To experiment with the UDP protocol and measure its reliability.
Client/Server communication happens a bit differently with UDP
as opposed to TCP. Instead of calling
connect to establish a long-term communication channel,
UDP programs specify the address each time they send data. Likewise,
they receive from any machine that sends data to them. There is no
The server program below waits for a client to send them a message. It then sends 10,000 digits of $\pi$ to that address, and finally a "-" character to mark the end of the transmission.
#!/usr/bin/python3 import socket # host (internal) IP address and port HOST = '10.142.0.3' PORT = 4040 # open the file and read our value of Pi f = open("pi.txt", "r") pi = f.read() # create our socket sock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM) # allow us to reuse an address for restarts sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1) # set the socket host and port number up sock.bind((HOST, PORT)) # wait for a client to connect to us with the "pi" message print("Waiting for a connection...") request, address = sock.recvfrom(1024) print("Sending pi to", address) # send the digits of pi one by one for digit in pi: sock.sendto(digit.encode(), address) sock.sendto(b"-", address) print("Done!") # close connections sock.close()
Notice there is no
accept call, and also the use of
recvfrom in place of the regular
For this lab, you will be testing how reliable UDP actually is - in a typical UDP transmission, how many packets are actually lost?
To do this, you will need to run the server above, as well as a client (provided below). You will then write a little code to compare the value of Pi received from the server to the actual value.
You will then run the programs a couple of times to see how many packets are lost and/or rearranged. You will turn in the results of this analysis.
wgetto download them directly.
Just the length is not enough to see if the value is correct - some digits may have been sent out of order. In order to measure the closeness of two sequences, we can use the Levenshtein distance, which will tell us the number of changes that were made between the two sequences. The algorithm is difficult to implement, so you can use this Python function to compute it (taken from Stack Overflow):
def MED_character(str1,str2): cost = 0 len1 = len(str1) len2 = len(str2) # output the length of other string in case the length of any of the string is zero if len1 == 0: return len2 if len2 == 0: return len1 # initializing a zero matrix accumulator = [[0 for x in range(len2)] for y in range(len1)] # initializing the base cases for i in range(0, len1): accumulator[i] = i; for i in range(0, len2): accumulator[i] = i; # we take the accumulator and iterate through it row by row. for i in range(1,len1): char1 = str1[i] for j in range(1,len2): char2 = str2[j] cost1 = 0 if char1 != char2: cost1 = 2 # cost for substitution accumulator[i][j]=min(accumulator[i-1][j]+1, accumulator[i][j-1]+1, accumulator[i-1][j-1] + cost1 ) cost = accumulator[len1 - 1][len2 - 1] return cost
Call this function on both versions of Pi and print the result of it. This function will take a little bit of time on such large strings (about 1 minute on my machine).
When you're finished, email your analysis to firstname.lastname@example.org. This will include your initial hypothesis, the result of the three executions of the client/server and also your answer to number 10.
Copyright © 2018 Ian Finlayson | Licensed under a Creative Commons Attribution 4.0 International License.