&B\end{bmatrix}, $$ $$ \begin{aligned}&\min_{\mathbf U}\ \tfrac12\mathbf U^\top\hl{H}{H}\mathbf U+(F\mathbf x_0)^\top\mathbf U\quad\text{s.t.}\quad G\mathbf U\le\mathbf w+E\mathbf x_0,\\ &H=2\big(S_u^\top\bar Q S_u+\bar R\big),\qquad F=2S_u^\top\bar QS_x, \end{aligned} $$

其中 $\bar Q=\mathrm{diag}(Q,\dots,Q,P_f)$，$\bar R=\mathrm{diag}(R,\dots,R)$。$H$ 正定（$R\succ0$ 即可），所以这是一个严格凸 QP，有唯一解。注意 $\mathbf x_0$ 只出现在线性项和约束右端：$H$ 与当前状态无关，线性定常系统可以离线算好、甚至离线分解。状态约束 $\mathbf x_k\in\mathcal X$ 经 $S_u$ 变成关于 $\mathbf U$ 的稠密不等式，这就是 6.8 节 FIG 4 里"墙"的那些约束行。

with $\bar Q=\mathrm{diag}(Q,\dots,Q,P_f)$ and $\bar R=\mathrm{diag}(R,\dots,R)$. $H$ is positive definite ($R\succ0$ suffices), so this is a strictly convex QP with a unique solution. Note that $\mathbf x_0$ appears only in the linear term and the right-hand side: $H$ does not depend on the current state, so for an LTI system it can be computed, and even factored, offline. State constraints $\mathbf x_k\in\mathcal X$ become dense inequalities in $\mathbf U$ through $S_u$: these are the "wall" rows of FIG 4 in Lesson 6.8.

无约束时就是 LQR。没有约束时 $\mathbf U^\star=-H^{-1}F\mathbf x_0$，第一块就是一个线性增益 $K_N$。若 $P_f=P$ 满足 Riccati 方程 $P=Q+A^\top PA-A^\top PB(R+B^\top PB)^{-1}B^\top PA$，则 6.12 节的反向递推从 $P_N=P$ 开始，每一步都得到同一个 $P$，于是 $K_N=K_\infty$ 对所有 $N$ 成立：这正是 FIG 2 里 $N=1$ 也与 LQR 相同的原因。

Without constraints it is LQR. Unconstrained, $\mathbf U^\star=-H^{-1}F\mathbf x_0$, whose first block is a linear gain $K_N$. If $P_f=P$ solves the Riccati equation $P=Q+A^\top PA-A^\top PB(R+B^\top PB)^{-1}B^\top PA$, the backward recursion of Lesson 6.12 started at $P_N=P$ returns the same $P$ at every step, so $K_N=K_\infty$ for every $N$: that is why FIG 2 matches LQR even at $N=1$.

稀疏形式与 RiccatiThe sparse form and Riccati

不消元，取 $\mathbf z=(\mathbf u_0,\mathbf x_1,\mathbf u_1,\dots,\mathbf x_N)$，动力学写成等式约束 $\mathbf x_{k+1}-A\mathbf x_k-B\mathbf u_k=0$，乘子为 $\boldsymbol\lambda_{k+1}$。拉格朗日函数的 KKT 条件（6.7 节）是一个线性方程组：

Without eliminating anything, take $\mathbf z=(\mathbf u_0,\mathbf x_1,\mathbf u_1,\dots,\mathbf x_N)$ and write the dynamics as equality constraints $\mathbf x_{k+1}-A\mathbf x_k-B\mathbf u_k=0$ with multipliers $\boldsymbol\lambda_{k+1}$. The KKT conditions of the Lagrangian (Lesson 6.7) are a linear system:

$$ \hl{KKT}{\begin{bmatrix}\mathcal H & \mathcal G^\top\\ \mathcal G & 0\end{bmatrix}}\begin{bmatrix}\mathbf z\\ \boldsymbol\lambda\end{bmatrix}=\begin{bmatrix}0\\ \mathbf b(\mathbf x_0)\end{bmatrix}, $$ $$ \mathcal H=2\,\mathrm{diag}(R,Q,R,Q,\dots,P_f),\quad \mathcal G=\begin{bmatrix}-B & I & & & \\ & -A & -B & I & \\ && & \ddots & \end{bmatrix}. $$

$\mathcal H$ 块对角，$\mathcal G$ 块双对角：重排变量后整个矩阵是带宽约 $2n+m$ 的带状矩阵。消元的顺序如果从 $k=N$ 往回做，得到的正是 Riccati 递推：每一步只处理一个 $(n+m)$ 维的小块，总代价 $O(N(n+m)^3)$。内点法每次迭代只是给 $\mathcal H$ 的对角加上约束的权重（第 $k$ 步的约束只影响第 $k$ 块），所以结构不变，每次迭代仍是一次 Riccati 递推；HPIPM、FORCES Pro 等 MPC 专用求解器就是这么做的。

$\mathcal H$ is block diagonal and $\mathcal G$ block bidiagonal: after reordering, the whole matrix is banded with bandwidth about $2n+m$. Eliminating from $k=N$ backwards yields exactly the Riccati recursion: each step handles one $(n+m)$-dimensional block, for a total cost $O(N(n+m)^3)$. An interior-point iteration only adds constraint weights to the diagonal of $\mathcal H$ (stage-$k$ constraints touch only block $k$), so the structure survives and each iteration is again one Riccati recursion; MPC-specific solvers such as HPIPM and FORCES Pro do exactly this.

