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- In cdma2000 reverse link, a user stream of 9.6kbps is sent to an encoder r = 14 (4 coded bit for every 1 information bit), K = 9. and then through a symbol repetition (2x) (repeats the encoded bits twice). What coded data rate after repetition is obtained? What Walsh code length should you use to spread that stream into a typical 1.2288Mcps CDMA channel?Consider two hosts S and R connected directly by a link. A link has a transmission speed of 107 bits/sec. It uses data packets of size 400 bytes. Data needed to transmitted is 400 bytes. Length of the link is 600 km and the propagation speed are 3 × 108 meter per second. T is the transmission delay and P is the propagation delay. The propagation delay in millisecond and transmission delay in microsecond respectively areFrames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25 ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). 1.1 What is the minimum number of bits (1) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
- Consider a full-duplex 512 kbps link with length 36,000Km connecting two hosts. The data frames carry a useful payload of size 6000 bits. Assume that both ACK and data frames have 400 bits of header information, and that ACK frames carry no data. The propagation speed along this link is 3×108 m/sec. a) What is the RTT on this link? b) What is the effective useful data throughput, in kbps, when using Stop-and-Wait?Consider two systems are connected with LAN cable. The length of the cable is 3 kilometers and the data transmitted at a rate of 30 MB/s. The signal speed in the cable is 2.5 x 108 m/s. What is the minimum frame size in bits in the CSMA/CD network?We have a 1000 m long 10 Mbps communication channel with ARQ protocol. The signal travels at a speed of 2 ∗ 10 ^ 8 ? / ?Assume that no frame is lost or damaged. Confirmation frames, control information are negligible in size.For what range of frame sizes do we get an efficiency of at least 70%?
- A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgment has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) ?Networking question: Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of the i. differential Manchester scheme using each of the given data streams, assuming that the last signal level has been positive. From thegraphs, guess the bandwidth for this scheme using the average number of changes in the signal level.In a selective repeat sliding window protocol the bandwidth of the link is 10 Mbps. The size of the frame is 10 KB and the one-way delay is 100 msec. If the efficiency of the protocol is 60 % then the minimum number of sequence bits that are required for the protocol is?
- Assume that there is a shared connection (for example, an Ethernet bus) with a speed of 1 Mbps. How much time (in microseconds) is required to send a frame with a length of 1000 bits over this link?Networking question: Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of thei. NRZ-Lii. NRZ-Iiii. Manchesteriv. differential Manchester schemeusing each of the given data streams, assuming that the last signal level has been positive. From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level.Consider the following scenario: Frames of 12000 bits are sent over a 400kbps channel using a satellite whose propagation delay is 270ms. Acknowledgements are always piggybacked onto data frames. (That is, in the case of stop-and-wait, the next frame cannot be transmitted until the entire frame containing the acknowledgement is received.) Five-bit sequence numbers are used. What is the maximum achievable channel utilization for (a) Stop-and-wait. (b) Protocol 5 (i.e., Go Back N). (c) Protocol 6 (i.e., Selective Repeat)