Computer Networks:
Switching and Queuing
Ivan Marsic
Rutgers University
Chapter 4 – Switching and Queuing Delay Models
Switching and Queuing Delay
Models
Chapter 4
Topic:
Packet Switching in Routers
 Router Architecture
 Forwarding Table Lookup
 Switching Fabric Design
 How Queuing Happens
Why Routers?
• (Recall Chapter 1)
• Avoid needing direct link between all hosts
(totally connected graph)
• Redundancy: If one path fails, find an
alternative path
Routing Delays
Source
Destination
Source
Router
Destination
transmission
time
propagation
time
Time
router
delay
detail
(a)
transmission
time (2nd link)
router
processing
time
(b)
Router Components
Input port
Output port
Input port
Output port
Network layer protocol
Network port
(bidirectional)
Link layer protocol
Link layer protocol
Switching
fabric
How Router Forwards Packets
2
3
Network Layer
Link & Physical
Layers
4
Services offered to incoming packets:
1
1
Receiving and storing packets
2
Forwarding decision for packets
3
Moving packets from input to output port
4
Transmission of packets
Services to Incoming Packets
1
Receiving and storing packets
Forwarding decision for packets
2
3
Moving packets from input to output port
4
Transmission of packets
Distribution of Protocol Layers
L1
L1
L1
Network
Key:
L3 = End-to-End Layer
L2 = Network Layer
L1 = Link Layer
L2
L1
L1
Router B
Router A
End system
L1
L3 L2 L1
2 L3
L1 L
L1 L2
L1
L1
1
L2 L
L1
L1
Router C
Note that router has a single (common) Network Layer protocol,
but each connection has a dedicate Link Layer protocol
End system
Forwarding Algorithms
Routing
function
Forwarding
algorithm
Unicast routing
Unicast routing
with Types of Service
Multicast routing
Longest prefix match
on destination address
Longest prefix match
on destination address
+ exact match on
Differentiated Services
field of IPv4 header
Longest match on
source address
+ exact match on
source address,
destination address,
and incoming interface
Router Architectures
CPU
packets
CPU
Memory
NFE
Processor
Memory
NFE
Processor
Line Card
#1
Line Card
#4
Line Card
#1
CPU
Memory
CPU
Memory
Line Card
#4
Line Card
#2
Line Card
#5
Line Card
#2
CPU
Memory
CPU
Memory
Line Card
#5
Line Card
#3
Line Card
#6
Line Card
#3
CPU
Memory
CPU
Memory
Line Card
#6
(a)
(b)
First generation router
Second generation router
CPU
Memory
packets
Line Card
#1
Fwd
Engine
Fwd
Engine
Line Card
#4
Line Card
#2
Fwd
Engine
Fwd
Engine
Line Card
#5
Line Card
#3
Fwd
Engine
Fwd
Engine
Line Card
#6
(c)
Third generation router
Switching via Memory / via Bus
Input port
CPU
Forwarding & Routing
processor
Memory
Network layer
Output port
First generation router
Link layer
Line card
Line card
packets
System bus
Input port
Output port
NFE processor
NFE processor
Routing processor
CPU
Memory
Network layer
packets
Link layer
Line card
Line card
Second generation router
System bus
Crossbar Switch Fabric
Input
N x N switching elements
allows N simultaneous packets
switched (in the best case
when all packets going to
different outputs)
Output
Goal: Reduce # Switching Elements
• System bus (in 1st and 2nd generation arch’s)
allows only one packet switched at a time
• Crossbar allows up to N packets switched at
a time
• Something in the middle? (+cheaper!)
Banyan Switch Fabric
1
Packet
with tag 1
Input port
tag
0
1
Input port
tag
Output port
tag
0
1
(b)
(a)
8x8 Banyan has only 12
switching elements
(while 8x8 crossbar
requires 64)
But, much greater
likelihood of collisions…
Input port
tag
Output port
tag
00
00
01
01
10
10
11
11
Output port
tag
000
000
001
001
010
010
011
011
100
100
101
101
110
110
111
111
Packet
with tag 100
100
(c)
Packet
with tag 001
001
Reducing Collisions
• (Show slide with a collision example)
• Collisions can be reduced if packets are
ordered on input ports by their output port
number
• The router cannot choose the ordering of
arriving packets, but we can insert a sorting
hardware between the input network ports
and the switching fabric …
Batcher Network
Comparator
Input
numbers
Output
numbers
7
5
5
Low
3
3
High
5
(b)
(a)
Input
list
9
3
L
4
H
5
4
4
1
7
4
L
H
L
3
5
2
4
6
2
3
3
4
5
2
6
5
1
3
7
2
5
6
6
1
7
7
7
5
7
1
4
5
9
H
Output
list
9
3
4
4
3
3
L
H
2
4
(c)
3
H
5
6
5
L
7
7
9
Batcher-Banyan Network
Batcher
sort
network
Trap
network
Shuffle
exchange
network
Banyan
network
Why Batcher-Banyan Network
Packets
with tags 1, 0, 0, 2
1
0
0
2
Empty input
Batcher sorter
0
1
0
0
1
0
2
0
0
0
1
1
2
Trap
network
Banyan network
0
0
1
1
2
2
This figure is meant to illustrate why a concentrator is needed,
because otherwise the gap in the input sequence will cause
collision in the Banyan, but the example does not work for a
4x4 network -- need an 8x8 network example!!!!
2
3
Topic:
Router Delays
& Queuing Models
 Where & Why Queuing Happens
 Little’s Law
 Queuing Models


