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Discussion #8 EE450, 10/13/2010 Sample Problems: -ARQ protocols (Sliding window)
Discussion #8EE450, 10/13/2010
Sample Problems: -ARQ protocols (Sliding window)
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Problem#1: Description
Draw a timeline diagram for the sliding window algorithm with SWS=RWS=4 frames for two given scenarios using the following assumptions:
1. The receiver sends a duplicate acknowledgement if it does not receive the expected frame. For example it sends dupack[2] when it expects to see frame[2] but receives frame[3] instead.
2. The receiver sends a cumulative acknowledgement after it receives all the outstanding frames. For example it sends ACK[5] when it receives the lost frame[2] after it already received frame[3], frame[4] and frame[5].
3. The timeout interval is about 2RTT.
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Problem#1: Two Scenarios
a. Frame[2] is lost. Retransmission takes place upon timeout (as usual).
b. Frame[2] is lost. Retransmission takes place either upon receipt of the first dupack or upon timeout. Does this scheme reduce the transaction time?
Note that some end-to-end protocols (e.g. variants of TCP) use a similar scheme for fast retransmission.
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Problem#1 Solution: Scenario (a)
Transmission time
Frame[2]
Ack[1]
time time
Sender Receiver
Frame[3]Frame[4]
Frame[5]
Frame[1]
Dupack[2]Dupack[2]Dupack[2]Timeout=2R
TTFrame[2]
Ack[5]
Frame[6]
Ack[6]
Ttrans + RTT
Ttrans + RTT
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Frame[2]
Ack[1]
time time
Problem#1 Solution: Scenario (b)
Sender Receiver
Frame[3]Frame[4]
Frame[5]
Frame[1]
Dupack[2]Dupack[2]Dupack[2]
Frame[2]
Ack[5]
Frame[6]
Ack[6]
Ttrans + RTT
Ttrans + RTT
Transmission time
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Comparing the two scenarios
(a) Delay1= 2Ttrans + 3RTT (b) Delay2= 4Ttrans + 2RTT
Frame[2]
Ack[1]
Frame[3]Frame[4]
Frame[5]
Frame[1]
Dupack[2]Dupack[2]Dupack[2]
Timeout=2RTT
Frame[2]
Ack[5]
Frame[6]
Ack[6]
Ttrans + RTT
Frame[2]
Ack[1]
Frame[3]Frame[4]
Frame[5]
Frame[1]
Dupack[2]Dupack[2]Dupack[2]
Frame[2]
Ack[5]
Frame[6]
Ack[6]
2Ttrans
2Ttrans + 2RTT
Transmission time
Transmission time
Ttrans
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Comparing the transaction time (transfer delay) in the two scenarios
RTT=2Tprop
Delay1= 2Ttrans + 3RTT= 2Ttrans + 6Tprop
Delay2= 4Ttrans + 2RTT=4Ttrans + 4Tprop
?
2Ttrans + 6Tprop > 4Ttrans + 4Tprop
?
2Tprop > 2Ttrans
If Tprop > Ttrans , in Scenario (b) the transaction time is reduced!
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Problem#2: Description
A is connected to B via an intermediate router R.
The A-R and R-B links each accept and transmit only one packet per second in each direction (so two packets take 2 seconds).
The two directions transmit independently. A sends to B using the sliding window
protocol with SWS=4.
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Problem#2: Questions?
a) For Time=0, 1, 2, 3, 4, 5 state what packets arrive at and leave each node, or label them on a timeline.
A R B1 pkt/sec 1 pkt/sec
b)What happens if the links have a propagation delay of 1 seconds, but accept immediately as many packets as are offered (i.e. latency= 1 second but bandwidth is infinite).
A R B
Tprop=1 sec
BW=∞ BW=∞
Tprop=1 sec
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Part (a) SolutionSWS=4 packets, RTT=4 sec
T=0 sec
A R B
Data[2]Data[1]
Data[0]
Data[3]
A R B
Data[3]Data[2]
Data[0]
0<T<1 sec
A R B
Data[3]Data[2]
Data[1]Data[0]
1<T<2 sec Data[1]
2<T<3 secA R B
Data[3]
A R BData[3]
3<T<4 sec A R BData[3]
4<T<5 sec
Data[1]Data[2]
Ack[0]
Data[2]
Data[0]
Ack[0]
Ack[0]
Data[2]
Data[1]
Ack[1]
Ack[1] Ack[2]
Part (a) Solution Ctd.Ignoring Ack Transmission Time
At T=2 sec Data[0] arrives at B
At T=3 sec Data[1] arrives at B
At T=4 sec Data[2] arrives at B
At T=5 sec Data[3] arrives at B, and Data[4] arrives at R
Data[4]
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Part (b) SolutionSWS=4 packets, infinite BW, RTT=4 sec
T=0
A R B
Data[2]Data[1]
Data[0]
Data[3]
T=0+sec
A R B
T=1 sec
Data[1]Data[0]Data[2]Data[3]
A R B
Data[1]Data[0]Data[2]Data[3]
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Part (b) Solution Ctd.Virtually No Transmission Time!
