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

Hi Tom,

Probably, this material would help you:


> 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 <>:
>> 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

Received on Tuesday, 24 January 2017 07:54:28 UTC