Re: Is a faithful HTTP/2 response scheduler necessarily O(n) in the worst case?

[Kazuho Oku <[e-mail]>]

Regarding your question, I am unaware of a scheduler design that is
better than O(n) for both dequeuing (i.e. selecting the stream to
send) and for rescheduling a stream (especially a stream with many
decendants).

While designing the scheduler of H2O (you are right in pointing out
the fact that it is O(depth)), I came up with two options. One was to
retain the vertical (i.e. parent-children) relationship between the
streams. The other was to squash the vertical relationships to
generate a one-dimensional list of streams ordered by priority.

By taking the latter approach, you could create a scheduler that
dequeues at O(1). But such scheduler would need to perform O(N)
operation when receiving a priority frame or a stream-level window
update (in this case N is number of direct and indirect decendants of
the reprioritized stream).

Considering this, we chose to implement the scheduler of H2O as O(1)
weight-wise, and O(n) depth-wise, but that the constant for O(n) would
be small enough so that it cannot be used an attack vector.

2017-01-24 9:39 GMT+09:00 Tom Bergan <[e-mail]>:
> I implemented the HTTP/2 response scheduler in Go's HTTP/2 server library.
> I'm trying to understand the worst-case behavior of that scheduler. I
> believe the worst-case behavior is necessarily O(n) operations per frame
> sent on the wire, where n is the number of streams in the dependency tree.
> Here is my rationale.
>
> The key operation is finding the highest-priority stream that is ready to
> send data.
>
> If we don't care about weights, and we don't care about balancing bandwidth
> usage across sibling nodes in a tree, then we can label each node with two
> fields: "ready" (true if the stream is ready to send data) and "depth" (the
> node's distance from the root of the tree). The scheduler must find a node
> with the smallest depth over all nodes with ready = true. It is fairly
> trivial to implement this in O(log n).
>
> Now, let's introduce weights. The scheduler must allocate bandwidth to all
> ready nodes, which happens recursively as follows:
>
>   func allocateBandwidth(node, bw) {
>     if (node.ready) {
>       node.bandwidthShare = bw
>       return
>     }
>     totalWeight = 0
>     for (n in node.children) {
>       if (n.ready || descendantIsReady(n)) {
>         totalWeight += n.weight
>       }
>     }
>     for (n in node.children) {
>       if (n.ready || descendantIsReady(n)) {
>         allocateBandwidth(n, bw * n.weight / totalWeight)
>       }
>     }
>   }
>   allocateBandwidth(root, 1.0)
>
> I believe the above definition is a direct translation of RFC 7540 Section
> 5.3.2 (also see this thread, which discussed bandwidth allocation). Given a
> complete bandwidth allocation, the server can translate each node's
> bandwidthShare to a number of tokens that are updated using a token bucket
> algorithm (or similar). The scheduler would then pick the node with the most
> available tokens. The scheduler looks something like:
>
>   func scheduleNextFrame() {
>     if (tree changed since last frame written) {
>       allocateBandwidth(root, 1.0)
>       assign tokens and build a priority queue containing all nodes with
> allocated bandwidth
>     }
>     node = priorityQueue.head()
>     node.consumeTokens() // consume up to frame size or flow-control limit
>     priorityQueue.update(node)
>     return node
>   }
>
> There are two steps in scheduleNextFrame. The first step updates the
> bandwidth allocation if the tree changed since the last frame was written.
> This is the most expensive step. I don't believe it's possible to implement
> allocateBandwidth using fewer than O(n) worst-case steps. For example, if
> the tree is mostly flat, meaning that most nodes are children of the same
> node, then the loop to compute totalWeight is O(n). I believe Firefox
> creates mostly-flat trees. Further, allocateBandwidth makes O(depth)
> recursive steps, where "depth" is the maximum depth of the dependency tree.
> If the tree is mostly-linear, then O(depth) becomes O(n). Chrome creates
> mostly-linear trees.
>
> The overall runtime depends on how often the tree changes. If the tree
> changes rarely, the scheduler is cheap. If the tree changes frequently, the
> scheduler is worst-case O(n). A tree has "changed" if it changed shape
> (nodes are added/moved/removed) or if any node's ready state has changed.
> Both kinds of changes can happen often in practice, suggesting that the
> overall scheduler is worst-case O(n). For example, consider a server that
> sends small responses (one or two DATA frames per response) -- each stream
> will be closed after one or two DATA frames, so on average, the tree will
> change shape every few frames. Further, it's possible for ready states to
> change more frequently than you might think. In Go's HTTP/2 implementation,
> a stream does not become ready until the application handler calls Write to
> buffer response data. After that buffered data is serialized on the wire,
> the stream transitions to "not ready" because the scheduler cannot know when
> the next Write call will happen. The stream will transition back to "ready"
> during the next Write call. Each Write call typically buffers about 32KB of
> data. This is a "push" model, where the application handler "pushes" data
> into the scheduler. I'm aware of one other HTTP/2 server that works
> similarly. I suspect that frequent ready/not-ready transitions are common to
> most HTTP/2 servers that use a "push" model. These servers will be more
> susceptible to the O(n) worst case.
>
> Questions:
>
> 1. Am I missing a clever implementation, or is it true that a faithful
> HTTP/2 scheduler necessarily requires O(n) operations per frame sent on the
> wire, in the worst case? I could not find much discussion of this question
> after a quick search. H2O claims to implement an O(1) scheduler, however,
> the code seems to be worst-case O(depth) or O(n) -- see here, here, and
> here.
>
> 2. If the above is correct, should I be concerned about the O(n) worst case?
> I doubt that a typical web browsing session will trigger O(n) behavior
> frequently, so I'm less concerned about the average case; I'm more concerned
> about pathological cases or possible DoS vectors. Also, think about cases
> where the "client" is actually a proxy server, meaning the HTTP/2 connection
> may have many more concurrent streams than a typical browsing session. For
> comparison, if you recall the predecessor to HTTP/2 (SPDY), a SPDY scheduler
> could be trivially implemented in O(1), since SPDY used just eight priority
> buckets.
>
> 3. If I should be concerned about an O(n) worst case, are there any
> suggested mitigations beyond setting SETTINGS_MAX_CONCURRENT_STREAMS to a
> smallish constant?



-- 
Kazuho Oku