Midterm Exam

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Midterm Exam

• March 29 (Thursday) 2 weeks from now
• April 3 (Tuesday)

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Performance of CSMA/CD
Derivation of maximum throughput of CSMA/CD

• Let A be the probability that some station can
  successfully transmit in a slot. We get
• In the above formula, A is maximized when P1/ N.
  Thus

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Performance of CSMA/CD

• Prob contention interval has a length of j
  slots
• Prob 1 successful attempt x Prob j-1
  unsuccessful attempts

The expected number of slots in a contention
interval is then calculated as
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Performance of CSMA/CD

• Now we can calculate the maximum efficiency of
  CSMA/CD with our usual formula

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Performance of CSMA/CD

• If N 100, A 0.9999 0.3697
• Efficiency 1 / (1 5.4a)
• If we have just one computer wanting to transmit,
  then we can get almost 100 efficiency.
• The above equation is when we have many computers
  wanting to transmit at the same time (which is
  the common case)

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Performance of CSMA/CD

• Given an Ethernet LAN with a 2.5 km long
• Assume average packet size is 620 bits
  (practical)
• Propagation delay 2500 / (2 x 108) 1.25 x
  10-5 sec.
• Transmission time of a packet 620 / (10 x 106)
  6.2 x 10-5 sec
• a 1.25 x 10-5 sec / 6.2 x 10-5 sec 0.2016
• Efficiency 1 / (1 5.4 x 0.2016) 48

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Performance of CSMA/CD

• In this case, the effective transmission rate of
  the 10-Mbps Ethernet is only 4.8 Mbps.
• If the Ethernet transmits TCP/IP frames, about 30
  bytes are for header.
• Hence only (620 240) / 620 61 of the frame
  bits is user data bits
• Therefore, the maximum rate at which the Ethernet
  network can transmit user data is 61 x 4.8 Mbps
  2.928 Mbps.
• This efficiency is very sensitive to the length
  of packets.

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Performance of CSMA/CD

• Suppose we have a 200 m, 1 Gbps shared Ethernet.
• We have the same average packet length of 620
  bits.
• Propagation delay 200 / (2 x 108) 1.0 x 10-6
  sec.
• Transmission time of a packet 620 / (1 x 109)
  0.62 x 10-6 sec
• a 1.0 x 10-6 sec / 0.62 x 10-6 sec 1.613
• Efficiency 1 / (1 5.4 x 1.613) 10.3
• That is why we find only switched Gbps Ethernet

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Average Delay of CSMA/CD

• Recall the efficiency of the Ethernet is
• Ttime / (Ttime 5.4 Propagation delay)

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Average Delay of CSMA/CD

• Before each successful transmission, 5.4 x
  propagation delay is wasted
• A cycle consists of 5.4 x propagation delay
  Ttime
• Given N nodes wanting to transmit, we assume that
  each node gets to transmit after N cycles
• Average time N x (5.4 x propagation delay
  Ttime)

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Average Delay of CSMA/CD

• Given an Ethernet LAN with a 2.5 km long
• Assume average packet size is 620 bits
  (practical)
• Assume, we have 40 computers connected
• Average time N x (5.4 x propagation delay
  Ttime)
• 40 x (5.4 x 1.25 x 10-5 sec 0.62 x 10-6 sec)
  2.72 10-3 sec
• This is reasonable (but not guaranteed) for voice
  applications (phone conversation)
• since this Ethernet can accommodate the bandwidth
  needed (40 x 64 kbps 2.5 Mbps).
• Unlike Video applications where it would be too
  much (40 x 1.5 Mbps 60 Mbps)

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Token Ring Priority Scheme Example

• (a) Station A has a frame with a priority of 4.
  Stations B and D have frames of priorities 5 and
  6, respectively.
• (1) A transmitted the frame (P4,R0)
• (2) B makes a reservation (P4,R5)
• (3) D found that its priority is higher than R
  (6gt5)
• (4) D also makes a reservation (P4,R6) to
  overwrite the old reservation
• (b) Station A received the frame
• (1) A issues a free token with a priority of 6
  (P6,R0) and keeps in mind that it upgrades the
  priority from 4 to 6 by pushing Sr4,Sx6 (stacks
  to store the old priority level and the new
  priority level, respectively)
• (2)B makes a second reservation (P6,R5) on this
  token
• (3) D gets this token (Pm P 6)
• (4) D transmits a frame with a priority of 6 and
  a reservation of 5.

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Token Ring Priority Scheme Example

• (c ) Station C has a frame with a priority of 7
• (1) The frame transmitted by D passing B (P6,
  R5)
• (2) B bypasses the frame. C found that this is a
  data frame and it has a higher priority (PmgtR).
• (3) C makes a reservation (P6,R7) to overwrite
  the old reservation. D begins to remove the
  transmitted frame
• (d) Station D received the frame
• (1) D issues a free token with a priority of 7
  (P7,R0), and keeps in mind that it upgrades the
  priority from 6 to 7 by pushing Sr6,Sx7.
• (2) A bypasses this frame
• (3) B makes a reservation again (P7,R5) on this
  token
• (4) C gets this token and transmits a frame with
  (P7,R5).

