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Kirill Smelkov
iproute2
Commits
9bac173f
Commit
9bac173f
authored
Oct 31, 2011
by
Michal Soltys
Committed by
Stephen Hemminger
Nov 02, 2011
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HFS manpage changes
Few minor changes and small additions.
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man/man7/tc-hfsc.7
man/man7/tc-hfsc.7
+65
-29
man/man8/tc-hfsc.8
man/man8/tc-hfsc.8
+1
-1
man/man8/tc-stab.8
man/man8/tc-stab.8
+28
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man/man7/tc-hfsc.7
View file @
9bac173f
.TH HFSC 7 "
25 February 2009
" iproute2 Linux
.TH HFSC 7 "
31 October 2011
" iproute2 Linux
.ce 1
\fBHIERARCHICAL FAIR SERVICE CURVE\fR
.
...
...
@@ -158,7 +158,7 @@ curve.
.IP "V()"
In linkshare criterion, arbitrates which packet to send next. Note that V() is
function of a virtual time \- see \fBLINKSHARE CRITERION\fR section for
details.
Virtual time \&'vt' corresponds to packets' heads
details. Virtual time \&'vt' corresponds to packets' heads
(vt\~=\~V^(\-1)(w)). Based on LS service curve.
.IP "F()"
An extension to linkshare criterion, used to limit at which speed linkshare
...
...
@@ -187,12 +187,12 @@ Interface 10mbit, two classes, both with two\-piece linear service curves:
.PP
Assume for a moment, that we only use D() for both finding eligible packets,
and choosing the most fitting one, thus eligible time would be computed as
D^(\-1)(w) and deadline time would be computed as D^(\-1)(w+l).
If the 2nd
D^(\-1)(w) and deadline time would be computed as D^(\-1)(w+l). If the 2nd
class starts sending packets 1 second after the 1st class, it's of course
impossible to guarantee 14mbit, as the interface capability is only 10mbit.
The only workaround in this scenario is to allow the 1st class to send the
packets earlier that would normally be allowed. That's where separate E() comes
to help.
Putting all the math aside (see HFSC paper for details), E() for RT
to help. Putting all the math aside (see HFSC paper for details), E() for RT
concave service curve is just like D(), but for the RT convex service curve \-
it's constructed using \fIonly\fR RT service curve's 2nd slope (in our example
\- 7mbit).
...
...
@@ -255,7 +255,7 @@ Such approach has its price though. The problem is analogous to what was
presented in previous section and is caused by non\-linearity of service
curves:
.IP 1) 4
either it's impossible to guarantee
both
service curves and satisfy fairness
either it's impossible to guarantee service curves and satisfy fairness
during certain time periods:
.RS 4
...
...
@@ -278,40 +278,40 @@ beyond of what the interface is capable of.
.RE
.IP 2) 4
and/or it's impossible to guarantee service curves of all classes at all
and/or it's impossible to guarantee service curves of all classes at the same
time [fairly or not]:
.RS 4
Even if we didn't use virtual time and allowed a session to be "punished",
there's a possibility that service curves of all classes couldn't be
guaranteed for a brief period. Consider following, a bit more complicated
example:
Root interface, classes A and B with concave and convex curve (summing up to
root), A1 & A2 (children of A), \fIboth\fR with concave curves summing up to A,
B1 & B2 (children of B), \fIboth\fR with convex curves summing up to B.
Assume that A2, B1 and B2 are constantly backlogged, and at some later point
A1 becomes backlogged. We can easily choose slopes, so that even if we
"punish" A2 for earlier excess bandwidth received, A1 will have no chance of
getting bandwidth corresponding to its first slope. Following from the above
example:
This is similar to the above case, but a bit more subtle. We will consider two
subtrees, arbitrated by their common (root here) parent:
.nf
R (root) -\ 10mbit
A \- 7mbit, then 3mbit
A1 \- 5mbit, then 2mbit
A2 \- 2mbit, then 1mbit
B \- 3mbit, then 7mbit
B1 \- 2mbit, then 5mbit
B2 \- 1mbit, then 2mbit
.fi
At the point when A1 starts sending, it should get 5mbit to not violate its
service curve. A2 gets punished and doesn't send at all, B1 and B2 both keep
sending at their 5mbit and 2mbit. But as you can see, we already are beyond
interface's capacity \- at 12mbit. A1 could get 3mbit at most. If we used
virtual times and kept fairness property, A1 and A2 would send at 3mbit
together with 5:2 ratio (so respectively at ~2.14mbit and ~0.86mbit).
R arbitrates between left subtree (A) and right (B). Assume that A2 and B are
constantly backlogged, and at some later point A1 becomes backlogged (when all
other classes are in their 2nd linear part).
What happens now ? B (choice made by R) will \fIalways\fR get 7 mbit as R is
only (obviously) concerned with the ratio between its direct children. Thus A
subtree gets 3mbit, but its children would want (at the point when A1 became
backlogged) 5mbit + 1mbit. That's of course impossible, as they can only get
3mbit due to interface limitation.
