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- A computer on 6 Mbps network is regulated by token bucket. The token bucket is filled at the rate of 1 Mbps. It is initially filled to capacity with 8 megabits. How long can the computer transmit at the fill 6 Mbps ?In GSM, a "TDMA frame" is composed of eight distinct time slots. Each GSM time slot is 577 s (microseconds) in duration and includes far more than a snapshot of each individual signal.Each time slot contains 148 bits. Just 114 of these 148 bits reflect speech or other info. The remainder are used for a number of different types of power. How many data bits are included in a single TDMA frame?A channel has a bit rate of 8 kbps and one-way propagation delay of 30 ms. The channel uses Stop and wait protocol. The transmission time of the acknowledgment frame is negligible. To get a channel efficiency of at least 60%, the minimum frame size should be byte.
- What is the delay of sending a data file from a source station to a destination station? Suppose that the length of the data file is 2 Mbytes and the bandwidth of the channel is 1600 bps, and the network link between the sender and receiver is 20000 Km and the speed of the light inside the link is 2x108 meters/second? Assume that the delay is composed of only transmission time and propagation time.In GSM, a "TDMA frame" is composed of eight distinct time slots. Each GSM time slot is 577 s (microseconds) in duration and includes far more than a snapshot of each individual signal.. Each time slot contains 148 bits. Just 114 of these 148 bits reflect speech or other info. The remainder are used for a number of different types of power. How many data bits are included in a single TDMA frame?A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 às to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channl is?
- We define data link layer communication efficiency as the ratio between a packet transmission time and the sum of the the packet transmission time and twice of its propagation time. A channel has a bit rate of 10 kbps and a propagation delay of 10,000 micro-second. For what range of frame sizes (in number of bit) does stop-and-wait give an efficiency of at most 90%?A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should beWe define data link layer communication efficiency as the ratio between a packet transmission time and the sum of the the packet transmission time-and twice of its, propagation time. A channel has a bit rate of 10 kbps and a propagation delay of 10,000 micro-second. For what range of frame sizes (in number of bit) does stop-and-wait give an efficiency of at most 60%?
- Suppose that we are sending a 30 Mb MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 × 108 meters/sec, and the distance between source and destination is 10,000 km. 1. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. What is the end-to-end delay?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?In selective repeat ARQ, packet size is 2000 bytes transmission time for one packet is 1ms. If distance between hosts is 10km and signal speed is 4ms per km (4ms/km) and frame sequence number are 6 bit long in frame format then the throughput (in Mbps) is