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RARD - Design and Analysis of Switching Systems - Solved Exam, Exams of Design and Analysis of Algorithms

Main points of this past exam are: Steady State Probabilities, Random Arrival, Random Departure, Queue Length, Arrival Probability, Departure Probability, Cell Discard Probability, Queue Carrying Traffic, Average Data Rate, Exceeding

Typology: Exams

2012/2013

Uploaded on 03/23/2013

sarath
sarath 🇮🇳

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Final Exam Solutions
May 3, 1999
1. (15 points) Give expressionsfor the steady state probabilities of a Random Arrival, Random Departure
(RARD) queue with a queue length of 20, cell arrival probability
:
2 and cell departure probability
:
5.
(
0
)=
:
5
1
,
:
2521
:
75
,
:
5
and
(
j
)=(
:
25
j
=:
5
)
(
0
)
.
Give an expression for the virtual cell discardprobability for a queue carrying traffic from 30 bursty
sources, each with a peak data rate of 8% of the link rate and an average data rate equal to 0.5% of
the link rate.
12 such sources can be on without exceeding the link rate and the probability that any given sources
is on is 1/16. So the virtual cell discard probability is
29
X
i
=
12
29
i
!
(
1
=
16
)(
15
=
16
)
29
,
i
(
:
08
(
i
+
1
)
,
1
)
=
(
:
08
(
i
+
1
))
The IBA model and virtual cell discard probability can be used to estimate the cell loss for bursty
sources. Which is more accurate in each of the following cases? Explain why.
50 bursty sources with peak rate equal to 5% of the link rate, an average rate equal to .5% of the
link rate all sendingdata to a queuelarge enough tohold one average size burst.
Virtual cell discard probability is more accurate in this case. In general, it produces accurate
results when the peak rate of the sources is much smaller than the link rate and the buffer is
small, as is true here. The IBA model works poorly in these situations.
10 bursty sources with peak rate equal to 50% of the link rate, an average rate equal to 5% of
the link rate all sending data to a queuelarge enough to hold 30 average size bursts.
Here the IBA model is better. It gives good results when the buffer is much larger than the
burst size and the peak rate is close to the link rate. The virtual cell discard probability is a poor
estimate for the actual cell discard probability in thiscase.
1
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Final Exam Solutions

May 3, 1999

  1. (15 points) Give expressions for the steady state probabilities of a Random Arrival, Random Departure (RARD) queue with a queue length of 20, cell arrival probability :2 and cell departure probability :5.

 ( 0 ) = ^1 : 25 : 215

: 75 :^5

 and  (j ) = (: 25 j^ =: 5 ) ( 0 ).

Give an expression for the virtual cell discard probability for a queue carrying traffic from 30 bursty sources, each with a peak data rate of 8% of the link rate and an average data rate equal to 0.5% of the link rate. 12 such sources can be on without exceeding the link rate and the probability that any given sources is on is 1/16. So the virtual cell discard probability is

X^29

i= 12

i

( 1 = 16 )( 15 = 16 )^29 i (: 08 (i + 1 ) 1 )=(: 08 (i + 1 ))

The IBA model and virtual cell discard probability can be used to estimate the cell loss for bursty sources. Which is more accurate in each of the following cases? Explain why.

 50 bursty sources with peak rate equal to 5% of the link rate, an average rate equal to .5% of the

link rate all sending data to a queue large enough to hold one average size burst. Virtual cell discard probability is more accurate in this case. In general, it produces accurate results when the peak rate of the sources is much smaller than the link rate and the buffer is small, as is true here. The IBA model works poorly in these situations.

 10 bursty sources with peak rate equal to 50% of the link rate, an average rate equal to 5% of

the link rate all sending data to a queue large enough to hold 30 average size bursts. Here the IBA model is better. It gives good results when the buffer is much larger than the burst size and the peak rate is close to the link rate. The virtual cell discard probability is a poor estimate for the actual cell discard probability in this case.

