Jonathan Nguy603 799 761Homework 41. These ACK packets do not need sequence numbers because the rdt3.0 protocol uses a stop-and-wait procedure, so all the packets must be in order when they are sent and ACK’d. 2. a) If the receiver is expecting the sequence k, and the window size is 3, the latest ACK is must’ve sent could have been k -1. Because the window size is 3, the earliest ACK that could be sent is k – 3 – 1, which is k – 4. With these two, we know that the possible ACK’s being sent back is between k – 4 and k – 1.b) If the receiver is expecting the sequence k next, that means it must have sent ACK’s up until k – 1. If none of those ACK’s have been received by the sender, the sender’s sequence numbers would be between k – N = k – 3 and k – 1. Next, if the sender received all of those ACK’s, that means that the next sequence numbers would be between k and k + N – 1= k + 2.3. a) The minimum possible latency = 2 RTT’s + size of object/transmission rate. The minimum W that achieves that is:RTT +S /RS/ RPlugging in 28 kbps to the first equation, we get:2∗.1+1000003500=28.77 secondsTo get the minimum W, we plug in the values to the second equation:.1+536 /(280008)536/(280008)=1.6So the minimum window size is 2.b) The minimum latency for 100 kbps is:2∗.1+1000001000008=2∗.1+8=8.2 secondsTo get the minimum window size:.1+536 /(1000008)536/(1000008)=3.33So the minimum window is 4.c) The minimum latency for 10 Mbps is:2∗.1+100000(100000008)=2∗.1+.08=.28 secondsTo get the minimum window size:.1+536 /(100000008)536/(100000008)=234.21So the minimum window is 235.4. a) SRTT = (.9 * (.9 * (.9 * SampleRTT1 + .1 * SampleRTT2) + .1 * SampleRTT3) + .1 * SampleRTT4)b) SRTTn = .9 * SRTTn-1 + .1 *
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