1. In Flow control on data link layer ARQ stands for…
- A. Asynchronous Request Repeat
- B. Automatic repeat request
- C. Automatic request repeat
- D. Asynchronous Repeat Request
- E. None of the above
Ans- B. Automatic repeat request
2. In Stop and Wait ARQ if sender wants to send 600 frames and link probability is 0.2. How many transmissions are require to send these 600 frames to receiver side.
- A. 750
- B. 850
- C. 700
- D. 600
- E. None of the above
Ans-A. 750
Here given values are
Total frames= 600
Link Probability= 0.2
Put these value in below formula to get total transmission
Total Number of Transmission= n(1/1-p)
=> 600(1/1-0.2) = 750
Read Go Back N ARQ MCQs
3. In Stop and Wait ARQ if sender side transmission time is ‘u’ and propagation time is ‘v’ then after what time sender send next packet, if size of packet is equal to size of acknowledgement packet and queuing and processing time is negligible.
- A. 2u+v
- B. 2u+2v
- C. u+v
- D. u+2v
- E. None of the above
Ans-B. 2u+2v
In Stop and Wait ARQ first of all sender takes time to put the frame to link which is called transmission time and this transmission time is given as ‘u’ and this data is propagate from sender to receiver in given time called propagation time which is given ‘v’.
after sending data from sender to receiver sender wait for acknowledgement of receiver
as per given question we have given that size of both sending data frame and acknowledgement is same
so sending of acknowledgment frame has taken equal transmission time and propagation time.
Total time for next sending frame is = sender frame transmission time + Propagation time + Receiver acknowledgment transmission time + Propagation time
Total time for next sending frame is = u+v+u+v =>2u+2v
4. Stop and wait ARQ is working on …
- A. Noiseless channel
- B. Noisy channel
- C. Both A and B
- D. TDM
- E. None of the above
Ans- B. Noisy channel
5. Stop and wait ARQ is used for…
- A. Flow control
- B. Error control
- C. Both A and B
- D. Neither Flow nor Error Control
- E. None of the above
Ans- C. Both A and B
6. In Stop and Wait ARQ a 2000 bit frame to be send from sender to receiver with link bandwidth is 2Mbps and this frame is send over 200km across a data link layer. find out efficiency in % of stop and wait ARQ , if velocity of link is given as 1*108 m/sec.
- A. 36.33 %
- B. 35.36 %
- C. 34.96 %
- D. 33.33 %
- E. None of the above
Ans-D. 33.33 %
Efficiency of Stop and Wait protocol = Useful Time/Total Time
Useful Time= Transmission time
Total Time = Transmission time + Propagation time
Efficiency = TT/TT+Tp
now first we calculate Transmission time which is length of frame/link bandwidth
TT=L/Bw
TT=2000/2*106=> 1 msec
now calculate Propagation time which is
Propagation Time = Distance / Velocity
Put the value Propagation Time = 200*103/ 108 =>2msec
Efficiency = (1/1+2)*100 =>33.33%
7. In Stop and Wait ARQ a 8192 bit frame to be send from sender to receiver with link bandwidth is 1024 Mbps and the round trip time is 16 μsec. find out efficiency(in %) and throughput of stop and wait ARQ.
- A. 50 % and 512 Mbps
- B. 55 % and 1024 Mbps
- C. 60 % and 2048 Mbps
- D. 45 % and 256 Mbps
- E. None of the above
Ans-A. 50 % and 512 Mbps
Efficiency of Stop and Wait protocol = Useful Time/Total Time(Round Trip Time)
Useful Time= Transmission time
Total Time = Transmission time + Propagation time or Total time = Round trip time(RTT)
Efficiency = TT / RTT
now first we calculate Transmission time which is length of frame/link bandwidth
TT=L/Bw
TT=8192/1024*106=> 8 μsec
RTT= 16 μsec
Now put the value of transmission time into efficiency formula
Efficiency = (8 μsec/16 μsec)*100 =>0.5*100=>50%
Throughput = Efficiency*Bandwidth
Throughput = 0.5*1024 Mbps=>512 Mbps
8. In Stop and Wait ARQ if sender sent 1000 frames than how many acknowledgement he already received.
- A. 1000
- B. 1100
- C. 999
- D. 1001
- E. None of the above
Ans-A. 1000
8. In Stop and Wait ARQ if sender wants to send 8 frames and every 3rd packet is lost than how many transmission are possible to send these whole frames.
- A. 9
- B. 11
- C. 10
- D. 12
- E. None of the above
Ans-A. 11
9. Stop and wait ARQ used …
- A. Commutative Acknowledgement
- B. Independent Acknowledgement
- C. Request Acknowledgement
- D. Both A and B
- E. None of the above
Ans- B. Independent Acknowledgement
10. In Stop and Wait ARQ a 2000 Bytes frame to be send from sender to receiver with link bandwidth is 160 Kbps. Acknowledgement frame size is 200 Bytes and link Transmission rate is 16 Kbps at receiver side. The propagation delay is 200msec then find throughput…
- A. 27.66 Kbps
- B. 26.66 Kbps
- C. 25.66 Kbps
- D. 24.66 Kbps
- E. None of the above
Ans- B. 26.66 Kbps
Throughput of Stop and Wait ARQ protocol = length of frame/Total Time(Round Trip Time)
Length of Frame = 2000 Bytes=>2000*8=16000 bits
Total Time = Transmission time(frame) + Propagation time or Total time +Transmission time(Acknowledgement) =
now first we calculate Transmission time of frame which is length of frame/link bandwidth
TT=L/Bw
TT(Frame)=16000/160*103=> 100msec
calculate Transmission time of Acknowledgement frame which is
length of Acknowledgement frame / link bandwidth
TT( Ack)=200*8/16*103=> 100msec
Propagation time = 200 msec
Now put the value of transmission time into Throughput formula
Throughput of Stop and Wait ARQ protocol = length of frame/Total Time(Round Trip Time)
=>16000*8/100+2*200+100 => 26.66
Throughput = 26.66 Kbps
11. In Stop and Wait ARQ a 2000 Bytes frame to be send from sender to receiver with link bandwidth is 106 bps. Acknowledgement frame size is negligible and the channel utilization is 25 % for Stop and Wait ARQ. The Propagation delay of frame is…
- A. 26 msec
- B. 28 msec
- C. 27 msec
- D. 24 msec
- E. None of the above
Ans- D. 24 msec