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Token Ring Priority Scheme Example

• (e) C transmitted the frame.
• (1) The frame transmitted by C passing A
  (P7,R5)
• (2) B bypasses the frame. C begins to remove this
  frame.
• (3) C issues a free token with (P7,R5)
• D found that the priority level in the incoming
  token (7) is upgraded by itself and downgrades
  the priority level from 7 to 6 by popping the Sx
  and Sr.
• (f) Station A received the downgraded token
• (1) A found that the priority level in the
  incoming token (6) is upgraded by itself and
  tries to downgraded the priority level from 6 to
  4. Nevertheless, it also found that the reserved
  priority level in the token is 5. As a result,
  the priority level is upgraded to 5 (Sx5,Sr4)
• (2) A issues a token with (P5,R0). B finally
  found a token for itself.
• (3) B uses this token to transmit a frame to C.

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Token Ring Priority Scheme Example
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(No Transcript)
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Performance of Token Rings

• Parameters and Assumption
• End-to-end propagation delay a
• Packet transmission time 1
• Number of stations N
• Assume that each station always has a packet
  waiting for transmission
• Note The ring is used either for data
  transmission or for passing the token

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Performance of Token Rings

• Define
• T1 Average time to transmit a frame. Per
  assumption, T1 1
• T2 Average time to pass the token
• Maximum Throughput
• Frame Time T1
• ------------------------------ ----------
• Time Frame Overhead T1 T2

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Effect of propagation delay

• Effect of propagation delay on throughput
• Case 1 a lt 1 (Packet longer than ring)
• T2 time to pass token to the next station a/N
• Case 2 a gt 1 (Packet shorter than ring)
• Note Sender finishes transmission after T1 1,
  but cannot release the token until the token
  returns
• T1T2 max(1, a) a/N

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Performance of Token Ring

• Given a Token Ring LAN with a 2.5 km long
• Assume average packet size is 620 bits
• Assume, we have 40 computers connected
• Assume it is a 10-Mbps LAN
• Length of Ring in bits 125 bits 40 (delay)
  bits 165 bits
• Efficiency Ttime / (Ttime propagation delay
  /N)
• Propagation delay 1.25 x 10-5 sec
• Ttime 6.2 x 10-5 sec
• Efficiency 99.5

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Average Delay of Token Ring

• Given a Token Ring LAN with a 2.5 km long
• Assume average packet size is 620 bits
• Assume, we have 40 computers connected
• Assume it is a 10-Mbps LAN
• Assume on average the token is half way around
  the ring.
• On average each computer will have to wait for 20
  other computers to finish
• Average delay 20 (Ttime propagation delay/40)
  1.24 10-3 sec
• Worse Case delay 40 (Ttime propagation
  delay/40) 2.48 10-3 sec

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Performance of Token Ring

• Given a Token Ring LAN with a 2.5 km long
• Assume average packet size is 620 bits
• Assume, we have 40 computers connected
• Assume it is a 100-Mbps LAN
• Length of Ring in bits 1250 bits 40 (delay)
  bits 1290 bits
• Efficiency Ttime / (propagation delay
  propagation delay /N)
• Propagation delay 1.25 x 10-5 sec
• Ttime 6.2 x 10-6 sec
• Efficiency 48.4

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dual attach node
c o n c e n t r a t o r
dual attach node
single attach nodes
FDDI Dual Ring Structure
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3
4
Bridge to Ethernet LAN
5
node 1
Concentrator
2
6
Concentrator
7
10
Bridge to token- ring LAN
8
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An example FDDI topology (only primary ring is
shown)
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FDDI MAC Operation

• When a token arrives each station follows this
  procedure
• THT TTRT TRT
• TRT 0
• Send Synchronous Data
• IF THT gt 0, enable THT and start sending
  Asynchronous data as long as THT gt 0

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FDDI MAC Example-1
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FDDI MAC Example-1
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FDDI MAC Example-2
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FDDI MAC Example-2
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FDDI MAC Example-3
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FDDI MAC Example-3
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FDDI MAC Example-Sync. (TTRT 80)
Maximum Throughput 80 / 84 95.23
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FDDI MAC Example-Asyn. (TTRT100)
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FDDI MAC Example(Asyn. TTRT100)
Time 404, station 1 gets the token
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Max Throughput

• Example n 4, TTRT 100, D 4
• Max. Throughput 384 / 404
• Max. Access DelayT(n-1)2D 308 (Station 4) in
  the above example