- Throughput
is the total amount of data
transmitted by a source
- Average Throughput
is the amount of data
transmitted by a source
per time unit
- Example:
- The throughtput in [0, 10] is 7 packets
- The average throughtput in [0, 10] is 0.7 packets/sec
- In this webpage, we will analyze the
behavior of the TCP protocol
when the
loss indications
are exclusively
Triple Duplicate ACKs
In other words: when TCP recovers exclusively from packets losses using the Triple Duplicate ACK recovery...
I.e.: only by fast recovery - click here
- In this case, the
recovery action is:
- Set CWND = CWND/2
(This is actually an approximation, TCP sets CWND to CWND/2 + 3, but if CWND is large, +3 is not going to do much) - Enter congestion avoidance phase
- Set CWND = CWND/2
- Nt
= number of packets transmitted in time interval
[0, t]
= TCP throughput - Bt = Nt/t
= average number of packets transmitted in time interval
[0, t]
= average TCP throughput in interval [0, t] - B = lim(t &rarr &infin)
Bt
= average TCP throughput over a very very long period of time
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- A TD Period (TDP)
is defined as a period between
two consecutive
Triple Duplicate ACK loss indications
- Graphically:
Notes:
- The figure
begins
with the previous
Triple Duplicate ACK period
Specifically, with the last round of transmission that contain a packet loss
NOTE:
- Due to the one lost, all lost packet loss assumption (see: click here ), the duplicate ACKs will be generated when the packets in the next transmission round are received
- The first few packets
in the last round
are acked normally
with new ACKs (green ACKs)
These new ACKs will slide sender window further and trigger new transmissions (these new transmissions will be part of the next transmission round)
- Some packet in the
last round is
lost
With the one lost, all lost assumption, the packets in the tail end of the last transmission round of the previous TD period will all be lost
- Then a new transmission
round in the
previous TD period begins...
When these new packets are received, they will trigger duplicate ACKs
- When the sending TCP
receives
3 duplicate ACKs, it will recover by:
- Set CWND = CWND/2
- Enter congestion avoidance
- TCP has now
started a
new TD period
This TD period continues until a packet loss occurs...
- The TD period is the interval between 2 consecutive Triple Duplicate Loss indications as denoted in the figure above.
- The figure
begins
with the previous
Triple Duplicate ACK period
- This figure shows what happens when TCP
recovers
by using Triple Duplicate ACKs:
- The TD period
repeats
indefinitely
Here are two consecutuve TD periods:
- Important note:
- The initial window size
in each TD period
will be different
- In fact: the initial window size of a TD period is a random variable !!!
- The initial window size
in each TD period
will be different
- Definitions and notations:
- p
= the
probability
that a packet is
lost
given that:
- It is the first packet
in the round
or
- All preceeding packets in the round are not lost
In more understandable English:
- p = the probability that a packet loss is the first loss inside a round of packet transmission
NOTE:
-
The packet loss probability
p
is a
parameter
in the TCP model
(and it is thus given and not computed)
-
Parameters are usually
obtained (determined)
by
empirical measurements
(e.g., count the number of packet losses...)
- It is the first packet
in the round
or
- B(p)
=
TCP's throughput
for a given
packet loss probability
p
Our mission: find an expression for B(p)....