对比：稠密形式一次 Cholesky 是 $(Nm)^3/3$，对 $N$ 三次方；稀疏是 $N$ 的一次方，但每步付出 $n^3$。所以 $Nm$ 小、$n$ 大（短时域、多状态少输入）时稠密划算，长时域时稀疏划算；介于其间的"部分缩并"（把每几步缩成一块）是很多实时求解器的默认选择。

Compare: one dense Cholesky costs $(Nm)^3/3$, cubic in $N$; the sparse form is linear in $N$ but pays $n^3$ per step. So the condensed form wins when $Nm$ is small and $n$ large (short horizons, many states, few inputs) and the sparse form wins for long horizons; "partial condensing" in between (condense every few steps into one block) is the default of many real-time solvers.

稳定性：最优代价是李雅普诺夫函数Stability: the optimal cost is a Lyapunov function

设终端集 $\mathcal X_f$ 与终端代价满足两个条件（Mayne 等，2000）：在 $\mathcal X_f$ 里有一个局部控制律 $\mathbf u=-K_f\mathbf x$，它满足约束、使 $\mathcal X_f$ 不变；并且

Suppose the terminal set $\mathcal X_f$ and terminal cost satisfy two conditions (Mayne et al., 2000): inside $\mathcal X_f$ there is a local control law $\mathbf u=-K_f\mathbf x$ that respects the constraints and keeps $\mathcal X_f$ invariant; and

$$ V_f\big((A-BK_f)\mathbf x\big)-V_f(\mathbf x)\le-\big(\mathbf x^\top Q\mathbf x+\mathbf x^\top K_f^\top RK_f\mathbf x\big)\qquad\forall\,\mathbf x\in\mathcal X_f . $$

取 $K_f=K_\infty$、$V_f=\mathbf x^\top P\mathbf x$，这个不等式取等号（这就是 Riccati 方程），$\mathcal X_f$ 取 LQR 不碰约束的一个不变椭圆或多面体。证明思路（"平移论证"）：设 $\mathbf x$ 处的最优计划是 $(\mathbf u_0^\star,\dots,\mathbf u_{N-1}^\star)$，终点 $\mathbf x_N^\star\in\mathcal X_f$。到了下一个状态 $\mathbf x^+=A\mathbf x+B\mathbf u_0^\star$，构造候选计划

With $K_f=K_\infty$ and $V_f=\mathbf x^\top P\mathbf x$ the inequality holds with equality (it is the Riccati equation), and $\mathcal X_f$ is an invariant ellipse or polytope on which LQR does not touch the constraints. Proof idea (the "shift argument"): let the optimal plan at $\mathbf x$ be $(\mathbf u_0^\star,\dots,\mathbf u_{N-1}^\star)$, ending at $\mathbf x_N^\star\in\mathcal X_f$. At the next state $\mathbf x^+=A\mathbf x+B\mathbf u_0^\star$, build the candidate plan

$$ \hl{warm}{\tilde{\mathbf U}=\big(\mathbf u_1^\star,\ \mathbf u_2^\star,\ \dots,\ \mathbf u_{N-1}^\star,\ -K_f\mathbf x_N^\star\big)} . $$

它满足所有约束（前 $N-1$ 步本来就满足，最后一步由 $\mathcal X_f$ 的不变性保证），所以 $\mathbf x^+$ 处的问题可行：这就是递归可行性。它的代价是原最优代价减去第一步的代价，再把终端代价换成"多走一步加新的终端代价"，由上面的不等式，后者不会变大：

It satisfies every constraint (the first $N-1$ steps already did, and the last is covered by the invariance of $\mathcal X_f$), so the problem at $\mathbf x^+$ is feasible: that is recursive feasibility. Its cost is the old optimal cost minus the first stage, with the terminal cost replaced by "one more stage plus the new terminal cost", which by the inequality above does not increase:

$$ V_N(\mathbf x^+)\ \le\ J(\mathbf x^+,\tilde{\mathbf U})\ \le\ V_N(\mathbf x)-\big(\mathbf x^\top Q\mathbf x+\mathbf u_0^{\star\top}R\,\mathbf u_0^\star\big). $$

