Test series

Question

Consider a 1MBPS hard disk is interfaced to the processor in cycle stealing mode of DMA whereby whenever 64B data is available in the buffer it is sent to the main memory. Processor word length is 64 bits. Machine cycle time is 2us. Percentage of CPU time consumed by DMA?

Comments

@nitish @Ashwin can u pls verify my approach?

Since transfer rate is 1MBPS it takes 1us to transfer 1 byte of data.

We have been given cycle stealing mode wherein DMA takes control of bus for every least addresable unit( Here given is word)

so to transfer 1 word i.e 8bytes of data total time taken is 8us. Out of this 2us is memory cycle time for which CPU remains blocked.

Since in cycle stealing we have pipelining , %CPU blocked= (2/8)*100=25%

 

And just 1 more thing to add, had this been burst transfer we would have taken time to transfer complete 64bytes of data. But since memory cycle time is for 1 word will we take memory cycle time as (64/8)*2=16us ?

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Out of this 2us is memory cycle time for which CPU remains blocked

@Kiran Karwa Can you please tell me the meaning of this line and what does the CPU do in the remaining 6sec?

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@vishal

memory cycle time is the time taken to read/write 1 word to/from memory. During this time the DMA gets hold of the system bus and hence CPU cant perform any memory related operation during this 2us. So we consider it blocked.

In the remaining 6us CPU can perform any memory related operation.