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%
A similar question is mentioned in d link below. Please kindly refer it for further reference:
https://gateoverflow.in/37355/dma-operation
I hope d explanation helps u. :)