In the left subtree \- we have the same situation as previously (fair split
between A1 and A2, but violated guarantees), but in the whole tree \- there's
no fairness (B got 7mbit, but A1 and A2 have to fit together in 3mbit) and
there's no guarantees for all classes (only B got what it wanted). Even if we
violated fairness in the A subtree and set A2's service curve to 0, A1 would
still not get the required bandwidth.
.RE
.
.SH "UPPERLIMIT CRITERION"
...
...
@@ -416,6 +416,19 @@ In the other words - LS criterion is meaningless in the above example.
You can quickly "workaround" it by making sure each leaf class has RT service
curve assigned (thus guaranteeing all of them will get some bandwidth), but it
doesn't make it any more valid.
Keep in mind - if you use nonlinear curves and irregularities explained above
happen \fIonly\fR in the first segment, then there's little wrong with
"overusing" RT curve a bit:
.nf
A \- ls 5.0mbit, rt 9mbit/30ms, then 1mbit
B \- ls 2.5mbit
C \- ls 2.5mbit
.fi
Here, the vt of A will "spike" in the initial period, but then A will never get more
than 1mbit, until B & C catch up. Then everything will be back to normal.
.
.SH "LINUX AND TIMER RESOLUTION"
.
...
...
@@ -434,7 +447,7 @@ If you have \&'tickless system' enabled, then the timer interrupt will trigger
as slowly as possible, but each time a scheduler throttles itself (or any
other part of the kernel needs better accuracy), the rate will be increased as
needed / possible. The ceiling is either \&'timer frequency' if \&'high
resolution timer support' is not available or not compiled in
. Otherwise
it's
resolution timer support' is not available or not compiled in
, or
it's
hardware dependent and can go \fIfar\fR beyond the highest \&'timer frequency'
setting available.
...
...
@@ -458,7 +471,7 @@ tc class add dev eth0 parent 1:0 classid 1:1 hfsc rt m2 10mbit
Assuming packet of ~1KB size and HZ=100, that averages to ~0.8mbit \- anything
beyond it (e.g. the above example with specified rate over 10x bigger) will
require appropriate queuing and cause bursts every ~10 ms.
As you can
require appropriate queuing and cause bursts every ~10 ms. As you can
imagine, any HFSC's RT guarantees will be seriously invalidated by that.
Aforementioned example is mainly important if you deal with old hardware \- as
it's particularly popular for home server chores. Even then, you can easily
...
...
@@ -510,6 +523,29 @@ curve there, and in such scenario HFSC simply doesn't throttle at all.
So, in rare case you need those speeds with only RT service curve, or with UL
service curve \- remember about drawbacks.
.
.SH "CAVEAT: RANDOM ONLINE EXAMPLES"
.
For reasons unknown (though well guessed), many examples you can google love to
overuse UL criterion and stuff it in every node possible. This makes no sense
and works against what HFSC tries to do (and does pretty damn well). Use UL
where it makes sense - on the uppermost node to match upstream router's uplink
capacity. Or - in special cases, such as testing (limit certain subtree to some
speed) or customers that must never get more than certain speed. In the last
case you can usually achieve the same by just using RT criterion without LS+UL
on leaf nodes.
As for router case - remember it's good to differentiate between "traffic to
router" (remote console, web config, etc.) and "outgoing traffic", so for
example:
.nf
tc qdisc add dev eth0 root handle 1:0 hfsc default 0x8002
tc class add dev eth0 parent 1:0 classid 1:999 hfsc rt m2 50mbit
tc class add dev eth0 parent 1:0 classid 1:1 hfsc ls m2 2mbit ul m2 2mbit
.fi
\&... so "internet" tree under 1:1 and "router itself" as 1:999
.
.SH "LAYER2 ADAPTATION"
.
Please refer to \fBtc\-stab\fR(8)
...
...
man/man8/tc-hfsc.8
View file @
9bac173f
.TH HFSC 8 "
25 February 2009
" iproute2 Linux
.TH HFSC 8 "
31 October 2011
" iproute2 Linux
.
.SH NAME
HFSC \- Hierarchical Fair Service Curve's control under linux
...
...
man/man8/tc-stab.8
View file @
9bac173f
.TH STAB 8 "
25 February 2009
" iproute2 Linux
.TH STAB 8 "
31 October 2011
" iproute2 Linux
.
.SH NAME
tc\-stab \- Generic size table manipulations
...
...
@@ -42,14 +42,14 @@ size is calculated only once \- when a qdisc enqueues the packet. Initial root
enqueue initializes it to the real packet's size.
