Switching
Lecture 1
Switch Architecture
Inter connection between N input links and N
output links
Switch Arbiter
O/p lines
I/p lines
2
1
2
1
3
4
3
4
5
5
O/p adapters
I/p adapters
Switch fabric
Switch Fabric
I/p lines
O/p lines
Cross bar
Switch Types
Single staged
Multi Staged
Input Queued
Packets stored at the input
Switch speed = line speed
Any combination of packets can be transferred across a
switch as long as the input and output terminals do not
overlap (Matching)
1-3, 2-1 allowed, 1-2, 3-2 not allowed, 1-4, 1-5 not
allowed
Output Queued
Packets stored at the output
Switch speed = N * line speed
(speedup = N)
Any N packets intended for an output can be transferred to
the output simultaneously, 1-2, 3-2 allowed
Combined Input Output Queued
Packets stored at both input and output
Speedup between 1 and N
Shared Memory Switch
Packets stored in the switch fabric; every output
line reads the paper as and when they are
transmitted
Switch Performance Metric
Throughput
Rate vector comprising of rate of transmission of
packets across each output (r1,….rN)
Delay Vector
Delay between packet transmission from input to
output
Input Queued Switch
rij is the average number of packets arriving at input
line i for output line j
Let an output line serve one packet each slot
i rij  1 for each j (output contention)
j rij  1 for each i (input contention)
Output Queued switch
j rij  1 for each i (input contention)
i rij  N for each j (output contention)?
Redundant constraint
Combined Input Output Queued
Switch
j rij  1 for each i (input contention)
i rij  k for each j (output contention)
speedup k
Shared memory switch
Constraints just like output queued switch
Less packet loss due to statistical multiplexing of
buffer memory
Output queued switch loses packet if any
single output buffer overflows, even if there
is additional space in other output buffers
Shared memory switches share all buffers
and will overflow only if the combined
buffers overflow
Scheduling for Input Queued
Switch
Choice of matchings determine whether any throughput
can be attained as long as the constraints are satisfied
FIFO scheduling
Every input maintains a queue of packets
Transfer the first packet of each queue subject to
matching constraints
HOL Contention
2
4
1
4
2
3
4
1
2
3
4
Only one packet can be transferred across the switch
4
2
1
4
2
3
4
1
2
3
4
Two packets can be transferred across the switch
Input constraints allow (jj rij  1 )/N close to 1
(rate of transfer of packets across the switch per
line)
Under FIFO scheduling this number is upper
bounded by 0.586
Penalty of HoL blocking
Proof
Lookahead scheduling
First, schedule among the head of line packets
in each input queue
If an input does not transfer packet but has a packet,
see whether the second packet can be scheduled
This can be generalized to considering the first w
packets in each queue
2
4
1
4
2
3
4
1
2
3
4
At the first cut 2-4 is scheduled, then look at the second
packet in queue 1 and schedule 1-2
Lookahead performance
(jj rij  1 )/N increases with increase in w,
but the limiting value for w is strictly less
than 1
Queueing in High-Performance Packet Switching
Hluchyj and Karol, IEEE JSAC December
88, Vol 6, No. 9
Achieving 100 % Throughput in
Input Queued Switch
100 % throughput is said to be attained if the
switch is able to sustain any arrival process as
long as the rate constraints are satisfied,
This will allow (jj rij  1 )/N close to 1
``Achieving 100 % Throughput in Input Queued Switch,’’
Mckeown et al, INFOCOM 95
``Stability properties of constrained queueing systems and
scheduling for maximum throughput in multihop radio
networks,’’ Tassiulas and Ephremides, IEEE Trans of
Automatic Control, Vol. 37, No. 12, pp. 1936-1949, 1992
Maximum weighted matching
based scheduling
Every input maintains separate logical queues for each
output
Weight of each queue is the queue length
Serve the packets from the queues which form a maximum
weighted matching
Weight of a matching is the sum of weights of
the queues in the matching
4
4
2
4
1
4
2
Schedules 1-4
3
4
1
2
3
4
Possible matchings are 1-2, 2-4,
weight = 2
1-4,
weight = 3
1-2,
weight = 1
2-4,
weight = 1
Maximum weighted matching based
scheduling gives priority to long queues,
and also tries to schedule a large number of
queues, but may not always schedule the
largest possible number of queues
Proof for optimality
Lecture 2
Proof for FIFO throughput
Queueing in High-Performance Packet Switching
Hluchyj and Karol, IEEE JSAC December
88, Vol 6, No. 9
Assumptions:
Input queues saturated, that is packets are always waiting at
the input queues
Inputs can be selected with equal probability for each output
Suppose a new packet reaches the HOL position of an input
queue, then its destination is a specific output queue with
equal probability as any other output queue (1/N)
Notations
Bim
Number of packets at the heads of the input
queue destined for the ith output in the mth
slot, but not selected for it.
