If you read "Nonlinear Model Predictive Control" by L. Grune and J. Pannek (and anywhere else), everyone says that the prediction horizon size $N$ must be larger or equal to $2$,$ N\geq2$.
Why?
Model Predictive Control: Why the horizon size, $N$, must be equal or larger than 2
control theorymodel predictive controloptimal controloptimization
Related Solutions
In addition to the answer of @Johan at point 5), note I use a different cost function, with which you are probably more familiar with
Define the cost function $\mathbf{V}(\mathbf{x},\mathbf{U})$ as
$$ \mathbf{V}(\mathbf{x},\mathbf{U}) := \mathbf{x}_N^\mathrm{T} \mathbf{P} \mathbf{x}_N + \sum_{i=0}^{N-1}\left( \mathbf{x}_i^\mathrm{T} \mathbf{Q} \mathbf{x}_i + \mathbf{u}_i^\mathrm{T} \mathbf{R} \mathbf{u}_i\right) $$
Herein $N$ is the control horizon, $\mathbf{P}$ is the terminal state weight, $\mathbf{Q}$ is the state weight and $\mathbf{R}$ is the input weight. Remark: $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$ are positive semidefinite. Furthermore, $\mathbf{U}$ is the stacked input vector, that is,
$$ \mathbf{U} := \begin{bmatrix} \mathbf{u}_0^T & \mathbf{u}_1^T & \ldots & \mathbf{u}_{N-1}^T \end{bmatrix}^T $$
The optimal control $\mathbf{U}^*$ input is given by
$$ \mathbf{U}^* = \arg \mathbf{V}^*(\mathbf{x},\mathbf{U}) = \arg \min_{\mathbf{U}} \mathbf{V}(\mathbf{x},\mathbf{U}) $$
The sum in the cost function can be rewritten in a matrix equation
$$ \begin{gathered} \mathbf{V}(\mathbf{x},\mathbf{U}) = \mathbf{x}_0^\mathrm{T} \mathbf{Q} \mathbf{x}_0 + \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix}^\mathrm{T} \begin{bmatrix} \mathbf{Q} & 0 & \cdots & 0 \\ 0 & \mathbf{Q} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{P} \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix} + \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_{N-1} \end{bmatrix}^\mathrm{T} \begin{bmatrix} \mathbf{R} & 0 & \cdots & 0 \\ 0 & \mathbf{R} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{R} \end{bmatrix} \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_{N-1} \end{bmatrix} \\ \Leftrightarrow \\ \mathbf{V}(\mathbf{x},\mathbf{U}) = \mathbf{x}^\mathrm{T}\mathbf{Q}\mathbf{x} + \mathbf{X}^\mathrm{T}\mathbf{\Omega}\mathbf{X} + \mathbf{U}^\mathrm{T}\mathbf{\Psi}\mathbf{U} \end{gathered} $$
Remark: $\mathbf{x} = \mathbf{x}_0$. Furthermore, $\mathbf{P} \succeq 0 \wedge \mathbf{Q} \succeq 0$ implies that $\mathbf{\Omega} \succeq 0$ and $\mathbf{R} \succ 0$ implies that $\mathbf{\Psi} \succ 0$. The cost function now still depends on the future state $\mathbf{X}$. To eliminate $\mathbf{X}$ the prediction matrices $\mathbf{\Phi}$ and $\mathbf{\Gamma}$ are computed such that the stacked state vector is a function of the initial state and the stacked input vector i.e. $\mathbf{X} = \mathbf{\Phi}\mathbf{x} + \mathbf{\Gamma}\mathbf{U}$.
$$ \mathbf{\Phi} = \begin{bmatrix} \mathbf{A} \\ \mathbf{A}^2 \\ \vdots \\ \mathbf{A}^N \end{bmatrix} \qquad \mathbf{\Gamma} = \begin{bmatrix} \mathbf{B} & 0 & \cdots & 0 \\ \mathbf{A}\mathbf{B} & \mathbf{B} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{A}^{N-1}\mathbf{B} & \mathbf{A}^{N-2}\mathbf{B} & \cdots & \mathbf{B} \end{bmatrix} $$
The cost function $\mathbf{V}(\mathbf{x},\mathbf{U})$ can now be written as a function of the initial state $\mathbf{x}$ and the stacked input vector $\mathbf{U}$
$$ \mathbf{V}(\mathbf{x},\mathbf{U}) = \frac{1}{2} \mathbf{U}^\mathrm{T} \mathbf{G} \mathbf{U} + \mathbf{U}^\mathrm{T} \mathbf{F}\mathbf{x} + \mathbf{x}^\mathrm{T} \left( \mathbf{Q} + \mathbf{\Phi}^\mathrm{T} \mathbf{\Omega} \mathbf{\Phi} \right) \mathbf{x} $$
Where $\mathbf{G} = 2\left(\mathbf{\Psi} + \mathbf{\Gamma}^\mathrm{T} \mathbf{\Omega} \mathbf{\Gamma} \right)$, $\mathbf{F} = 2 \mathbf{\Gamma}^\mathrm{T} \mathbf{\Omega} \mathbf{\Phi}$ and $\mathbf{G} \succ 0$. The last term of the newly derived cost function is independent on $\mathbf{U}$. Therefore, it can be neglected when computing the optimal input sequence $\mathbf{U}^*(\mathbf{x})$
Now lets add some input and state constraints
$$ \mathbf{u}_\mathrm{min} \leq \mathbf{u}_k \leq \mathbf{u}_\mathrm{max} \wedge \mathbf{x}_\mathrm{min} \leq \mathbf{x}_k \leq \mathbf{x}_\mathrm{max} $$
This can be rewritten to
$$ \begin{bmatrix} 0 \\ 0 \\ -\mathbf{I}_n \\ \mathbf{I}_n \end{bmatrix} \mathbf{x}_k + \begin{bmatrix} -\mathbf{I}_m \\ \mathbf{I}_m \\ 0 \\ 0 \end{bmatrix} \mathbf{u}_k \leq \begin{bmatrix} -\mathbf{u}_\mathrm{min} \\ \mathbf{u}_\mathrm{max} \\ -\mathbf{x}_\mathrm{min} \\ \mathbf{x}_\mathrm{max} \end{bmatrix} \Leftrightarrow \mathbf{M}_k\mathbf{x}_k + \mathbf{E}_k\mathbf{u}_k \leq \mathbf{b}_k $$
Now note that this is only valid for $k = 0,1 \ldots, N-1$ so not up to the control horizon $N$. As such we add a terminal constraint for $N$ on the state.
