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### b)Repeat the calculation for cycle-stealing mode

For Burst Mode :-.

### For how long would the device tie up the bus when transferring a block of 128 bytes

Bus has requested and release only one time for whole block of data transfer

Bus is tie up with DMA during data transfer to the memory

So total time =Bus request+release+n*cycle time

250+250 +128*500=64500 ns

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For Cycle Stealing Mode:-

DMA returns the bus after a word is transfer. The DMA effectively steals cycles
from the processor in order to transfer the byte, so single byte transfer is also known as cycle stealing.

Bus is requested  and release for every Byte of transfer So,

total Time= 128*(250+250+500) ==>128000 ns

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Can anyone please check this solution

$data\ transfer\ rate=50KB/sec$

$\mathit{bus \ cycle\ =500ns}$

$trasfer\ of\ bus\ control=250ns$

$data\ transfer\ 1\ Byte\ at\ a\ time$

$block\ size=128\ Byte$

$\mathit{Burst\ Mode)}$

$transfer\ time=\frac{block\ size}{data\ transfer\ rate}$

$\frac{128*8}{50*10^{3}*8}=2.56ms$

$actual\ transfer\ time=time\ to\ transfer\ bus\ control\ at\ beginning\ and\ end+transfer\ time$

$=250ns+2.56ms+250ns$

$=2.56ms+500ns$$=2.56ms+500ns(500ns \ is \ negligible)$

$\cong 2.56ms$

$\mathit{Cycle\ stealing\ mode)}$

$where\ only\ 1Byte\ is\ transfer\ not\ the\ entire\ data$

$time\ to\ transfer\ 1Byte=250+500+250=1000ns$

$= 1microsec$

$total\ transfer\ time=128*1microsec$

$=128microsec$
by

Hii @afroze,

Why are taking disk transfer time in Burst Mode ??

Data is already buffered in i/o buffer before the bus has granted to the DMA

In burst mode data should be transferred at io device transfer rate.