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Question

An I/O device transfers data at a rate of 10MB/s over a 100MB/S bus. The data is transferred in 4KB blocks. If the processor operates at 500MHz, and it takes a total of 5000 cycles to handle  each DMA request, find the fraction of CPU time handling the data transfer with and without DMA.

Answer 1

Given, Data Transfer rate =10 MB/s , size of data block = 4 KB, Processor operates at 500 MHz and each DMA requires 5000 cycles
Since processor is operating at 500 MHz we have a cycle time of 1/(500 x $10^6$) = 2 x $10^{-9}$ s = 2 ns

With DMA –
By default we consider Burst mode,
hence, % of CPU spent handling the data = Y/(X+Y) where X = data transfer time and Y = memory cycle time
X = 4 KB / 10 MB = 0.390625 x $10^{-3}$ s = 390.625 $\mu$s
Each DMA requires 5000 cycles = 5000 x 2 x $10^{-9}$ = 10 x $10^{-6}$ = 10 $\mu$s
Therefore, % of CPU idle/blocked or handling the data = Y/(X+Y) = 10 x $10^{-6}$ / ( 390.625 + 10 ) x $10^{-6}$ = 0.0249609 = 2.4961 % $\approx$ 2.5 %

Without DMA –
The processor here copies the data into memory as it is sent over the bus. Since the I/O device sends data at a rate of 10MB/s over the 100MB/s bus, 10 % of each second is spent transferring data. Thus 10% of the CPU time is spent copying data to memory.

Comments

@Sachin Mittal 1 sir, If there is anything wrong with the answer I have added please point it out.