$V_N$ 每步至少下降一个正的阶段代价，又有下界 0，所以它是闭环的李雅普诺夫函数（13.2 节），原点渐近稳定。FIG 4 的终端约束 $v_N\le0$ 是这个论证的一个简化版：$\mathcal X_f=\{v\le0\}$ 在 $u=0$ 下不变，只证了可行性，没要求代价下降。没有终端成分时，这两个结论都可能失败：FIG 2 的发散与 FIG 4 的不可行就是反例。时域足够长时它们通常又会成立（终端成分可以去掉），但"多长才够"依赖于问题，事先很难算。

$V_N$ drops by at least one positive stage cost per step and is bounded below by 0, so it is a Lyapunov function for the closed loop (Lesson 13.2) and the origin is asymptotically stable. FIG 4's terminal constraint $v_N\le0$ is a simplified version of the argument: $\mathcal X_f=\{v\le0\}$ is invariant under $u=0$, which proves feasibility but asks for no cost decrease. Without terminal ingredients both conclusions can fail: the divergence in FIG 2 and the infeasibility in FIG 4 are counterexamples. With a long enough horizon they usually hold again (the terminal ingredients can be dropped), but how long is "long enough" depends on the problem and is hard to compute in advance.

软约束Soft constraints

状态约束硬性要求时，模型误差或扰动随时可能让问题不可行（6.8 节的误区）。实践中把它们写成软约束：$\mathbf c(\mathbf x_k)\le\boldsymbol\sigma_k$，$\boldsymbol\sigma_k\ge0$，代价加 $\rho_1\mathbf 1^\top\boldsymbol\sigma+\rho_2\|\boldsymbol\sigma\|^2$。线性项是"精确罚"：只要 $\rho_1$ 大于对应硬约束的拉格朗日乘子，原问题可行时的解与硬约束完全相同，只在不可行时才允许最小的违反。FIG 7 里道路边界和障碍物就是这样处理的。

Hard state constraints can make the problem infeasible at any moment because of model error or disturbances (the pitfall in Lesson 6.8). In practice they are made soft: $\mathbf c(\mathbf x_k)\le\boldsymbol\sigma_k$, $\boldsymbol\sigma_k\ge0$, with $\rho_1\mathbf 1^\top\boldsymbol\sigma+\rho_2\|\boldsymbol\sigma\|^2$ added to the cost. The linear term is an "exact penalty": as long as $\rho_1$ exceeds the Lagrange multiplier of the corresponding hard constraint, the solution equals the hard-constrained one whenever that is feasible, and only when it is not is the smallest violation allowed. FIG 7 handles road edges and the obstacle this way.

非线性 MPC 与实时迭代；MPPINonlinear MPC and real-time iterations; MPPI

模型非线性 $\mathbf x_{k+1}=\mathbf f(\mathbf x_k,\mathbf u_k)$ 时，问题是一个非线性规划，通常用 SQP（6.7 节）：在当前猜测处线性化动力学，$\delta\mathbf x_{k+1}=A_k\delta\mathbf x_k+B_k\delta\mathbf u_k$，$A_k=\partial\mathbf f/\partial\mathbf x$，$B_k=\partial\mathbf f/\partial\mathbf u$，用 Gauss-Newton 近似代价的 Hessian（6.6 节），解一个 QP 得到修正量。实时迭代（RTI，Diehl 等，2005）只做一次 SQP 迭代：上一周期的解平移之后已经很接近最优，一次迭代就够；并且线性化可以在测到新状态之前准备好，测量到达后只剩解一个 QP。时变的 $A_k,B_k$ 意味着每个周期都要重新缩并或重新做 Riccati，FIG 5 的"每步缩并"曲线就是这种情况。

With a nonlinear model $\mathbf x_{k+1}=\mathbf f(\mathbf x_k,\mathbf u_k)$ the problem is a nonlinear program, usually solved by SQP (Lesson 6.7): linearize the dynamics at the current guess, $\delta\mathbf x_{k+1}=A_k\delta\mathbf x_k+B_k\delta\mathbf u_k$ with $A_k=\partial\mathbf f/\partial\mathbf x$, $B_k=\partial\mathbf f/\partial\mathbf u$, approximate the cost Hessian by Gauss-Newton (Lesson 6.6), and solve one QP for the correction. The real-time iteration (RTI, Diehl et al., 2005) performs only one SQP iteration per cycle: the shifted previous solution is already close to optimal, so one iteration suffices; and the linearization can be prepared before the new state is measured, leaving only one QP once the measurement arrives. Time-varying $A_k,B_k$ mean re-condensing or redoing Riccati every cycle: the "condense every step" curve of FIG 5.

MPPI（6.11 节）解的是同一个"滚动时域"问题，只是不求导：在平移后的上一计划附近采 $K$ 条噪声序列，各自用模型往前仿真得到代价 $S_i$，按 $w_i\propto\exp(-(S_i-S_{\min})/\lambda)$ 加权平均。它能直接用不连续的代价（撞了罚一个大数）和任意不可导的模型，代价是需要很多次仿真（通常在 GPU 上并行），而且只能把约束写成罚函数。

MPPI (Lesson 6.11) solves the same receding-horizon problem without derivatives: sample $K$ noisy sequences around the shifted previous plan, simulate each forward with the model to get a cost $S_i$, and average with weights $w_i\propto\exp(-(S_i-S_{\min})/\lambda)$. It can use discontinuous costs (a big penalty for a hit) and any non-differentiable model directly; the price is many simulations (usually parallel on a GPU), and constraints can only enter as penalties.

