Disks

Question

A certain hard disk rotates at 6000 rpm. It has 1 KB per sector and  64 sectors/track and has a seek time of 5 msec. Find the time required to read:

i. 800 successive sectors?

ii. 800 random sectors?

Comments

Is it :
For successive :
[5msec +5msec]_1sector + [799*5msec]_799sector+[125msec]_transfer=4130msec
For Random
10msec*800+125=8125msec?

---

For the random sectors your answer is correct but for successive it is given as 134.8msec.

---

can u please in detail how are u carrying out the calc... for both successive and random

---

Please can you tell in detail how this calculation is done? why seek time is added twice 5+5 = 10msec

---

to access some data on disk , disk have to first reach that desired track and then sector.

so to reach desired track the time consumed is called SEEK TIME. so reach desired sector then time consumed is called Rotational latency. and now it only have to transfer data(data transfer time)...

seek time to access 1 sector or n successive sector is same as disk head once have to move desired track then just read , rotational latency would also be same as let suppose you have to access sector 5 and your head is at sector 2 then RL would be to reach 2 to 5 and then to reach 5 to 6 no RL so in that manner to access successive RL=0 , data transfer time is directly proportional to data size...in one rotation , one TRACK data can be transferred ....

so to question no 1 )

RL=5ms ST=5ms  DTT=125ms

so total time to access 800 successive sectors is = 5+5+125=135ms

question 2) here random sector accessing so everytime disk head have to seek desired track and then desired sector then DTT so just multiply 800 to ST and RL. but transer time would be same as last as it only depends on data size , which is still same...

8000+125=8125ms...

---

Seek time =5ms given

Rotational latency =time to complete 1/2 rotation

60sec-------->>>6000rotation

so 1/2 rotation ------>>>5ms

Data transfer time (DTT)= time to transfer data

60sec------->>>>6000 rotation (means 6000 tracks)

so 10ms to transfer data of 1 track, which is 64*10³B

10ms--------64*10³B

800 sector data size= 800*10³B

so to transfer 800 sector data DTT=(10ms/64*10³ )*800*10³ = 125ms

Answer 1

Avg rotational latency = 5 msec // Time to reach the start of that track

Avg. seek time = 5 msec // Time to move onto particular track

For Successive sectors: 
= avg. seek time + time to get to starting sector(or avg rotational latency ) + rotational latency for req. no of sectors

= 5ms + 5ms + [  (12 tracks * 10msec rotational latency) + (time to read 32 sectors  of 13^th track)  ]

= 10 + 120 + 5

= 135msec

For Random sectors: 

To read one sector  = avg. seek time + time to get to starting of sector(or avg rotational latency ) + rotational latency for one  sector

= 5ms + 5ms + 10/64

Total time to read 800 sectors = 800 * [ 5ms + 5ms + 10/64 ] = 8125 msec

Comments

@Shubhanshu @Habibkhan

Please check the solution.

---

Hello shivam

your answer is correct as per my knowledge.