Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7. The total time for the data transfer (in milliseconds) from the hard disk is ________. (rounded off to one decimal place)

### 1. Disk Parameters

• Rotational speed = $15000\text{ rpm}$

• Time for one full rotation = $\frac{60}{15000}\text{ s} = 4\text{ ms}$

• Average rotational latency ($t_{\text{rot}}$) = $\frac{4\text{ ms}}{2} = 2\text{ ms}$

• Transfer time for one sector ($t_{\text{tr}}$) = $\frac{\text{Rotation time}}{\text{Sectors per track}} = \frac{4\text{ ms}}{400} = 0.01\text{ ms}$

### 2. Seek Time The head moves in the order: $\text{Track } 0 \to 5 \to 12 \to 7$.

• Track $0 \to 5$: $|5 - 0| \times 1\text{ ms} = 5\text{ ms}$

• Track $5 \to 12$: $|12 - 5| \times 1\text{ ms} = 7\text{ ms}$

• Track $12 \to 7$: $|7 - 12| \times 1\text{ ms} = 5\text{ ms}$

### 3. Rotational Latency and Transfer Time Data is transferred from $10$ randomly located sectors on each of the $3$ tracks, giving a total of:

For each randomly located sector:

For 30 sectors:

### 4. Total Time