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1. The Prob lem

Consider an error-free 64kbps satellite channel used to send 512-byte data frames in one direction, with

very short acknowledgements coming back the other way. What is the maximum throughput for window

sizes of 1, 7, 15 and 127? We assume that the earth-satellite propagation time is 270 msec.

2. The Solut ion

2.1 Elementary QuantitiesWhat we know:

the bandwidth: b=64kbits/second the frame length: l=512 bytes=4096 bits the propagation time: tprop=270msec

We can also compute the time required to transmit one frame on the channel:

the transmission time: 5.621024*64

4096

b

lttrans msec.

2.2 The Analytical SolutionLet us consider a window size of N and see that happens:

t=0

t=ttrans

t=N* ttrans

t=tprop

t= ttrans+tprop

t= ttrans+2*tprop

After this, all the frames are acknowledged one after the other, and the process re-starts. So, one

can see that the on the line there are always N or N-1 frames in transit. The sender sends N frames during

the time interval t= ttrans+2*tprop. Thus the throughput is:

proptrans ttlNthroughput*2

*

However, there is a condition to be met in order for things to work the way they have been

described: when the sender receives the first acknowledge, it should have send its entire window, that is,

proptranstrans tttN *2* ,

or equivalently:

S

R

ACK1 ACK2 ACKN

S

R

N 2

S

R

N-1N 1

S

R

N-1N 1

S

R

1

S

R

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trans

prop

t

tN *21

Hence, the general formula for the throughput is given by:

trans

prop

trans

prop

proptrans

t

tNifb

t

tNif

tt

lN

throughput

*21

*21

*2

*

2.3. The Numerical Solution

In our case, the condition becomes: N 9, that is, the top throughput is reached for N=9.Hence, the numeric solution for our problem is:

N Throughput Efficiency

1 6798 bits/second 10.37%

7 47586 bits/second 72.61%

15 64 kbits/second 100%

127 64 kbits/second 100%

3. Conclus ion

We can conclude that when using a sliding window protocol on an error-free channel, the throughput

grows linearly with the size of the window until the bandwidth of the transmission line is reached.

Test questions

1. A channel has a bit rate of 3.8kbits/seconds and a propagation delay of 20milllseconds. For whatrange of frame size does stop and wait gives an efficiency of 60%

2. On February 22, 2008 pakistan telecom authority issue an order to Pakistan ISPs to block acces tothree P address belonging to youtube 208.65.153.238, 208.65.153.253, 208.65.153.251 one

operator noted that these address were belonging to the same /24 prefix

a. What should have been done by youtube to avoid this problemb. Explain what kind of solution would you propose to improve the security of inter-domain routing

throughput

windows size

b

trans

prop

r

t*21

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3. A. explain the different between broadcast and collision domainB. what is congestion and why does it occur

4. a Give the xtics of an ideal LAN

b Explain how MPLS have overcome the problem found in the conventional IP networks and how are

routes discover or selected

Channel has data rate of 4 Kbps and propagation delayof 20mswhat range of frame size do stop and wait givean efficiency of atleast 50?

Answer:

You need to use two formulas - firstly a= propagation delay/transmission time. We'lldenote this by tprop and ttrans.

tpropis given in the question, it is 20 x 10-3

ttransis Distance/Data Rate

so ttrans = L / 4 x 103

ttrans = 80/L

(becasue the powers of ten cancel each other out so you just multiply 20 x 4.we use powers of ten because we need the same format as the propogation delay

a = 80/L

2. Use the formula 1/1+2a >= .5 = 1/1+ 160/L>=.5

First 1>= .5(1+160/L)

second 1>= .5 + 80/L

third 80/L >=1x .5

fourth .5 >= 80/LL >= 80/.5

L >= 160