COSC 6377 Fall 2002

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COSC 6377 Fall 2002

• Homework 1
• Solution

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• 1. (LG Chapter 1 Problem 16 Modified)
• The propagation delay is the time required for
  the energy of a signal to propagate from one
  point to another. The speed of light in space is
  3.0 x 108 meters/second.
• a. (5 points)
• Find the propagation delay for a signal
  traversing up and down to a geo-stationary
  satellite. Assume the round-trip distance from
  earth surface to the satellite and back to earth
  surface is 2.0 x 36,000 km.
• b. (5 points) How many bits are in transit
  during the propagation delay in the case
  described in (a). if bits are entering the
  network at the following transmission speed 10
  megabits/second. (note 1 megabits 1.0 x 106
  bits)
• Answer
• a. (2.0 x 36,000 x 103 ) / (3.0 x 108 ) 0.24
  seconds
• b. 0.24 seconds x 1.0 x 106 bits/seconds
  240000 bits

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• 2. (LG Chapter 1 Problem 22 Modified)
• The official standards of the Internet community
  are published as a Request for Comment, or RFC.
  Use your Web browser to access the IETF Web page,
  http//www.ietf.org.
• a. (4 points)
• Find and retrieve the RFC titled Internet
  Official Protocol Standards. This RFC has number
  3000. This RFC gives the state of
  standardization of the various Internet
  protocols. What is the state and standard
  number of the following protocols IP, UDP, TCP,
  ARP, DNS, FTP?
• b. (3 points)
• Find and retrieve the RFC titled Assigned
  Numbers. This RFC, number 1700, contains all the
  numbers and constants that are used in Internet
  Protocols. What are the port numbers for Telnet,
  FTP, and http?
• c. (3 points)
• What is the state of RFC 1300?
• Answer
• a.
• b. Telnet 23 FTP 20/21 HTTP 80
• c. Informational

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• 3. (a) (LG Chapter 2 Problem 19 Modified) (5
  points)
• Suppose an application layer entity wants to
  send an L-byte message to its peer process, using
  an existing TCP connection. The TCP segment
  consists of the message plus 20 bytes of header.
  The segment is encapsulated into an IP packet
  that has an additional 20 bytes of header. The
  IP packet in turn goes inside an Ethernet frame
  that has 18 bytes overhead, which consists of
  header and trailer. What percentage of the
  transmitted bits in the physical layer
  corresponding to message information if L 10
  bytes? 1000 bytes?
• (b) (LG Chapter 2 Problem 20 Modified) (5
  point)
• Suppose that the TCP entity, that has a TCP
  overhead of 20 bytes, receives a 1.5 megabyte
  file from the application layer and that the IP
  layer is willing to carry blocks of maximum size
  1500 bytes (including both IP overhead and IP
  message). Calculate the amount of overhead,
  which equals to the total TCP overhead and total
  IP overhead, incurred from segmenting the file
  into packet-sized units.
• Answer
• (a) For L 10 bytes, efficiency 10/68
  0.147 14.7
• For L 1000 bytes, efficiency 1000/1058
  0.945 94.5
• (b) Available bytes for application layer file
  1500 -20 -20 1460 bytes
• 1.5 MBytes / 1460 bytes 1027.39 This
  implies that 1028 blocks IP packets are needed to
  transfer the file.
• 40 bytes 1028 41120 bytes

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• 4. (a) (LG Chapter 3 Problem 11 Modified) (5
  points)
• Consider an analog repeater system in which the
  signal has power sx2, and each stage adds noise
  with power sn2. For simplicity assume that each
  repeater recovers the original signal without
  distortion but that the noise accumulates. Also,
  assume sx2 100 sn2. Find the SNR after 10
  repeater links. Write the expression in
  decibels SNR dB 10 log10SNR. (Note the
  accumulated noise for 10 repeaters is 10 sn2)
• (b) (LG Chapter 3 Problem 20) (5 points)
• Suppose that a low-pass communication system has
  1 MHz bandwidth. What bit rate is attainable
  using 8-level pulses? What is the Shannon
  capacity of this channel if the SNR is 99. (Note
  log2x (log10x) / (log102), and log102 0.3)
• Answer
• (a) SNR dB 10 log10SNR. 10 log10 (sx2 / 10
  sn2) 10 log10 (100 sn2 / 10 sn2 )
• 10 log10 (10) 10
• (b) Nyquist rate 106 x 2 x 3 6 x 106 bits
  /seconds
• Shannon capacity 106 x log2 (1 99) 106 x
  log10(100) / log102 106 x 2 / 0.3
• 6.67 x 106 bits /seconds

