Reaching the double exponential upper bound in Fourier-Motzkin elimination One round of Fourier-Motzkin elimination may yield $n^2/4$ inequalities where $n$ is the original number of linear inequalities, whence an upper bound of $n^{2^d}/2^{2^{d+1}-2}$ for $d$ rounds of elimination without removal of redundant constraints.

I'm wondering whether it is possible to meet this upper bound.

(Note: this is a distinct question from that of counting the number of true faces in a projection, that is, after removal of redundant constraints, for which McMullen's upper bound provides a tighter bound.)
I think this upper bound is tight. As an example, consider the following system

\begin{align*}
+x_0 &amp; +\frac{1}{2} x_1  +\frac{1}{4} x_2 \le 0 \\
+x_0 &amp; +\frac{1}{2} x_1 +\frac{1}{4} x_2  \le 0 \\
+x_0 &amp; +\frac{1}{2} x_1  +\frac{1}{4} x_2 \le 0 \\
+x_0 &amp; +\frac{1}{2} x_1 +\frac{1}{4} x_2  \le 0 \\
-x_0 &amp; +\frac{1}{2} x_1  +\frac{1}{4} x_2 \le 0 \\
-x_0 &amp; +\frac{1}{2} x_1 +\frac{1}{4} x_2  \le 0 \\
-x_0 &amp; -\frac{3}{2} x_1  +\frac{1}{4} x_2 \le 0 \\
-x_0 &amp; -\frac{3}{2} x_1 -\frac{7}{4} x_2  \le 0 \\
\end{align*}

It has 8 inequalities. Though some of them are identical, it is not a problem since we do not remove redundant constraints. One can also add dummy variables to make them different. Reducing $x_0$, we obtain system with 16 inequalities. Then, reducing $x_1$, we obtain system with $64$ inequalities.

More generally, let $S_{n}^k$ denote the following system of inequalities over $x_0, x_1, \ldots, x_{k - 1}$. Here we assume that $k \le n, n\ge 2$. There will be $2^n$ inequalities in $S_{n}^ k$. Take any $m \in\{1, 2, \ldots, 2^n\}$.  Let $m^{th}$ inequality of $S_{n}^ k$ be
$$\alpha^m_0 x_0 + \ldots + \alpha^m_{k - 1} x_{k - 1} \le 0,$$
where
$$\alpha^m_i = \begin{cases} 2^{-i} &amp; \mbox{if $m \le 2^n - 2^{n - i - 1}$}, \\ -2 + 2^{-i} &amp; \mbox{otherwise.} \end{cases}$$

Here we use a restriction mentioned above : provided $k\le n$, we have that $2^{n - i - 1}$ is an integer. 

For example, system above is $S_{3}^3$. It is easy to verify that if we eliminate $x_0$ from $S_{n} ^k$, we obtain $S_{2n - 2} ^{k - 1}$. Indeed, when we eliminate $x_0$, we add up all the inequalities from the first half with all the inequalities from the second half.  Thus we get $2^{n - 1} \cdot 2^{n - 1} = 2^{2n - 2}$ inequalities. What happens with $x_i$ for $i &gt; 0$? In the first $$2^{n - 1} \cdot (2^{n - 1} - 2^{n - i - 1}) = 2^{2n - 2} - 2^{2n - 2 - i} = 2^{2n - 2} - 2^{(2n - 2) - (i - 1) - 1}$$ inequalities a coefficient before $x_i$ will be $2^{-i} + 2^{-i} = 2^{-(i - 1)}$. In the last $2^{-i} + (-2 + 2^{-i}) =  2^{(2n - 2) - (i - 1) - 1}$ inequalities a coefficient before $x_i$ will be $2^{-i} + (-2 + 2^{-i}) = -2 + 2^{-(i - 1)}$. This is $S_{2n - 2}^{k - 1}$, but over variables $x_1, \ldots, x_{k - 1}$. Since, $2n - 2 \ge n - 1 \ge k - 1$, a restriction on the number of variables is also preserved.

Further, if we eliminate $x_1$, we obtain $S_{4n - 6}^{k - 2}$, and so on. After $d$ steps we will have $S^{k - d}_{2^d n - (2^{d + 1} - 2)}$. The latter has exactly
$$2^{2^d n - (2^{d + 1} - 2)} = N^{2^d}/ 2^{2^{d + 1} - 2}$$
inequalities, where $N = 2^n$ is the number of inequalities of the initial system.