Each qdisc can use different size table, but the adjusted size is stored in
area shared by whole qdisc hierarchy attached to the interface
(technically
,
it's stored in skb). The effect is, that if you have such setup, the last qdisc
with a stab in a chain "wins". For example, consider HFSC with simple pfifo
attached to one of its leaf classes. If that pfifo qdisc has stab defined, it
will override lengths calculated during HFSC's enqueue, and in turn, whenever
HFSC tries to dequeue a packet, it will use potentially invalid size in its
calculations. Normal setups will usually include stab defined only on root
qdisc, but further overriding gives
extra flexibility for less usual setups.
area shared by whole qdisc hierarchy attached to the interface
. The effect is
,
that if you have such setup, the last qdisc with a stab in a chain "wins". For
example, consider HFSC with simple pfifo attached to one of its leaf classes.
If that pfifo qdisc has stab defined, it will override lengths calculated
during HFSC's enqueue, and in turn, whenever HFSC tries to dequeue a packet, it
will use potentially invalid size in its calculations. Normal setups will
usually include stab defined only on root qdisc, but further overriding gives
extra flexibility for less usual setups.
Initial size table is calculated by \fBtc\fR tool using \fBmtu\fR and
\fBtsize\fR parameters. The algorithm sets each slot's size to the smallest
...
...
@@ -59,18 +59,16 @@ table will usually support more than is required by \fBmtu\fR.
For example, with \fBmtu\fR\~=\~1500 and \fBtsize\fR\~=\~128, a table with 128
slots will be created, where slot 0 will correspond to sizes 0\-16, slot 1 to
17\~\-\~32, \&..., slot 127 to 2033\~\-\~2048. Note, that the sizes
are shifted 1 byte (normally you would expect 0\~\-\~15, 16\~\-\~31, \&...,
2032\~\-\~2047). Sizes assigned to each slot depend on \fBlinklayer\fR parameter.
17\~\-\~32, \&..., slot 127 to 2033\~\-\~2048. Sizes assigned to each slot
depend on \fBlinklayer\fR parameter.
Stab calculation is also safe for an unusual case, when a size assigned to a
slot would be larger than 2^16\-1 (you will lose the accuracy though).
During kernel part of packet size adjustment, \fBoverhead\fR will be added to
original size, and after subtracting 1 (to land in the proper slot \- see above
about shifting by 1 byte) slot will be calculated. If the size would cause
overflow, more than 1 slot will be used to get the final size. It of course will
affect accuracy, but it's only a guard against unusual situations.
original size, and then slot will be calculated. If the size would cause
overflow, more than 1 slot will be used to get the final size. It of course
will affect accuracy, but it's only a guard against unusual situations.
Currently there're two methods of creating values stored in the size table \-
ethernet and atm (adsl):
...
...
@@ -82,8 +80,8 @@ This is basically 1\-1 mapping, so following our example from above
and so on, up to slot 127 with 2048. Note, that \fBmpu\fR\~>\~0 must be
specified, and slots that would get less than specified by \fBmpu\fR, will get
\fBmpu\fR instead. If you don't specify \fBmpu\fR, the size table will not be
created at all
, although any \fBoverhead\fR value will be respected during
calculations.
created at all
(it wouldn't make any difference), although any \fBoverhead\fR
value will be respected during
calculations.
.IP "atm, adsl"
.br
ATM linklayer consists of 53 byte cells, where each of them provides 48 bytes
...
...
@@ -127,7 +125,7 @@ IPoA in LLC case requires SNAP, instead of LLC\-NLPID (see rfc2684) \- this is
the reason, why it actually takes more space than PPPoA.
.IP \(bu
In rare cases, FCS might be preserved on protocols that include ethernet frame
(Bridged and PPPoE).
In such situation, any ethernet specific padding
(Bridged and PPPoE). In such situation, any ethernet specific padding
guaranteeing 64 bytes long frame size has to be included as well (see rfc2684).
In the other words, it also guarantees that any packet you send will take
minimum 2 atm cells. You should set \fBmpu\fR accordingly for that.
...
...
@@ -136,11 +134,20 @@ When size table is consulted, and you're shaping traffic for the sake of
another modem/router, ethernet header (without padding) will already be added
to initial packet's length. You should compensate for that by subtracting 14
from the above overheads in such case. If you're shaping directly on the router
(for example, with speedtouch usb modem) using ppp daemon,
layer2 header will
not be added yet
.
(for example, with speedtouch usb modem) using ppp daemon,
you're using raw ip
interface without underlying layer2, so nothing will be added
.
For more thorough explanations, please see \fB[1]\fR and \fB[2]\fR.
.
.SH "ETHERNET CARDS CONSIDERATIONS"
.
It's often forgotten, that modern network cards (even cheap ones on desktop
motherboards) and/or their drivers often support different offloading
mechanisms. In context of traffic shaping, 'tso' and 'gso' might cause
undesirable effects, due to massive tcp segments being considered during
traffic shaping (including stab calculations). For slow uplink interfaces,
it's good to use \fBethtool\fR to turn off offloading features.
.
.SH "SEE ALSO"
.
\fBtc\fR(8), \fBtc\-hfsc\fR(7), \fBtc\-hfsc\fR(8),
...
...
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