- p
= the
probability
that a packet is
lost
given that:
- Notations:
- Yi
= number of packets sent
in the
ith
TD period
( Yi includes packets that were lost)
- Ai
= duration
of the
ith
TD period
- Wi = window size at the end of the ith TD period
- Yi
= number of packets sent
in the
ith
TD period
- Illustrations:
- NOTE:
the following holds in the above figure
- The initial window size
of TD period 2 is equal to
W1/2
I.e.: w2 = (W1 / 2)
- Also in the figure above: w3 = (W2 / 2)
- The initial window size
of TD period 2 is equal to
W1/2
- B(p) can be
expressed as:
Number of packets sent in n TD period B = --------------------------------------- , or: Duration of n TD period
Y1 + Y2 + ... + Yn B = -------------------- , or: A1 + A2 + ... + An
+- -+ | Y1 + Y2 + ... + Yn | | ------------------ | | n | +- -+ B = -------------------- , or: +- -+ | A1 + A2 + ... + An | | ------------------ | | n | +- -+
Avg. number of packets sent in ONE TD period B = ---------------------------------------------- Avg. duration of ONE TD periodThe first important equation: E[ Y ] B = -------- ............. (1) E[ A ] - NOTE:
- E[Y]
= the expected value
(or "average") of the random variable
Y
- E[A] = the expected value (or "average") of the random variable A
- E[Y]
= the expected value
(or "average") of the random variable
Y
- Remember our objective:
- Find a relationship for the throughput B of TCP
- We just saw that:
E[ Y ] B = -------- E[ A ] - So our mission is now:
- Find an expression for E[Y]
- Find an expression for E[A]
- Find a relationship for the throughput B of TCP
- To find E[Y]
and E[A],
we must analyze
a TD period
carefully
- Let us look at
an arbitrary
TD period
- Let us pick and study the ith TD period
The figure illustrate the behavior at the start of the ith TD period
(We will look at the behavior at the end of the TD period later)
Behavior at the start of the ith TD period:
Notes:
- The preceeding
TD period was
the (i-1)th
TD period
Assume that the (i-1)th TD period ended with this CWND value:
- CWND = Wi-1
- TCP recovers using
Triple Duplicate ACKs
(i.e., Fast Recovery)
and starts a
new
ith TD period
The starting CWND size in this ith TD period is equal to:
- W = Wi-1 / 2
- Inside the ith
TD period,
as long as there are
no packet losses ,
the
congestion window size CWND
will increase as follows:
- CWND = CWND + 1/b
for each transmission round
- The number of packets
transmitted in
one transmission round
is equal to
the
CWND value
at the start of the round.
Therefore:
- # packets transmitted in round 1 = Wi-1 / 2
- # packets transmitted in round 2 = Wi-1 / 2 + 1/b
- # packets transmitted in round 3 = Wi-1 / 2 + 2/b
- And so on... (upto and including the last round in the ith TD period)
- Let us now look at the
end of
the
ith TD period...
Behavior at the end of the ith TD period:
Notes:
- Assume that:
- The first packet loss occured inside the round Xi of transmission.
In that case, there are Xi complete transmission rounds inside the ith TD period
(There is a partial round following round Xi )
(See above figure)
- Let
αi
= the index of the
first packet
that was lost
(And by assumption, all packets following this packets in round Xi are also lost)
Note:
- αi
is an absolute index
- E.g., if there are 10 packets in the first transmission period, and packet #5 in the second transmission period is lost, then: αi = 15
- αi
is an absolute index
- The congestion window used in
the
round
Xi
is equal to:
Wi-1 Xi W in round Xi = ----- + --- 2 b
- There are
no more new ACKs
following the
packet loss
in the
round Xi !!!
Hence, congestion window will not increase anymore inside the ith TD period
Therefore, Wi (= the value of the CWND at the end of the ith TD period) is:
Wi-1 Xi Wi = ----- + --- 2 b
- There is a partial round
of transmissions
following
round Xi
The number of packets transmitted in this partial round is equal to αi - 1 We can also reason as follows:
- After the packet
αi ,
(first packet that was lost),
the sender will transmit:
Wi - 1
packets
- After the packet
αi ,
(first packet that was lost),
the sender will transmit:
- Assume that:
- Consider the
ith TD period:
- Recall that
αi
is the
(absolute) index of the
first packet that was
lost
in the
ith TD period
Recall that after the packet αi , the sender will transmit Wi - 1 more packets.
- Therefore, the
total number of packets
Yi
transmitted in the
ith TD period
is equal to:
- Yi = αi + (Wi - 1) ............................... (2)
- From equaltion (2) and the
Theory of Probability, we have that:
E[Y] = E[ α ] + E[W] - 1 ............................... (2a)
(The notation E[...] means "expected value". For laymen: you can think of it as "average").
