Lecture 9
CS3516:
These slides are generated from those made
available by the authors of our text.
Computer
Networking: A Top
Down Approach
6th edition
Jim Kurose, Keith Ross
Addison-Wesley
March 2012
Introduction 1-1
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-2
TCP flow control
application may
remove data from
TCP socket buffers ….
… slower than TCP
receiver is delivering
(sender is sending)
application
process
application
TCP
code
IP
code
flow control
receiver controls sender, so
sender won’t overflow
receiver’s buffer by transmitting
too much, too fast
OS
TCP socket
receiver buffers
from sender
receiver protocol stack
Transport Layer 3-3
to application process
TCP flow control
RcvBuffer

receiver “advertises” free
buffer space by including
rwnd value in TCP header
of receiver-to-sender
segments
 RcvBuffer size set via
socket options (typical default
is 4096 bytes)
 many operating systems
autoadjust RcvBuffer


rwnd
buffered data
free buffer space
TCP segment payloads
receiver-side buffering
sender limits amount of
unacked (“in-flight”) data to
receiver’s rwnd value
guarantees receive buffer
will not overflow
Transport Layer 3-4
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-5
Connection Management
before exchanging data, sender/receiver “handshake”:


agree to establish connection (each knowing the other willing
to establish connection)
agree on connection parameters
application
connection state: ESTAB
connection variables:
seq # client-to-server
server-to-client
rcvBuffer size
at server,client
network
Socket clientSocket =
newSocket("hostname","port
number");
application
connection state: ESTAB
connection Variables:
seq # client-to-server
server-to-client
rcvBuffer size
at server,client
network
Socket connectionSocket =
welcomeSocket.accept();
Transport Layer 3-6
the well-known “two-army problem”
The red armies
together can defeat the
blue army, but
separately the blue
army wins.
Red army
Blue army
Red army
Q: how can the 2 red armies agree on an attack time?
 Fact: the one who sends a message does not know
whether the message is delivered
 Basic rule: one cannot send an ACK to acknowledge an
ACK – it goes on forever.
4//26/05
CS118 7
TCP 3-way handshake
client state
server state
LISTEN
LISTEN
choose init seq num, x
send TCP SYN msg
SYNSENT
received SYNACK(x)
indicates server is live;
ESTAB
send ACK for SYNACK;
this segment may contain
client-to-server data
SYNbit=1, Seq=x
choose init seq num, y
send TCP SYNACK
SYN RCVD
msg, acking SYN
SYNbit=1, Seq=y
ACKbit=1; ACKnum=x+1
ACKbit=1, ACKnum=y+1
received ACK(y)
indicates client is live
ESTAB
Syn
Ack
Transport Layer 3-8
TCP: closing a connection

client, server each close their side of connection
 send TCP segment with FIN bit = 1

respond to received FIN with ACK
 on receiving FIN, ACK can be combined with own
FIN

simultaneous FIN exchanges can be handled
Transport Layer 3-9
TCP: closing a connection
client state
server state
ESTAB
ESTAB
clientSocket.close()
FIN_WAIT_1
FIN_WAIT_2
can no longer
send but can
receive data
FINbit=1, seq=x
CLOSE_WAIT
ACKbit=1; ACKnum=x+1
wait for server
close
FINbit=1, seq=y
TIMED_WAIT
timed wait
for 2*max
segment lifetime
CLOSED
can still
send data
LAST_ACK
can no longer
send data
ACKbit=1; ACKnum=y+1
CLOSED
Fin
Ack
Transport Layer 3-10
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-11
Principles of congestion control
congestion:




informally: “too many sources sending too much
data too fast for network to handle”
different from flow control!
manifestations:
 lost packets (buffer overflow at routers)
 long delays (queueing in router buffers)
a top-10 problem!
Transport Layer 3-12
Causes/costs of congestion: scenario 1


lout
Host A
unlimited shared
output link buffers
Host B
R/2
delay

two senders, two
receivers
one router, infinite
buffers
output link capacity: R
no retransmission
throughput:
lout

original data: lin

lin R/2
maximum per-connection
throughput: R/2

lin R/2
large delays as arrival rate, lin,
approaches capacity
Transport Layer 3-13
Causes/costs of congestion: scenario 2