7机器人中的它：非线性 MPC 跟踪路径In robotics: nonlinear MPC for path tracking

自动驾驶的横向控制是 MPC 的经典应用。车用 9.1 节的运动学自行车模型：状态 $(x,y,\theta,\delta)$，$\dot x=v\cos\theta$，$\dot y=v\sin\theta$，$\dot\theta=\tfrac vL\tan\delta$，控制量是转向角速度 $\dot\delta=\omega$，$|\omega|\le0.4$ rad/s，$|\delta|\le0.55$ rad，轴距 $L=2.7$ m，车速 $v$ 恒定。路面宽 7 m，前方有一个可以拖动的红色障碍物。三个控制器同时跑：

非线性 MPC（蓝）：代价是横向误差、航向误差、转角和转向角速度的平方和；道路边界和障碍物写成横向误差的软约束 $e_{\rm lo}(s_k)\le e_k\le e_{\rm hi}(s_k)$。每 0.1 s 做一次实时迭代（一次 Gauss-Newton 线性化 + 一个 QP，用本页的内点法）。
MPPI（琥珀）：同一个模型、同一个代价（约束以同样的罚函数出现），$K$ 条采样，噪声 $\sigma=0.2$ rad/s，温度 $\lambda=1$。
纯追踪（紫，9.1 节）：预瞄距离 $\max(4,\,0.5v)$ m，同样的转向速率上限。它只看路径，看不见障碍。

Lateral control of a car is a classic MPC application. The car is the kinematic bicycle of Lesson 9.1: state $(x,y,\theta,\delta)$, $\dot x=v\cos\theta$, $\dot y=v\sin\theta$, $\dot\theta=\tfrac vL\tan\delta$, with the steering rate $\dot\delta=\omega$ as input, $|\omega|\le0.4$ rad/s, $|\delta|\le0.55$ rad, wheelbase $L=2.7$ m and constant speed $v$. The road is 7 m wide, and a draggable red obstacle sits ahead. Three controllers run at once:

Nonlinear MPC (blue): the cost is the sum of squares of lateral error, heading error, steering angle and steering rate; road edges and the obstacle are soft constraints on the lateral error, $e_{\rm lo}(s_k)\le e_k\le e_{\rm hi}(s_k)$. Every 0.1 s it does one real-time iteration (one Gauss-Newton linearization plus one QP, solved by this page's interior-point method).
MPPI (amber): the same model and the same cost (constraints appear through the same penalties), $K$ samples, noise $\sigma=0.2$ rad/s, temperature $\lambda=1$.
Pure pursuit (purple, Lesson 9.1): look-ahead $\max(4,\,0.5v)$ m with the same steering-rate limit. It sees the path but not the obstacle.

FIG 7非线性 MPC、MPPI 与纯追踪：同一条路，同一个障碍Nonlinear MPC, MPPI and pure pursuit: one road, one obstacle

全景：拖动红色障碍OVERVIEW · drag the red obstacle

跟随视角CHASE VIEW

右上角选择要看哪个控制器的预测：非线性 MPC 的预测（蓝色虚线带点）、MPPI 的加权平均计划（琥珀色虚线）和其中 32 条采样（琥珀色淡线），或纯追踪的预瞄点（紫点）。三辆车从障碍物前 45 m 同时出发，绕赛道一直开下去；每经过一次障碍记录一次"绕过 / 碰到"（以车后轴中点加 1 m 车身半宽和车头车尾长度判断）。ms/步是你的浏览器里每个控制周期的实测计算时间。改动任何参数都会重新开始。 The selector at the top right picks whose prediction to show: nonlinear MPC's prediction (dashed blue with dots), MPPI's weighted plan (dashed amber) and 32 of its samples (faint amber), or pure pursuit's look-ahead point (purple dot). The three cars start together 45 m before the obstacle and keep lapping the track; each pass by the obstacle is recorded as "cleared" or "hit" (judged from the rear-axle midpoint with a 1 m half-width and the car's front and rear overhang). ms/step is the computation time per control cycle measured in your browser. Changing any parameter restarts the run.