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• 5. (LG Chapter 3 Problem 41)
• Let g(x) x3 x2 1. Consider the information
  sequence 1001.
• (a) (5 points)
• Find the codeword corresponding to the preceding
  information sequence.
• (b) (5 points)
• Suppose that the codeword has a transmission
  error in the first bit. What does the receiver
  obtain when it does its error checking? (meaning
  the value of the check bits)
• Answer
• (a) codeword 1001110
• (b) The value of the check bits is 101

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• 6. (a) (LG Chapter 3 Problem 43) (5 points)
• Suppose a header consists of four 16-bit words
  (11111111 1111111), (11111111 00000000),
  (11110000 11110000), (11000000 11000000). Find
  the Internet checksum for this code.
• Answer
• Define the four words as b1, b2, b3, and b4
  respectively.
• x (b1 b2 b3 b4 ) (216 - 1) (11
  10110000 10101111) (11111111 11111111)
• 10110000 10110010
• Therefore, the internet checksum b5 -x
  01001111 01001101
• (b) (LG Chapter 3 Problem 45) (5 points)
• Take any binary polynomial of degree 7 that has
  an even number of nonzero coefficients. Show by
  longhand division that the polynomial is
  divisible by x 1.
• Answer For example, let i(x) x7 x5 x2
  1, then i(x) (x 1) (x6 x5 x 1)
• (Note to TA The above answer is just an
  examples. There are numerous other valid
  examples. You need to check each answer to see
  whether it is correct.)

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• 7. (LG Chapter 3 Problem 53) (10 points)
• A (6,3) linear code has check bits given by
• b4 b1 b2
• b5 b1 b3
• b6 b2 b3
• a. Find the set of all codewords.
• Answer Each codewords consists of the
  information bits and their corresponding check
  bits.

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• 8. (a) (LG Chapter 6 Problem 46) (5 points)
• Construct the Walsh orthogonal spreading
  sequence of length 16.
• Answer
• 00000000 00000000
• 01010101 01010101
• 00110011 00110011
• 01100110 01100110
• 00001111 00001111
• 01011010 01011010
• 00111100 00111100
• 01101001 01101001
• 00000000 11111111
• 01010101 10101010
• 00110011 11001100
• 01100110 10011001
• 00001111 11110000
• 01011010 10100101
• 00111100 11000011
• 01101001 10010110

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• 8. (b) (LG Chapter 6 Problem 47) (5 points)
• Decode the sum signal in Figure 6.42 using the
  Walsh sequence for channel 4. What do you get?
  Explain why.
• Answer
• Note For channel 4
• The sequence is (-1,1,1,-1)
• Each value of the correlator output pulse is
  equal to the value of the sum signal pulse
  multiplied by the corresponding value of the
  sequence pulse. Therefore the correlator output
  are
• (-1,-1,-1,3) (1,1,-3,1) (-1,-1,3,-1)
• Each integrator output is the sum of its
  corresponding sequence pulse values. Therefore
  the integrator output values are (0) (0) (0).
• We obtain 0 for every information bit. This is
  because no channel 4 signal is presented in the
  sum signal, and the Walsh sequence of channel 4
  is orthogonal to other channel signals.

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• 9. (LG Chapter 6 Problem 50) (10 points)
• On bridge forwarding table build-up
• Answer
• Bridge forwarding table for B1
• Bridge forwarding table for B2

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• 10. (a) (LG Chapter 5 Problem 47) (5 points)
• Perform bit destuffing for the following
  sequence 11101111101111101111110
• Answer
• 1110111111111101111110
• (b) (LG Chapter 5 Problem 56) (5 points)
• The PPP byte stuffing method uses an escape
  character defined by 0x7D (01111101). When the
  flag, which is 0x7E, is observed inside the
  frame, the escape character is placed in front of
  the flag and the flag is exclusive-ORed with
  0x20. That is, 0x7E is encoded as 0x7D 0x5E. An
  escape character itself (0x7D) is encoded as 0x7D
  0x5D.
• What are the contents of the following received
  sequence of bytes after byte destuffing
• 0x7D 0x5E 0xFE 0x24 0x7D 0x5D 0x7D 0x5D 0x62
  0x7D 0x5E
• Answer
• 0x7E 0xFE 0x24 0x7D 0x7D 0x62 0x7E