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127 notes > lecture 7 > Flashcards

lecture 7 Flashcards

1
Q

3 types of multiple-access protocols

A
  1. random access protocols
  2. controlled access protocols
  3. channelization protocols
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2
Q

what protocols does wifi use

A

CSMA

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3
Q

about contention (random access) protocols (2)

A
  • no station is superior to another station

- no station permits/does not permit another station to send

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4
Q

two types of aloha protocols

A
  • pure aloha

- slotted aloha

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5
Q

throughput in pure aloha =

A

G * e^-2G

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6
Q

the maximum throughput in pure aloha is when G =

A
  • 0.5

- S = 0.184

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7
Q

how does pure aloha protocol work (3)

A
  • transmit at any time
  • detect collision after sending
  • if collision occurs, wait a random time and retry
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8
Q

checksum error in pure aloha

A
  • if the first bit of a frame collides with the last bit of another frame
  • both will have to be retransmitted
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9
Q

how does slotted aloha protocol work

A
  • divide time into intervals (slots), one for each frame
  • stations agree upon time intervals
  • users transmit only at beginning of a time slot
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10
Q

throughput in slotted aloha =

A

G * e^-G

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11
Q

maximum throughput in slotted aloha is when G =

A
  • 1

- S = 0.368

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12
Q

graph for aloha protocols

A
  • throughput per time frame on the y-axis

- G (attempts per packet time) on the x-axis

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13
Q

how do CSMA protocols improve channel utilization

A

listen to the medium/stations before sending

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14
Q

1-presistent

A

the station transmits with a probability of 1 whenever it finds the channel idle

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15
Q

p-presistent

A
  • if empty, sends with probability p

- defers with probability q = 1 - p

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16
Q

what is channelization

A

a multiple access method in which the available bandwidth of a link is shared in time, frequency or through code, between diff stations

17
Q

in frequency division multiple access (FDMA)

A

the available bandwidth of the common channel is divided into bands that are separated by guard bands

18
Q

in time division multiple access (TDMA)

A

the bandwidth is just one channel that is timeshared between different stations

19
Q

in code division multiple access (CDMA)

A
  • one channel carries all transmissions simultaneously
  • code attached to each station
  • code is decrypted when it reaches destination
20
Q

IEEE 802.3 =

A

ethernet

21
Q

IEEE 802.11 =

A

internet

22
Q

a network interface card (NIC) can be in one of three states

A
  • transmitting
  • receiving
  • cycling between transmitting and receiving
23
Q

what is the period of silence a NIC waits before it can transmit

A

9.6 us (microsec)

24
Q

why is the period of silence necessary

A
  • to allow the receiver to cycle from transmitting mode to receiving mode
  • this prevents the frame from being lost
25
Q

in CSMA/CD, to detect the collision, the sender must still be

A

in transmitting mode

26
Q

which layer does the NIC work under

A

data link

27
Q

to guarantee that sender is still in transmission state when a collision happens

A

the transmission delay should be twice the propagation delay
- T = 2 * P + extra delays if any

28
Q

data rate of basic/standard ethernet

A

10Mbps

29
Q

how to find min frame size

A
  1. RTT = 2 * propagation time
  2. propagation time = distance/speed + (n)delay of each repeater
  3. transmission delay = packet size/ 10Mbps = RTT
30
Q

what is the minimum frame size according to IEE standards

A
  • 46 bytes/368 bits in pure data portion
  • 64 bytes/512 bits including overheads (because we are in the data link layer)
  • 18 bits of added overhead
31
Q

to find how many bytes of padding is needed

A

46 - present byte value = amount of padding in BYTES

32
Q

binary exponential backoff

A
  • wait 0 to (2^i) - 1

- i = number of collisions

33
Q

the more collisions we have

A

the lower the probability of sending

34
Q

what is the maximum frame size according to IEE standards

A

1518 bytes/12,144 bits

35
Q

why do we need maximum and minimum frame size (2 pts each)

A

minimum:
- to ensure sender is still in transmitting mode
- for the transmission delay to equal twice the propagation delay, otherwise it will equal less and RTT wont be achieved
maximum:
- to restrict large chunk of data being sent without any fragmentation
- retransmission of the frame in the case of errors (even 1 bit) would take less time

127 notes (11 decks)

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