percentage of time of cpu blockage during dma transfer closed

Question

Consider a disk drive with following specifications: 16 surfaces, 512 tracks/surface, 512 sectors/sector, 1KB/sector, rot. speed 3000rpm. The disk is operated in cycle stealing mode whereby whenever one byte word is ready it is sent to memory; similarly for writing the disk surface reads a 4 byte word from the memory in each dma cycle. Memory cycle time is 40ns. The max. percentage of time that cpu gets blocked during dma operation:

10%  25%  40% 50%

Comments

Duplicate : https://gateoverflow.in/1393/gate2005_70

Answer 1

Let x be preparation time and y be transfer time 

Then % of time CPU Blocked = $\displaystyle \frac{y}{x+y}*100$

Total byte to be transfer = 
1 byte word is ready it is sent to memory + 4 byte word for writing in each dma cycle
= 5 Byte 
X =  In one rotation $\frac{60}{3000}$sec  it reads 512 KB
         so how much time  for 5 B ?

= 195ns(approx)

Y = 5*memory cycle time 
      5*40 = 200ns 

Therefore     $\displaystyle \frac{y}{x+y}*100 = \frac{200}{200+195}*100 = 50 \%$

Comments

I didn't understand how did u calculated total byte to be transfered = 5B ?
What is that 1 byte word is ready it is sent to memory  ?