In an 802.11b wireless LAN, suppose there are only two nodes, a source and a destination; the source node has a large amount of data to send to the destination node. The nodes are separated by 750 meters, and have powerful enough radios to communicate at this range. No other nodes operate in the area. Assume the following: Signal propagation speed is 3×108 meters/sec. DIFS, which is the first component of delay between noticing the medium being idle and beginning a new RTS transmission, is 50msec. The backoff time, which is the additional delay right after DIFS in case the source has additional packages to send, is chosen randomly for each packet. When there is no contention, it is uniformly likely to be anywhere from 0 to 31 slots long, where one slot is 20msec. SIFS has a value of 10msec. The frame header bits take a 200msec per packet to transmit; the Data field consists of 1100 bytes. Data is transmitted at an 11 Mbps rate. An ACK, RTS, and CTS each have a 200msec transmission time. a) Find the time between the moment the source node decided to transmit a file, to the moment the ACK of the first frame is received (do not include the expected backoff time). Show your detailed work b) Find the expected backoff time of the source (as described above) before transmitting subsequent packages. Show your detailed work

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In an 802.11b wireless LAN, suppose there are only two nodes, a source and a destination; the source node has a large amount of data to send to the destination node. The nodes are separated by 750 meters, and have powerful enough radios to communicate at this range. No other nodes operate in the area. Assume the following:

Signal propagation speed is 3×108 meters/sec.

DIFS, which is the first component of delay between noticing the medium being idle and beginning a new RTS transmission, is 50msec.

The backoff time, which is the additional delay right after DIFS in case the source has additional packages to send, is chosen randomly for each packet. When there is no contention, it is uniformly likely to be anywhere from 0 to 31 slots long, where one slot is 20msec.

SIFS has a value of 10msec.

The frame header bits take a 200msec per packet to transmit; the Data field consists of 1100 bytes.

Data is transmitted at an 11 Mbps rate.

An ACK, RTS, and CTS each have a 200msec transmission time.

a) Find the time between the moment the source node decided to transmit a file, to the moment the ACK of the first frame is received (do not include the expected backoff time). Show your detailed work

b) Find the expected backoff time of the source (as described above) before transmitting subsequent packages. Show your detailed work

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b) To find the expected backoff time of the source before transmitting subsequent
packages, we can use the binary exponential backoff algorithm.
Initially, the source node selects a random backoff time between 0 and 31 slots. If the
medium is active when the backoff timer reaches 0, the node doubles the backoff time
and selects a new random value between 0 and the new maximum backoff time. This
process is repeated until the node successfully transmits its packet.
E[N]
Σ(η 0 → ∞)n
(pd ≤)
Σ(η = 0 → ∞).x²
Σ(n = 0→∞)x"
Σ(η = 0 ->> ∞)nx"
E[N] = {(n = 0 →31)n * (¹)
=
=
=
n-1
=
=
1
1-x
*
1
X
(1-x)²
2
2
(1-x)²
n
1
2
31
1 * (-²) ¹ + 2 * (¹) + +31* (¹)
2
E[N] ≈ 7.99999996
n
The expected backoff time is the expected number of slots times the slot time:
E[T] = E[N] * 20µsec = 160µsec
Transcribed Image Text:b) To find the expected backoff time of the source before transmitting subsequent packages, we can use the binary exponential backoff algorithm. Initially, the source node selects a random backoff time between 0 and 31 slots. If the medium is active when the backoff timer reaches 0, the node doubles the backoff time and selects a new random value between 0 and the new maximum backoff time. This process is repeated until the node successfully transmits its packet. E[N] Σ(η 0 → ∞)n (pd ≤) Σ(η = 0 → ∞).x² Σ(n = 0→∞)x" Σ(η = 0 ->> ∞)nx" E[N] = {(n = 0 →31)n * (¹) = = = n-1 = = 1 1-x * 1 X (1-x)² 2 2 (1-x)² n 1 2 31 1 * (-²) ¹ + 2 * (¹) + +31* (¹) 2 E[N] ≈ 7.99999996 n The expected backoff time is the expected number of slots times the slot time: E[T] = E[N] * 20µsec = 160µsec
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