- Recall:
- ni
=
the number of
TD periods
in one Reno cycle
Note that the series of the TD period ends in a Timeout recovery
Situation:
- ni
=
the number of
TD periods
in one Reno cycle
- Categorizing the
ni
TD periods:
- ni - 1
TD periods will end in
Triple Duplicate ACKs recovery
- The last one TD periods will end in Timeout recovery
- ni - 1
TD periods will end in
Triple Duplicate ACKs recovery
- Define:
- Q = the probability that a TD period is ended by a Timeout recovery
Assumption:
- Each TD period is independent of each other
- E[n] can be computed as follows:
E[n] = average number of TD periods in one Reno cycle
E[n] = 1 * P[ 1st TD period ends in Timeout recovery ] + 2 * P[ 2nd TD period ends in Timeout recovery ] + 3 * P[ 3rd TD period ends in Timeout recovery ] + 4 * P[ 4th TD period ends in Timeout recovery ] + ...
E[n] = 1 * Q + 2 * (1-Q) * Q + 3 * (1-Q)2 * Q + 4 * (1-Q)3 * Q + ...
E[n] = Q * ( 1 + 2 * (1-Q) + 3 * (1-Q)2 + 4 * (1-Q)3 + ... )
- Using
maple, we can compute the
sum:
1 + 2 * (1-Q) + 3 * (1-Q)2 + 4 * (1-Q)3 + ... maple: sum( k* (1-Q)^(k-1), k=1..infinity); answer: 1 ---- 2 Q
- Hence:
- Q = the probability that a TD period is ended by a Timeout recovery
1 E[n] = Q * --- Q2 1 = --- . . . . . . . . . . . . . . . . . . . . . . (1) Q
- Warning:
Fasten your seat belt - computing
Q
is long and tedious....
Before we discuss how to compute Q, we need to compute 2 probability values
- The probability that a
TD period
will
end is
a Timeout
depends on
- The size of
the congestion window
Example:
If CWND ≤ 3, TCP Reno can never recover using Triple Duplicate ACKs and the TD period will certainly end in timeout
- Which packet in the
TD period is
lost
Example:
If the packet loss happens at the first, second or third packet, TCP Reno can not recover using Triple Duplicate ACKs and the TD period will also end in timeout
- The size of
the congestion window
- To express the
probabilities
for situations where a
TD period ends in
Timeout, we
need the following expressions:
- A(W, k) =
the probability that the first
k
packets of W packets in a TD period
are ACKed (i.e., not lost)
when
given that there is a packet loss
in the window
- C(n, m) = the probability that the first m packets of n packets in a transmission are ACKed (i.e., not lost)
- A(W, k) =
the probability that the first
k
packets of W packets in a TD period
are ACKed (i.e., not lost)
when
given that there is a packet loss
in the window
- The following probability
is needed to compute
Q :
Definition:
- A(W, k)
= the probability that the
first k packets
of W packets
in a TD period
are ACKed (i.e., not lost)
when
given that there is
a packet loss
in the window
- Note:
- We are gievn that there is a packet loss because the TD period is ended (in a timeout)
- Recall the
packet lost assumption:
- When packet j in a TD period is lost, then all packets j+1, j+2, ..., W are also lost !
- A(W, k)
= the probability that the
first k packets
of W packets
in a TD period
are ACKed (i.e., not lost)
when
given that there is
a packet loss
in the window
- Illustrating the event where:
-
first k packets
of W packets
in a TD period
are ACKed (i.e., not lost)
when
given that there is
a packet loss
in the window
-
A(W, k)
is computed as a
Conditional Probability
as follows:
A(W, k) = Prob [ first k packets not lossed | loss in W packets ]
Re-write the conditional probability using the rule:
Prob [ (first k packets not lossed) AND (loss in W packets) ] A(W, k) = -------------------------------------------------------------- Prob[ loss in W packets ] Prob [ (first k packets not lossed) AND (loss in W packets) ] = -------------------------------------------------------------- 1 - Prob[ NO losses in W packets ] Prob [ (first k packets not lossed) AND (loss in W packets) ] = -------------------------------------------------------------- 1 - (1 - p)W
The packet loss must have happen at packet k+1 to get "first l packets not lossed" Therefore: Prob [ (first k packets not lossed) AND (packet k+1 lost) ] A(W, k) = ------------------------------------------------------------- 1 - (1 - p)W (1 - p)k * p = ---------------- 1 - (1 - p)W
In conclusion: (1 - p)k * p A(W, k) = ---------------- . . . . . . . . . . . . . . . . . . (2) 1 - (1 - p)W
- The following probability
is also necessary to compute
Q :
Definition:
- C(n, m)
= the probability that the
first m packets
of n packets in
a transmission
are ACKed (i.e., not lost)
- Note:
- C(n, m) is a unconditional probability
- C(n, m)
= the probability that the
first m packets
of n packets in
a transmission
are ACKed (i.e., not lost)
- Illustrating the event where
first m packets
of n packets in
a transmission
are ACKed (i.e., not lost):
- We must
distinguish
2 different cases:
- m < n
- m = n
- Case 1: m < n
- first m packets of n packets in a transmission are ACKed
means:
- The first m packets were ACKed
- Packet m+1 is lost.
