##### DISK Numerical Ex.(2)

**1. If the Disk has average rotational speed of 3000 rpm and seek latency of 16 ms find the Average access time.**

Solution:-

Average Access Time or Latency = Seek latency + rotational delay + data transfer delay

**Rotational latency:** If it is not given in question then we can get as

** 1/2 (one rotation time)**

**So, **

3000 rotation in 1 minute

1 roatation in = (1/3000 ) * 60 sec

= 1/50 sec.

So rotational latency = 1/2 *(1/50)= 100 sec. = 10 m-sec.

**Average Access Latency = 16 + 10 = 26 m-sec**. // here data transfer delay is not given , so neglect it.

**2. For the following data are given to calculate Average Access time for transferring 250 bytes from disk.**

** No. of tracks = 500**

** No of sectors per track = 100**

** Bytes per sector = 500 B**

** Rotation time =600 rpm**

Time taken to move from one track to adjacent track is 1 msec.

**Solution:**

**Average Access time = avg seek time + rotational delay + data transfer delay**

let's divide the question into three parts:

**1. Calculation of Avg .seek time: **

They are saying that no. of tracks = 500 and time taken to move from one track to adjacent track is 1 msec.

Avg. time to move from track_{0} to track _{499} = { (500 + 0) / 2} * 1msec

= **250 msec.**

**2. Calculation of Rotational delay= 1/2(one rotation time)**

600 rotation in 1 minute

1 roatation in (1/600) * 60 sec. = 100 m-sec

Rotational delay = 100/2 = **50 msec.**

**3. Data tranfser rate : **

Capacity of track = 100 * 500 = 50000 B

50000 B in 1 rotation or

50000 B in 100 msec

250 B in (100 / 50000) * 250 = 0.5 msec.

Answer is

So Total latency = 250 + 50 + 0.5 =** 300.5 msec **