- Recall that
αi
is the
(absolute) index of the
first packet that was
lost
in the
ith TD period
- We have assumed that
the
length of
each
(transmission) round
is constant (= RTT)
(see: click here)
- The number of
(complete) rounds
of transmission
in the
ith TD period
is:
- Xi
And there is one partial round of transmissions.
Therefore:
- The number of (complete and partial) rounds of transmission in the ith TD period = Xi + 1
- The duration
of the
ith TD period
is equal then to:
- Ai = (Xi + 1) × RTT ............................... (3)
- With equation (3) and the
Theory of Probability, we have that:
- E[A] = (E[X] + 1) × RTT ............................... (3a)
- We have expressed
E[Y]
in terms of
E[X]
But we do not know anything about E[X] .... yet...
- We can derive a relationship
between:
- E[X] (number of complete rounds in a TD period) and
- E[W] (the window size at the end of a TD period)
- Relationship between E[X] and E[W]:
- The window size at the
end of
ith TD period
is equal to:
(see: click here)
- Wi
- From studying
the anatomy of
the ith TD period:
The window size at the end of the the ith TD period is also equal to:
- Wi-1 / 2 + Xi / b
- Therefore, we have:
Wi-1 Xi Wi = ----- + ----- .......... (4) 2 b
- The window size at the
end of
ith TD period
is equal to:
(see: click here)
- Using equation (4) and the
Theory of Probability, we have that:
E[W] E[X] E[W] = ----- + ----- 2 b
E[W] E[X] <==> ----- = ----- 2 b2 <==> E[W] = --- E[X] ............. (4a) b Or: b <==> E[X] = --- E[W] ............. (4b) 2
===========================================================
- When working out a problem, always keep in mind:
- What's the goal (Where are you going).
- What have you achieved so far (Where are you now).
This will help you decide what to do next.
- Goal:
- find an expression for the throughput of TCP (B(p)
- Subgoal 1: we found:
E[ Y ] B = -------- ............. (1) E[ A ]We need to find an expression for:
- E[Y]
- E[A]
- Subgoal 2: we found 3 equations:
E[Y] = E[ α ] + E[W] - 1 . . . . . . . . . . . . . . . . (2a) E[A] = (E[X] + 1) × RTT . . . . . . . . . . . . . . . . (3a)
b E[X] = --- E[W] . . . . . . . (4b) 2Substitute E[X] = (b/2) × E[W] in equation (3a), and we get:
E[Y] = E[α] + E[W] - 1 . . . . . . . . . . . . . . . (2a) 2 E[A] = (--- E[W] + 1) × RTT . . . . . . . . . . . . . (5) b
And therefore: E[α] + E[W] - 1 B = -------------------------- . . . . . . . . . . . . (6) 2 (--- E[W] + 1) × RTT b
- This formula still have (new) unknowns:
- E[α], and
- E[W]...
- New subgoal: find an expression for:
- E[α]
- E[W]
- NOTE:
-
It may look like we did not make any progress,
but looks are deceptive.
The fact is: E[α] and E[W] are a lot easier to calculate !
- Recall the definition of
α:
- α =
the index
of the first packet
that is lost
in a TD period
(α is an absolute index, counting every packet inside a TD period.
I.e., it counts across different transmission rounds)
- α =
the index
of the first packet
that is lost
in a TD period
- In other words:
Note:
- The
first
α - 1
packets
in the TD period are
not lost,
and
- The next packet (packet α) is lost
(Those that have had Theory of Probability will immediately see that this is the classic "coin toss" experiment)
- The
first
α - 1
packets
in the TD period are
not lost,
and
-
Recall that:
-
p
= probability
that the
first packet
in the TD period is lost
or
that a packet is lost when all previous packets
are not lost
See: click here
-
p
= probability
that the
first packet
in the TD period is lost
or
that a packet is lost when all previous packets
are not lost
- Then:
Prob[ α = k] = Prob[ the first packet lost is packet #k ] = Prob[ first k-1 packets not lost and the next packet is lost ] = Prob[ first k-1 packets not lost ] × P[ next packet is lost ] <==> Prob[ α = k] = (1 - p)k-1 × p . . . . . . . . . . . . . . . . . . . (7)We have found the probability density function for the random variable α
We can use the probability density function to compute the expected value E[α]
- Computing the
expected value
E[α]:
By definition: &infin --- \ E[α] = > k × Prob[ α = k] / --- k=1
&infin --- \ <==> E[α] = > k × (1 - p)k-1 × p / --- k=1
&infin --- \ <==> E[α] = p > k × (1 - p)k-1 / --- k=1
Using Maple: > sum( k * (1-p)^(k-1), k=1..infinity); 1 ---- 2 p
1 <==> E[α] = p x --- p21 E[α] = --- . . . . . . . . . . (8) p
- One down, one more to go....