one router, finite buffers
sender retransmission of timed-out packet
 application-layer input = application-layer output: lin =
lout
 transport-layer input includes retransmissions : l‘in lin
lin : original data
l'in: original data, plus
lout
retransmitted data
Host A
Host B
finite shared output
link buffers
Transport Layer 3-14
Causes/costs of congestion: scenario 2
lout
idealization: perfect
knowledge
 sender sends only when
router buffers available
R/2
lin : original data
l'in: original data, plus
copy
lin
R/2
lout
retransmitted data
A
Host B
free buffer space!
finite shared output
link buffers
Transport Layer 3-15
Causes/costs of congestion: scenario 2
Idealization: known loss

packets can be lost,
dropped at router due
to full buffers
sender only resends if
packet known to be lost
lin : original data
l'in: original data, plus
copy
lout
retransmitted data
A
no buffer space!
Host B
Transport Layer 3-16
Causes/costs of congestion: scenario 2

packets can be lost,
dropped at router due
to full buffers
sender only resends if
packet known to be lost
R/2
when sending at R/2,
some packets are
retransmissions but
asymptotic goodput
is still R/2 (why?)
lout
Idealization: known loss
lin : original data
l'in: original data, plus
lin
R/2
lout
retransmitted data
A
free buffer space!
Host B
Transport Layer 3-17
Causes/costs of congestion: scenario 2


packets can be lost, dropped
at router due to full buffers
sender times out prematurely,
sending two copies, both of
which are delivered
R/2
lin
l'in
timeout
copy
A
when sending at R/2,
some packets are
retransmissions
including duplicated
that are delivered!
lout
Realistic: duplicates
lin
R/2
lout
free buffer space!
Host B
Transport Layer 3-18
Causes/costs of congestion: scenario 2


packets can be lost, dropped
at router due to full buffers
sender times out prematurely,
sending two copies, both of
which are delivered
R/2
when sending at R/2,
some packets are
retransmissions
including duplicated
that are delivered!
lout
Realistic: duplicates
lin
R/2
“costs” of congestion:


more work (retrans) for given “goodput”
unneeded retransmissions: link carries multiple copies of pkt
 decreasing goodput
Transport Layer 3-19
Causes/costs of congestion: scenario 3



four senders
multihop paths
timeout/retransmit
Host A
Q: what happens as lin and lin’
increase ?
A: as red lin’ increases, all arriving
blue pkts at upper queue are
dropped, blue throughput g 0
lin : original data
l'in: original data, plus
lout
Host B
retransmitted data
finite shared output
link buffers
Host D
Host C
Transport Layer 3-20
Causes/costs of congestion: scenario 3
lout
C/2
lin’
C/2
another “cost” of congestion:
 when packet dropped, any “upstream
transmission capacity used for that packet was
wasted!
Transport Layer 3-21
Approaches towards congestion control
two broad approaches towards congestion control:
end-end congestion
control:



no explicit feedback
from network
congestion inferred
from end-system
observed loss, delay
approach taken by
TCP
network-assisted
congestion control:

routers provide
feedback to end systems
 single bit indicating
congestion (SNA,
DECbit, TCP/IP ECN,
ATM)
 explicit rate for
sender to send at
Transport Layer 3-22
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
4.5 routing algorithms
link state
distance vector
hierarchical routing
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-23
TCP congestion control: additive increase
multiplicative decrease
approach: sender increases transmission rate (window
size), probing for usable bandwidth, until loss occurs
 additive increase: increase cwnd by 1 MSS every
RTT until loss detected
 multiplicative decrease: cut cwnd in half after loss
AIMD saw tooth
behavior: probing
for bandwidth
cwnd: TCP sender
congestion window size

additively increase window size …
…. until loss occurs (then cut window in half)
time
Transport Layer 3-24
TCP Congestion Control: details
sender sequence number space
cwnd
last byte
ACKed

sent, notyet ACKed
(“inflight”)
last byte
sent
sender limits transmission:
TCP sending rate:
 roughly: send cwnd
bytes, wait RTT for
ACKS, then send
more bytes
rate
~
~
cwnd
RTT
bytes/sec
LastByteSent< cwnd
LastByteAcked

cwnd is dynamic, function
of perceived network
congestion
Transport Layer 3-25
TCP Slow Start
when connection begins,
increase rate
exponentially until first
loss event:
Host B
RTT