Aim
Number of packets moving to the heads of
the input queue destined for the ith output in
the mth slot
Fm

Number of packets transmitted through the
switch in the mth slot (expected value: F)
Expected number of packets passing through
the switch per slot per output line (F/N)
Bim = max(0, Bim-1 + Aim –1)
Eqn 1
Aim is binomial (Fm-1, 1/N)
For large N, binomial (Fm-1, 1/N) is Poisson(F/N)
Thus the dynamics of Eqn 1 resembles a queueing system
(M/D/1) where Aim is the arrival process, Poisson(), Bim is
the number of packets in the queue.
From standard M/D/1 result, expected value of Bim is
2/2(1- )
Fm-1 = N - i Bim-1
Dividing both sides by N, and taking expectations
we have  = 1 – expected value of Bi
 = 1 - 2/2(1- )
Solving for ,  = 2 - 2
Proof of optimality of the
maximum weighted matching
algorithm for input queues
Preview of Markov Process
• A sequence of random variables X1 , X2,….,Xn
,….. such that
– Xi+1 is independent of X1,….Xi-1 given Xi
– Pr(Xi+1 = ai+1/ Xi = ai, …., X1 = a1) = Pr(Xi+1 = ai+1/ Xi =
ai )
• Discrete time discrete state markov chain
• So a markov chain evolution can be specified by
– Initial states
– Transition probabilities Pr(Xi+1 = ai+1/ Xi = ai ) = pi,i+1
0.5
1
0
2
1.0
0.5
0.5
0.5
State A communicates with state B if there is a positive probability
path from A to B
A set of states is closed if all states in the set communicate
with each other, and no state in the set communicates to any
state outside the set, e.g., {0, 1}
A state a is open if it communicates to some other state
which does not communicate to a, e.g., {2}
A state has a period d if the process can only return to this state
in intervals which are multiples of d
More precisely, di is the g.c.d. of {k: pi ik > 0},
Where pi ik is Pr(Xk = i/ X0 = i)
0.5
1
0
d0= 1
0.5
0.5
0.5
1.0
1
0
1.0
d0 = 2
Period of any two communicating states are equal
A state is aperiodic if its period is 1
Consider a function of the states f(x)
Ef(x) = x p(x)f(x)
Consider the input queueing system
Let the system have random arrivals, i.e., the
number of arrivals for the different sessions is
random
Arrival process for each input-output pair is i.i.d (independent
and identically distributed)
Number of arrivals of i-j in slot t is independent of the
number of arrivals of any other pair in past or future slots,
and also independent of the number of arrivals of any other
pair in the same slot
Probability that a packet arrives for i-j in any slot t is qij
Under this assumption the queue length process at the nodes
(values of the Bij(t) s for all pairs i-j) constitute a markov process
Under maximum weighted matching, this markov chain
consists of one closed set with periodicity 1 (check it!)
Let f(x) be the total number of packets in the input queues in
state x
Ef(x) = x p(x)f(x)
A scheduling strategy is said to be optimum if it attains
finite expected queue length as long as any other strategy
attains it.
