is it $2.78 \%$?

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A channel has bit rate of $1 \hspace{0.1cm} Mbps$ and propagation delay of $270 \hspace{0.1cm} msec$ .Frame size is of $125 \hspace{0.1cm} bytes$ . Acknowledgement is always piggibacked onto data frames .Four bit sequence no. Is used .ignore header size,then what is maximum achievable channel utilization for GO-BACK-N?

- $1.48\%$
- $0.18\%$
- $2.95\%$
- $2.78\%$

0

so first know what is channel utilization:- Channel utilization means how much fraction of channel your are using which is available.

we have given a frame of size 125B or 1000bits. Using n=4, using GO-BACK N we can send atmost 15 frame.

so, during round trip time ( 2*propogation time i.e tp )=270*2=540msec we are sending only 15 frame whose transmission time is 1msec (1000bits/1Mbps).

Total time used = 15*1msec=15msec

Total available time=540msec

utilization = Total time used/ total available time

15/540=0.02777 => approx. 2.78 %

we have given a frame of size 125B or 1000bits. Using n=4, using GO-BACK N we can send atmost 15 frame.

so, during round trip time ( 2*propogation time i.e tp )=270*2=540msec we are sending only 15 frame whose transmission time is 1msec (1000bits/1Mbps).

Total time used = 15*1msec=15msec

Total available time=540msec

utilization = Total time used/ total available time

15/540=0.02777 => approx. 2.78 %

0 votes

Transmission time $T_{t}=\frac{125 \times 8}{10^{6}}\text{sec=1 msec}$

Propagation time=$T_{p}=270 msec$

Window size$=2^{n}-1,\text{n=No.of bits in seq no. }=15$

Efficiency=$\frac{\text{window size}}{1+2a} ,a=\frac{T_p}{T_t}$

$\frac{15}{1+2 \times 540} \times 100\approx 2.78 \%$

Propagation time=$T_{p}=270 msec$

Window size$=2^{n}-1,\text{n=No.of bits in seq no. }=15$

Efficiency=$\frac{\text{window size}}{1+2a} ,a=\frac{T_p}{T_t}$

$\frac{15}{1+2 \times 540} \times 100\approx 2.78 \%$

0 votes

so first know what is channel utilization:- Channel utilization means how much fraction of channel your are using which is available.

we have given a frame of size 125B or 1000bits. Using n=4, using GO-BACK N we can send atmost 15 frame.

so, during round trip time ( 2*propogation time i.e tp )=270*2=540msec we are sending only 15 frame whose transmission time is 1msec (1000bits/1Mbps).

considering ack is piggybacked:- total frame now 16 in duration of 541 msec.

Total time used = 16*1msec=15msec

Total available time=541msec

utilization = Total time used/ total available time

16/541=0.02957 => approx. 2.95 %

we have given a frame of size 125B or 1000bits. Using n=4, using GO-BACK N we can send atmost 15 frame.

so, during round trip time ( 2*propogation time i.e tp )=270*2=540msec we are sending only 15 frame whose transmission time is 1msec (1000bits/1Mbps).

considering ack is piggybacked:- total frame now 16 in duration of 541 msec.

Total time used = 16*1msec=15msec

Total available time=541msec

utilization = Total time used/ total available time

16/541=0.02957 => approx. 2.95 %

0 votes

option (D)

as we know in case of Go Back N ,N+1=2^n,where N is window size,n is number of bits for sequence number field

n=4 hence N=15

we know that link utilization or channel utilization or sender utilization=N/1+2a,where a=PT/TT

a=270*10^-3*1*10^6/125*8=270

efficiency=N/1+2a=15/1+2*270=.002772=2.772%(nearly 2.78%)

as we know in case of Go Back N ,N+1=2^n,where N is window size,n is number of bits for sequence number field

n=4 hence N=15

we know that link utilization or channel utilization or sender utilization=N/1+2a,where a=PT/TT

a=270*10^-3*1*10^6/125*8=270

efficiency=N/1+2a=15/1+2*270=.002772=2.772%(nearly 2.78%)

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