40 40 Consider the figure below, which plots the evolution of TCP's congestion window at the beginning of each time unit. In this abstract model (NOT TCP Reno or Tahoe), TCP sends a "flight" of packets of size cwnd at the beginning of each time unit. The result of sending packets is that either (i) all packets are ACKed at the end of the time unit; (ii) there is a timeout for the first packet; (iii) there is a triple duplicate ACK for the first packet. The initial value of cwnd is 1 and the initial value of ssthresh is 8. 35 30 25 25 Congestion window size (in segments) 20 15 15 10 10 5 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Time unit (in RTT)
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Given evolution of TCP's congestion window, provide following: a. List the times, at which TCP is in slow start, congestion avoidance and fast recovery states. b. List the times, at which the first packet in the sent flight of packets is lost and indicate whether the packet loss is detected via timeout, or by triple duplicate ACKs. c. List the times at which the value of ssthresh changes and provide the new value of ssthresh.
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- Consider a node called A that sends packets to an adjacent node called B. To control the packet flow to node B, node A employs a credit manager scheme in which parameters C=3(credits), C_max=4(credits) and \tao=6(msec), respectively. node A has an infinite buffer to temporarily store packets. When a packet arrives at node A, node A stores the packet at the bottom of the buffer. Node A also has a single server (transmitter). As soon as the server becomes idle, the server picks up a packet at the head of the buffer, if any, and serves the packet for packet transmission time T_p as far as there remains a credit. (The packet transmission time T_p = 4 (msec).) We observed the arrival times of the first 12 packets, denoted by aA(1)⋯A(12), which were as follows: n A(n) 1 0.5 msec 2 1.0 3 1.5 4 2.0 5 4.5 6 5.5 7 6.5 8 12.5 9 13.0 10 13.5 11 14.0 12 14.6 1. Let R(n) denote the departure time of the n th packet from the buffer for n∈{1,2,...}. Find R(n) for…22. A datagram subnet allows routers to drop packets whenever they need to. The probability of a router discarding a packet is p. Consider the case of a source host connected to the source router, which is connected to the destination router, and then to the destination host. If either of the routers discards a packet, the source host eventually times out and tries again. If both host-router and router-router lines are counted as hops, what is the mean number of a. (a) hops a packet makes per transmission? b. (b) transmissions a packet makes? (c) hops required per received packet?22. A datagram subnet allows routers to drop packets whenever they need to. The probability of a router discarding a packet is p. Consider the case of a source host connected to the source router, which is connected to the destination router, and then to the destination host. If either of the routers discards a packet, the source host eventually times out and tries again. If both host-router and router-router lines are counted as hops, what is the mean number of a. (a) hops a packet makes per transmission? b. (b) transmissions a packet makes? c. (c) hops required per received packet?
