TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A conservative routing matrix preserves total flux

Statement

Let $P$ be a conservative nonnegative routing matrix from Conservative nonnegative network advection routing, and let $q^{i n} \in R^{n_{i n}}$ be entrywise nonnegative. Then $q^{o u t} = P q^{i n}$ is entrywise nonnegative and $\sum_{j = 1}^{n_{o u t}} q_{j}^{o u t} = \sum_{i = 1}^{n_{i n}} q_{i}^{i n} .$

Facts & Assumptions

Given: A finite incoming flux vector $q^{i n} \geq 0$ , a nonnegative routing matrix $P$ , and the unit incoming column-sum condition.

[F1]

$P_{j i} \geq 0$ and $\sum_{j = 1}^{n_{o u t}} P_{j i} = 1$ for each incoming index $i$ (Conservative nonnegative network advection routing).

Proof

technique · direct finite-sum calculation

Given: The nonnegative incoming flux vector and the routing matrix of the Statement.

1.1F1givenalgebra

For every outgoing index $j$ , $q_{j}^{o u t} = \sum_{i} P_{j i} q_{i}^{i n}$ is a sum of nonnegative terms by [F1] and the assumption $q^{i n} \geq 0$ . Hence $q^{o u t} \geq 0$ .

2.1F1step 1.1algebra∎

All sums are finite, so they may be interchanged: $\sum_{j} q_{j}^{o u t} = \sum_{j} \sum_{i} P_{j i} q_{i}^{i n} = \sum_{i} (\sum_{j} P_{j i}) q_{i}^{i n} = \sum_{i} q_{i}^{i n},$ where the final equality uses the unit column sums of [F1].

Depends on

Conservative nonnegative network advection routing

Used by

Density continuity can violate conservative advection balance Counterexample

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ward Whitt, Continuous-Time Markov Chains, full lecture notes (2006) (standard reference, not scraped)