lecture 7

Question 1

3 types of multiple-access protocols

Answer 1

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

Question 2

what protocols does wifi use

Answer 2

CSMA

Question 3

about contention (random access) protocols (2)

Answer 3

• no station is superior to another station

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

Question 4

two types of aloha protocols

Answer 4

• pure aloha

- slotted aloha

Question 5

throughput in pure aloha =

Answer 5

G * e^-2G

Question 6

the maximum throughput in pure aloha is when G =

Answer 6

• 0.5

- S = 0.184

Question 7

how does pure aloha protocol work (3)

Answer 7

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

Question 8

checksum error in pure aloha

Answer 8

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

Question 9

how does slotted aloha protocol work

Answer 9

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

Question 10

throughput in slotted aloha =

Answer 10

G * e^-G

Question 11

maximum throughput in slotted aloha is when G =

Answer 11

• 1

- S = 0.368

Question 12

graph for aloha protocols

Answer 12

• throughput per time frame on the y-axis

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

Question 13

how do CSMA protocols improve channel utilization

Answer 13

listen to the medium/stations before sending

Question 14

1-presistent

Answer 14

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

Question 15

p-presistent

Answer 15

• if empty, sends with probability p

- defers with probability q = 1 - p

Question 16

what is channelization

Answer 16

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

Question 17

in frequency division multiple access (FDMA)

Answer 17

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

Question 18

in time division multiple access (TDMA)

Answer 18

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

Question 19

in code division multiple access (CDMA)

Answer 19

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

Question 20

IEEE 802.3 =

Answer 20

ethernet

Question 21

IEEE 802.11 =

Answer 21

internet

Question 22

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

Answer 22

• transmitting
• receiving
• cycling between transmitting and receiving

Question 23

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

Answer 23

9.6 us (microsec)

Question 24

why is the period of silence necessary

Answer 24

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

Question 25

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

Answer 25

in transmitting mode

Question 26

which layer does the NIC work under

Answer 26

data link

Question 27

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

Answer 27

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

Question 28

data rate of basic/standard ethernet

Answer 28

10Mbps

Question 29

how to find min frame size

Answer 29

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

Question 30

what is the minimum frame size according to IEE standards

Answer 30

- 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

Question 31

to find how many bytes of padding is needed

Answer 31

46 - present byte value = amount of padding in BYTES

Question 32

binary exponential backoff

Answer 32

- wait 0 to (2^i) - 1 | - i = number of collisions

Question 33

the more collisions we have

Answer 33

the lower the probability of sending

Question 34

what is the maximum frame size according to IEE standards

Answer 34

1518 bytes/12,144 bits

Question 35

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

Answer 35

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