In a multi commodity network, we define a demand to be a vector $d \in \mathbb{R}^{k}$, where $k$ is the number of pairs of sinks. That is, $k = \binom{S}{2}$, where $S$ is the set of sinks (aka terminals).

For example, if $\left| S \right| = 3$ we have $\binom{3}{2} = 3$ possible pairs of sinks. A demand can be of the type $\left< 1, 0,\frac{1}{2}\right>$. It translates to "Need flow $1$ for the first couple" and "need $0$ flow for the second couple" and "need $\frac{1}{2}$ for the third couple". All $3$ conditions are to be satisfied simultaneously.

For a given flow network a demand is said to be feasible if we can find a flow function to achieve it, while maintaining the capacity constraints and all normal flow rules.

We define the set of all feasible demands as $D = \left\{ d \in \mathbb{R}^{k} | d \text{ is feasible}\right\}$.

I have recently came across a statement which says $D$ can be seen as a polytope in $\mathbb{R}^{k}$, yet I find this statement a bit puzzling. Is there a proof for this, or even perhaps an intuition behind it?

(Perhaps, it is easy to show for the example of $\left| S \right| = 3$)

Note: full definitions can be seen here.


1 Answer 1


It is a polytope because it can be expressed as a system of linear inequalities. See https://en.wikipedia.org/wiki/Multi-commodity_flow_problem. In particular, let $d$ be a demand and $f$ a flow; there is a system of linear inequalities on $d,f$ that capture the condition that $f$ is a legal flow that achieves $d$. It follows that the set

$$S = \{(d,s) \mid f \text{ is a legal flow that achieves } d\}$$

is a convex polytope. Moreover, the set $D$ has the form $D = \{d \mid (d,s) \in S\}$, and the projection of any polytope onto fewer dimensions is itself a polytope. Therefore, $D$ is a polytope.

If that sounds like a lot to absorb, here is an elementary proof that $D$ is convex, which might help give some partial intuition:

Suppose $d_1,d_2$ are two feasible demands. Let $f_1$ be a legal flow that achieves demand $d_1$, and $f_2$ a legal flow that achieves $d_2$. Let $\lambda$ be any real number with $0 \le \lambda \le 1$. Define $d_3 = \lambda d_1+ (1-\lambda) d_2$ and $f_3 = \lambda f_1 + (1-\lambda) f_2$. It is easy to verify that $f_3$ is a legal flow, and that it achieves demand $d_3$. It follows that $d_1,d_2 \in D$, then $\lambda d_1 + (1-\lambda) d_2 \in D$, for all $\lambda$ with $0 \le lambda \le 1$. This proves that the region $D$ is convex.


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