-  Assume a simple network of two directly adjacent routers. At time T=0 seconds both sent updates (A -> B "can reach 10.0.0.0 with cost 1", B -> A "can reach 20.0.0.0 with cost 1"). At time T=10 second router B exploded and, as a sad consequence, it stopped sending or receiving any packets. Select all statements that are true: - If at T=15s A sends a packet to 20.0.0.1, A will immediately drop the packet - If at T=15s A sends a packet to 20.0.0.1, it will not reach the destination - If at T=125s A sends a packet to 20.0.0.1, A will immediately drop the packet - At time T=125s node A has no knowledge about the route to 20.0.0.0/24 - If at T=185s A sends a packet to 20.0.0.1, A will immediately drop the packet - At time T=185s node A has no knowledge about the route to 20.0.0.0/24 - If at T=305s A sends a packet to 20.0.0.1, A will immediately drop the packet - At time T=305s node A has no knowledge about the route to 20.0.0.0/24Below is a nested MPLS network by MTN. With the assumption that Customer 6 (C6) is connected to Router 6 (R6), execute a Dijkstra Algorithm on a Link State routing to determine the shortest path for Customer 6 to transmit packets to every other Customer/Destination connected to the rest of the Routers on the network. Tabulate the executed iteration for the Dijkstra Algorithm using the table below. Iteration Nodes R1 R2 R3 R4 R5 R7 R8 R9 R10 0 [6] 1 [6,3] 2 [6,3,2] 3 [6,3,2,1] 4 [6,3,2,1,4] 5 [6,3,2,1,4,5] 6 [6,3,2,1,4,5,7]…Below is a nested MPLS network by MTN. With the assumption that Customer 6 (C6) is connected to Router 6 (R6), execute a Dijkstra Algorithm on a Link State routing to determine the shortest path for Customer 6 to transmit packets to every other Customer/Destination connected to the rest of the Routers on the network. Tabulate the executed iteration for the Dijkstra Algorithm using the table below. Iteration Nodes R1 R2 R3 R4 R5 R7 R8 R9 R10 0 [6] 1 [6,3] 2 [6,3,2] 3 [6,3,2,1] 4 [6,3,2,1,4] 5 [6,3,2,1,4,5] 6 [6,3,2,1,4,5,7]…
- Consider the figure below in which a TCP sender and receiver communicate over a connection in which the segments can be lost. The TCP sender wants to send a total of 10 segments to the receiver and sends an initial window of 5 segments at t = 1, 2, 3, 4, and 5, respectively. Suppose the initial value of the sequence number is 151 and every segment sent to the receiver each contains 612 bytes. The delay between the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t = 8, and an ACK for this segment arrives at t = 15. As shown in the figure, 1 of the 5 segments is lost between the sender and the receiver, but none of the ACKs are lost. Assume there are no timeouts and any out of order segments received are thrown out.Below are the two questions, so make sure to answer each part carefully: Why does TCP implement congestion control if it already has flow control to manage the sender's window? Consider our recent reading [Chiu+89] Analysis of the Increase and Decrease algorithms for congestion avoidance in computer networks. How does this article showcases that TCP is fair?1. Below is a nested MPLS network by MTN. With the assumption that Customer 6 (C6) is connected to Router 6 (R6), execute a Dijkstra Algorithm on a Link State routing to determine the shortest path for Customer 6 to transmit packets to every other Customer/Destination connected to the rest of the Routers on the network. (a) i. Tabulate the executed iteration for the Dijkstra Algorithm using the table below. Iteration Nodes R1 R2 R3 R4 R5 R7 R8 R9 R10 0 [6] 1 [6,3] 2 [6,3,2] 3 [6,3,2,1] 4 [6,3,2,1,4] 5 [6,3,2,1,4,5] 6 [6,3,2,1,4,5,7]…
- P1. Consider the network below. a. Show the forwarding table in router A, such that all traffic destined to host H3 is forwarded through interface 3. b. Can you write down a forwarding table in router A, such that all traffic from H1 destined to host H3 is forwarded through interface 3, while all traffic from H2 destined to host H3 is forwarded through interface 4? (Hint: This is a trick question.)Question 3 Consider a satellite channel running TCP/IP as shown in the figure below. Due to the long round trip time (RTT) and high bit error rate (BER) the conventional TCP congestion mechanism doesn’t work well in a satellite transmission system. 3.1 Assume traditional TCP/IP is used and that the congestion window of the TCP sender is 10k bytes, and it is in congestion avoidance (CA) state with the threshold variable equal to 8k bytes. Now, the sender receives 3 duplicated acknowledgements (ACKs). As a result, explain the effect on the congestion window size and the threshold.Consider the figure below in which a TCP sender and receiver communicate over a connection in which the sender->receiver segments may be lost. The TCP sender sends an initial window of 5 segments. Suppose the initial value of the sender->receiver sequence number is 362 and the first 5 segments each contain 638 bytes. The delay between the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t=8. As shown in the figure below, 3 of the 5 segment(s) are lost between the segment and receiver.