Chapter 5 - 5.2 - The Components of a Computer System - Practice Problems - Page 244: 2

The seek time depends on the number of tracks over which the read head must move. This could range from $0,$ if the arm does not need to move, to a worst case of the arm having to move from the far inside track to the far outside track, a total of 49 tracks. The average, as stated in the problem, is a move across 20 tracks. The best-case rotational delay is $0,$ whereas the worst case is one complete revolution. The rotational speed is $2,400$ rev/min $=40$ rev/sec $=25$ msec/rev. On the average, we will wait about one-half of a revolution. Finally, the transfer time is the same in all cases, the time it takes for one sector $(1 / 20 \text { of a track) to rotate }$ under the read/write head, which is 1$/ 20$ rev $\times 25$ msec/rev $=1.25$ msec. Putting all this together in a table produces the following values for the time (in msec) required for each task: \begin{array}{lll}{} & {\text { Best Case }} & {\text { Average Case }} & {\text { Worst Case }} \\ {\text { Seek time }} & {0.0} & {20 \times 0.4=8.0} & {49 \times 0.4=19.6} \\ {\text { Latency }} & {0.0} & {0.5 \times 25=12.5} & {1 \times 25=25.0} \\ {\text { Transfer }} & {1.25} & {1.25} & {1.25} \\ {\text { Total }} & {1.25} & {21.75} & {45.85}\end{array}

The seek time depends on the number of tracks over which the read head must move. This could range from $0,$ if the arm does not need to move, to a worst case of the arm having to move from the far inside track to the far outside track, a total of 49 tracks. The average, as stated in the problem, is a move across 20 tracks. The best-case rotational delay is $0,$ whereas the worst case is one complete revolution. The rotational speed is $2,400$ rev/min $=40$ rev/sec $=25$ msec/rev. On the average, we will wait about one-half of a revolution. Finally, the transfer time is the same in all cases, the time it takes for one sector $(1 / 20 \text { of a track) to rotate }$ under the read/write head, which is 1$/ 20$ rev $\times 25$ msec/rev $=1.25$ msec. Putting all this together in a table produces the following values for the time (in msec) required for each task: \begin{array}{lll}{} & {\text { Best Case }} & {\text { Average Case }} & {\text { Worst Case }} \\ {\text { Seek time }} & {0.0} & {20 \times 0.4=8.0} & {49 \times 0.4=19.6} \\ {\text { Latency }} & {0.0} & {0.5 \times 25=12.5} & {1 \times 25=25.0} \\ {\text { Transfer }} & {1.25} & {1.25} & {1.25} \\ {\text { Total }} & {1.25} & {21.75} & {45.85}\end{array}