试一试Try this

把障碍物拖到道路中心线的右侧（横向偏移 $-1.5\le e<0$ m，读数里有 $e$），车速调到 12 m/s 或更高：非线性 MPC 改从左边绕过去，且一次都不碰到。Drag the obstacle to the right of the centre line (lateral offset $-1.5\le e<0$ m, shown in the readout) and set the speed to 12 m/s or more: nonlinear MPC now goes around on the left without touching it.
把预测步数减到 $N\le10$，找到非线性 MPC 绕不过去（碰到障碍或冲出路面）的时域。这时的预瞄距离是多少米？转向角速度只有 0.4 rad/s，要横移 2 m 需要提前多久开始打方向？Reduce the prediction steps to $N\le10$ and find a horizon at which nonlinear MPC can no longer get around (it hits the obstacle or leaves the road). What is the preview distance then? With a steering rate of only 0.4 rad/s, how early must it start turning to move 2 m sideways?

真实系统里In real systems

起源：MPC 最早来自炼油和化工的过程控制（动态矩阵控制 DMC 等，Qin 与 Badgwell 的综述统计了数千套工业应用），那里采样周期是分钟级的。机器人把它推到了毫秒级。四足：MIT Cheetah 3 的凸 MPC（Di Carlo 等，2018）把机身看作单刚体，绕当前偏航角线性化姿态动力学，以未来若干步的足底力为决策变量、摩擦锥线性化成金字塔，每个周期解一个 QP；计算出的足底力再交给更高频的腿部控制（13.7 节的全身控制会接着做）。ETH 的 OCS2 库在 ANYmal 等机器人上做非线性 MPC。自动驾驶：Autoware 的横向控制器提供一个线性 MPC 实现，默认用带一阶转向延迟的运动学自行车模型；赛车 MPC 用 MPCC（轮廓控制，Liniger 等，2015）同时优化路径进度。四旋翼：非线性 MPC 用全刚体模型直接输出推力与角速度指令。求解器：OSQP（ADMM，Stellato 等，2020）、qpOASES（有效集，热启动极好，Ferreau 等，2014）、HPIPM（稀疏内点法，Riccati 结构）、acados（Verschueren 等，2022：在 HPIPM 之上实现 RTI，自动生成 C 代码）、CasADi（建模与自动微分）；MPPI 则常在 GPU 上运行。显式 MPC：对很小的线性问题，可以离线算出分段仿射的 $\kappa_N(\mathbf x)$（Bemporad 等，2002），在线只查表，适合没有余力解 QP 的嵌入式芯片。

Origins: MPC grew out of process control in refineries and chemical plants (dynamic matrix control, DMC, and others; Qin and Badgwell's survey counts thousands of industrial installations), with sampling periods of minutes. Robotics pushed it to milliseconds. Quadrupeds: the convex MPC of MIT Cheetah 3 (Di Carlo et al., 2018) treats the body as a single rigid body, linearizes the attitude dynamics about the current yaw, uses the foot forces over the coming steps as decision variables with friction cones linearized into pyramids, and solves one QP per cycle; the resulting foot forces go to a faster leg-level controller (Lesson 13.7's whole-body control continues from there). ETH's OCS2 library runs nonlinear MPC on ANYmal and other robots. Autonomous driving: Autoware's lateral controller offers a linear MPC implementation whose default model is the kinematic bicycle with a first-order steering delay; racing MPC uses MPCC (model predictive contouring control, Liniger et al., 2015) to optimize progress along the path at the same time. Quadrotors: nonlinear MPC on the full rigid-body model outputs thrust and body-rate commands directly. Solvers: OSQP (ADMM, Stellato et al., 2020), qpOASES (active set, excellent warm starts, Ferreau et al., 2014), HPIPM (sparse interior point with Riccati structure), acados (Verschueren et al., 2022: RTI on top of HPIPM with generated C code), CasADi (modelling and automatic differentiation); MPPI usually runs on a GPU. Explicit MPC: for very small linear problems the piecewise-affine $\kappa_N(\mathbf x)$ can be computed offline (Bemporad et al., 2002) so that the online work is a table lookup, which suits embedded chips with no room for a QP solver.

8常见误区Common pitfalls

误区 1：时域越长越好，终端代价可有可无Pitfall 1: the longer the horizon the better, and the terminal cost is optional

长时域的代价是计算（稠密形式对 $N$ 三次方增长）和对模型误差更敏感的远期预测。没有终端代价时，短时域可以直接让闭环不稳定（FIG 2：$N\le7$ 时摆会倒）。正确的顺序是：先放 $P_f=P$，让时域只需覆盖"约束会起作用的那段未来"，例如 FIG 4 里刹车所需的距离；再按计算预算选 $N$。

A long horizon costs computation (cubic in $N$ for the condensed form) and relies on far-future predictions that are more sensitive to model error. Without a terminal cost a short horizon can make the closed loop outright unstable (FIG 2: the pendulum falls for $N\le7$). The right order: first set $P_f=P$, so that the horizon only has to cover "the part of the future where constraints act", such as the braking distance in FIG 4; then choose $N$ to fit the computation budget.

