###### Operating-Systems

October 11, 2023###### Operating-Systems

October 11, 2023# Operating-Systems

Question 13 |

Consider a system with 3 processes that share 4 instances of the same resource type. Each process can request a maximum of K instances. Resource instances can be requested and released only one at a time. The largest value of K that will always avoid deadlock is _________.

2 | |

3 | |

4 | |

5 |

Question 13 Explanation:

No. of process = 3

No. of resources = 4

Let’s distribute each process one less than maximum demands i.e., (k-1) resources.

So, for three processes, 3(k – 1) resources.

For deadlock avoidance provide an additional resource to any one of the process.

∴ Total resources required to avoid deadlock in any case is 3(k – 1) + 1 = 3k – 2

Now this 3k – 2 should be less than equal to available no. of resources, i.e.,

3k – 2 ≤ 4

k ≤ 2

So maximum value of k = 2

No. of resources = 4

Let’s distribute each process one less than maximum demands i.e., (k-1) resources.

So, for three processes, 3(k – 1) resources.

For deadlock avoidance provide an additional resource to any one of the process.

∴ Total resources required to avoid deadlock in any case is 3(k – 1) + 1 = 3k – 2

Now this 3k – 2 should be less than equal to available no. of resources, i.e.,

3k – 2 ≤ 4

k ≤ 2

So maximum value of k = 2

Correct Answer: A

Question 13 Explanation:

No. of process = 3

No. of resources = 4

Let’s distribute each process one less than maximum demands i.e., (k-1) resources.

So, for three processes, 3(k – 1) resources.

For deadlock avoidance provide an additional resource to any one of the process.

∴ Total resources required to avoid deadlock in any case is 3(k – 1) + 1 = 3k – 2

Now this 3k – 2 should be less than equal to available no. of resources, i.e.,

3k – 2 ≤ 4

k ≤ 2

So maximum value of k = 2

No. of resources = 4

Let’s distribute each process one less than maximum demands i.e., (k-1) resources.

So, for three processes, 3(k – 1) resources.

For deadlock avoidance provide an additional resource to any one of the process.

∴ Total resources required to avoid deadlock in any case is 3(k – 1) + 1 = 3k – 2

Now this 3k – 2 should be less than equal to available no. of resources, i.e.,

3k – 2 ≤ 4

k ≤ 2

So maximum value of k = 2

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