#mca4

Huawei OSN 8800 APE Related Boards Price

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Huawei OSN 8800 MCA4 TN11MCA4 Function

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For real pictures of related product, visit the HongTelecom Gallery.

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Sketch the Manchester and differentiate Manchester encodings for the bit stream: 100111010101

Solution:

Question3. A bit stream 101010101101 transmitted using standard CRC method. The generator polynomial is x 4 +1. Show the actual bit string transmitted. Suppose the fourth bit from the left is inverted during transmission. Show that this error is detected at the receiver’s end.

Solution: Our generator G(x) = x4 + 1 encoded as 10101. Because the generator polynomial is of the degree three we append three zeros to the lower end of the frame to be transmitted. Hence after appending the 4 zeros the bit stream is 1010101011010000. On dividing the message by generator after appending three zeros to the frame we get a remainder of 0011. We do modulo 2 subtraction thereafter of the remainder from the bit stream with the three zeros appended. The actual frame transmitted is 1010101011010011. See below.

What is the cause of thrashing? How does the system detect thrashing? Once it detects thrashing, what can the system do to eliminate this problem?

Solution:

Thrashing is caused by under allocation of the minimum numberof pages required by a process, forcing it to continuously page fault.The system can detect thrashing by evaluating the level of CPU utilizationas compared to the level of multiprogramming. It can be eliminatedby reducing the level of multiprogramming.

Question 1: On a disk with 1000 cylinders, numbers 0 to 999, compute the number of tracks the disk arm must move to satisfy all the requests in the disk queue. Assume the last request serviced was at track 345 and head is moving to track 0. The queue in FIFO order contains requests for the following tracks: 123, 874,692, 475, 105, 376. Perform the computation for the following disk scheduling algorithms: (a) FIFO (b) SSTF (c) SCAN (d) LOOK 

Solution:

           (a)   FIFO - 2013

           The tracks traveled to will be 345,123,874,692,475,105            

      and 376 making the total distance 222+751+182+217+370+271=2013.

           (b)   SSTF-1298

         The tracks traveled to will be 345,376,475 ,692,874, 123 and 105 making

      the total distance 529+769=1298

           (c)   SCAN- 1219

                          The tracks traveled to will be  345,123,105,0,376,475,692 and 874  

                    making the total distance 345+874 =1219

            (d)  LOOK-1009

       The tracks traveled to will be  345,123,105,376,475, 692 and 874

                     making the total distance 240+769=1009