If the system is markov with an aperiodic closed set, and
other open states, and
Suppose we can find a scalar function V(x) of vector x such
that E(V(B(t + 1))- V(B(t)) / B(t) = x)  0 for all states x,
except a finite number of states and V(x) 0 for all states x,
(negative drift condition) Meyn and Tweedie, Book
Then the probability distribution of the markov chain
converges to a probability distribution with finite expectation
Assume that the expected arrival rates satisfy the
rate constraints for the input queued switch,
we will show that the negative drift condition
is satisfied
The function V(x) we choose is ixi2
Notations:
B(t)
Vector of queue lengths in slot t (column vector)
A(t)
Vector of arrivals in slot t
B(t)
Vector of departures in slot t
B(t+1) = B(t) + A(t+1) – D(t+1)
V(B(t + 1))- V(B(t)) = (B(t + 1) - B(t))T (B(t + 1) + B(t))
= (A(t+1) - D(t+1))T (A(t+1) + D(t+1) + B(t))
= (A(t+1) - D(t+1))T (A(t+1) + D(t+1)) + (A(t+1) - D(t+1))T B(t)
E[V(B(t + 1))- V(B(t))/ B(t) = x] =E[(A(t+1) - D(t+1))T (A(t+1) +
D(t+1))/ B(t) = x] + E[(A(t+1) - D(t+1))T B(t)/ B(t) = x]
The first term in the summation can be upper bounded
by a positive constant since the expected arrivals in slot
t + 1 are finite and independent of the queue lengths in
slot t, the expected departure from each pair is at most
1 and at least 0
E[(A(t+1))T B(t)/ B(t) = x] = aTx
a = i i Mi
(using rate constraints) where M is a
matching vector and i i  1,
i  0
aTx = i i weight of matching i under x
 (i i ) weight of maximum weight matching
under x
E[(D(t+1))T B(t)/ B(t) = x] = weight of maximum weight matching
under x
E[(A(t+1) - D(t+1))T B(t)/ B(t) = x]  ((i i ) - 1) weight
of maximum weight matching under x
Depending on state x, this can become as highly negative as
desired
Thus the negative drift condition holds except for a finite
number of states.
Performance Difference Between
Input and Output Queues
Note that input queued switch has additional rate
constraints at the output (as compared to output
queued switches)
For output line stability these constraints will
eventually arise in output queued switches as well.
Thus throughput wise under maximum
weighted matching input and output queued
switches perform similarly.
Delay performances can be different because of the
nature of constraints.
Can an input queued switch emulate an output
queued switch?
Lecture 12, TCOM 799, Fall 01
Lecture 13
Computationally Simple Algorithms
for Maximum possible throughput in
input queued switches
Implementational complexity
Maximum weighted matching will require the arbiter
to know the instantaneous queue lengths and then
computing the maximum weighted matching.
Will there be a loss in throughput if the
scheduling is as per the maximum weighted
matching in a previous slot?
Not, as long as the delay is finite and the
maximum number of arrivals are upper
bounded in a slot
The proof for throughput optimality uses properties of
maximum weight matching in computing the following value
E[(D(t+1))T B(t)/ B(t) = x] = weight of maximum weight matching
under x
If D(t+1) is not a maximum weight matching, then
(D(t+1))T B(t) differs from the weight matching by at
most a constant provided the maximum number of
arrivals in any slot is finite.
The result follows.
Computational complexity
A maximum weighted matching can now be
computed once in a certain interval, and used
throughout the interval without any loss in
throughput.
O notation
Maximum weighted matchings can be computed in
bipartite graphs in O(N3 log N)
Other low complexity matchings
Maximum size matchings
Low throughput for nonuniform arrival rates
Computation can become simpler for maximum weight
matchings with weight no longer the queue lengths
•"A Practical Scheduling Algorithm to Achieve 100%
Throughput in Input-Queued Switches."
Adisak Mekkittikul, and Nick McKeown
IEEE Infocom 98, Vol 2, pp. 792-799, April 1998, San
Francisco.
Choose a maximum weight matching where the
Weight of a pair is now the sum of the queue lengths at
the input and the output ports for the pair.
This is also a maximum size matching.
Simpler algorithms for computing maximum size
matchings can be used to compute this maximum
weight matching in O(N2.5 )
Proof for optimality
Consider a symmetric matrix T (N2 X N2) such that
maxmatching M MTB(t)
constant .
is not upper bounded by a
Then the scheduling policy which schedules the
matching M which maximizes the above every
slot is also throughput optimal
T can be chosen suitably to reduce the complexity of
computing such a optimal matching
Examples of T
Identity matrix
Maximum weight matching with weight = queue
length
Tij = 2 if i = j
1 if i/N = j/N
1 if i mod N = j mod N
0 otherwise
Maximum weight matching with weight of a pair is now the sum of
the queue lengths at the input and the output ports for the pair
Can there be a T to represent maximum size matching?
