Midterm Exam: General Guidelines

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Midterm Exam General Guidelines

• April 3 300 420 LTF (Same as lecture)
• 5-6 questions (including 1-2 multiple choice
  questions)
• The other questions are calculation questions
• Closed book/notes. Bring calculator
• Come and ask questions

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Problem 1

• Given a channel with an intended capacity of 20
  Mbps. The bandwidth of the channel is 3 MHz. What
  signal-to-noise-ratio (SNR) is required in order
  to achieve this capacity?
• Solution
• The maximum number of bits/sec H log2(1 S/N),
• So 20 M 3 M log2(1 S/N).
• We can compute that S/N must be 100.6,
• so 10log10(S/N) 20 dB.

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Problem 2

• There are 200 computers in a lab which are
  attached to an Ethernet (10 Mbps) with a coaxial
  cable of 1500 m. The packets are 800 bits long.
  The propagation delay is 2108 m/sec. On
  average, how many packets can each computer send
  per second?
• Note the efficiency of the Ethernet is equal to
  1/(12a/A), where a propagation
  delay/transmission delay and A (1-1/N)N-1,
  where N is the number of computers.

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Solution

• The efficiency of the Ethernet is 1/(15.4a)
  where a 0.093.
• Efficiency 68.
• Consequently, the packet transmission rate per
  computer is equal to (0.68 x 10 x 106)/(200 x
  800) 44 packets/sec.

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Problem 3

• Using the CRC error detection method, let M
  10100001, and P 1001.
• What is the value of FCS (frame check sequence) ?
• Solution
• Divide 101000010000/1001 you will get
• a remainder of 111 which is the FCS

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Problem 4

• A group of N stations share a 56 kbps pure ALOHA
  LAN. Each station outputs a 1000-bit packet at an
  average of one packet every 100 sec. What is the
  maximum useful value of N?
• Solution
• With pure ALOHA, the usable maximum bandwidth is
  0.18 56 kbps 10.1 kbps. Each station requires
  10 bps, so N 10100/10 1010 stations.

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Problem 5

• A channel has a data rate of 10 Mbps and a
  propagation delay of 5 microsec/km. The distance
  between the sending and the receiving nodes is 50
  km. Nodes exchange fixed-size frames of 1000
  bits. What is the size of the window needed for a
  sliding-window protocol to have the utilization
  at 50? What is the size of the sequence number
  field?
• Solution
• a propagation delay / transmission delay 2.5
• W/(12a) 50
• Solve this equation, we can get that W3, which
  requires a sequence number field size of 2 bits.

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Problem 6

• Given an FDDI local network with 5 stations.
  Assume that TTRT has been set up to be equal to
  28 seconds where 20 seconds are equally allocated
  to the stations for synchronous transmission, and
  8 seconds are allocated for asynchronous
  transmission.
• Assume all stations are fully loaded with
  synchronous and asynchronous traffic, and the
  propagation delay from one station to the next is
  1 second.
• If station 1 gets the token at time t4, when
  will station 4 have a chance to transmit its
  asynchronous traffic?

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Solution
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Problem 7

• Answer the following questions pertaining to DLL
• a) For a given link, assume a bit error rate of
  10-3 (bit errors are independent and identical)
  and a frame size of 125 bytes, what is the
  probability of a frame error?
• Prframe error 1 - (1 - 10-3)1258 0.6323
• b) Assume a 1 Mbps link of 1000-kilometers in
  length with 5 nanoseconds per meter propagation
  delay. Assume negligible time for ACK frames and
  processing delay. The sender always has data to
  send. For Stop-and-Wait (SAW), what is the
  minimum frame size required to achieve an
  utilization of no less than 75?
• tpr 106 5x10-9 5 millisec
• tfr (size / 106) which is (size / 103) for
  millisec (size is in bits)
• U tfr / (2tpr tfr)
• 0.75 (size / 1000) / (25 (size / 1000)) gt
  size 3750 bytes

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Problem 7

• c) Assume a 1 Mbps link of 1000-kilometers in
  length with 5 nanoseconds per meter propagation
  delay. Assume constant 250-byte data frames and
  negligible time for ACK frames and processing
  delay. The sender always has data to send. For
  sliding window, what is the minimum possible
  window size to achieve an utilization of no less
  than 75?
• tpr 106 5x10-9 5 millisec
• tfr (250 8) / 106 2 millisec
• U (Ntfr) / (2tpr tfr)
• 0.75 (N 2) / (25 2) gt N 4.5 and taking
  the ceiling gt N 5

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Problem 8

• A Token ring consisting of N computers, each of
  which is always ready to transmit packets. When a
  station acquires a token, it transmits for T
  units of time.
• Assume the propagation delay is smaller than the
  frame transmission time. The propagation delay is
  P units of time.
• Derive an expression for the token rotation time,
  the time it takes for the token t make one full
  circle around the ring
• Solution
• Token rotation time N x (T P/N)

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Problem 9

• A token ring transmits at a rate of 5 Mbps and
  has a length of 800 meters. The free token us 24
  bits long. The propagation speed is 2 108 m/sec.
• A) How many meters of cable is equivalent to a
  1-bit delay?
• B) For proper operation of the ring, how bit
  delays should be inserted by the powered
  stations?
• Solution
• A) 1/5Mbps d / 2 108 m/sec d 40 m.
• 800/40 20 bits.
• We must add another additional 4 bits by the
  stations.