G22.2262: Data Communications and Network Design

Homework 2.

 

 

Due Date: Thursday, December 9, 1999

 

This homework has 4 questions. Question 1 is worth 85 points, question 2 is worth 9 points, and questions 3 and 4 are worth 3 points each.

  1. The ARP implementation is very straightforward. The ARP code can be called from the IP layer or from the Ethernet layer. For this assignment, you must write a program that implements ARP!

    2. You will assume you have an IP address of 198.11.109.34 and Ethernet address of 08:00:20:98:A1:35.

    Your program should take a single parameter, and that parameter is the name of a file. The file contains lines of the following form:

    IP:IP_Address:Broadcast_flag:Data
    or
    Ether:Data

    Each line represents data passed down to ARP from IP or up to ARP from Ethernet. After reading each line of the file, you must show the ARP table and any packets that ARP tells Ethernet to send. You must assume that every line arrives 15 seconds apart, and you must delete entries from the ARP table if they have not been accessed IP in 20 minutes.

     

    For example, the file can contain the following lines:

    IP:198.11.109.08:0:IPdatagram1
    IP:198.11.109.15:0:IPdatagram2
    Ether:0001 0800 06 04 0002 08 00 20 11 F1 A6 198.11.109.08 08 00 20 98 A1 35 198.11.109.34
    IP:198.11.109.255:1:IPdatagram3
    IP:198.11.109.15:0:IPdatagram4
    IP:198.11.109.15:0:IPdatagram2
    Ether:0001 0800 06 04 0002 08 00 20 15 19 87 198.11.109.15 08 00 20 98 A1 35 198.11.109.34
    IP:198.11.109.08:0:IPdatagram1
    IP:198.11.109.08:0:IPdatagram5
    Ether:0001 0800 06 04 0001 08 00 20 61 73 A4 198.11.109.18 00 00 00 00 00 00 198.11.109.34
    Ether:0001 0800 06 04 0001 08 00 20 A1 C3 4F 198.11.109.19 00 00 00 00 00 00 198.11.109.32
    IP:198.11.109.19:0:IPdatagram6
    Ether:0001 0800 06 04 0001 08 00 20 A1 C3 4F 198.11.109.19 00 00 00 00 00 00 198.11.109.32
    Ether:0001 0800 06 04 0002 08 00 20 A1 C3 4F 198.11.109.19 08 00 20 98 A1 35 198.11.109.34

     

    Your program should be invoked as follows:

     

    $ myarp filename

    Your program should print the following output:

     

    Arp table:
    198.11.109.08 0
    Tell Ethernet to send:
    FF FF FF FF FF FF 08 00 20 98 A1 35 0806
    0001 0800 06 04 0001 08 00 20 98 A1 35 198.11.109.34 00 00 00 00 00 00 198.11.109.08
     
    Arp table:
    198.11.109.08 0
    198.11.109.15 0
    Tell Ethernet to send:
    FF FF FF FF FF FF 08 00 20 98 A1 35 0806
    0001 0800 06 04 0001 08 00 20 98 A1 35 198.11.109.34 00 00 00 00 00 00 198.11.109.15
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 0
    Tell Ethernet to send:
    08 00 20 11 F1 A6 08 00 20 98 A1 35 0800
    IPdatagram1
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 0
    Tell Ethernet to send:
    FF FF FF FF FF FF 08 00 20 98 A1 35 0800
    IPdatagram3
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 0
    Tell Ethernet to send:
    Nothing
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 0
    Tell Ethernet to send:
    FF FF FF FF FF FF 08 00 20 98 A1 35 0806
    0001 0800 06 04 0001 08 00 20 98 A1 35 198.11.109.34 00 00 00 00 00 00 198.11.109.15
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    Tell Ethernet to send:
    08 00 20 15 19 87 08 00 20 98 A1 35 0800
    IPdatagram2
    08 00 20 15 19 87 08 00 20 98 A1 35 0800
    IPdatagram4
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    Tell Ethernet to send:
    08 00 20 11 F1 A6 08 00 20 98 A1 35 0800
    IPdatagram1
     
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    Tell Ethernet to send:
    08 00 20 11 F1 A6 08 00 20 98 A1 35 0800
    IPdatagram5
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    198.11.109.18 08 00 20 61 73 A4
    Tell Ethernet to send:
    08 00 20 61 73 A4 08 00 20 98 A1 35 0806
    0001 0800 06 04 0002 08 00 20 98 A1 35 198.11.109.34 08 00 20 61 73 A4 198.11.109.18
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    198.11.109.18 08 00 20 61 73 A4
    Tell Ethernet to send:
    Nothing
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    198.11.109.18 08 00 20 61 73 A4
    198.11.109.19 0
    Tell Ethernet to send:
    FF FF FF FF FF FF 08 00 20 98 A1 35 0806
    0001 0800 06 04 0001 08 00 20 98 A1 35 198.11.109.34 00 00 00 00 00 00 198.11.109.19
     
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    198.11.109.18 08 00 20 61 73 A4
    198.11.109.19 08 00 20 A1 C3 4F
    Tell Ethernet to send:
    80 00 20 A1 C3 4F 08 00 20 98 A1 35 0800
    IPdatagram6
     
    Arp Table:
    198.11.109.08 08 00 20 11 F1 A6
    198.11.109.15 08 00 20 15 19 87
    198.11.109.18 08 00 20 61 73 A4
    198.11.109.19 08 00 20 A1 C3 4F
    Tell Ethernet to send:
    Nothing

     

     

    Here are some notes:

    You must assume every line in the file comes in 15 seconds apart. So, if an entry is not accessed in 20 minutes, you must delete it from the ARP table. You should reset the timer every time it is accessed by an "IP" line, but not by an "Ether" line.

    The the ARP table is completely empty, you should display lines that say:

    Arp Table:
    Empty

    If you are waiting for an ARP reply from machine X, and IP asks you to send a datagram to X, only send another ARP request if the IP datagram is in the queue for machine X.

    Have your output look exactly like that in the example!

     

  2. There is a series of documents called "Request for Comments" that describe the TCP/IP protocols. Please read RFC 1868, "ARP Extension - UNARP". How would you modify your program in question 1 to implement UNARP? Be specific!

  3. Let's assume the sender's ethernet address in the ARP data gets corrupt during transmission in such a way that the corruption passes the Ethernet CRC. Would that cause any problems? Explain.

  4. Suppose a machine on the 198.11.109.0 network wanted to send a packet to a machine with an IP address of 199.13.45.34. If the machine with IP address 199.13.45.34 did not exist, explain how sending the packet generates excess broadcast traffic on the 199.13.45.0 network