Question

Consider a device operating on 8MBPS speed and transferring the data to memory
using cycle stealing mode of DMA. if it takes 250 nanoseconds to transfer 16 bytes data
to memory when it is ready/ prepared. Then the percentage of time the CPU is blocked
due to DMA is ?

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Cycle Stealing Mode - 1 byte is transferred at a time. Then BR (Bus Request) is sent and BG (Bus Grant) is given this process is repeated until entire data is sent. So, for 64 bytes we need 64 cycles of CPU time assuming BR and BG is done in one machine cycle. Each cycle being 2 micro second, totally it takes 128 micro seconds.

Data transfer time = Data/Speed + CPU Time = 64*8 bits/1 Mbps + 128 = 640 microseconds

So, % of CPU cycles = 128/640 * 100 = 20%.

B) In Burst Mode entire data is sent together and CPU is not given BUS access during this time. But CPU is free to perform non-memory operations.