CS 4594 Broadband

Switch Analysis

Diagram of Buffered Switch

N inputsN outputs

buffer

buffer

buffer

buffer

buffer

buffer

buffer

buffer

buffer

General Queuing Assumptions

For this NxN switch:

Assume that at each clock cycle, cells arrive evenly on each input, say with probability p.

Assume the probability of each incoming cell on any input going to any particular output is the same.

Then the probability of a cell arriving at input i and going to output j is p/N.

What is the probability that m cells arrive at output j at the same time?

Diagram of Input Buffered Switch

N inputsN outputs

buffer

buffer

buffer

Reference

• Chao, Lam, Oki, Broadband Packet Switching Technologies, Wiley, 2001.

• De Prycker, Asynchronous Transfer Mode, 3rd Edition, Prentice Hall, 1995.

Assumptions for Input-Buffered

• N inputs and N outputs

• Queues are FIFOs

• Of all the cells destined for a particular output, only one cell can pass through the switch at a time

• As soon as a cell is taken from the head of the line (HOL) a cell replaces it (saturation)

Assumptions for Input Buffered Switch

• N = number of inputs.• Ai[m] = number of cells arriving at the heads of

the queues during the mth time slot and destined for output i.

• Bi[m]= number of remaining cells at input buffers destined for output i in the mth time slot.

• Bi[m]= max (0, Bi[m-1]+ Ai[m] –1)• F[m] = number of cells transported through switch

during mth time slot

Equations for Input Queue

kmFki NNk

mFkmA

]1[)/11()/1(

]1[]][Pr[

N

i

i mBNmF1

]1[]1[

N

i

imAmF

1

]1[

1)– [m]A 1]-[mB (0,max ][B iii m

Results for Input Queue

NFB /1

NasBi

,)1(2

)(

0

20

1)1(1

1 N

NB

NB

N

i

i

NF /Poisson Rate:

Steady state:

Results for Input Queue

)1(2

)(1

0

20

0

220

Input Queuing

For saturated switch with an infinite number of inputs Pmax = 0.586

Diagram of Output Buffered Switch

N inputsN outputs

buffer

buffer

buffer

Queuing Assumptions

The probability of a cell arriving at input i and going to output j is p/N.

What is the probability that m cells arrive at output j at the same time?

The Answer

Probability i cells arrive at a certain output queue

= Comb(i, N) (p/N)i (1- p/N)N-i

Here Comb(i, N) is the combination number of how many ways a group of i items can be chosen from a set of N elements.

Markov Chains

• Markov Chain: Description of a system in terms of a set of states. The system satisfies the Markov assumption:

The future state of the system depends only upon its current state.

• Each state is assigned a probability and transition from one state to another is determined by a fixed set of transition probabilities.

Assumptions for Output Buffered• Ak = Prob[A = k] = Comb(k,N) (p/N)k (1-p/N)N-k

• Let Q[m] be the number of cells in the ith queue at the end of the mth time slot

• Q[m] = min(max(0, Q[m-1] + Am – 1), b)

0 1 2 3A0

A1A1

A0 A0 A0

A2 A2A2

A3 A3

A1

A4

A3

A2

A1

A4

A3

Markov Chain for Output Buffered

• Ak = Prob[A = k] = Comb(k,N) (p/N)k (1-p/N)N-k

• P[i,j]=Pr[Q[m]=j | Q[m-1]=i]

0 1 2 3A0

A1A1

A0 A0 A0

A2 A2A2

A3 A3

A1

A4

A3

A2

A1

A4

A3

A4

Assumptions for Output BufferedP[0,0]=A0+A1

P[0,1]=A2

P[0,2]=A3

…P[0,N]=AN+AN+1+AN+2+…+AM

P[1,0]=A0

P[1,1]=A1

P[1,2]=A2

…P[1,N]=AN-1+AN+AN+1+…+AM

P[2,0]=0P[2,1]=A0

P[2,2]=A1

…P[2,N]=AN-2+AN-1+AN+…+AM

……P[N,0]=0…P[N,N-1]=A0

P[N,N]=A1+A2+…+AM

0 1 2 3

A0

A1 A1

A0 A0 A0

A2A2 A2

A3A3

A1

A4

A3

A2

A1

A4

A3

A4

A4

•P[i,j]=Pr[Q[m]=j | Q[m-1]=i]

In and Out

in out equation

A0*Q0 = A0*Q0

0 (A0+A1)*Q0 + A0*Q1 (A0+A1+…)*Q0 A1*Q0 + A0*Q1 + A0*Q0 = Q0

1 A2*Q0 + A1*Q1 + A0*Q2 (A0+A1+…)*Q1 A2*Q0 + A1*Q1 + A0*Q2 = Q1

2 A3*Q0 + A2*Q1 + A1*Q2 + A0*Q3 (A0+A1+…)*Q2 A3*Q0 + A2*Q1 + A1*Q2 + A0*Q3 = Q2

3 A4*Q0 + A3*Q1 + A2*Q2 + A1*Q3 + A0*Q4 (A0+A1+…)*Q3 A4*Q0 + A3*Q1 + A2*Q2 + A1*Q3 + A0*Q4 = Q3

… ... … …

b Ab+1*Q0 + (Ab+Ab+1)*Q1 + …+ (A2+…+Ab+1)*Qb-1 + (A1+…+ Ab+1)*Qb

(A0+A1+…)*Qb Ab+1*Q0 + Ab*Q1 + …+A2*Qb-1 + A1*Qb + A0*Qb-1

+ R= Qb

zzQQARzzQAzQzA b )()()( 001

00

Solving

zzQQARzzQAzQzA b )()()( 001

00

zzQQAzQAzQzA )()()( 0000

)1(*)(*))(( 00 zQAzQzzA

zzA

zQAzQ

)(

)1(*)( 00

Generating Function

Let

xi = Comb(i, N) (p/N)i (1- p/N)N-i

Then

xi zi , for i = 0, 1, 2, . . ., N

= (z (p/N) + (1-p/N))N

Average Queue Size

It can be shown that the average queue size is:

Q = (N-1)/N p2 /(2 (1-p))

And the average waiting time is

W = (N-1)/N p /(2 (1-p))

Graph of Results for N=2, 4, 8, 16

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Central Queuing

• Central Queues behave like output queues according to de Precker