Proof for optimality
The function V(x) we choose is XTX
V(B(t + 1))- V(B(t)) = (B(t + 1) - B(t))T T (B(t + 1) + B(t))
= (A(t+1) - D(t+1))T T(A(t+1) + D(t+1) + 2B(t))
= (A(t+1) - D(t+1))T T(A(t+1) + D(t+1)) +2 (A(t+1) - D(t+1))T TB(t)
E[V(B(t + 1))- V(B(t))/ B(t) = x] =E[(A(t+1) - D(t+1))T T(A(t+1) +
D(t+1))/ B(t) = x] + 2E[(A(t+1) - D(t+1))T TB(t)/ B(t) = x]
The first term in the summation can be upper bounded
by a positive constant since the expected arrivals in slot
t + 1 are finite and independent of the queue lengths in
slot t, the expected departure from each pair is at most
1 and at least 0
E[(A(t+1))T TB(t)/ B(t) = x] = aT Tx
a = i i Mi
(using rate constraints) where M is a
matching vector and i i  1,
i  0
aT Tx = i i Mi Tx
 (i i ) maxmatching M MTx
E[(D(t+1))T TB(t)/ B(t) = x] = maxmatching M MTx
E[(A(t+1) - D(t+1))T T B(t)/ B(t) = x]  ((i i ) - 1)
maxmatching M MTx
This can be as negative as desired for suitable x,
result holds
Linear complexity algorithms
•“Linear complexity algorithms for maximum throughput in
radio networks and input queued switches"
Leandros Tassiulas
IEEE Infocom 98, Vol 2,April 1998, San Francisco.
Randomized algorithms
Choose the schedule randomly with a probability
distribution which depends on the queue lengths
Maximum weighted matching chooses the matching
M which attains maxmatching M Mx when the queue
length vector is x
The randomized algorithm chooses schedules with a
certain probability, the distribution is such that the
probability of choosing the maximum weighted
matching is at least , where  can be pre-selected as
any number between 0 and 1.
Example choice:
include each pair independently with probability 0.5
if the choice is not a matching dont include any edge
probability of choosing any matching is 2-(N*N)
Let this schedule be I
Let I(t) = I if B(t)I B(t)I(t-1)
= I(t-1) otherwise
The policy will be to schedule I(t) in each slot
Markov process representation is Y(t) = (B(t), I(t))
B(t+1) = B(t) + A(t+1) – I(t)
Proof of optimality
Let IB(t) be the maximum weighted matching vector for queue
lengths B(t)
V(Y) = V1(Y) + V2(Y)
V1(Y) = ibi2
V2(Y) = ((IB-I)T B)2
V1(Y(t + 1))- V1(Y(t)) = (B(t + 1) - B(t))T (B(t + 1) + B(t))
= (A(t+1) - I(t))T (A(t+1) + I(t) + 2B(t))
= (A(t+1) - I(t))T (A(t+1) + I(t)) +2 (A(t+1) - I(t))T B(t)
(A(t+1) - I(t))T B(t) = (A(t+1) – IB(t) )T B(t) +
(IB(t) – I(t))T B(t)
E[(A(t+1) - IB(t) )T B(t)/ Y(t) = Y ]  ((i i ) - 1) wt of maximum
Matching under B(t)
 (1/N)(i i ) - 1) V1(Y)
E[(IB(t) - I(t+1))T B(t)/ Y(t) = Y] =  V2(Y)
E[(A(t+1) - I(t))T (A(t+1) + I(t))/ Y(t) ] = constant
E[V1(Y(t + 1))- V1(Y(t))/ Y(t) ]  2(1/N)((i i ) - 1) V1(Y) +
2 V2(Y) + constant
E[V2(Y(t + 1))/ Y(t) = Y] = 0. P[I(t+1) = IB(t+1) ] +
E[V2(Y(t + 1))/ Y(t)=Y, IB(t+1)  I(t+1) ] P[I(t+1)  IB(t+1) ]
(1-) E[((IB(t+1) – I(t+1))T B(t+1))2 / Y(t), IB(t+1)  I(t) ]
E[((IB(t+1) – I(t+1))T B(t+1))2 / Y(t), IB(t+1)  I(t+1) ] =
E[((IB(t+1) – I(t+1))T (B(t) + A(t+1) – I(t))2 / Y(t), IB(t+1) 
I(t+1) ]
= E[((IB(t+1))T B(t) - (I(t+1))T B(t) + (IB(t+1) – I(t+1))T (A(t+1) –