T=2 sec
T=2+ sec
T=1+ secA R B
Data[1]Data[0]Data[2]Data[3]
A R B
A R B
Ack[2]Ack[3]Ack[1]Ack[0]
Data[1]Data[2]
Data[3]
Data[0]
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T=3 sec
T=3+ sec
A R B
Ack[2]Ack[3]Ack[1]Ack[0]
A R B
Ack[2]Ack[3]Ack[1]Ack[0]
T=4 sec A R B
Ack[2]Ack[3]Ack[1]Ack[0]
Part (b) Solution Ctd.
Data[6]Data[5]
Data[4]
Data[7]
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Part (b) Solution Ctd.
T=4+sec
A R B
Data[5]Data[4]Data[6]Data[7]
T=5 sec
A R B
Data[5]Data[4]Data[6]Data[7]
A R B
Data[5]Data[4]Data[6]Data[7]
T=5+sec
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Problem#3: Description
A is connected to B via an intermediate router R.
The A-R link is instantaneous but the R-B link transmit only one packet per second (so two packets take 2 seconds).
The two directions transmit independently.
A sends to B using the sliding window protocol with SWS=4.
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Problem#3 : Questions?
a) For Time=0, 1, 2, 3, 4, 5 state what packets arrive at and are sent from each node.
b) How large does the queue at R grow?
A R B
Tprop=0 sec
BW=Infinity 1 pkt/sec
Buffer
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Part (a) SolutionSWS=4 packets, RTT=2 sec
T=0 sec
A R B
Data[2]Data[1]
Data[0]
Data[3]
A R BData[0]
A R B
0<T<1 sec
Data[1]Data[0]T=0+sec
Data[3]Data[2]
Data[2]Data[1]
Data[3]
1<T<2 secA R B
A R B
Data[1]
Ack[0]
Data[0]
Ack[0]
Data[1]
Part (a) Solution Ctd.Ignoring Ack Transmission Time
Data[3]Data[2]
Data[3]
Ack[1]
T=2 secData[2]
At T=1 sec Data[0] arrives at B
A R B
A R B
Data[2]
Ack[1]
Data[3]
Data[1]
Ack[1]
Data[2]
Ack[2]
Part (a) Solution Ctd.
Data[4]
Ack[0]
3<T<4 sec
T=2+sec Data[3]Data[4]
At T=3 sec Data[2] arrives at B, and Ack[1] arrives at R
A R B
A R B
Data[3]
Ack[2]
Data[4]
Data[2]
Ack[2]
Data[3]
Ack[3]
Part (a) Solution Ctd.
Data[5]
Ack[1]
4<T<5 sec
T=3+sec Data[4]Data[5]
At T=4 sec Data[3] arrives at B, and Ack[2] arrives at R
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A R BData[4]
Ack[3]
Data[3]
Part (a) Solution Ctd.
Ack[2]
T=4+sec Data[5]Data[6]
The steady-state queue size at R is two Data Frames.
Part (b) Solution
At T=5 sec Data[4] arrives at B, and Ack[3] arrives at R
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Problem#4: Description
Consider the same topology and situation as in problem #2.
Assume that the router has a buffer size of 1, i.e. it can hold one packet (or frame) in to addition to the one it is sending (in each direction).
Assume that A has a timeout interval of 4 sec and SWS=4.
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Problem#4 : Questions?
From Time=0, show what happens at each second until
all four packets successfully arrive at B.
A R B
Tprop=0 sec
BW=∞ 1 pkt/sec
Buffer size=1
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SolutionSWS=4 packets, RTT=2 sec
T=0 sec
A R B
Data[2]Data[1]
Data[0]
Data[3]
A R B
Data[2]Data[1]
Data[3] Dropped
A R BData[0]
0<T<1 secData[1]
Data[0]
T=0+sec
T=2 sec
Solution Ctd.Ignoring Ack Transmission Time
A R B
Data[1]
Ack[1]Ack[0]
A R B
Ack[0]T=2+sec
Data[4]
Ack[1]
1<T<2 sec A R BData[1]
Data[0]
Ack[0]
Solution Ctd.Ignoring Ack Transmission Time
A R B
Ack[1]T=3+sec
Data[5]
A R B
Ack[1]
T=4A R B
Data[4]
Data[3]Data[2]
At T=4 Timeout for Data[2] and Data[3]. A starts retransmitting them.
B receives an out of order frame, it does not send an Ack.
Buffered at B.
Data[4]
T=3 sec
At T=4 sec Data[5] also arrives at B.
