a) .Consider two resources are available and value of semaphore is 2.  Now consider operations P() and V() and their execution sequence P(), P(), V(), P(), P(), V(), P(), P(), P(), V(), V(), P(). Answer the followings after the completion of above functions:

i.Number of the process which is sleeping on P()?

ii. What will be the value of semaphore?

iii. Number of available resources?

b) Find the value of logical address and limit register. The value of the base register is 400K and the value of the physical address is 532k.

Transcribed Image Text:

Interrupt 800K Partition 4 700K Limit Register Comparator Partition 3 400K Partition 2 Base Register Adder 200K Partition 1 100K Memory Management Unit (MMU) Operating System Logical Address Physical Address Interrupt 800K Partition 4 700K Limit Register Comparator Partition 3 400K Partition 2 Base Register Adder 200K Partition 1 100K Memory Management Unit (MMU) Operating System Logical Address Physical Address

Answer rating: 100% (QA)

Solution A The code for P operation is P ... View the full answer