- A simplification:
- We can simplify
the expression for
E[Y] in
Equation (2a):
E[Y] = E[ α ] + E[W] - 1 . . . . . . . . . . . . . . . . . . . . (2a) using Equation (8) that we have just discovered above :
E[α] = 1/p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (8)
- Simplified expressin for E[Y]:
E[Y] = (1/p) + E[W] - 1 = E[W] + (1/p) - 1
= E[W] + (1 - p)/p . . . . . . . . . . . . . (9)
There are two unknowns (E[Y] and E[W]) in the equation....
In order to find a solution, we need to find one more relationship between E[Y] and E[W]
- We can simplify
the expression for
E[Y] in
Equation (2a):
-
The other relationship between E[Y] and E[W] can be obtained by counting the number of packets inside a TD period.
The following figure shows the CWND in each transmission round in the ith TD period :
Note:
- Notice that:
- The CWND does
not increase by 1 (one)
after each round of transmissions
This is due to the use of delayed ACKs !!!!
- The CWND will increase by 1 (one) after b rounds of transmissions
- The CWND does
not increase by 1 (one)
after each round of transmissions
- Hence, TCP will transmit:
- Wi-1/2 in round 1
- Wi-1/2 in round 2
- ...
- Wi-1/2 in round
b-1
- Wi-1/2 + 1 in round b
- Wi-1/2 + 1 in round b + 1
- ...
- Wi-1/2 + 1 in round
2b -1
- Wi-1/2 + 2 in round 2b
- And so on...
- Notice that:
- The total number of packets
transmitted inside
ith TD period
is then equal to:
<-------- b terms --------> Wi-1 Wi-1 Wi-1 Y = ----- + ----- + ... + ----- 2 2 2 Wi-1 Wi-1 Wi-1 + ( ----- + 1 ) + ( ----- + 1 ) + ... + ( ----- + 1 ) 2 2 2 Wi-1 Xi-1 Wi-1 Xi-1 Wi-1 Xi-1 + ( ----- + ----- ) + ( ----- + ----- ) + ... + ( ----- + ----- ) 2 b 2 b 2 b + βi ( βi is the number of packet in the partial round)
Wi-1 Wi-1 Wi-1 Xi-1 <==> Y = b ----- + b ( ----- + 1 ) + ... + b ( ----- + ----- ) + βi 2 2 2 b <------------------- Xi/b terms ------------------>
Xi Wi-1 Xi-1 <==> Y = b ----- ----- + b ( 0 + 1 + 2 + .... + ----- ) + βi b 2 b
XiWi-1 Xi-1 <==> Y = ------- + b ( 0 + 1 + 2 + .... + ----- ) + βi 2 b <---- Xi/b terms ---->
XiWi-1 Xi Xi-1 <==> Y = ------- + b (----) ( average{ 0, ----- } ) + βi 2 b b
XiWi-1 Xi Xi-1 <==> Y = ------- + ----- b ( ------ ) + βi 2 b 2 b
XiWi-1 Xi Xi - 1 <==> Y = ------- + ----- -------- + βi 2 b 2 This is almost equal Formula (9) in Padhye's paper ( click here) I think he has a typo in his paper... Or I has some round-off error above In any case, the difference is very small...
