A channel has a bit rate of 4 kbps and a propaga tion delay of 20 msec. For what range of frame sizes does stop-and-wait given an efficiency of at least 50 percent? Note:- Final Answer L= 5.33× 10-7 kbps Give step by step answer
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Topic :- Data link Layer
A channel has a bit rate of 4 kbps and a propaga tion delay of 20 msec. For what range of frame sizes does stop-and-wait given an efficiency of at least 50 percent?
Note:- Final Answer L= 5.33× 10-7 kbps
Give step by step answer
Step by step
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- Consider TDM-based circuit switching with 8000 frames per second and 32 time slots per frame. During each time slot 8 bits are sent. What is the shortest time possible (in seconds) to transmit a 1 Mbyte file over the given link?A channel has a bit rate of 4 kbps and a propagation delay of 20 msec For what range of frame sizes does stop-and-wait given an efficiency of at least 50 percent?A planet located far from Earth is around 9 x 1010 meters away. If a stop-and-wait protocol is employed for frame transmission on a 64 Mbps point-to-point link, what is the channel utilization? Assume that the speed of light is 3 108 m/s and that the frame size is 32 KB.
- For the following stream of digital signal which is being transmitted by bit rate 5 Kbps: 101100000000110 a) Draw the waveform for all line codes you studied. b) Determine if the line code is self synchronized or not. c) Determine if the code can be used with ac coupling or not d) Determine the Bandwidth of the line code e) Which code is used in magnetic recording f) Which code is used in LANConsider a full-duplex 256 Kbps satellite link with a 240 millisecond end-to-end delay. The data frames carry a useful payload of size 12,000 bits. Assume that both ACK and data frames have 320 bits of header information, and that ACK frames carry no data. What is the effective data throughput When using Stop and Wait?Suppose a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The distance from Earth to Mars (when they are closest together) is approx- imately 55 Gm, and data travels over the link at the speed of light-3 × 108 m/s. (a) Calculate the minimum RTT for the link. (b) Calculate the delay x bandwidth product for the link. (c) A camera on the rover takes pictures of its surroundings and sends these to Earth. How quickly after a picture is taken can it reach Mission Control on Earth? Assume that each image is 5 Mb in size.
- Suppose a 128-kbps point-to-point link is set up between the Earth and a rover on Mars. The distance from the Earth to Mars (when they are closest together) is approximately 55 Gm, and data travels over the link at the speed of light-3 × 108 m/s. (a) Calculate the minimum RTT for the link. X (b) Calculate the delay x bandwidth product for the link. (c) A camera on the rover takes pictures of its surroundings and sends these to Earth. How quickly after a picture is taken can it reach Mission Control on Earth? Assume that each image is 5 Mb in size.A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?Consider that the link capacity of a channel is 512 Kbps and round – trip delay time is 1000ms find the bandwidth delay product.
- Data communication Suppose that the spectrum of a channel is between 8 MHz and 10 MHz, and the intended capacity is 8 Mbps. 1= With multiple signaling level, how many levels are required to obtain this capacity and how many bits can be transferred with each signal element? Constant values= c=3x^10°m/s, k=1,38x^10231/KA line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 kHz. What is the maximum data rate supported by this line? If the bandwidth of the channel is 5 kbps, how long does it take to send a frame of 100,000 bits out of this device? What is the maximum data rate of a channel with a bandwidth of 200 kHz if we use four levels of digital signaling?III. The figure below shows synchronous TDM, the data rate for each input connection is 1 Mbps. If 1 bit at a time is multiplexed, Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, and (d) the output frame rate. ... 1 1 1 1 1 1 Mbps 0 0 0 0 0 1 Mbps Frames 01010001110100010101 ... MUX 1 0 1 0 1 1 Mbps *** 0 0 1 0 0 1 Mbps