Computer Networks - TDC 463
September 27, 1999
Homework Set #2
Due October 4, 1999 by the start of class

1) Browse RFC 894, what value is placed into the type field for an Ethernet
   frame to designate IP is the next protocol encapsulated in the next
   higher layer?

   Answer: 0x0800

2) Using Ethernet, you can take two computers and connect them directly
   together using a cross-over cable.  No hub or switch is needed.
   Can you do the same thing in token ring or do you need a token ring
   hub?  How or why?

   Answer: No.  Token ring requires a "insertion" process and phantom
           voltage which each station would not be able to provide for
           the other.

3) List three Ethernet switch products from three different vendors.
   Please provide a link to each of their product's page.

   Answer: Cisco Catalyst 5xxx series of switches
             http://www.cisco.com/warp/public/cc/cisco/mkt/switch/cat/c5000/index.shtml
           3Com's Corebuilder series of switches
             http://www.3com.com/nsc/switchjump/cb_switches.html
           Cabletron's SmartSwitch 6000
             http://www.cabletron.com/switching/6000/

4) Answer Tanenbaum 4.30.

   Answer: At 4Mb/s, a station can transmit 40,000 bits or 5000 bytes in 10
           milliseconds.  This is an upper bound on frame length.  From this
           amount, some overhead bytes must be subtracted, giving a slighly
           lower limit for the data portion.

5) What is the primarily, potential problem of using the cut-through
   technique in LAN switches?

   Answer: Cut through does not check the frame for errors.  Bad frames will
           be at the least, processed unnecessarily and steal bandwidth.

6) If a Ethernet station collides 4 times in trying to transmit a single
   frame, how long might it be before the next attempt?

   Answer: 2^4 = 16.  0 to 16 slot times potential.  16 * 512 = 8192 bit
           times.  16 * 51.2 microseconds = 819.2 microseconds.

   or      based on the book's algorithm (2)^4 - 1 = 15 or 15 * 512 which is
           7680 bit times or 768 microseconds.

7) If you put a network analyzer on a token ring network and saw that
   all frames had the monitor bit in the access control field were set to
   zero (off), what do you know about the active monitor station in
   relation to the network analyzer?

   Answer: The active monitor is downstream from the analyzer.

8) Answer Tanenbaum 4.36.

   Answer: The maximum wire length would need to be 1/10 as long.

9) What is the lowest category grade of cabling that can be used for
   100BASE-T?

   Answer: Category 3 using all 4 pairs.

10) If when you send a frame on a shared LAN and each station has to
    check the frame to see if it contains it's destination address
    anyway, why not just send all frames to a broadcast address?  Why
    use MAC layer addresses at all?

    Answer: Determining the MAC layer address is done on the network
            adapter (hardware) which is relatively low cost.  If all
            stations had to pass every packet up the protocol stack,
            issuing interrupts to the CPU and using software, performance
            suffers.  In addition, when you have to cross data link
            networks via a bridge (or router), you would cause all
            networks to have to receive and process the frame which would
            typically be unnecessary.  Also, how would you map a global
            network layer address to a data link interface?