Closed
Bug 1106030
Opened 10 years ago
Closed 9 years ago
Kibana query to report active weekly, bi-weekly or monthly desktop client users
Categories
(Cloud Services :: Operations: Metrics/Monitoring, task)
Tracking
(Not tracked)
RESOLVED
WONTFIX
People
(Reporter: RT, Unassigned)
References
Details
User Story
A "Recurring user" is a desktop client user who has participated in at least one call or have been in a room with another user, per time period, 2 periods in a row.
No description provided.
Reporter | ||
Updated•10 years ago
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User Story: (updated)
Summary: Loop dashboard "Recurring callers" needs to include room usage → Kibana query to report active weekly, bi-weekly or monthly desktop client users
Comment 1•10 years ago
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Romain, how long do they need to have been in a room with another user for the call to count as "successful"?
Flags: needinfo?(rtestard)
Reporter | ||
Comment 2•10 years ago
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I think we want here to report calls technically successful where both peers have the opportunity to communicate if they want to (we then have another bug for call duration distribution) - I mean by that no minimal duration. That said I am not sure which events you are basing your reports on and if these events are a true reflection of "2 users in a room concurrently". There may be timeout values that we need to take into account to consider that both peers have had a chance to be in a room - Mark/Alexis is there such timeout we should consider to be guaranteed that if 2 peers have been together in a room for X seconds it means that technically the call was successful?
Flags: needinfo?(standard8)
Flags: needinfo?(rtestard)
Flags: needinfo?(alexis+bugs)
Comment 3•10 years ago
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(In reply to Romain Testard [:RT] from comment #2) > That said I am not sure which events you are basing your reports on and if > these events are a true reflection of "2 users in a room concurrently". > There may be timeout values that we need to take into account to consider > that both peers have had a chance to be in a room - Mark/Alexis is there > such timeout we should consider to be guaranteed that if 2 peers have been > together in a room for X seconds it means that technically the call was > successful? Currently the messages we get are: - Room Join - Room Refresh (iirc every 5 mins * 0.9 - the factor is inserted by the client to ensure we hit the refresh time) - Room Leave These will be received one set per user. If there's no room leave after 5 minutes from a join or refresh, then the user crashed or had some other issue. We don't know at which part in that 5 minutes it happened. Obviously to determine if two people are in the room at the same time, we need to analyse individual room tokens for when the room join -> room leave period for two users overlaps.
Flags: needinfo?(standard8)
Reporter | ||
Comment 4•10 years ago
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(In reply to Mark Banner (:standard8) from comment #3) > (In reply to Romain Testard [:RT] from comment #2) > > That said I am not sure which events you are basing your reports on and if > > these events are a true reflection of "2 users in a room concurrently". > > There may be timeout values that we need to take into account to consider > > that both peers have had a chance to be in a room - Mark/Alexis is there > > such timeout we should consider to be guaranteed that if 2 peers have been > > together in a room for X seconds it means that technically the call was > > successful? > > Currently the messages we get are: > > - Room Join > - Room Refresh (iirc every 5 mins * 0.9 - the factor is inserted by the > client to ensure we hit the refresh time) > - Room Leave > > These will be received one set per user. > > If there's no room leave after 5 minutes from a join or refresh, then the > user crashed or had some other issue. We don't know at which part in that 5 > minutes it happened. > > Obviously to determine if two people are in the room at the same time, we > need to analyse individual room tokens for when the room join -> room leave > period for two users overlaps. Thanks. A = Count of overlaps when 2 peers have joined a room before the other peer leaves B = Count of instances where 2 peers are in a room and one or 2 peers never left it OR one or 2 peers never generated a refresh Valid count of room conversations = A - B Sounds right?
Comment 5•10 years ago
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On the server side, we know normally when a user joins and when a user leaves. This should be enough to find out how long 2 users where around. Katie, do you agree on that?
Flags: needinfo?(alexis+bugs) → needinfo?(kparlante)
Comment 6•9 years ago
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(In reply to Alexis Metaireau (:alexis) from comment #5) > On the server side, we know normally when a user joins and when a user > leaves. This should be enough to find out how long 2 users where around. > > Katie, do you agree on that? Yes, we can track when a user joins/leaves, but we also need to count the case where we never see a leave event. In that case, perhaps we want to compare join/last refresh. We might be able to do this fairly easily with current logs using the heka sessionization stuff. For this case (recurring users) we're using the "participants" logging ( https://bugzilla.mozilla.org/show_bug.cgi?id=1124371). I do think we should be measuring the times ourselves as well (separate bug).
Flags: needinfo?(kparlante)
Comment 7•9 years ago
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Here's a bug for doing something to collect/process that timing information: https://bugzilla.mozilla.org/show_bug.cgi?id=1126801
Comment 8•9 years ago
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Reopen if necessary.
Status: NEW → RESOLVED
Closed: 9 years ago
Resolution: --- → WONTFIX
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Description
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