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

[Tatsuhiro Tsujikawa <[e-mail]>]

Hi,

On Wed, Jan 25, 2017 at 7:22 AM, Tom Bergan <[e-mail]> wrote:

> Thanks for the feedback. Sounds like the worst-case time really is O(n).
>
> > kazuhooku@
> > the constant for O(n) would be small enough so that it cannot be used an
> attack vector
>
> Don't you need to ensure that n is small, otherwise even O(n) with a small
> constant factor can lead to degraded performance? e.g.,
> if SETTINGS_MAX_CONCURRENT_STREAMS = 20,000, I may end up computing a sum
> of 20,000 numbers every few frames, which is not terribly slow, but not
> terribly fast either.
>
>
​If large n is concern, server can limit SETTINGS_MAX_CONCURRENT_STREAMS to
the reasonable number.  Usually we use 100, and it is pretty fast.

Best regards,
Tatsuhiro Tsujikawa​



> > kazu@
> > http://www.mew.org/~kazu/material/2015-http2-priority2.pdf
> > http://www.mew.org/~kazu/doc/paper/http2-haskell-2016.pdf
>
> Thanks. IIUC, the algorithms described in both links are still at least
> O(depth), which can be O(n) for dependency trees generated by certain
> clients such as Chrome.
>
> > stefan.eissing@
> > The very same problem exists for stream processing in order to generate
> response data.
>
> What did you mean by "stream processing"?
>
> On Tue, Jan 24, 2017 at 12:26 AM, Stefan Eissing <
> [e-mail]> wrote:
>
>> The very same problem exists for stream processing in order to generate
>> response data.
>>
>> > Am 24.01.2017 um 08:53 schrieb Kazu Yamamoto (山本和彦) <[e-mail]>:
>> >
>> > Hi Tom,
>> >
>> > Probably, this material would help you:
>> >
>> > http://www.mew.org/~kazu/material/2015-http2-priority2.pdf
>> > http://www.mew.org/~kazu/doc/paper/http2-haskell-2016.pdf
>> >
>> > --Kazu
>> >
>> >> 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
>> >>
>> >
>>
>>
>>
>