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Elza [17]
2 years ago
8

Two communicating devices are using a single-bit even parity check for error detection. The transmitter sends the byte 10101010

and, because of channel noise, the receiver gets the byte 10011010. Will the receiver detect the error
Computers and Technology
1 answer:
icang [17]2 years ago
7 0

Answer:

The receiver will not detect the error.

Explanation:

The byte sent by transmitter: 10101010

The byte received by receiver due to channel noise: 10011010

If you see the bold part of the both sent and received bytes you can see that the number of bits changed is 2.

The two communicating devices are using a single-bit even parity check. Here there are two changed bits so this error will not be detected as this single bit even parity check scheme has a limit and it detects the error when the value of changed bit is odd but here it is even.

This parity scheme basically works well with the odd number of bit errors.

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Considering the following algorithm, which of the following requirements are satisfied?
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Answer:

b) Bounded Waiting

Explanation:

int currentThread = 1;

bool thread1Access = true;

bool thread2Access = true;

thread1 { thread2 {

While (true) {

                   While (true)

                                   {

                     while(thread2Access == true)

                                       {

                                      while(thread1Access == true)

                                       {

                                            If (currentThread == 2) {

                                              If (currentThread == 1)

                                                {        

                                                  thread1Access = false; thread2Access = false;

                                                  While (currentThread == 2);

                                                 While (currentThread == 1);

                                                  thread1Access = true; thread2Access = true;

} }

/* start of critical section */ /* start of critical section */

currentThread = 2 currentThread = 1

… ...

/* end of critical section */ /* end of critical section */

thread1Access = false; thread2Access = false;

… ...

} }

} }

} }

It can be seen that in all the instances, both threads are programmed to share same resource at the same time, and hence this is the bounded waiting. For Mutual exclusion, two threads cannot share one resource at one time. They must share simultaneously. Also there should be no deadlock. For Progress each thread should have exclusive access to all the resources. Thus its definitely the not the Progress. And hence its Bounded waiting.

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