Discussion #9 EE450
Sample Problems – CSMA/CD
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Problem#1: Description
n Two nodes A and B on the same 10Mbps Ethernet Segment.
n The propagation delay between them is equivalent to 225 bit times (225*bit duration).
n Both nodes start to transmit at the same time , t=0.
n Upon detecting the collision, each node transmit a
jamming signal equivalent to 48 bit times.
n Node “A” will retransmit immediately after it senses the medium is idle (not after it detects a collision).
n Station B will schedule its retransmission 51.2 microsec after it senses the medium is idle (not after it detects a collision).
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Problem#1 : Questions?
n Construct a timeline diagram to indicate all the events involved in the question.
n At what time will node “A” start retransmission?
n At what time will the frame from “A” be completely delivered to “B”?
n Will there be a collision the second time?
n What is the effective throughput for station “A” assuming that the frame length is the minimum allowed which is 512 bits?
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Problem#1: Solution
BW=10Mbps
AB
Tprop=225 bit times
1 bit time = 1 bit / 10 Mbps = 0.1 microsec Tprop = 225 bit times = 22.5 microsec
T= 0 sec A Data[A]
Data[B] B
Tprop=22.5 microsec
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Collision
T= 0 + 22.5/2
= 11.25 microsec
Collision
Data[A]
AB T=11.25 microsec T=11.25 microsec
Data[B]
Collision is detected Collision is detected
T= 11.25 + 22.5/2 =22.5 microsec
AB T=11.25 microsec T=11.25 microsec
A transmitting data station that detects another signal while transmitting a frame, stops transmitting that frame, transmits a jam signal, and then waits for a random time interval (known as “backoff delay” and determined using the truncated binary exponential backoff algorithm) before trying to send that frame again.
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Jamming Signal
T= 22.5 +4.8
= 27.3 microsec
A Jam[B] B TJam =48 bit times= 4.8 microsec
B’s Jamming signal is transmitted.
T= 27.3 + 22.5
= 49.8 microsec
Jam[B]
AB Tprop=22.5 microsec
The last bit of B’s Jamming signal is received at A.
A now senses the medium as idle so it starts retransmission. B schedules its retransmission for 51.2 microsec later,
i.e. at T= 49.8 + 51.2 = 101 microsec
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A’s Retransmission
T= 49.8 + 22.5
= 72.3 microsec
Data[A]
AB Tprop=22.5 microsec
The first bit of A’s retransmitted frame is received at B at T= 72.3 microsec.
T= 72.3 + 51.2
= 123.5 microsec A
Data[A]
B
Tprop=22.5 microsec
Frame size =512 bits , Frame transmission time= 512/10 Mbps=51.2 microsec
The last bit of A’s retransmitted frame is received at B at T= 123.5 microsec.
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A second Collision?
• There won’t be a second collision, because at T=101 microsec, B senses the medium to be busy so it doesn’t transmit and reschedules its retransmission.
• Throughput for A :
Frame size / Transfer time = 512 bits/123.5 microsec = 4.15 Mbps
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Problem#2: Description
n Consider a 100Mbps 100BaseT Ethernet.
n Assume propagation speed is 1.8 ×108 m/sec
n Assume a frame length of 72 bytes and no repeaters.
n In order to have an efficiency of 0.5, what should be the
maximum distance between the nodes?
n Does this maximum distance ensure that a transmitting node A will be able to detect whether any other node transmitted while A was transmitting? Why or why not?
n How does your maximum distance compare with the actual 100 Mbps standard?
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Problem#2: Solution
n We want 1/(1 + 5a) = 0.5 or
n Equivalently a = 0.2 = tprop / ttrans (I)
n tprop = d /(1.8 ×108 ) m/sec (II)
n ttrans = Frame size/BW= (576 bits ) /(108 bits/sec ) = 5.76μ
sec (III)
n Substitute (II) & (III) in (I)
n Solve for d and we obtain d = 207 meters.
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Problem#2: Solution
n For the 100 Mbps Ethernet standard, the maximum distance between two hosts is 200 m.
n For transmitting station A to detect whether any other station transmitted during A ‘s interval, ttrans must be greater than 2tprop.
n Therefore 2×207 m/1.8 ×108 m/sec = 2.30μ sec.
n Because 5.76 > 2.30, A will detect B ‘s signal before the end of its transmission.
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