A. Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms o
ID: 3863476 • Letter: A
Question
A. Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms on average.
A packet’s link and physical layer headers are always set at the same bitrate and take a
total of 125us to transmit. If a packet is sent with a link layer payload of 1000 bytes at
a bitrate of 1Mbps, what overhead do the physical and link layers introduce? Calculate
overhead as the fraction of the complete packet time taken up by backoff and link/physical layer headers.
.
B. What is the overhead if the packet sent at 1 Mbps is 30 bytes long?
.
C. What is the overhead if a 1500 byte payload is sent at 48Mbps?
Explanation / Answer
A)
Given,
Backoff time is = 1ms and
Link and physical layer headers transmission time is = 125us = 0.125 ms.
Length of the payload is = 1000 bytes = 8000 bits.
Bit Rate for the transmission is = 1Mbps.
Time taken for the transmission of payload is = Length of payload / Bit Rate
= 8 * 103 / 106
= 8 * 10-3
= 8 msecs
Hence, Time taken for the transmission of payload is = 8 msecs.
Overhead time = Backoff time + Link and physical layer headers transmission time.
= 1+0.125 ms
Hence,Overhead time = 1.125 ms.
A Complete Packet is Payload + Overhead.
Now, Complete packet time is = Time taken for the transmission of payload + Overhead time
= 8+1.125 ms
= 9.125
Hence, Complete packet time = 9.125 ms.
Given, Overhead is the fraction of the complete packet time taken up by backoff and
link/physical layer headers.
So Overhead is = overhead time / Complete packet time
= 1.125 / 9.125
= 0.123
Hence Overhead is 0.123 and it is 12.3% of total packet.
B)
Length of the packet is = 30 bytes = 240 bits and
Bit Rate for the transmission is = 1Mbps.
Now, Complete packet time is = Length of packet / Bit Rate
= 240 / 106
= 240 us
Hence, Complete packet time = 240 us.
From above Overhead time = 1.125 ms = 1125 us.
Overhead is = overhead time / Complete packet time
= 1125 / 240
= 4.6875
Hence Overhead is = 4.6875.
C)
Given,
Backoff time is = 1ms and
Link and physical layer headers transmission time is = 125us = 0.125 ms.
Length of the payload is = 1500 bytes = 12000 bits.
Bit Rate for the transmission is = 48Mbps.
Time taken for the transmission of payload is = Length of payload / Bit Rate
= 12 * 103 / 48 *106
= 0.25 * 10-3
= 0.25 msecs
Hence, Time taken for the transmission of payload is = 0.25 msecs.
Overhead time = Backoff time + Link and physical layer headers transmission time.
= 1+0.125 ms
Hence,Overhead time = 1.125 ms.
A Complete Packet is Payload + Overhead.
Now, Complete packet time is = Time taken for the transmission of payload + Overhead time
= 0.25+1.125 ms
= 9.125
Hence, Complete packet time = 1.375 ms.
Given, Overhead is the fraction of the complete packet time taken up by backoff and
link/physical layer headers.
So Overhead is = overhead time / Complete packet time
= 1.125 / 1.375
= 0.82
Hence Overhead is 0.82 and it is 82% of total packet.
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