Source Destination (i) Consider the transfer of a file containing 4 billion characters from one computer to another via an intermediate point (see figure). Transfer consists of a sequence of cycles in a connection- oriented communication transmission method. All channels are the same in terms of propagation speed and characteristics. The throughput of the channel refers to the number of sequences required to transfer 4 billion characters. Each character in its digital form requires 1 byte. The computers (source -intermediate point) are D=2500m apart, whereas the intermediate byte. The computers (source -intermediate point) are D=2500m apart, whereas the intermediate point to destination is 5000m away. Each terminal generates a data rate of b=2Mb/s, with a packet sizes=32 Bytes, which includes 10 Bytes of overhead (header). The propagation speed on the bus is 400 m/sec on all channels. Find the total elapsed time using throughput and total Capacity measures over the links for a bus topology, having each frame acknowledged with an 88-bit frame before the next frame is sent. (12 Marks) (ii) Now consider the transfer of a file containing 4 billion characters from one computer to another. Transfer consists of a sequence of cycles in an ATM transmission mode/method. The throughput of the channel refers to the number of sequences required to transfer 4 billion characters. Each character in its digital form requires 1 byte. The computers (source- intermediate point) are D=2500m apart, whereas the intermediate point to destination is 5000m away, and each terminal generates a data rate of b=2Mb/s. The propagation speed on the bus is 400 m/sec. Each ATM frame acknowledges a 1000 Bytes frame before the next frame is sent (every 1MB it acknowledges the correct receipt (sending ACK)). Find the total elapsed time using throughput and total Capacity measures over the links for a bus topology, having each frame acknowledged with an 88-bit frame before the next frame is sent. (8 Marks) (iii) Compare the (i) and (ii) and draw constructive conclusions (substantiated) for the total elapsed time using throughput and total Capacity measures. (3 Marks) 3. Bit transmission errors between computers sometimes occur, where one computer sends a zero but other computer receives a 1(or vice versa). Because of this, the computer sending a message repeats each bit three times, so a 0 is sent as 000 and a 1 as 111. The receiving computer "decodes" each triplet by majority rule: whichever number 0 orl, appears more often in a triplet is declared to be the intended bit. For example, both 000 and 100 are decoded as 0, while 101 and 011 are decoded as 1. Suppose that 6% of bits are switched (0 to 1 or 1 to 0) during transmission between two particular computers, and these errors occur independently during transmission. a) Find the probability that a triplet is decoded incorrectly by the receiving computer. b) Using your answer to part (a), explain how using triplets reduces communication errors. c) How does your answer to part (a) change if each bit is repeated 5 times (instead of 3)? d) Imagine 25 kilobit message (i.e., one requiring 25000 bits to send) what is the expected number of errors if there is no bit repetition implemented? If each bit is repeated 3 times?