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+12 votes

A single processor system has three resource types $X, Y$ and $Z$, which are shared by three processes. There are $5$ units of each resource type. Consider the following scenario, where the column **alloc **denotes the number of units of each resource type allocated to each process, and the column **request **denotes the number of units of each resource type requested by a process in order to complete execution. Which of these processes will finish **LAST**?

alloc | request | |||||
---|---|---|---|---|---|---|

$X$ | $Y$ | $Z$ | $X$ | $Y$ | $Z$ | |

P0 | $1$ | $2$ | $1$ | $1$ | $0$ | $3$ |

P1 | $2$ | $0$ | $1$ | $0$ | $1$ | $2$ |

P2 | $2$ | $2$ | $1$ | $1$ | $2$ | $0$ |

- $P0$
- $P1$
- $P2$
- None of the above, since the system is in a deadlock

+24 votes

Best answer

The answer is (**C**).

$X$ | $Y$ | $Z$ |

$0$ | $1$ | $2$ |

Now, $P1$ will execute first, As it meets the needs.

After completion, The available resources are updated.

$X$ | $Y$ | $Z$ |

$2$ | $1$ | $3$ |

Now $P0$ will complete the execution, as it meets the needs.

After completion of $P0$ the table is updated and then $P2$ completes the execution.

Thus $P2$ completes the execution in the last.

0

please explain me how you get "Available Resources" , even if it is not mention in the question or may be i am not able to fetch that particular point so please help me [email protected]_keeda

+2

Calculate the total allocated resources. 5 instances of X are allocated 4 of Y and 3 of Z.

So the available or left resources are x=5-5=0 :::: y=5-4=1::::z=5-3=2.

Hope it helps.

So the available or left resources are x=5-5=0 :::: y=5-4=1::::z=5-3=2.

Hope it helps.

+1

@ Shubham Aggarwal for this particular question what is other sequence other than P1 -> P0 -> P2 ?

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