Host A
 initially cwnd = 1 MSS
 double cwnd every RTT
 done by incrementing
cwnd for every ACK
received

summary: initial rate is
slow but ramps up
exponentially fast
time
Transport Layer 3-26
TCP: detecting, reacting to loss



loss indicated by timeout:
 cwnd set to 1 MSS;
 window then grows exponentially (as in slow start)
to threshold, then grows linearly
loss indicated by 3 duplicate ACKs: TCP RENO
 dup ACKs indicate network capable of delivering
some segments
 cwnd is cut in half window then grows linearly
TCP Tahoe always sets cwnd to 1 (timeout or 3
duplicate acks)
Transport Layer 3-27
TCP: switching from slow start to CA
Q: when should the
exponential
increase switch to
linear?
A: when cwnd gets
to 1/2 of its value
before timeout.
Implementation:


variable ssthresh
on loss event, ssthresh
is set to 1/2 of cwnd just
before loss event
Transport Layer 3-28
Summary: TCP Congestion Control
duplicate ACK
dupACKcount++
L
cwnd = 1 MSS
ssthresh = 64 KB
dupACKcount = 0
slow
start
timeout
ssthresh = cwnd/2
cwnd = 1 MSS
dupACKcount = 0
retransmit missing segment
dupACKcount == 3
ssthresh= cwnd/2
cwnd = ssthresh + 3
retransmit missing segment
New
ACK!
new ACK
cwnd = cwnd+MSS
dupACKcount = 0
transmit new segment(s), as allowed
cwnd > ssthresh
L
timeout
ssthresh = cwnd/2
cwnd = 1 MSS
dupACKcount = 0
retransmit missing segment
timeout
ssthresh = cwnd/2
cwnd = 1
dupACKcount = 0
retransmit missing segment
New
ACK!
new ACK
cwnd = cwnd + MSS (MSS/cwnd)
dupACKcount = 0
transmit new segment(s), as allowed
.
congestion
avoidance
duplicate ACK
dupACKcount++
New
ACK!
New ACK
cwnd = ssthresh
dupACKcount = 0
dupACKcount == 3
ssthresh= cwnd/2
cwnd = ssthresh + 3
retransmit missing segment
fast
recovery
duplicate ACK
cwnd = cwnd + MSS
transmit new segment(s), as allowed
Transport Layer 3-29
TCP throughput

avg. TCP thruput as function of window size, RTT?
 ignore slow start, assume always data to send

W: window size (measured in bytes) where loss occurs
 avg. window size (# in-flight bytes) is ¾ W
 avg. thruput is 3/4W per RTT
avg TCP thruput =
3 W
bytes/sec
4 RTT
W
W/2
Transport Layer 3-30
TCP Futures: TCP over “long, fat pipes”



example: 1500 byte segments, 100ms RTT, want
10 Gbps throughput
requires W = 83,333 in-flight segments
throughput in terms of segment loss probability, L
[Mathis 1997]:
. MSS
1.22
TCP throughput =
RTT L
➜ to achieve 10 Gbps throughput, need a loss rate of L
= 2·10-10 – a very small loss rate!

new versions of TCP for high-speed
Transport Layer 3-31
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
4.5 routing algorithms
link state
distance vector
hierarchical routing
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-32
Interplay between routing, forwarding
routing algorithm
local forwarding table
dest address output link
address-range 1
address-range 2
address-range 3
address-range 4
routing algorithm determin
end-end-path through netw
forwarding table determin
local forwarding at this ro
3
2
2
1
IP destination address in
arriving packet’s header
1
3 2
Network Layer 4-33
Graph abstraction
5
u
2
2
1
graph: G = (N,E)
v
x
3
w
5
z
3 1
1
y
2
N = set of routers = { u, v, w, x, y, z }
E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w
aside: graph abstraction is useful in other network contexts
P2P, where N is set of peers and E is set of TCP connectio
Network Layer 4-34
Graph abstraction: costs
5
u
2
v
2
1
x
3
w
c(x,x’) = cost of link (x,x’)
e.g., c(w,z) = 5
5
z
3 1
1
y
2
cost could always be 1, or
inversely related to bandwidth,
or inversely related to
congestion
cost of path (x1, x2, x3,…, xp) = c(x1,x2) + c(x2,x3) + … + c(xp-1,xp)
key question: what is the least-cost path between u and z ?
routing algorithm: algorithm that finds that least cost path
Network Layer 4-35
Routing algorithm classification
Q: global or decentralized
information?
global:
 all routers have complete
topology, link cost info
 “link state” algorithms
decentralized:
 router knows physicallyconnected neighbors, link
costs to neighbors
 iterative process of
computation, exchange of
info with neighbors
 “distance vector” algorithms
Q: static or dynamic?
static:
 routes change slowly over
time
dynamic:
 routes change more
quickly
 periodic update
 in response to link
cost changes
Network Layer 4-36
Lecture 9 outline
3.5 connection-oriented
transport: TCP