When m < n, C(n, m) is equal to:
C(n, m) = Prob [ (first m packets NOT lossed) AND (packet m+1 lost) ] = (1 - p)m * p
Therefore: C(n, m) = (1 - p)m * p (for m < n) . . . . . . (3a)
- Case 2: m = n
- first m packets of m packets in a transmission are ACKed
means:
- All m packets were ACKed (i.e., not lost)
When m = n, C(n, m) is equal to:
C(n, m) = Prob [ (all m packets NOT lossed) ]
C(n, m) = (1 - p)n (m = n) . . . . . . (3b)
- OK, we are ready to compute
Q
Recall the definition of Q:
- Q = the probability that a TD period is ended by a Timeout recovery
- There is only one reason
why TCP Reno will
end up in timeout:
- The number of duplicate ACKs received by Reno is < 3
- Several events can
result in the above
"reason"
- In order to compute
Q ,
we must
enumerate
all events
that
leads to a
timeout recovery
- There are 3 events that will
lead to Reno
receiving
< 3 duplicate ACKs:
- The congestion window CWND is
≤ 3
- The congestion window CWND is
≥ 4,
but:
- the packet loss happens very early in the transmission round
- The congestion window CWND is
≥ 4, and the
packet loss happens
late
in the transmission round,
but:
- the packet loss happens very early in the RE-transmission round
We will illustrate the cases in more detail....
- The congestion window CWND is
≤ 3
- Case 1: ( W <= 3 )
Notes:
- Sender will send at most
3 packets
- One of the
3 packets is
lost
So the receiver will receive at most 2 packets
- Therefore, the receiver will
send at most
2 duplicate ACKS
- Since Reno needs 3 duplicate ACKs to recover with fast recovery, the sender will timeout.
- Sender will send at most
3 packets
- Case 2: ( W >= 4 )
and (packet 1, 2 or 3
in the transmission round was
lost)
Notes:
- Suppose
any one
of the packet 1, 2 or 3
is lost
(Remember that Padhye assumes that when a packet in a cycle is lost, all packets after that packet will also be lost).
- This will cause the receiver to send back
at most 2 new ACKs
These new ACKs will free up (at most) 2 window sequence numbers for new packets in the next round of transmission
- When the receiver receives these
new packets,
it will send
(at most) 2 duplicate ACKs
- Since Reno needs 3 duplicate ACKs to recover with fast recovery, the sender will timeout.
- Suppose
any one
of the packet 1, 2 or 3
is lost
- Case 3: ( W >= 4 )
and (packet 1, 2 and 3
in the transmission round were
not lost), but
(packet 1, 2 or 3
in the
RE-transmission round
was lost)
It should be obvious now that Reno will receive < 3 duplicate ACKs in this case also...
- Observation (from the above 3 cases):
- Q
= the probability
that a TD period
is ended by
a Timeout recovery
- Q depends on the congestion window size (W)
- Q
= the probability
that a TD period
is ended by
a Timeout recovery
- Define:
- Q(W) = the probability that a TD Period with initial congestion window size W will end in Timeout recovery
Note:
-
Q(W)
will
split the
outcome space
into
separate (smaller) cases
(or "conditions"
- All cases will
collectively
form the entire outcome space
- This property allow
us to use the
Law of Total Probability
on
Q(W)
to compute the
the probability
Q .
(Law of Total Probability: see click here)
- We will be using:
to determine Q(W) .
- Deriving an expression for
Q(W):
Q(W) = the probability that a TD period with CWND = W ends in Timeout
Case 1: W ≤ 3
Q(W) = 1 (if W <= 3) . . . . . . . . . . . . . . (4a)
(Because Reno will certainly ends in timeout....)
Case 2: W ≥ 4
- Each of the events is independent from one another
- Recall: A(W, k) =
the probability that the first
k
packets of W packets in a TD period
are ACKed (i.e., not lost)
when
given that there is a packet loss
in the window
- Q(W) = probability that a TD period with CWND = W ends in Timeout
Because it ends in Timeout, it is given there is a packet loss !!!