I(t)))2 / Y(t), IB(t+1)  I(t+1) ]
(IB(t+1))T B(t)  (IB(t))T B(t)
(I(t+1))T B(t) I(t)T B(t)
E[((IB(t+1))T B(t) - (I(t+1))T B(t) + (IB(t+1) – I(t+1))T
(A(t+1) – I(t)))2 / Y(t), IB(t+1)  I(t+1) ]
 E[((IB(t))T B(t) - (I(t))T B(t) + (IB(t+1) – I(t+1))T (A(t+1)
– I(t)))2 / Y(t), IB(t+1)  I(t+1) ]
= E [((IB(t) – I(t))T B(t))2 / Y(t), IB(t+1)  I(t+1) ] +E [ (IB(t+1) –
I(t+1))T (A(t+1) – I(t)))2 / Y(t), IB(t+1)  I(t+1) ]
+ 2E[((IB(t) – I(t))T B(t)) (IB(t+1) – I(t+1))T (A(t+1) – I(t))) /
Y(t), IB(t+1)  I(t+1) ]
E [((IB(t) – I(t))T B(t))2 / Y(t), IB(t+1)  I(t+1) ] =
((IB(t) – I(t))T B(t))2
= V2(Y)
E [ (IB(t+1) – I(t+1))T (A(t+1) – I(t)))2 / Y(t), IB(t+1)
 I(t+1) ]  constant
E[((IB(t) – I(t))T B(t)) (IB(t+1) – I(t+1))T (A(t+1) –
I(t))) / Y(t), IB(t+1)  I(t+1) ]  constant ((IB(t) – I(t))T
B(t))
= constant  V2(Y)
E[((IB(t+1) – I(t+1))T B(t+1))2 / Y(t), IB(t+1)  I(t+1) ] 
V2(Y) + constant + constant  V2(Y)
E[V2(Y(t + 1))/ Y(t), IB(t+1)  I(t+1) ] P[I(t+1)  IB(t+1) ]
(1-) (V2(Y) + constant + constant  V2(Y))
E[V2(Y(t + 1))/ Y(t) = Y] (1-) (V2(Y) + constant +
constant  V2(Y))
E[V2(Y(t + 1)) - V2(Y(t )) / Y(t) = Y] -V2(Y) + (1-)
(constant + constant  V2(Y))
E[V(Y(t + 1))- V(Y(t))/ Y(t) ]  2(1/N)((i i ) - 1) V1(Y) V2(Y) + (2-const.) V2(Y) + constant
Shared Memory Switches
Resource management
constraints
Packets are stored in the switch fabric
As soon as packets arrive at the input, they are read
in the switch memory and stored there.
Each output line serves packets intended for it as
and when it is available.
There are N logical queues
Output lines serve packets independent of each other.
Scheduling among the outputs is not an issue
Different queues share the same physical memory.
So memory management rather than scheduling is
the issue
How does memory management
affect performance?
Performance metric: Throughput or packet drop
Let N = 2
Suppose the entire switch memory consists of packets
for output 1
Packets cleared from the memory at rate 1 per slot.
If the memory had packets for both outputs, packets
would be cleared at rate 2
Load balancing reduces packet drop!
Memory can be managed to balance the load
Memory management options
When a packet arrives:
(a) It can be accepted
(b) It can be rejected
(c) It can be accepted while dropping some other
packet (pushout)
The objective is to choose the optimal course of action so
as to minimize packet drops
Some architectures do not allow pushout!
Optimal memory management in
presence of pushout
Optimal Buffer Sharing
I. Cidon, L. Georgiadis, R. Guerin, A. Khamisy
IEEE JSAC, Vol. 13, No. 7, September 1995
Optimal Strategy for N = 2
Optimal memory management strategy accepts packets
whenever the buffer has space without any push outs
If the buffer is full, then the arrival for output port j is accepted
pushing out one for the other port, if the number of packets for
port j is below a certain threshold, j = 1,2
The thresholds for different ports can be different, but their
sum equals the total memory B
Can the queue lengths for both ports be above their respective
thresholds?