segment structure
reliable data transfer
flow control
connection management
4.5 routing algorithms
link state
distance vector
hierarchical routing
3.6 principles of congestion
control
3.7 TCP congestion control
Transport Layer 3-37
A Link-State Routing Algorithm
Dijkstra’s algorithm


net topology, link costs
known to all nodes
 accomplished via “link state
broadcast”
 all nodes have same info

computes least cost paths
from one node (‘source”)
to all other nodes

 gives forwarding table for
that node

notation:
 c(x,y): link cost from
iterative: after k
iterations, know least cost
path to k dest.’s

node x to y; = ∞ if not
direct neighbors
D(v): current value of
cost of path from source
to dest. v
p(v): predecessor node
along path from source to
v
N': set of nodes whose
least cost path definitively
known
Network Layer 4-38
Dijsktra’s Algorithm
1 Initialization:
2 N' = {u}
3 for all nodes v
4
if v adjacent to u
5
then D(v) = c(u,v)
6
else D(v) = ∞
7
8 Loop
9 find w not in N' such that D(w) is a minimum
10 add w to N'
11 update D(v) for all v adjacent to w and not in N' :
12
D(v) = min( D(v), D(w) + c(w,v) )
13 /* new cost to v is either old cost to v or known
14 shortest path cost to w plus cost from w to v */
15 until all nodes in N'
Network Layer 4-39
Dijkstra’s algorithm: example
D(v) D(w) D(x) D(y) D(z)
Step
N'
u
0
uw
1
uwx
2
3 uwxv
4 uwxvy
5 uwxvyz
p(v)
p(w)
p(x)
p(y)
7,u 3,u 5,u ∞
5,u 11,w
6,w
11,w
6,w
10,v
p(z)
∞
∞
14,x
14,x
12,y
9
notes:


construct shortest path tree by
tracing predecessor nodes
ties can exist (can be broken
arbitrarily)
x
5
4
7
8
3
u
w
y
3
7
2
z
4
v
Network Layer 4-40
Dijkstra’s algorithm: another example
Step
0
1
2
3
4
5
N'
u
ux
uxy
uxyv
uxyvw
uxyvwz
D(v),p(v) D(w),p(w)
2,u
5,u
2,u
4,x
2,u
3,y
3,y
D(x),p(x)
1,u
D(y),p(y)
∞
2,x
D(z),p(z)
∞
∞
4,y
4,y
4,y
5
u
2
v
2
1
x
3
w
5
z
3 1
1
y
2
Network Layer 4-41
Dijkstra’s algorithm: example (2)
resulting shortest-path tree from u:
v
w
u
z
x
y
resulting forwarding table in u:
destination link
v (u,v)
x (u,x)
y (u,x)
w (u,x)
z (u,x)
Network Layer 4-42
Dijkstra’s algorithm, discussion
algorithm complexity: n nodes



each iteration: need to check all nodes, w, not in N
n(n+1)/2 comparisons: O(n2)
more efficient implementations possible: O(nlogn)
oscillations possible:

e.g., support link cost equals amount of carried traffic:
A
1
D
1
B
0
0
0
1+e
C
e
e
initially
2+e
D
0
C
0
B
1+e 1
0
1
A
0
given these costs,
find new routing….
resulting in new costs
D
A
0
1
C
2+e
B
0
1+e
2+e
D
A
0
B
1+e 1
0
C
0
given these costs,
given these costs,
find new routing….
find new routing….
resulting in new costs resulting in new costs
Network Layer 4-43
The End is Near!