- Therefore: Prob[ packet k in window (W) is lost ] = A(W, k)
- Recall:
C(n, m)
= the probability that the
first m packets
of n packets in
a transmission
are ACKed (not lost)
- Hence:
-
Prob[ packet 1, 2 or 3 of m packets lost] =
C(1, m) +
C(2, m) +
C(3, m)
Q(W) = probability that a TD period with CWND = W ends in Timeout = Prob[ (packet 1 in window lost) OR (packet 2 in window lost) OR (packet 3 in window lost) OR (packet 4 in window lost AND (packet 1, 2 or 3 in retrans lost) OR (packet 5 in window lost AND (packet 1, 2 or 3 in retrans lost) OR ... OR (packet W in window lost AND (packet 1, 2 or 3 in retrans lost) ]
Therefore: Q(W) = Prob[ packet 1 in window lost ] + Prob[ packet 2 in window lost ] + Prob[ packet 3 in window lost ] + Prob[ packet 4 in window lost ] * Prob[ packet 1, 2 or 3 of 3 packets lost] + Prob[ packet 5 in window lost ] * Prob[ packet 1, 2 or 3 of 4 packets lost] + ... + Prob[ packet W in window lost ] * Prob[ packet 1, 2 or 3 of W-1 packets lost] ]
Hence: Q(W) = A(W, 0) + A(W, 1) + A(W, 2) + A(W, 3) * Prob[ packet 1, 2 or 3 of 3 packets lost] + A(W, 4) * Prob[ packet 1, 2 or 3 of 4 packets lost] + ... + A(W, W-1) * Prob[ packet 1, 2 or 3 of W-1 packets lost] ]
We have: Q(W) = A(W, 0) + A(W, 1) + A(W, 2) + A(W, 3) * ( C(3,0) + C(3,1) + C(3,2) ) + A(W, 4) * ( C(4,0) + C(4,1) + C(4,2) ) + ... + A(W, W-1) * ( C(W-1,0) + C(W-1,1) + C(W-1,2) )
In summary: Q(W) = A(W, 0) + A(W, 1) + A(W, 2) W-1 2 ---- +- ---- -+ \ | \ | + > | A(W, k) * > C(k, m) | . . . . . . . (4b) / | / | ---- +- ---- -+ k=3 m=0
- Computing Q(W)
We use Maple to obtain a simplified expression for Q(W) as follows:
- NOTE: This is Equation (23) in the paper
Define A(W,k) (see: click here ) > A := (1-p)^k*p/ (1 - (1-p)^W); k (1 - p) p A := ------------ W 1 - (1 - p)
First, compute: A(W, 0) + A(W, 1) + A(W, 2) > Q1 := subs(k=0, A) + subs(k=1, A) + subs(k=2, A); > Q2 := simplify(Q1); 2 p (3 - 3 p + p ) Q2 := - ---------------- W -1 + (1 - p)
Next, compute: W 2 ---- +- ---- -+ \ | \ | > | A(W, k) * > C(k, m) | / | / | ---- +- ---- -+ k=3 m=0 > C := p*(1-p)^m; > Q3 := A * ( subs(m=0, C)+subs(m=1, C)+subs(m=2, C) ); > Q4 := sum( Q3, k=3..W-1 ); > Q5 := simplify(Q4); 2 W W 2 3 p (3 - 3 p + p ) ((1 - p) - (1 - p) p - 1 + 3 p - 3 p + p ) Q5 := -------------------------------------------------------------- W -1 + (1 - p)
Finally: > Q := Q2 + Q5; 2 2 W 2 3 p (3 - 3 p + p ) p (3 - 3 p + p ) ((1 - p) - 1 + 3 p - 3 p + p ) Q := - ---------------- + ------------------------------------------------- W W -1 + (1 - p) -1 + (1 - p) 2 W W 2 3 p (3 - 3 p + p ) ((1 - p) - (1 - p) p - 2 + 3 p - 3 p + p ) Q(W) = -------------------------------------------------------------- . . (5) W -1 + (1 - p)
- We can also verify that the limit of Q(W)
for p -> 0 is equal
to 3/W :
> limit(Q, p=0, left); 3/W
- Recall that:
- Q = the
probability that a
TD period ends in
Timeout
- Q(W) = the probability that a TD Period with a congestion window size W will end in Timeout
- Q = the
probability that a
TD period ends in
Timeout
- Q
can be computed from
Q(W)
using
the Law of Total Probability :
Q = Q(1) * Prob[ W = 1 ] + Q(2) * Prob[ W = 2 ] + Q(3) * Prob[ W = 3 ] + Q(4) * Prob[ W = 4 ] + .... (upto infinity) . . . . . . . . . . . (6)
-
Major problem :
- Computing Prob[ W = w ] for w = 1, 2, ... is very difficult...
