Delay Bandwidth product
Delay bandwidth product: The number of bits the sender must
transmit before the first bit arrives at the receiver.
The bits in the pipe are said to be “in flight”
If the sender does not fill the pipe (i.e. if he does not send “delay X
bandwidth” worth of data before it waits for an Ack signal), he will
not fully utilize the network.

TCP data throughput : An Example
Assume that data is to travel between src A and dst B,
and that each segment contains the same number of
bytes.
Assume that the time required to send one segment from A to B can be divided
into 4 equal parts (starting with 0th time interval).
Neglect the processing time at A and B.
TCP begins data transmission with a slow-start.
TCP data throughput:
Travel of the first segment:
At the beginning of t = 0th interval, the sender starts
sending the first segment.
At the beginning of t = 1st interval, the segment goes
through x.
Similarly at the beginning of t =2nd and 3rd interval the
segment goes through the points y and z respectively.
At the end of t = 3rd interval, it has been received at B.
At the beginning of t = 4th interval, B will send the Ack.
A
x
y
z
TCP: Travel of first Ack:
At the end of t = 4, 5, 6 the Ack reaches the points z, y and x respectively.
At the end of t = 7, Ack1 reaches A.
A can now send the 2nd segment. The TCP slow-start procedure
would now allow the 3rd segments also to be sent, immediately
after the 2nd, even if Ack2 is yet to be received.
At the beginning of t = 8, A sends the 2nd segment.
At the beginning of t = 9, A sends the 3rd segments.
Similarly when Ack2 and Ack3 are received by A, its congestion window is of 4 segments.
Motion of the ACKs and segments is now explained in the table in the next few slides.
B
TCP: Movement of Segments and Acks
Beginning
of
End of
A sends
B sends
t = 11
B receives the
2nd segment
t = 12
Ack2
t = 13
Ack3
t = 15
t = 16
Comments
B receives the
3rd segment at
the end of t=12
Ack 2 reaches
A
Seg4
t = 16
Ack 3 reaches
A
TCP: Movement of Segments and Acks
(continued)
Beginning End of
of
A sends
t
t
t
t
t
t
t
Seg5
Seg6
Seg7
=
=
=
=
=
=
=
17
18
19
20
21
22
23
B sends
Comments
Ack 4
Ack 5
Ack 6
Ack7
t = 23
Ack 4 reaches A
TCP: Movement of Segments and Acks
(continued)
On receiving Ack 4, Ack 5, Ack 6, Ack 7, A should be able to send 8 segments.
Beginnin End of
g of
t = 24
t = 25
t = 26
A sends
B sends
Comments
Seg8
Seg9
Seg10
TCP: Movement of Segments and Acks
(continued)
Beginnin End of
g of
A sends
t = 27
Seg11
t = 28
Seg12
Ack 8
t = 29
Seg13
Ack 9
t = 30
Seg14
Ack 10
t = 31
Seg15
Ack 11
t=31
B sends
Comments
A receives Ack8
TCP: Movement of Segments and Ack
(continued)
Beginning End of
of
A sends
B sends
t = 32
Seg16
Ack12
A receives
Ack9
t = 32
t = 33
Comments
Seg17
Ack13
t = 33
Bandwidth-delay product:
the TCP example
Thus at t = 31 and all successive times the pipe between the
sender and the receiver is full.
At each unit of time, a segment is received by B and a new
segment is sent by A. The number of data segment, at any time, in
the pipe is equal to the number of Ack segments in the pipe. This
is the ideal steady state of the connection.
Hence the sender should have 8 segments unacknowledged at any
time for maximum throughput.
Question: How big should the advertised window of the receiver be?
Bandwidth-delay product:
Capacity of a pipe in bits = bandwidth (bits/sec)*RTT (sec).
A receives
Ack10
the TCP example:
T1: Raw bit rate = 1.544Mbps
One bit out of 193 is used for SYNC,
So net bit rate = 1.536 Mbps.
Similarly
T3: Raw bit rate = 44.736Mbps
Net bit rate = 44.21Mbps
For RTT = 60ms,
Bandwidth-delay product for T1
= 1.536*60*103
= 92160 bits
= 11520 bytes
the TCP example: (continued):
Bandwidth-delay product for T3
= 44.21*60*103
= 2652600 bits
= 331575 bytes.
This is larger than the TCP window advertisement can be, so TCP window scale
option has to be used.
The capacity of the pipe can be affected by a change in
either the bandwidth or the delay.