PUERTA | GATE-CS-2015 (Conjunto 2) | Pregunta 65

Considere un disco típico que gira a 15000 rotaciones por minuto (RPM) y tiene una tasa de transferencia de 50 × 10 ⁶ bytes/seg. Si el tiempo promedio de búsqueda del disco es el doble del retraso de rotación promedio y el tiempo de transferencia del controlador es 10 veces el tiempo de transferencia del disco, el tiempo promedio (en milisegundos) para leer o escribir un sector de 512 bytes del disco es _____________
(A) 6.1

Respuesta: (A)
Explicación:

Disk latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead
Seek Time? Depends no. tracks the arm moves and seek speed of disk
Rotation Time? depends on rotational speed and how far the sector is from the head 
Transfer Time? depends on data rate (bandwidth) of disk (bit density) and the size of request

Disk latency = Seek Time + Rotation Time + 
                        Transfer Time + Controller Overhead

Average Rotational Time = (0.5)/(15000 / 60) = 2 milliseconds
[On average half rotation is made]

It is given that the average seek time is twice the average rotational delay
So Avg. Seek Time =  2 * 2 = 4 milliseconds.

Transfer Time = 512 / (50 × 106 bytes/sec)
              = 10.24 microseconds

Given that controller time is 10 times the average transfer time
Controller Overhead = 10 * 10.24 microseconds
                    = 0.1 milliseconds

Disk latency = Seek Time + Rotation Time + 
                           Transfer Time + Controller Overhead
             = 4 + 2 + 10.24 * 10-3 + 0.1 milliseconds
             = 6.1 milliseconds

Consulte http://cse.unl.edu/~jiang/cse430/Lecture%20Notes/reference-ppt-slides/Disk_Storage_Systems_2.ppt

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