  1. (10 points) Consider the IP routing table shown below.

prefix next hop

  • 3 10* 5 010* 8 1011* 11 00101* 13 100101 9 101100* 12 010110* 7 0100110* 4 1011011* 10 10101111* 7 01001101* 6

What would be the next hop for a packet whose address started with the bits 1011001? 12 Show a binary trie that could be used to represent this routing table.







 







































 





















 













  1. (15 points) Consider an ATM switch using a Benes network B 512 ; 8 with static routing, together with binary cell replication and recycling for multicast. If all ports are used for both external traffic and for recycling, what speed advantage is needed to ensure that there is no blocking. Assume that 20% of the outgoing traffic is multicast and that the external links operate at 2.4 Gb/s and the maximum virtual circuit rate is 150 Mb/s.

For this network to be nonblocking for unicast, we need a speed advantage of ( 3 ( 1 +  )=d)( 1 + (d

1 )(k 1 )) + B =. Here d = 8 , k = 3 ,  = : 2 and B = = 1 = 16 so the required speed advantage is

Suppose that the maximum speed advantage that can be implemented in the technology you are using is 4. If you implement multicast by dedicating some of the 512 ports for recycling, how many external ports can the system have?

h =  +( 1 = )(B = ) n = (: 2 =(: 2 + 4 : 0625 )) 512  25 , so we can have 512 25 = 487 external

ports in this system.

6. (5 points) The table below shows the values of a;b (x) for all values of x from 0 to ab 1. What are

the values for a and b?

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 a;b (x) 0 2 4 6 8 10 12 14 1 3 5 7 9 11 13 15

Well the product of a and b is clearly 16, so the possible choices are a = 2 , b = 8 , a = 4 , b = 4 and a = 8 , b = 2. The last of these is the correct choice.

  1. (15 points) Give an expression the blocking probability for the following networks using Lee’s method. Assume each external link and internal link can carry just one virtual circuit at a time. Let p denote the probability that each network input is busy. C (^203) ; 4 ; 6

( 1 ( 1 2 p= 3 )^2 )^6

X 4 ; 3 (X 2 ; 3  X 3 ; 4 ) X 3 ; 2

If p  3 = 4 we have [ 1 ( 1 4 p= 3 )( 1 2 p= 3 )( 1 8 p= 9 )]^3. Otherwise, the blocking probability is

1 (according to Lee’s method). B 27 ; 3

1 ( 1 p)^2

1 ( 1 ( 1 p)^2 )^3

  1. (10 points) Assume you are using B 27 ; 3 as a contend-and-repeat network in which the first two stages do random traffic distribution. Give an expression for the maximum throughput for this network.

n

h

1 ( 1 = 3 )( 1 ( 2 = 3 )^3 )

 3 io^3

  1. (10 points) The figure shown below shows a copy network and related components that can be used in conjunction with a sorting network. The input cells at left are shown with the fanout and multicast index fields (MI). The letters are just labels for identifying the different cells. Show the cells that make it to the fifos just before the final sort & route section, after two operational steps. Write the letters for the cells in the appropriate places on the figure.

0, IDQRXW   

  



DGGHU FRQF DGGHU

FRS\ UDQJH

FRS\

                    

URXWLQJ WDEOHV

          

              

VRUW

URXWH

G

F

J F

J E

I E

H E

H D

N D      

 

D E G H I K

J

F ^2











k

q

Z

U

p

Assume the system operates as described in class. That is, the fanout sums are computed top to bottom and the copy-range block computes the copy network outputs directly, without any regard for positional unfairness.

  1. (10 points) The following diagram shows a sorting network with 16 inputs and outputs. Draw boxes around the different merge networks that make up the sorting network and label them M 16 , M 8 , M 4 , M 2. For each of the M 4 networks put an arrow to indicate the direction in which its sorting elements sort the values (the arrow should point from largest values to smallest values). Assume that the overall sorting network delivers the smallest values to the top outputs and the largest values to the bottom outputs.

0

0

0

0

0

0

0

0