M / M / 1, M / M / 1/ m, M / G / 1
Networks of Queues
Delay Components in Forwarding
Time
First bit received
Reception delay =
Input port
Fwd decision
queuing delay
t xI
Last bit received
Forwarding decision delay = tf
Fabric traversal
queuing delay
Switch fabric traversal delay = ts
Switch fabric
Transmission
queuing delay
First bit transmitted
Output port
Transmission delay =
t xO
Last bit transmitted
Road Intersection Analogy
Head of Line
Blocking
Car experiencing
head-of-line
blocking
Where & Why Queuing Happens
• At input ports, Head Of Line queuing
• At output ports, if output link is “too
narrow” (low data rate) for incoming traffic
• Inside switch fabric, if collision occurred
An Input-queued Switch
Input ports
Switch fabric
Arbiter
Output ports
General Service Model
Server
System
Arriving
customers
in
Departing
customers
Interarrival
time = A2  A1
out
Input
sequence
C1
C2
Ci
Time
A1
Output
sequence
C3
A2
A3
C1
Service
time = X1
Ai
C2
X2
Waiting
time = W 3
C3
X3
Ci
Xi
Simple Queuing Model
System
Queue
Arriving packets
Queued packets
Server
Serviced packet
Source
Source
Arrival rate
=  packets per second
Service rate
=  packets per second
Departing
packets
Delay Time
Total delay of customer i = Waiting + Service time
Ti = W i + Xi
Server
Service rate 
Arrival rate 
customers
unit of time
customers
unit of time
NQ
Customers in queue
Customer
in service
N
Customers in system
Why Queuing Happens?
C3
Time
11
:00
C2
10
:41
10
:00
Time
11
:00
Server
C1
10
:29
C3
10
:00
10
:0
10 5
:0
10 9
:13
C1
C2
Departure times sequence
10
:17
Arrival times sequence
Arrival Sequences
Sequence 1
(arrivals)
(departures)
Sequence 2
t (s)
0
1
2
3
t (s)
Server
Server
4
0
1
Service
delay
2
3
4
Queuing
delay
Sequence 3
Sequence 4
t (s)
0
1
2
3
4
t (s)
0
1
2
3
4
Cumulative
Number of arrivals, A(t)
Number of departures, B(t)
Birth and Death Processes
A(t)
T2
N(t)
B(t)
T1
Time
N(t)

Time
Little’s Law
• Average number of packets in the system =
arrival rate  average time that packet
spends in the system
• N=T
• Problem
– We would like to know more, such as what are
the probabilities of finding different number of
customers on arrival, etc.
Intuition for the Balance Principle
Room 0
Room n – 1
Room n
Room n + 1
t(n+1  n)
t(n  n+1)
 t(n  n+1)  t(n+1  n) 1
See: Global Balance Equations
Transition Probability Diagram
1  
1    
1    

0

1

1    

2

1    
1    

n
n1

n1

M/G/1 Example
A2 = 9:30
A4 = 10:10
A1 = 9:05
(a)
A3 = 9:55
A6 = 11:05
A5 = 10:45
X1 = 30
X2 = 10
X3 = 75
X4 = 15
X5 = 20
X6 = 25
1
2
3
4
5
6
9:00 9:05
9:35
9:45
9:55
11:10
Customer 5
service time = 15 min
Customer 6
arrives at 11:05
(c)
Residual service time r()
(b)
11:25
11:45
Time
Customer 4
service time = 20 min
Server
Customers 4 and 5
waiting in queue
Service time for customer 3
X3 = 75 min
75
Customer 6
arrives at 11:05
Time 
5
9:55
11:10
Ni = 2
Customer 3 in service;
Residual time = 5 min
Residual service time r()
Expected Residual Time
X1
Time 
X1
X2
XM(t)
t