If the service rates are equal, then the threshold is lower
than B/2 for the one with the higher arrival rate
Optimal strategy for arbitrary N
Known for identical arrival and transmission rates for all ports.
Accept a packet if the buffer is not full.
If the buffer is full, reject any arrival for the largest
queue, accept arrivals for any other packet while
dropping packets from the largest queue
Proofs involve markov decision process
Lecture 4
Optimal policy when push out is
not allowed
Sharing memory optimally
G. Foschini, B. Gopinath,
IEEE transactions on Communications, Vol 31, No. 3,
March 1983
Load balancing still reduces the packet drop.
Since push out is not allowed, load can be balanced by
rejecting new arrivals of overloaded queues even when
the buffer has space
Optimal Strategy for N = 2
Accept a packet for port j if
(a) Buffer has space
(b) Number of packets waiting for port j is less
than a threshold mj
The policy reserves B – m1 memory units for
output 2 and B – m2 memory units for output 1
Optimal Strategy for N = 3
Accept a packet for port j if
(a) Buffer has space
(b) Number of packets waiting for port j is less
than a threshold mj
(c) ij xij  mij even after accepting the packet
The policy reserves memory units for individual
outputs as also the combinations
Proof for optimality
In general, every policy will have a set of states 
and will admit packets only if it does not move out of
the set.
The objective will be to find the set which
reduces the blocking.
The system can be modeled by a continuous time
markov chain if arrivals are Poisson and departures
exponential, with the state vector consisting of queue
lengths of individual queues
System Model
Queue j has Poisson arrivals at the rate j exponential
service at rate j
Utilization j = j / j
Steady state distribution (a,…. ) = 1a…. N../ a..
1a…. N….
Reversibility
Lecture 8
Optimal memory management in
presence of pushout
Queue j has Poisson arrivals at the rate j exponential service at rate
j
Loss minimization is equivalent to throughput maximization
Whenever a packet is served, system gets a reward of 1 unit.
The objective is to maximize the total reward
System state: queue length vector (x1 x2 ….. xN)
Control action: memory management
ej A vector with 1 in the ith position and 0 in the rest
I(x) = 0 if x = 0
= 1 otherwise
J(x1 x2 ….. xN) = j I(xj  0) +  i max(J(x), J(x + ei), J(x + ei ej))
if i xi  B
= j I(xj  0) +  i max(J(x), J(x + ei - ej)) if
 i xi = B
Optimal Strategy for N = 2
Properties for the cost function J(x, y)
Monotonicity and Boundedness in x:
0  J(x+1, y) – J(x, y)  1, 0  x  B - 1
Monotonicity and Boundedness in y:
0  J(x, y+1) – J(x, y)  1, 0  y  B - 1
Concavity along x:
J(x+1, y) – J(x, y)  J(x, y) - J(x-1, y), 1  x  B - 1
Concavity along y:
J(x, y+1) – J(x, y)  J(x, y) - J(x-1, y), 1  x  B - 1
Concavity along the line x + y = b, 2  b  B :
J(x+1, y-1) – J(x, y)  J(x, y) - J(x-1, y+1), 1  x  B – 1, 1
 y B–1
Derivation of Optimal strategy
from the properties
Monotonicity and boundedness implies that the optimal
decision is to accept a packet for a port if the buffer has
space
The concavity property on the straight line x + y = B
implies that the function J(x, y) has either a unique
maxima on the line or two consecutive maximas. These
correspond to the replacement thresholds.
Nature of Thresholds
If 1  2 1 = 2 J(x, y) J(y, x), y  x
Approximate computation of
thresholds
J(x, y)  f(x,y) = d – c11x – c22y
d = (1 + 2)/(1- )
cj = j/(1 - (1 - j (1- j )))
j roots of quadratic equation
The maximum of f(x, B-x) gives the thresholds.
Optimal Strategy for N for
symmetric arrival rates
Properties for the cost function J(x, y)
Monotonicity and Boundedness :
0  J(x+ ej ) – J(x)  1,
Symmetry
J(x) = J(y), where y is a permutation of x
Balancing:
J(x)  J(x + ei - ej) if ith component of x is less than the
jth component
Drop from the longest queue