Operating-Systems
October 5, 2023Computer-Networks
October 5, 2023Operating-Systems
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Question 8
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The storage area of a disk has innermost diameter of 10 cm and outermost diameter of 20 cm. The maximum storage density of the disk is 1400 bits/cm. The disk rotates at a speed of 4200 RPM. The main memory of a computer has 64-bit word length and 1µs cycle time. If cycle stealing is used for data transfer from the disk, the percentage of memory cycles stolen for transferring one word is
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0.5%
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1%
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5%
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10%
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Question 8 Explanation:
y μs is cycle time.
x μs is data transfer time.
% of time CPU idle is,
y/x × 100
Maximum storage density is given, so consider innermost track to get the capacity
= 2 × 3.14 × 5 × 1700 bits
= 3.14 × 14000 bits
Rotational latency = 60/4200 s = 1/70 s
Therefore, to read 64 bits, time required
(106 × 64)/(70 × 3.14 × 17000) μs = 20.8 μs
As memory cycle time given is 1μs,
Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%
x μs is data transfer time.
% of time CPU idle is,
y/x × 100
Maximum storage density is given, so consider innermost track to get the capacity
= 2 × 3.14 × 5 × 1700 bits
= 3.14 × 14000 bits
Rotational latency = 60/4200 s = 1/70 s
Therefore, to read 64 bits, time required
(106 × 64)/(70 × 3.14 × 17000) μs = 20.8 μs
As memory cycle time given is 1μs,
Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%
Correct Answer: C
Question 8 Explanation:
y μs is cycle time.
x μs is data transfer time.
% of time CPU idle is,
y/x × 100
Maximum storage density is given, so consider innermost track to get the capacity
= 2 × 3.14 × 5 × 1700 bits
= 3.14 × 14000 bits
Rotational latency = 60/4200 s = 1/70 s
Therefore, to read 64 bits, time required
(106 × 64)/(70 × 3.14 × 17000) μs = 20.8 μs
As memory cycle time given is 1μs,
Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%
x μs is data transfer time.
% of time CPU idle is,
y/x × 100
Maximum storage density is given, so consider innermost track to get the capacity
= 2 × 3.14 × 5 × 1700 bits
= 3.14 × 14000 bits
Rotational latency = 60/4200 s = 1/70 s
Therefore, to read 64 bits, time required
(106 × 64)/(70 × 3.14 × 17000) μs = 20.8 μs
As memory cycle time given is 1μs,
Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%
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