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Question

DMA is transferring characters to the processor, from a device transmitting at 16000 bits per second. Assume DMA is using cycle stealing, If the Processor needs access to main memory once every microsecond, find how much percentage will the processor be slowed down due to DMA activity?

a. 0.20

b 0.25

c.0.30

d.0.35

Comments

answer A...

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     16000 bits------------------>1 sec

=>2000 characters----------->1 sec

therefore 1 character transmit in--------->1/2000=0.0005 sec

how to proceed further?

If the Processor needs access to main memory once every microsecond, find how much percentage will the processor be slowed down due to DMA activity?

What they are asking....Please explain.....

Answer 1

Answer is A).

DMA transmits 16000 bits per sec .

we take it as 2000 characters per sec by assuming data bus width = 1 byte  and runs for every cycle for every byte .

1 byte character is transmitted in ---> 1/2000 = 500 microsec.

Hence , a byte is transferred , one in every 500 microsec .

If the Processor needs access to main memory once every microsecond , so 1 access in 1 micro second , means 10⁶  accesses in 1 sec .

This means DMA steals the bus in every 500th cycle , if processor performs 10⁶  accesses in 1 sec .

So , percentage the processor be slowed down due to DMA activity

= 1/500 * 100 

= 0.20 

Comments

ANS : A ) 20%

Virtually similar question from stalling : 

Solution : 

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S, transfer time is basically time required to make the device interface ready to send the data to CPU, right?

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How can you conclude cycle time from the frequency of memory access?

Why is it neccesary for every cycle to have a memory fetch which seems a presumption here

Answer 2

Now, I require 500 micro sec for 1 byte transfer which corresponds to 1 steal cycle.
So, in 1 sec , how many steal cycles ? = 1000000/500
Now, processor needs to access Main memory every 1 micro sec i.e. 1 cycle/ 1us
So, total cycles =  1000000 cycles in 1 sec.
Out of these total cycles, I am stealing or using additional cycles = 1000000/500
SO, percentage = (1000000/500) / 1000000  * 100 = 0.20

Comments

how do you conclude that a cycle = 1 micro second?