Answer:
a. 11 blocks / track
b. 1400 bytes / track
c. 1819 track
d. 2,545,200 bytes / file
e. 18190 ms = 18.19 s
f. 16.842 s
Explanation:
a. D = Number of blocks per track
E = B * L = 1600 bytes / block
The number of blocks per track must be an integer. The INT function finds the integer part of a number.
D = INT( R / E ) = INT(11.875) = 11 blocks / track
b. F = Waste per track = R - D * E = (19000 bytes / track) - (11 blocks / track) * (1600 bytes / block) = 1400 bytes / track
This is a wastage of about 7.4%.
c. H = Number of tracks required to store the entire file.
G = Number of records per track = (11 blocks / track) * (10 records / block) = 110 records / track
The CEIL operator is the Ceiling operator. It is similar to the INT operator, except that it takes the next higher integer if the argument is not an exact integer.
H = Number of tracks = CEIL[ N / G ] = 1819 tracks
d. J = Total waste to store the entire file = (H - 1) * F = 2,545,200 bytes / file
e. K = Time to write all of the blocks. (Use period of revolution; ignore the time it takes to move to the next track.)
Assume it takes one revolution to record one track.
K = H * T = (1819 tracks) * (10 ms / track) = 18190 ms = 18.19 s
f. M = Time to write all of the records if they are not blocked. (Use period of revolution; ignore the time it takes to move to the next track.)
Assuming we treat the disk like a tape. The total file is 32,000,000 bytes long, which easily can fit in a PC RAM. We can declare a buffer to be the length of the track, and worry about parsing the file back into records when we reread the data.
In this case, we use CEIL(N * L / R) = 1685 tracks. This is 4 surfaces.
M = (N * L / R) * T = 16.842 s
g. P = Optimal blocking factor to minimize waste = INT(R / L) = 118 records / block
The wastage is Q = R - P * L = 120 bytes / track
h. What would be the answer to (e) if the time it takes to move to the next track were 5 ms?
Since we used 1819 tracks, we used tracks on 4 surfaces. When we change surfaces, we do not need to move the arm to another track. Assume also that we do not need to move the arm to position it to the first record in the file.
U = K + (1819 - 4)* 5 ms = 18190 ms + 9075 ms = 27265 ms = 27.265 s
i. What would be the answer to (f) if the time it takes to move to the next track were 5 ms?
In this
V = M + [CEIL(N * L / R) - 4] * 5 ms = 16842 ms + [1685 - 4] * 5 ms = 16842 + 8504 = 25247 ms = 25.247 s
