Operating-Systems
October 10, 2023Nielit Scentist-B [02-12-2018]
October 10, 2023Operating-Systems
| Question 420 |
A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is__
| 6 | |
| 7 | |
| 8 | |
| 9 |
Question 420 Explanation:
Three programs and each requires 3 tapes.
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
Correct Answer: B
Question 420 Explanation:
Three programs and each requires 3 tapes.
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
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