- But luckily:
E[W] = 1 * Prob[ W = 1 ] + 2 * Prob[ W = 2 ] + 3 * Prob[ W = 3 ] + 4 * Prob[ W = 4 ] + .... (upto infinity)
E[f(W)] = f(1) * Prob[ W = 1 ] + f(2) * Prob[ W = 2 ] + f(3) * Prob[ W = 3 ] + f(4) * Prob[ W = 4 ] + .... (upto infinity)
Therefore: Q = Q(1) * Prob[ W = 1 ] + Q(2) * Prob[ W = 2 ] + Q(3) * Prob[ W = 3 ] + Q(4) * Prob[ W = 4 ] + .... (upto infinity) . . . . . . . . . . . (6) = E[ Q(W) ] !!!So luckily:
- Equation (6) is also the expression for the E[ Q(W)] or the average value of Q(W)
- Why do I say luckily ?
- Because there is a standard trick in Probability Theory to approximate an average !!!
Approximation trick:
E[ f(W) ] ≅ f( E[W] ) . . . . . . . . . . . (7)
Clarification:
- Equation (7) says that E[ f(W) ] can be approximated by the function value of the function f(x) at the point x = E[W]
- Example:
(Prob[ W = w ]) ( Q(W) ) | Probability | Probability that | that TCP uses | TCP using CWND = w w | CWND = w | ends in Timeout -------------+------------------+--------------------- 1 | 0.1 | 0.1 2 | 0.4 | 0.3 3 | 0.3 | 0.6 4 | 0.2 | 0.9 - Then E[ Q(W) ]
is equal to:
Q = Q(1)*P[W=1] + Q(2)*P[W=2] + Q(3)*P[W=3] + Q(4)*P[W=4] Q = 0.1*0.1 + 0.3*0.4 + 0.6*0.3 + 0.9*0.2 Q = 0.01 + 0.12 + 0.18 + 0.18 Q = 0.49
(But we can't do it this way, because the value Prob[W=w] are too hard to obtain)
- The approximation
of
E(Q(W)] (= Q)
is to use the
function value
of function Q(x)
at the point
x = E[W]:
Q = E[Q(W)] ≅ Q( E[W] ) . . . . . . . . . . . (7)
How to use the approximation trick:
1. Find the function Q(x) that has the following function values: Q(1) = 0.1 Q(2) = 0.3 Q(3) = 0.6 Q(4) = 0.9 The common way is to use polynomial interpolation With Maple: > f := interp([1,2,3,4], [0.1, 0.3, 0.6, 0.9], w); > Q := unapply(f, w); 3 2 -0.01666666667 w + 0.1500000000 w - 0.1333333333 w + 0.1 Double check: > Q(1); 0.1 > Q(2); 0.3000000000 > Q(3); 0.6000000000 > Q(4); 0.8999999998
2. Find E[W]: Prob[W=1] = 0.1 Prob[W=2] = 0.4 Prob[W=3] = 0.3 Prob[W=4] = 0.2 E[W] = 1 * 0.1 + 2 * 0.4 + 3 * 0.3 + 4 * 0.2 E[W] = 0.1 + 0.8 + 0.9 + 0.8 E[W] = 2.6
3. The approximation of E[Q(W)] is equal to the function value of Q(x) at x = E[W] = 2.6 3 2 Q(w) = -0.01666666667 w + 0.1500000000 w - 0.1333333333 w + 0.1 > Q(2.6); 0.4744000000 E[Q(W)] = 0.4744 (approximately, actual: 0.49)
- To compute
Q( E[W] ) ,
we must:
- Find the function Q(x)
- Find average window size E[W]
- We already obtained
Q(W)
as a function in W
in
Equation 5
(see: click here)
2 W W 2 3 p (3 - 3 p + p ) ((1 - p) - (1 - p) p - 2 + 3 p - 3 p + p ) Q(W) = -------------------------------------------------------------- ....... (5) W -1 + (1 - p) - In the analysis of recovery by
Triple Duplicate ACKs,
we have already found
the expression for E[W]
(See:
click here):
-------------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) . . . . . . . . (13) 3b \/ 3bp 3b This was formula (13) in Padhye's paper
- Now substitute E[W]
(Equation 13)
for W in the
Equation (5)
and you will obtain the
approximation expression
for
Q...
It is very messy....
Even Padhye did not venture to write the result out, so I will skip it also :-)
(BTW, we are on page 6 of Padhye's paper: click here )
- We have just finished "computing"
E[n]
because:
1 E[n] = --- . . . . . . . . . . . . . . . . . . . . . . (1) Q Q = E[Q(W)] ≅ Q( E[W] ) . . . . . . . . . . (7)