###### Operating-Systems

October 5, 2023###### Computer-Networks

October 5, 2023# Operating-Systems

Question 8 |

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

0.5% | |

1% | |

5% | |

10% |

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

(10

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

(10

^{6}× 64)/(70 × 3.14 × 17000) μs = 20.8 μsAs 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

(10

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

(10

^{6}× 64)/(70 × 3.14 × 17000) μs = 20.8 μsAs memory cycle time given is 1μs,

Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%

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