Data[5]Buffered at B.
Data[4]
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Solution Ctd.Ignoring Ack Transmission Time
A R B4<T<5 sec
A R BData[3]
Data[2]
T=4+sec
At T=4+sec, R has started transmitting Data[2] while Data[3] is enqueued.
Data[2]Data[3]Data[5]Buffered at B.
Data[4]
Data[5]Buffered at B.
Data[4]
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A R B
Solution Ctd.
T=5+sec
5<T<6 sec
Data[3] Data[2]
A R B
Data[2]Data[3]
Ack[2]
T=6 secA R B
Data[5]Buffered at B.
Data[4]
Data[5]Buffered at B.
Data[4]
Data[5]Buffered at B.
Data[4]
B receives Data[3] and can send a Cumulative Ack for Data[3],Data[4] and Data[5]
B received Data[2] which was in-order, so it sends an Ack[2].
Data[3]
Ack[2]
Solution Ctd.
A R B
T=6+sec A R B
Ack[2] Ack[5]
Data[6]
Cumulative
6<T<7 sec A R B
Ack[5]Cumulative
Data[6]
T=7+secAck[5]
Cumulative
Data[6]Data[9]Data[8] Data[7]
Ack[6]
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Problem#5: Description
Suppose we run: A sliding Window Protocol (or
algorithm) With SWS=5, RWS=3 And no out of order arrivals
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Problem#5 : Questions?
a) What is the smallest value for MaxSeqNum?
Hint: Assume that it suffices to find the smallest MaxSeqNum such that if Data[MaxSeqNum] is in the receive window, Data[0] can no longer arrive.
b) Give an example showing that MaxSeqNum-1 is not sufficient.
c) State a general rule for the minimum MaxSeqNum in terms of SWS and RWS.
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Part (a) Solution
The smallest working value for MaxSeqNum is SWS+RWS= 5 +3 =8.
The earliest possible receive window is Data[6], Data[7], Data[8].
Data[5] was delivered and Acknowledged. Since SWS=5, all Data[0]s were sent
before Data[5] and by the “no out of order” hypothesis, Data[0] can no longer arrive.
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Part (b) Solution MaxSeqNum-1=7 is not enough because:MaxSeqNum-1=7 is not enough because:
When the receiver is expecting Data[7], an old Data[0] can still arrive.
0 and 7 are indistinguishable in mod (MaxSeqNum-1) so receiver can not tell which one has arrived.
Consider this example:Consider this example: Sender sends Data[0] through Data[4] and they all
arrive so receiver sends Ack[4] but it’s slow. The receive window is now Data[5],Data[6] and
Data[7]. Sender times out and retransmits Data[0]. Receiver accepts it as Data[7].
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Part (c) Solution
A general rule for minimum
MaxSeqNum in terms of SWS and
RWS:
MaxSeqNum >= SWS+ RWS
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Problem#6: Description
Suppose we run: A sliding Window Protocol (or algorithm) With SWS=RWS=3 And no out of order arrivals And infinite-precision sequence
numbers
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Problem#6 : Questions?
a) If Data[6] is in the receive window,
Data[0] (or in general any older data) can
no longer arrive (hence that MaxSeqNum
= 6 would have sufficed).
b) Show that if Ack[6] may be sent ( or more
literally, that DATA[5] is in the sending
window) then ACK[2] (or earlier) cannot
be received.
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Reminder: Let ACK[N] denote the acknowledgment of all data packets less than N.
Note that data below the sending window (that is, <LAR) is never sent again.
Hence – because out-of-order arrival is disallowed – if DATA[N] arrives at the receiver then nothing at or before DATA[N-3] can arrive later.
Similarly for ACKs, if ACK[N] arrives then (because ACKs are cumulative) no ACK before ACK[N] can arrive later.
Problem #6: Hints
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Part (a) Solution
If DATA[6] is in the receive window,
then the earliest that window can be
is DATA[4]-DATA[6].
This implies ACK[4] was sent, thus
DATA[1]-DATA[3] were received, and
thus DATA[0] can no longer arrive.
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Part (b) Solution
If ACK[6] may be sent, then the lowest
the sending window can be is
DATA[3]...DATA[5].
This means that ACK[3] must have
been received.
Once an ACK is received, no smaller
ACK can ever be received later.
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Problem#7
Design a sliding window ARQ protocol for:A 1Mbps point-to-point link to the moonWith a one-way latency of 1.25secAnd a frame size of 1KB
Minimum number of bits needed for the sequence number?
Solution:
RTT=2 × 1.25sec=2.5 sec BW=1Mbps=125 KBps
Window size=BW × RTT=125KB/sec × 2.5 sec=312 KB or 312 frames
Sequence number space= 2 × window size= 2 × 312=624
bits required for sequence number space= log2 624=9.285
~10 bits
