Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > A system has n resources R0,...,Rn-1,and k pr...
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A system has n resources R0,...,Rn-1,and k processes P0,....Pk-1.The implementation of the resource request logic of each process Pi is as follows:
if (i % 2 == 0) {
if (i < n) request Ri
if (i+2 < n) request Ri+2
}
else {
if (i < n) request Rn-i
if (i+2 < n) request Rn-i-2
}
if (i < n) request Ri
if (i+2 < n) request Ri+2
}
else {
if (i < n) request Rn-i
if (i+2 < n) request Rn-i-2
}
In which one of the following situations is a deadlock possible?
- a)n=40, k=26
- b)n=21, k=12
- c)n=20, k=10
- d)n=41, k=19
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A system has n resources R0,...,Rn-1,and k processes P0,....Pk-1.The i...
Option B is answer
No. of resources, n = 21
No. of processes, k = 12
No. of processes, k = 12
Processes {P0, P1....P11} make the following Resource requests:
{R0, R20, R2, R18, R4, R16, R6, R14, R8, R12, R10, R10}
{R0, R20, R2, R18, R4, R16, R6, R14, R8, R12, R10, R10}
For example P0 will request R0 (0%2 is = 0 and 0< n=21).
Similarly, P10 will request R10.
P11 will request R10 as n - i = 21 - 11 = 10.
As different processes are requesting the same resource, deadlock
may occur.
may occur.
Most Upvoted Answer
A system has n resources R0,...,Rn-1,and k processes P0,....Pk-1.The i...
The implementation of the resource request logic of each process Pi can be described as follows:
1. If i is an even number (i % 2 == 0):
- Process Pi requests resource Ri.
- If resource Ri is available, the process is granted the resource and continues its execution.
- If resource Ri is not available, the process waits until it becomes available.
2. If i is an odd number (i % 2 != 0):
- Process Pi requests resource R(i-1).
- If resource R(i-1) is available, the process is granted the resource and continues its execution.
- If resource R(i-1) is not available, the process waits until it becomes available.
This logic ensures that even-numbered processes request the resource with the same index, while odd-numbered processes request the resource with the index one less than their own.
1. If i is an even number (i % 2 == 0):
- Process Pi requests resource Ri.
- If resource Ri is available, the process is granted the resource and continues its execution.
- If resource Ri is not available, the process waits until it becomes available.
2. If i is an odd number (i % 2 != 0):
- Process Pi requests resource R(i-1).
- If resource R(i-1) is available, the process is granted the resource and continues its execution.
- If resource R(i-1) is not available, the process waits until it becomes available.
This logic ensures that even-numbered processes request the resource with the same index, while odd-numbered processes request the resource with the index one less than their own.
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A system has n resources R0,...,Rn-1,and k processes P0,....Pk-1.The implementation of the resource request logic of each process Pi is as follows:if (i % 2 == 0) { if (i < n) request Ri if (i+2 < n) request Ri+2}else { if (i < n) request Rn-i if (i+2 < n) request Rn-i-2}In which one of the following situations is a deadlock possible?a)n=40, k=26b)n=21, k=12c)n=20, k=10d)n=41, k=19Correct answer is option 'B'. Can you explain this answer?
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