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Recent activity by Prashant Sharma 1
2
answers
1
GATE CSE 1998 | Question: 1.16
In serial communication employing $8$ data bits, a parity bit and $2$ stop bits, the minimum band rate required to sustain a transfer rate of $300$ characters per second is $2400$ band $19200$ band $4800$ band $1200$ band
In serial communication employing $8$ data bits, a parity bit and $2$ stop bits, the minimum band rate required to sustain a transfer rate of $300$ characters per second ...
8.9k
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commented
Jun 27, 2016
Computer Networks
gate1998
computer-networks
communication
serial-communication
normal
out-of-gate-syllabus
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1
answer
2
MadeEasy Test Series: Computer Networks - Flow Control Methods
Bandwidth of a link is 1000 Mbps and round trip time is given as 250 μ sec. If frame size is 500 bits, the utilization (in percentage) of channel when STOP and WAIT ARQ is used is _______.
Bandwidth of a link is 1000 Mbps and round trip time is given as 250 μ sec. If frame size is 500 bits, the utilization (in percentage) of channel when STOP and WAIT ARQ ...
1.4k
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commented
Jun 24, 2016
Computer Networks
made-easy-test-series
computer-networks
flow-control-methods
stop-and-wait
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1
answer
3
Stop and wait protocol
R=transmission rate S=signal speed D=Distance between sender and receiver T=time to create one frame F=number of bits in a frame N=number of data bits in a frame A=number of bits in an acknowledgement F>N Q.1 Calculate the channel utilization My take: (F/R) / (F/R + 2*D/S + A/R ... .2 Calculate the effective data rate. My take: ( (F/R) / (F/R + 2*D/S + A/R + 2*T) ) * (N/F)
R=transmission rateS=signal speedD=Distance between sender and receiver T=time to create one frameF=number of bits in a frameN=number of data bits in a frameA=number of b...
1.1k
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edited
Jun 24, 2016
Unknown Category
stop-and-wait
sliding-window
computer-networks
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1
answer
4
Efficiency in ethernet
Consider a 10 Mbps Ethernet LAN that has stations attached to a 2.5 km long coaxial cable.Given that the transmission speed is 2.3*10^8 m/s, the packet size is 128 bytes out of which 30 bytes are overhead,find the effective transmission rate ... for the calculation of the effective transmission rate I should consider just the date bytes(not the overhead) i.e 128-30=98 bytes.
Consider a 10 Mbps Ethernet LAN that has stations attached to a 2.5 km long coaxial cable.Given that the transmission speed is 2.3*10^8 m/s, the packet size is 128 bytes ...
4.5k
views
edited
Jun 23, 2016
Computer Networks
ethernet
computer-networks
csma-cd
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