Sliding Window Protocol Question 1

Question

Consider  1 Mbps error-free line , The maximum frame size is 1000 bits .New packets are generated about 1 second apart . Timeout interval is 10m/sec if the special acknowledgement timer were eliminated unnecessary timout would occur . How any times average msg would have been transimitted ?

a) Only once

b) Twice

C) thrice

d) None

Comments

Ans D) ?

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What is the logic ?

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What is the meaning of Timeout interval is 10m/sec?

Answer 1

• Transmission delay (Tt) = L / B = 1000 bits / 10^6 bits per sec = 1 msec.
• After the packet is put on the link, the time out timer is started which is 10 msec long.
• The next packet is transmitted after 1 sec = 1000 msec.
• If no acknowledgment is received within 10 msec, the packet will be retransmitted.
• We have been asked how many times the average message be transmitted i.e. how many retransmissions are possible.
• Retransmission occurs or not depends on the propagation delay (Tp).
• If Tp is more, time out will occur and retransmission will take place but if Tp is less, then there will be no time out.
• Since propagation delay (Tp) is not given in the question, therefore we can not say anything.

Answer 2

I am still not sure what you meant by "time interval is 10m/sec". If I am correct, it is a typing error. That should be 10msec (milli second).

10^6 bits in 1 sec

therefore bits that can be transferred before the timeout is = 10^6 * 10/1000 = 10000 bits

and avg  msg size is (0+1000)/2=500 bits 

number of frames (or msgs) that can be transferred are 10000/500=20

D) NONE

Comments

timeout interval strats after sending the packet from sender to reciever..what is meant by these many bits can be sent in timout interval??

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in the above question he asked to calculate number of average retransmissions of a frame and he gave bandwidth 10⁶                

how did u calculated average message size i didnot get u 

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explaination given like this :
Suppose A sent B a frame that arrived correctly, but there was no reverse
traffic. After a while A would time out and retransmit. B would notice that
the sequence number was incorrect, since the sequence number is below
FrameExpected. Consequently, it would send a NAK, which carries an
acknowledgement number. Each frame would be sent exactly two times.

Answer 3

Given BW=1Mbps   and frame size =1000 bits

in 1 second -----> 10^6 bits  We can transfer that much bits

and a timeout occurs  10msec( typo in question) so in 10 mesc------> 10^4 bits we can transfer that much bits

(10^4)/10^3 = 10  frame can be transmitted

Option :D" NONE

Answer 4

I think to make sure transmissions occur RTT>Timeout.RTT=2*Tp.As no information is given regarding Tp,so we cannot say how many transmissions will occur in avg.

Answer 5

Suppose A sent B a frame that arrived correctly, but there was no reverse traffic.

After a while A would time out and retransmit. B would notice that the sequence number was incorrect, since it would be below FrameExpected.

 

Consequently, it would send a NAK, which carries an acknowledgement number. Each frame would be sent exactly two times.