Performance: Throughput. Consider now the (new!) network shown below, with a new sender (E) sending to D over a new additional TCP session. The links again have transmission rates of R₁ = 100 Mbps (i.e., 100 x 106 bits per second and R₂ = 150 Mbps. What is the maximum end-to-end throughput achieved by the new E-to-D session, assuming all senders are sending to receivers at the maximum rate possible? B R₂₁ 25 Mbps 100 Mbps 75 Mbps 50 Mbps 150 Mbps E R₁ с
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- Consider four Internet hosts, each with a TCP session. These four TCP sessions share a common bottleneck link - all packet loss on the end-to-end paths for these four sessions occurs at just this one link. The bottleneck link has a transmission rate of R. The round trip times, RTT, for all fours hosts to their destinations are approximately the same. No other sessions are currently using this link. The four sessions have been running for a long time. i) What is the approximate throughput of each of these four TCP sessions? Explain your answer briefly. ii) What is the approximate size of the TCP window at each of these hosts? Explain briefly how you arrived at this answer.Consider the scenario below where 4 TCP senders are connected to 4 receivers. The servers transmit to the receiving hosts at the fastest rate possible (i.e., at the rate at which the bottleneck link between a server and its destination is operating at 100% utilization, and is fairly shared among the connections passing through that link). R =1 Gbps and Rc is 300 Mbps and Rs is 400 Mbps. And that all four senders have data to send, What is the minimum value of Re that will ensure that the connections to Host-1 and Host-2 are not bottlenecked at links with capacity Rc or Re?TCP a. Consider two TCP connections, one between Hosts A (sender) and B (receiver), and another between Hosts C (sender) and D (receiver). The RTT between A and B is half that of the RTT between C and D. Suppose that the senders' (A's and C's) congestion window sizes are identical. Is their throughput (number of segments transmitted per second) the same? Explain. b. Now suppose that the average RTT between A and B, and C and D are identical. The RTT between A and B is constant (never varies), but the RTT between C and D varies considerably. Will the TCP timer values of the two connections differ, and if so, how are they different, and why are they different? Give one reason why TCP uses a three-way (SYN, SYNACK, ACK) handshake rather than a two-way handshake to initiate a connection. a.
- In this problem, we consider sending real-time voice from Host A to Host B over a packet switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte packets. There is one link between Hosts A and B; its transmission rate is 2 Mbps and its propagation delay is 10 msec. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?Consider a router buffer preceding an outbound link. In this problem, you will use Little’s formula, a famous formula from queuing theory. Let N denote the average number of packets in the buffer plus the packet being transmitted. Let a denote the rate of packets arriving at the link. Let d denote the average total delay (i.e., the queuing delay plus the transmission delay) experienced by a packet. Little’s formula is N=a⋅d . Suppose that on average, the buffer contains 10 packets, and the average packet queuing delay is 10 msec. The link’s transmission rate is 100 packets/sec. Using Little’s formula, what is the average packet arrival rate, assuming there is no packet loss?Consider two hosts, A and B that are connected by a transmissions link of2.1 Mbps. Assume that packets are of length 2.0 Kb (Kilobits) and the length of the link is 100Km. a. What is the propagation delay from A to B, that is the amount of time from when the first bit of the packet is transmitted at A, until it is received at B? b. What is the transmission time of the packet at A (the time from when the first bit of the packet is sent into the wire and the time at which the last bit is sent into the wire). c. Suppose now that that length of the link is doubled. What is the propagation delay from A to B now and what is the transmission time? d. Now suppose that node C is connected to node B also by a 2 Mbps, 100 Km link. How long does it take from when the first bit is transmitted by A to when the last bit is received at C, assuming B operates in a store-and -forward manner?
- In this problem, we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte packets and when a packet is formed, it is send to host B. There is one link between Hosts A and B; its transmission rate is 2 Mbps and its propagation delay is 10 msec. What's the elapsed time since when the first bit is formed (from the original analog signal at Host A) until the bit is received at host B?Consider a long-lived TCP session with an end- to-end bandwidth 500 Mb/s. The session starts with a sequence number of 3465. The minimum time (in seconds) before this sequence number can be used again isConsider that only a single TCP (Reno) connection uses one 10Mbps link which does not buffer any data. Suppose that this link is the only congested link between the sending and receiving hosts. Assume that the TCP sender has a huge file to send to the receiver, and the receiver's receive buffer is much larger than the congestion window. We also make the following assumptions: each TCP segment size is 1,500 bytes; the two-way propagation delay of this connection is 150 msec; and this TCP connection is always in congestion avoidance phase, that is, ignore slow start. A. What is the maximum window size (in segments) that this TCP connection can achieve? B. What is the average window size (in segments) and average throughput (in bps) Of this TCP connection? C. How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?
- Consider a long-lived TCP session with an end- to-end bandwidth of 1 Gbps (= 10⁹ bits-per- second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again isMaximum end-end throughput (e). Consider the scenario below where 4 TCP senders are connected to 4 receivers. Servers 1 - 3 transmit to the receiving hosts at the fastest rate possible (i.e., at the rate at which the bottleneck link between a server and its destination is operating at 100% utilization, and is fairly shared among the connections passing through that link). Server 4 has nothing to send, so its sending rate is zero. Suppose that R = 1 Gbps, Rc is 300 Mbps and Rs is 400 Mbps. What is the TCP throughput achieved on each of the 3 sctive-sending connections? A. 4R B. Rc C. R/4…Computer Networks Consider a packet of length L that begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let d, s, and R denotes the length, propagation speed, and the transmission rate of link i, for i=1,2,3 . The packet switch delays each packet by d . Assuming no queuing delays, in terms of d, s , R, (i=1,2,3), and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation speed on all three links is the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay? In the above problem, suppose R1=R2=R3=R and dproc=0. Further, suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it…