****Please show all work needed to find answers******* Q1. TCP and Throughput a.
ID: 3815226 • Letter: #
Question
****Please show all work needed to find answers*******
Q1. TCP and Throughput
a. For a 10 Mbits/s (bps) Internet and a maximum packet size of 1,518bytes, what is the transmission time?
b. What kind of throughput do we expect with a typical RTT (Round Trip Time) 40 msec, and a window size of 64 Kbytes?
c. What is the optimum window size for a 10 Gbps (G is 109) link between Chicago and New York with 25 msec round trip latency?
Q2. Parity bit error checking
a. Derive the probability of error for 3-bit packet using a parity bit as a method of error checking. List all the cases that the parity bit won’t be able to detect an error when it’s in fact occurring vs. the cases that the parity bit algorithm would detect an error, and compute out of all the possible combinations, what is the probability of parity bit error checking failing?
b. Extend this analysis to an n-bit packet and find the probability of an error in parity checking as n tends to infinity.
Q3. Checksum
a. Explain how come the bigger the divisor in a checksum the better the check? Show with an example.
Q4. Cyclic Redundancy Check (CRC)
a. For a CRC generator of 1011 find the CRC check for the sequence 10111101
b. CRC is claimed to detect bursts of errors up to n, where n is the number of bits in the generator. Give an example of 4-bit burst (consecutive errors) in the above sequence and how the CRC check is able to detect it.
Explanation / Answer
Q1. TCP and Throughput
a) For a 10 Mbits/s (bps) Internet and a maximum packet size of 1,518bytes, what is the transmission time?
Answer
Given, Bandwidth of an internet (bps) = 10 Mbits/sec and
Maximum packet size = 1518 bytes
Transmission time --> Time taken for a packet to transmit from souce to destination.
We know that,
Packet transmission time = Packet size / Bandwidth
= (1518 * 8 bit) / (10 * 10^6 bits/s) [ Since 1 byte = 8 bits and 1 Mbit = 10^6 bits ]
= 12144 / 10^7 bits/sec
= 1.2144 milli seconds [Since 1 milli/sec = 10^(-3) sec]
Here "^" indicates power for example 10^6 is 10 power 6.
b) What kind of throughput do we expect with a typical RTT (Round Trip Time) 40 msec, and a window size of 64 Kbytes?
Answer
Given, Round Trip Time = 40msec
Window size = 64 Kbytes
Throughput --> Number of instructions executed in time
we know that
Throughput = Size of Window / Round Trip Time
= 64 Kbytes / 40msec
= 64 * 10^3 bytes / 40 * 10^(-6) sec [Since 1 kbytes = 10^3 bytes and 1 msec = 10^(-6) ]
= (64/4) * 10^8 bytes/sec
= 16 * 10^8 bytes/sec
= 1600 Mega bytes/sec [Since 1 mega byte = 10^6 bytes]
c. What is the optimum window size for a 10 Gbps (G is 109) link between Chicago and New York with 25 msec round trip latency?
Answer
Given, Throughtput = 10 Gbps
Round trip latency = 25msec
Optimun window size = ?
Optimun window size = Throughtput * Round trip latency
= 10 Gbps * 25 msec
= (10 * 10^9 bps) * (25 * 10^(-6) sec) [Since 1 Giga byte = 10^9 bytes and 1 msec = 10^(-6)]
= 10^10 * 25 * 10^(-6) bytes/sec
= 25 * 10^4 bytes/sec
= 250 Kbps [Since 1 kilo byte = 10^3 bytes]
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