$$ \begin{bmatrix} -\mathbf{I}_{n} \\ \mathbf{I}_{n} \\ \end{bmatrix} \mathbf{x}_N \leq \begin{bmatrix} -\mathbf{x}_\mathrm{min} \\ \mathbf{x}_\mathrm{max} \\ \end{bmatrix} \Rightarrow \mathbf{M}_N\mathbf{x}_N \leq \mathbf{b}_N $$
All constraints can now be combined to
$$ \begin{gathered} \begin{bmatrix} \mathbf{M}_0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \mathbf{x}_0 + \begin{bmatrix} 0 & \cdots & 0 \\ \mathbf{M}_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \mathbf{M}_N \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix} + \begin{bmatrix} \mathbf{E}_0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \mathbf{E}_{N-1} \\ 0 & \cdots & 0 \end{bmatrix} \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_N \end{bmatrix} \leq \begin{bmatrix} \mathbf{b}_0 \\ \mathbf{b}_1 \\ \vdots \\ \mathbf{b}_N \end{bmatrix} \\ \Leftrightarrow \\ \mathbf{\mathcal{D}}\mathbf{x} + \mathbf{\mathcal{M}}\mathbf{X} + \mathbf{\mathcal{E}}\mathbf{U} \leq \mathbf{c} \end{gathered} $$
Again this equation still depends on $\mathbf{X}$, which we eliminate using the same procedure as before. As a result, the above equation can be written as
$$\mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x}$$
With $\mathbf{J} = \mathbf{\mathcal{M}}\mathbf{\Gamma} + \mathbf{\mathcal{E}}$ and $\mathbf{W} = -\mathbf{\mathcal{D}} - \mathbf{\mathcal{M}}\mathbf{\Phi}$. Now the constraints only depend on the stacked input vector and the initial state. The constrained MPC problem can then be formulated as
$$ \arg \underset{\mathbf{U}}{\min}\; \mathbf{V}(\mathbf{x},\mathbf{U}) \quad \text{s.t.} \quad \mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x} $$
Now if you use a QP solver you will have the following QP problem
$$ \begin{aligned} \mathbf{U}^*(\mathbf{x}) = \arg \underset{\mathbf{U}}{\min}\;\;& \frac{1}{2}\mathbf{U}^\mathrm{T} \mathbf{G} \mathbf{U} + \mathbf{U}^\mathrm{T} \mathbf{F} \mathbf{x} \\[1ex] \text{subject to:}\;\;&\mathbf{x}_0 = \mathbf{x} \hspace{3cm} \text{Measurement} \nonumber \\ &\mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x} \hspace{1.82cm} \text{Constraints} \nonumber \\ &\mathbf{Q} \succ 0, \mathbf{R} \succ 0 \hspace{1.99cm} \text{Cost weights} \end{aligned} $$
Now if you want to solve these kind of typical MPC problems, I would strongly advice to use the MPT toolbox, http://people.ee.ethz.ch/~mpt/3/, an example http://control.ee.ethz.ch/~mpt/3/UI/RegulationProblem. Furthermore, the slides from this workshop may help you as well https://www.kirp.chtf.stuba.sk/pc11/data/workshops/mpc/
Yes, effectively the same thing. Also called receding horizon control.
Best Answer
The $N\geq 2$ condition in the referenced book is due to a slightly non-standard notation in the definition of the problem, using the objective $\sum_{k=0}^{N-1} \ell(x_k,u_k)$, which still makes sense for $N=1$ but typically would lead to the trivial solution $u = 0$ as only the current input and no future state is penalized in the objective. It would still be a well-defined problem, but perhaps a bit silly.
A more common form (in papers with a theoretical bias) is $\sum_{k=0}^{N-1} \ell(x_k,u_k) + \Psi(x_{N})$ for some terminal penalty $\Psi$, and then $N=1$ would make sense as the one-step ahead predictive control setup. Another form which would make $N=1$ reasonable is $\sum_{k=0}^{N-1} \ell(x_{k+1},u_k)$, as it typically makes no sense to add the current state to the objective.