Question

Consider a device which operates with 20KBPS operating speed. The device is
operating on program control mode of 10 and it has to transfer data of 1008 from it.
The data is transferred byte wise. Size of status register is 2Bytes. Total time needed
to perform the data transfer is microseconds?
(Note: The status is checked only once, in the beginning)​

Answer 1

Answer:

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Answer 2

Answer:

5100 microsecond

Explanation:

as device operates with 20KB/s

i.e 20KB data is transferred in 1 sec

-> 1 B data will be transferred in 0.5 * 10^-4 sec

as its given in question 100 B is transferred

therefore , 100B data will be transferred in   0.5 * 10^-2 sec

now , size of status is 2 bytes and device operates with 1 B per 0.5 * 10^-4 sec (solved above)

transfer time = 2* 0.5*10^10-4 sec = 10^-4 sec

programmed IO time = 100B data read time from IO + transfer time

                                   = (0.5 * 10^-2) + (10^-4)

                                   = 51*10^-4 sec = 5100 microsec