误区 2：所有约束都写成硬约束Pitfall 2: making every constraint hard

输入约束（电机力矩、转向速率）是物理事实，硬约束天然可行。状态约束（位置、速度、车道）一旦遇上模型误差、测量噪声或扰动，硬约束就可能让 QP 不可行，控制器在最需要输出的时刻没有输出。状态约束应当写成带精确罚的软约束，并给"求解失败"准备一个后备动作（例如刹车或沿用平移后的上一计划）。

Input constraints (motor torque, steering rate) are physical facts, and hard constraints on them are always feasible. State constraints (position, speed, lanes) meet model error, measurement noise or disturbances, and as hard constraints they can make the QP infeasible, leaving the controller with no output exactly when it matters. Make state constraints soft with an exact penalty, and prepare a fallback for "the solver failed" (brake, or reuse the shifted previous plan).

误区 3：MPC 自带鲁棒性Pitfall 3: MPC is robust by construction

重新测量、重新求解确实给了 MPC 反馈（FIG 1），但第 6 节的稳定性和可行性证明都假设模型精确。模型误差下，"平移的上一计划"不再可行，名义 MPC 可能违反约束甚至失稳。需要保证时，要用鲁棒 MPC（例如管道 MPC：把约束收紧一个由扰动界算出的余量）或随机 MPC（机会约束）。另外，跟踪常值参考而模型有偏差时，名义 MPC 会留下稳态误差，要像 13.1 节的积分项那样加一个扰动估计（偏移消除）。

Measuring and re-solving does give MPC feedback (FIG 1), but the stability and feasibility proofs of Section 6 assume an exact model. With model error, "the shifted previous plan" is no longer feasible, and nominal MPC can violate constraints or even lose stability. When guarantees are needed, use robust MPC (e.g. tube MPC, which tightens the constraints by a margin computed from the disturbance bound) or stochastic MPC (chance constraints). Also, when tracking a constant reference with a biased model, nominal MPC leaves a steady-state error; add a disturbance estimate (offset-free MPC), the counterpart of the integral term in Lesson 13.1.

误区 4：忽略计算延迟Pitfall 4: ignoring the computation delay

求解要花时间：在 $t$ 时刻测到的状态，算出的输入要到 $t+\Delta$ 才生效。如果 $\Delta$ 与采样周期相当，优化里应该从"预测的 $t+\Delta$ 时刻的状态"出发，或者把延迟写进模型。RTI 把线性化挪到测量之前（准备阶段），正是为了把测量到输出的延迟压到只剩一个 QP。另一方面，求解时间不稳定也很危险：要设迭代上限（FIG 6），并确保上限内交出的解至少可行。

Solving takes time: the state measured at $t$ yields an input that takes effect at $t+\Delta$. If $\Delta$ is comparable to the sampling period, the optimization should start from the predicted state at $t+\Delta$, or the delay should be part of the model. RTI moves the linearization before the measurement (the preparation phase) precisely so that only one QP separates measurement from output. Unpredictable solve times are dangerous too: cap the iterations (FIG 6) and make sure the solution delivered at the cap is at least feasible.

误区 5：模型越精细越好Pitfall 5: the more detailed the model, the better

MPC 只能和它的模型一样好，但模型也决定了能不能及时解完。FIG 7 的运动学自行车模型假设轮胎不侧滑，侧向加速度接近 $\mu g$ 时就不对了（9.1 节的误区）。FIG 7 本身就越过了这条线：赛道最急的弯半径约 12.5 m，光是沿路径开过去就需要 $v^2/R$：10 m/s 时约 8 m/s²，12 m/s 时约 11.5 m/s²，14 m/s 时约 16 m/s²，而干燥路面上轮胎最多给出约 $\mu g\approx9$ m/s²。所以 12 到 14 m/s 的结果只说明控制器在这个模型里的行为，真车会侧滑，这时应换成动力学自行车模型；可每多一个状态，稠密 QP 就更大、非线性更强。四足的凸 MPC 故意用单刚体模型，正是为了每个周期都能解一个凸 QP。选模型时要问：控制器需要预测的那几秒里，哪些效应真的重要？

MPC is only as good as its model, but the model also decides whether it can be solved in time. FIG 7's kinematic bicycle assumes the tires do not slip sideways and becomes wrong as lateral acceleration approaches $\mu g$ (the pitfall of Lesson 9.1). FIG 7 itself crosses that line: the tightest bend of the track has a radius of about 12.5 m, so merely following the path there takes $v^2/R$, about 8 m/s² at 10 m/s, 11.5 m/s² at 12 m/s and 16 m/s² at 14 m/s, while tires on dry asphalt deliver at most about $\mu g\approx9$ m/s². So the results at 12 to 14 m/s only describe how the controllers behave in this model; a real car would slide, and you would switch to a dynamic bicycle model. Yet every extra state makes the QP larger and the problem more nonlinear. The quadruped convex MPC deliberately uses a single-rigid-body model precisely so that every cycle is a convex QP. When choosing the model, ask which effects really matter over the few seconds the controller predicts.