XiWi-1 Xi Xi - 1 <==> Y = ------- + ----- -------- + βi 2 2 b
Xi Xi - 1 <==> Y = --- ( Wi-1 + -------- ) + βi 2 b
Xi Wi-1 Wi-1 Xi 1 <==> Y = --- ( ----- + ----- + --- - --- ) + βi 2 2 2 b bUsing Equation (4), we get: Xi Wi-1 1 <==> Y = --- ( ----- + Wi - --- ) + βi . . . . . . . . (10) 2 2 b This is almost Formula (10) in Padhye's paper ( click here) According to Padhye: Xi Wi-1 Y = --- ( ----- + Wi - 1 ) + βi 2 2 The difference is very small !!!
Note:
- I will continue the discussion using Padhye's results
- From Equation (10) and using Theory of Probability, we can compute
the expected value of Y:
E[X] E[W] E[Y] = ------ ( ------ + E[W] - 1 ) + E[β] 2 2
Use Equation (4b) to replace E[X] and we have: b 3 <==> E[Y] = --- E[W] ( --- E[W] - 1 ) + E[β] 4 2
- Estimating E[β]
- There is
no way
to compute E[β]
exactly
Paghye estimated E[β]
- E[β] = average number of packets transmitted in
the last (chopped off) transmission round
- A good estimate
for E[β] is
half the CWND, so
Padhye used:
- E[β] = E[W]/2
- There is
no way
to compute E[β]
exactly
- This will give us the
second relationship
between E[Y] and E[W]:
b 3 E[W] ==> E[Y] = --- E[W] ( --- E[W] - 1 ) + ------ . . . . . . . (12) 4 2 2
- Use Equation (9)
and
Equation (12), and we will obtain:
1 - p E[Y] = ------ + E[W] . . . . . . (9) (See: click here) p b 3 E[W] E[Y] = --- E[W] ( --- E[W] - 1 ) + ------ . . . . . . . (12) 4 2 2
1 - p b 3 E[W] ------- + E[W] = --- E[W] ( --- E[W] - 1 ) + ------ p 4 2 2When you solve this quadratic equation in E[W], (homework problem), you will get:
-------------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) . . . . . . . . (13) 3b \/ 3bp 3b This is formula (13) in Padhye's paper
- We can finally compute the average throughput of TCP...
- First, we have the throughput formula in Equation (5):
E[α] + E[W] - 1 B = -------------------------- . . . . . . . . . . . . (5) 2 (--- E[W] + 1) × RTT b - We found an expression for
E[α]
in Equation (7):
1 E[α] = --- . . . . . . . . . . (7) p - And we have just found an expression for E[W]
in Equation (13):
-------------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) . . . . . . . . (13) 3b \/ 3bp 3b
- Substitute (7) and (12) into Equation (5), and work a long long time,
you will get a formula for
the throughput of TCP Reno
when the loss indications are ONLY Triple Duplicate ACKS:
- For
small value of p we
can ignore the o(1/sqrt(p)) term and get the following
very well-known
formula for TCP throughput:
------- 1 / 3 B(p) = ----- \ / ----- RTT \/ 2bp
- Equation 9:
average number packets sent in one TD period:
E[Y] = average number packets sent in one TD period 1 - p E[Y] = ------ + E[W] . . . . . . (9) (See: click here) pUse Equation (13) to replace E[W] in this equation, and we will have:
- p = probability of packet loss
(See: click here )
- b = number of packets received before sending an ACK
---------------------- / 1-p 2+b / 8(1-p) 2+b 2 E[Y] = ----- + ----- + \ / -------- + ( ----- ) . . . . (14) p 3b \/ 3bp 3b
- p = probability of packet loss
(See: click here )
- Equation 5:
average duration of one TD period
E[A] = average duration of one TD period 2 E[A] = ( --- E[W] + 1) * RTT . . . . . (5) (See: click here) bUse Equation (13) to replace E[W] in this equation, and we will have:
-- ------------------------- --- | 2+b / 2 b (1-p) 2+b 2 | E[A] = RTT * | --- + \ / ----------- + ( ----- ) + 1 | . . . (15) | 6 \/ 3p 6 | -- ---
- These 2 equation will be useful in a sebsequent analysis....
- A similar result has already obtained by
Mahdavi and Floyd when p is small and
b = 1
(See reference [6] in their paper: click here)