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SY CE/CSE ISE even 2026

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S1 = 1, S2 = 2

Processes:

  • P1: wait(S1), wait(S2)

  • P2: wait(S2), wait(S1)

  • P3: wait(S2)

Maximum processes in deadlock:

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Consider:

P1: wait(A); wait(B);

P2: wait(B); wait(A);

This leads to:

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Consider contiguous allocation of physical memory to processes using variable partitioning scheme. Suppose there are 88 holes in the memory of sizes 20 straight K straight B comma 4 text end text straight K straight B20KB,4 KB25 text end text straight K straight B comma 18 text end text straight K straight B comma 7 text end text straight K straight B comma 9 text end text straight K straight B comma 15 text end text straight K straight B25 KB,18 KB,7 KB,9 KB,15 KB, and 1212 KB. Assume that no two holes are adjacent. Two processes text P1 end textP1 of size 1616 KB and text P2 end textP2 of size 99 KB arrive in that order, and they are allocated memory using the best-fit technique. After allocating space to text P1 end textP1 and text P2 end textP2, the number of holes of size less than 88 KB is straight _ straight _ straight _ straight _ straight _ straight _ straight _. (answer in integer)

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What are the advantages of multithreading?

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Four processes have maximum requirements of 6, 4, 5, and 3 resources respectively.

Find the maximum number of resources that can lead to deadlock.

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Semaphore S = 2.

Processes P1, P2, P3, P4 execute

wait(S) simultaneously.

After all operations, how many processes are blocked?

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What distinguishes time-sharing from multiprogramming?
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