9小练习Check yourself

1. MPC 每个周期解出一整串最优输入 $\mathbf u_0^\star,\dots,\mathbf u_{N-1}^\star$，却只执行 $\mathbf u_0^\star$。主要原因是什么？1. Every cycle MPC computes a whole optimal sequence $\mathbf u_0^\star,\dots,\mathbf u_{N-1}^\star$ but applies only $\mathbf u_0^\star$. What is the main reason?

选项 C。如果模型完美、没有扰动，执行整串计划和滚动执行的结果相同；现实里两者都不成立。FIG 1 的开环模式里，模型误差 ×1.2 就让小车停在离目标几十厘米处，而滚动时域每步都从真实状态重新规划，误差被消除。Option C. With a perfect model and no disturbances, executing the whole plan and executing it in receding fashion give the same result; in reality neither holds. In FIG 1's open-loop mode a ×1.2 model error leaves the cart tens of centimetres from the target, while the receding horizon re-plans from the true state each step and removes the error.

2. 线性系统、无约束（或约束都不起作用），终端代价取 Riccati 方程的解 $P$。时域 $N$ 从 1 增加到 50，MPC 的控制律会怎样？2. A linear system, no constraints (or none active), and the terminal cost is the Riccati solution $P$. As the horizon $N$ grows from 1 to 50, what happens to the MPC law?

选项 B。$\mathbf x_N^\top P\mathbf x_N$ 恰好是从 $\mathbf x_N$ 出发的无限时域最优代价，所以有限时域问题与无限时域问题的最优解相同（贝尔曼原理）；Riccati 递推从 $P_N=P$ 出发每一步都得到 $P$，增益恒为 $K_\infty$。FIG 2 里 $N=1$ 时差别小于 $10^{-6}$。选项 A 描述的是 $P_f=0$ 或 $P_f=Q$ 的情形。Option B. $\mathbf x_N^\top P\mathbf x_N$ is exactly the infinite-horizon optimal cost from $\mathbf x_N$, so the finite- and infinite-horizon problems share the optimal solution (Bellman's principle); the Riccati recursion started at $P_N=P$ returns $P$ at every step, so the gain is always $K_\infty$. In FIG 2 the difference at $N=1$ is below $10^{-6}$. Option A describes $P_f=0$ or $P_f=Q$.

3. FIG 4 里，没有终端约束、$N=10$ 的 MPC 从安全的起点出发，几秒后变得不可行。最根本的原因是什么？3. In FIG 4, MPC with $N=10$ and no terminal constraint starts from a safe state and becomes infeasible a few seconds later. What is the root cause?

选项 A。起点在可挽救集里（选项 D 错：满力刹车能停住，更不用说一开始就慢慢走），QP 是凸的且每步都精确求解（B、C 错）。问题出在 MPC 自己的选择：它进入了"此刻可行、却已注定"的红色区域。加终端约束 $v_N\le0$，或者把时域加长到覆盖刹车距离，都能恢复递归可行性。Option A. The start is in the safe set (D is wrong: full braking would stop it, let alone moving slowly from the beginning), and the QP is convex and solved exactly every step (B and C are wrong). The trouble is MPC's own choice: it entered the red region that is "feasible now but already doomed". A terminal constraint $v_N\le0$, or a horizon long enough to cover the braking distance, restores recursive feasibility.

4. 一个状态维数 $n=12$、输入维数 $m=4$ 的线性 MPC，时域从 $N=20$ 加到 $N=80$。稠密（缩并）形式每次 Cholesky 分解的运算量大约变成原来的几倍？稀疏 Riccati 形式呢？4. A linear MPC with state dimension $n=12$ and input dimension $m=4$ has its horizon raised from $N=20$ to $N=80$. By roughly what factor does one Cholesky factorization of the condensed form grow? And the sparse Riccati form?

选项 D。稠密 Hessian 是 $Nm\times Nm$，Cholesky 是 $(Nm)^3/3$，$N$ 变成 4 倍，运算量变成 $4^3=64$ 倍。Riccati 每步处理固定大小的块，总量与 $N$ 成正比，4 倍。所以长时域几乎总是用稀疏形式（或部分缩并）。Option D. The condensed Hessian is $Nm\times Nm$ and Cholesky costs $(Nm)^3/3$; a 4× longer horizon costs $4^3=64$ times more. Riccati handles a fixed-size block per step, so its total is proportional to $N$: 4×. That is why long horizons almost always use the sparse form (or partial condensing).

5. 关于 MPPI 与基于梯度的非线性 MPC（SQP/RTI），哪个说法是对的？5. Which statement about MPPI versus gradient-based nonlinear MPC (SQP/RTI) is correct?

