Chapter 1:

6. The answer is in the book. 

7. Propagation delay is 2 * 10^3 m/(2 * 10^8 m/sec) = 1 * 10^-5s. For
   100-byte packets, 100 B/1 * 10^-5s is 10MB/sec, or 80Mb/sec. For 
   512-byte packets, this rises to 409.6Mb/sec.

8. Propagation delay is 5 x 10^4m/(2 x 10^8s) = 2.5 x 10^-4s.  For 100-byte
   packets, 100/2.5 x 10^-4s = 400KB/s = 3.2 Mbps.  For 512-byte packets
   512/2.5 x 10^-4s, = ~2MB/s = ~16 Mbps.

15. (a) The minimum RTT is 2 * 385,000,000m / 3 * 10^8 m/sec = 2.57 sec. 

    (b) The delay * bandwidth product is 2.57 sec*100Mb/sec = 257Mb = 32 MB. 

    (c) This represents the amount of data the sender can send before it 
	would be possible to receive a response. 

    (d) We require at least one RTT before the picture could begin arriving 
	at the ground (TCP would take two RTTs). Assuming bandwidth delay 
	only, it would then take 25MB/100Mbps = 200Mb/100Mbps = 2.0 sec to 
	finish sending, for a total time of 2.0 + 2.57 = 4.57 sec until the 
	last picture bit arrives on earth.

17. (a) Delay-sensitive; the messages exchanged are short. 

    (b) Bandwidth-sensitive, particularly for large files. (Technically 
	this does presume that the underlying protocol uses a large message 
	size or window size; stop-and-wait transmission (as in Section 2.5 
	of the text) with a small message size would be delay-sensitive.)

    (c) Delay-sensitive; directories are typically of modest size. 

    (d) Delay-sensitive; a file s attributes are typically much smaller 
	than the file itself (even on NT filesystems).

28. (a) 640 × 480 × 3 × 30 bytes/sec = 27.6 MB/sec 

    (b) 160 × 120 × 1 × 5 = 96KB/sec 

    (c) 650MB/75min = 8.7 MB/min = 144 KB/sec 

    (d) (assume 3 bytes/pixel) 8 × 10 × 72 × 72 × 3  pixels = 1.24MB
	At 14,400 bits/sec, this would take 691 seconds (ignoring overhead 
	for framing and acknowledgements).

Chapter 2:

1. See figure 2.10 for similar result

3. The answer is in the book. 

7. The answer is in the book. 

11. Suppose an undetectable three-bit error occurs. The three bad bits must
    be spread among one, two, or three rows. If these bits occupy two or
    three rows, then some row must have exactly one bad bit, which would be
    detected by the parity bit for that row. But if the three bits are all
    in one row, then that row must again have a parity error (as must each
    of the three columns containing the bad bits).

19. The answer is in the book. 

43. (a) Assuming 48 bits of jam signal was still used, the minimum packet
        size would be 4640 + 48 bits = 586 bytes.  

    (b) This packet size is considerably larger than many higher-level 
        packet sizes, resulting in considerable wasted bandwidth.  

    (c) The minimum packet size could be smaller if maximum collision
        domain diameter were reduced, and if sundry other tolerances were 
        tightened up.

46. The observation is that if the hosts are not perfectly synchronized the
    preamble of the colliding packet will interrupt clock recovery.