Made easy book

Question

Q(1) Consider a disk I/O transfer, in which 1500 bytes are to be transferred, but number of bytes on a track is 1000, and rotation speed of disk is 1500 rps but the average time required to move the disk arm to required track is 15milisec, then what will be total access time?

(a) 16.33 ms (b) 15.33 ms (c) 14.33 (d) none

Please answer ASAP

Comments

Is it A)??

---

Yeah, but how?

---

@Shubhanshu What are you considering ! I am getting 15+0.33+0.67 for first 1000 bytes=16ms
Then for next 500 15+0.33+0.22=15.55 are you not considering seek and latency for 500 please verify and correct me.!

Answer 1

Here we should know : 

a) Total access time = Seek Time + Rotational latency + Transfer time

b) Transfer time is decided by :    Number of bytes / Transfer rate where Transfer rate is found using the fact that the data equivalent to track capacity is transferred in 1 rotation.

As data to be transferred  =   1500 B which is greater than track capacity which is given as 1000 B , so all data is not in the same track , so the disk arm needs to move at least once , so at least 1 seek time is involved.

Given seek time      =      15 ms

Average rotational latency    =  1/2 * 1 Rotation time   =    0.5 rotation time

For transfer time , we know in 1 rotation time , number of bytes that can be transferred  =   1000 B

                  So number of rotation times needed for transfer of 1500 B                          =   1.5 rotation times

Hence rotational latency + transfer time      =        2 rotation times.

Now given ,

          Number of rotations in  1 s              =        1500

          Hence time taken for 1 rotation        =        1 / 1500 s        =    1000 / 1500 ms

          Hence time taken for 2 rotations      =         2000 / 1500 ms

                                                               =        1.33 ms

          Hence rotational latency + transfer time     =    1.33 ms

          Thus ,

          Total delay                                    =         Seek time + Rotational latency + Transfer time

                                                              =         15  +  1.33

                                                              =         16.33 ms

   Hence A) should be the correct option .

Comments

In all the question I'd see that we take Rotational latency as half of the rotation time. But nowhere I got the reason, Why we do this??

---

The reason is simple : It is termed as average rotational latency if u see in books . So this may vary from 0 to 1 rotation time. Hence taking mean we get 1/2 rotation time.

Answer 2

b=1500 bytes  r=1500

N=1000 bytes

T_s=15ms

T=1500/1000*1500=2=1ms

Rotational Latency= 1/(2*1500)=0.33ms

T_a= 15+1 + 0.33=16.33ms

Source: William Stallings :)