选项 B。MPPI 同样每个周期预测 $N$ 步、执行第一步、平移计划（6.11 节）。它以采样代替求导，所以代价和模型可以不可导；约束只能通过罚项进入，满足程度取决于采样是否找到可行区域。选项 D 错：RTI 每周期只做一次 SQP 迭代，靠热启动在实践中工作良好，而"可行即稳定"的结论说明次优解也可以稳定。Option B. MPPI also predicts $N$ steps each cycle, applies the first and shifts the plan (Lesson 6.11). It replaces derivatives by sampling, so the cost and model need not be differentiable; constraints can only enter through penalties, and how well they are met depends on whether the samples find the feasible region. Option D is wrong: RTI does a single SQP iteration per cycle and works well in practice thanks to the warm start, and "feasibility implies stability" shows suboptimal solutions can be stabilizing.

10延伸阅读Further reading

J. B. Rawlings, D. Q. Mayne & M. M. Diehl, Model Predictive Control: Theory, Computation, and Design, 2nd ed., Nob Hill Publishing, 2017 — MPC 的标准教材：稳定性、递归可行性、鲁棒与数值方法The standard text: stability, recursive feasibility, robustness and numerical methods
F. Borrelli, A. Bemporad & M. Morari, Predictive Control for Linear and Hybrid Systems, Cambridge University Press, 2017 — 线性 MPC、不变集与显式 MPCLinear MPC, invariant sets and explicit MPC
D. Q. Mayne, J. B. Rawlings, C. V. Rao & P. O. M. Scokaert, Constrained Model Predictive Control: Stability and Optimality, Automatica 36(6), 2000 — 终端代价与终端集条件的经典综述The classic survey of terminal-cost and terminal-set conditions
P. O. M. Scokaert, D. Q. Mayne & J. B. Rawlings, Suboptimal Model Predictive Control (Feasibility Implies Stability), IEEE Trans. Automatic Control 44(3), 1999 — 次优解为什么也能稳定Why suboptimal solutions can still stabilize
S. J. Qin & T. A. Badgwell, A Survey of Industrial Model Predictive Control Technology, Control Engineering Practice 11(7), 2003 — MPC 在工业中的历史与现状History and practice of MPC in industry
M. Diehl, H. G. Bock & J. P. Schlöder, A Real-Time Iteration Scheme for Nonlinear Optimization in Optimal Feedback Control, SIAM J. Control and Optimization 43(5), 2005 — 实时迭代（RTI）The real-time iteration (RTI)
A. Bemporad, M. Morari, V. Dua & E. N. Pistikopoulos, The Explicit Linear Quadratic Regulator for Constrained Systems, Automatica 38(1), 2002 — 显式 MPCExplicit MPC
B. Stellato, G. Banjac, P. Goulart, A. Bemporad & S. Boyd, OSQP: An Operator Splitting Solver for Quadratic Programs, Mathematical Programming Computation 12, 2020 — FIG 6 的 ADMM 来源The ADMM behind FIG 6
R. Verschueren et al., acados: a Modular Open-Source Framework for Fast Embedded Optimal Control, Mathematical Programming Computation 14, 2022 — RTI、HPIPM 与代码生成RTI, HPIPM and code generation
J. Di Carlo, P. M. Wensing, B. Katz, G. Bledt & S. Kim, Dynamic Locomotion in the MIT Cheetah 3 Through Convex Model-Predictive Control, IROS 2018 — 四足凸 MPCConvex MPC for a quadruped
G. Williams, A. Aldrich & E. A. Theodorou, Model Predictive Path Integral Control: From Theory to Parallel Computation, J. Guidance, Control, and Dynamics 40(2), 2017 — MPPIMPPI
A. Liniger, A. Domahidi & M. Morari, Optimization-Based Autonomous Racing of 1:43 Scale RC Cars, Optimal Control Applications and Methods 36(5), 2015 — 赛车的轮廓控制 MPC（MPCC）Model predictive contouring control (MPCC) for racing

1为什么机器人需要它Why robots need this

2直觉：看得见几秒，就只为这几秒做打算Intuition: plan for the seconds you can see

3动手玩：时域之外，和约束之内Play: beyond the horizon, and inside the constraints

时域之外的代价：终端代价 \(P_f\)The cost beyond the horizon: the terminal cost \(P_f\)

约束之内：MPC 与"截断的 LQR"Inside the constraints: MPC versus "clipped LQR"

4第二个视角：MPC 会把自己逼进死胡同A second view: MPC can corner itself

5第三个视角：几毫秒内解完A third view: solving it within milliseconds

同一个 QP，两种写法：稠密（缩并）与稀疏One QP, two ways to write it: condensed and sparse

实时预算与热启动Real-time budgets and warm starts

6再看数学The math

问题The problem

缩并成稠密 QPCondensing into a dense QP

稀疏形式与 RiccatiThe sparse form and Riccati

稳定性：最优代价是李雅普诺夫函数Stability: the optimal cost is a Lyapunov function

软约束Soft constraints

非线性 MPC 与实时迭代；MPPINonlinear MPC and real-time iterations; MPPI

7机器人中的它：非线性 MPC 跟踪路径In robotics: nonlinear MPC for path tracking

8常见误区Common pitfalls

9小练习Check yourself

10延伸阅读Further reading