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Consider the input/output (I/O) request made at different instants of time directed at a hypothetical disk having $200$ tracks as given in the following table $:$

$\begin{array}{|l|l|l||l|} \hline \textbf{Serial No.} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} \\\hline \textbf{TranckNo.} & \text{12} & \text{85} & \text{40} & \text{100} & \text{75} \\\hline \textbf{Time of Arrival} & \text{65} & \text{80} & \text{110} & \text{100} & \text{175} \\\hline \end{array}$

Assume that $:$

Current head position is at track no. $65$

Direction of last movement is towards higher number of tracks.

Current clock time is $160$ milliseconds

Head movement time per track is $1$ millisecond

$``\text{look}”$ is a variant of $``\text{SCAN}”$ disk-arm scheduling algorithm. In this algorithm, if no more I/O requests are left in current direction, the disk head reveres its direction. The seek time in Shortest Seek First (SSF) and $``\text{look}”$ disk-arm scheduling algorithms respectively are :

  1. $144$ and $123$ milliseconds
  2. $143$ and $123$ milliseconds
  3. $149$ and $124$ milliseconds
  4. $256$ and $186$ milliseconds
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ans is B   143 ms  and  123 ms

at clock time 160 millisecond only request number 1 2 3 4 are there to be serviced 

request number 5 (with track no 75) will be ready after 15 millisecond 

                              starting head position 65

now for SSTF          12    40    75   85    100

 it will serve 85(since 75 is not come yet)  first then  75 then 100 then 40 and finally 12

 so total movements =20+10+25+60+28=143 total time 143x1 ms=143ms

 now for look          12    40    75   85    100   direction towards highest no of tracks 

 it will serve 85(since 75 is not come yet)  first then  100 then 75 then 40 and finally 12

 so total movements =20+15+25+35+28=123 total time 123x1 ms=123ms

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