FOR FREE MATERIALS
- Introduction: OS
- Booting Process
- Command Interpreter
- Dual mode of operation
- System Calls
- Difference between system call and library call
- Interrupt and Trap
- Introduction
- Scheduling Criteria, Terms and Concepts
- Different Type Scheduling Algorithms
- Longest Job First
- Longest Remaining Time First
- Highest Response Ratio Next (HRRN)
- Priority Scheduling
- Least Completed Next (LCN) Scheduling
- Multi-Level Queue Scheduling
- Multilevel Feedback Queue Scheduling
- Guaranteed Scheduling
- Fair Share Scheduling
- Lottery Scheduling
- Multi-Processor Scheduling
- Deadline Scheduling
- Rate-Monotonic Scheduling
- POSIX Real-Time Scheduling
- Priority Inversion
- Proportional Share Scheduling
- Comparison table of Scheduling Algorithm
- Algorithm Evaluation for Scheduling
- Important concept of MM
- Memory Representation as Address
- Memory layout for a Process
- Logical and Physical Address Space
- Necessity of Memory Management
- Introduction
- Concept of Non-contiguous memory allocation
- Basic concept of Paging
- Address Translation
- Process of Paging Method
- Calculation of Offset for paging
- Calculation of Logical Address Bit and number of Pages
- Calculation of Physical Address Bit and number of Frames
- Calculation of Page Table Size
- Organization of MMU
- Why Page size power of 2
- Paging Address Calculation: example1
- Paging Numeric Problem: Example2
- Paging Address Calculation: Example3
- Paging Address Calculation: Question1
- Some Important Question 2
- Paging Address Translation: Question3
- UGC NET Paging Question
- GATE Paging Question
- Number of information at Page Entry
- Exams Question on Page Entry Information
- Exam Question: Page Entry Information
- Calculates Page Table Entry Size
- Calculates Page Size
- Calculation of Optimal Page Size
- Question on Optimal Page Size
- Hashed Page Table with Example
- Inverted Page Table with Gate Question
- Question on Inverted Page Table
- Multi-Level Paging
- Multi-level paging: Example1
- Multi-level paging: Example2
- Allocation of Frames
- Subject Mock
Counting semaphore: Example
Example 1: Counting Semaphore
Let capacity of critical section is 3 i.e. S.value = 3 (four process can enter Critical Section).
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1. First P1 try to enter into its critical section, it initiated Wait() function and before it go to critical section, run line 1 (S.value = S.value - 1) , now S.value = 2. Then run line 2 (if (S.value < 0), not satisfied so, process go to critical section and continue.
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2. Now P2 come, first run line 1 (S.value = S.value - 1), S.value = 1 and enter into critical section because line 2 (if (S.value < 0)) not satisfied.
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3. Now P3 come, first run line 1 (S.value = S.value - 1), S.value = 0 and enter into critical section because line 2 (if(S.value < 0) not satisfied.
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4. Now P4 come, first-run line 1 (S.value = S.value-1), S.value = -1 and then line 2 (if (S.value < 0) is satisfied. Not allowed the process to enter into critical section (critical section is full). Put the process P4 at the suspended queue list and sleep (block) the process P4.
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# In counting semaphore increment and decrement signified by wait and signal operation.
But it can be identified two more ways up and down or V and P.
Q. At a particular time, the value of a counting semaphore is 10, it will become 7 after
a. 3V
b. 3P
c. 5V & 2P
d. 13p & 10V
Which one is correct:
A. a, b
B. b, c
C. b, d
D. d
Answer:
Counting semaphore = 10. After which operation it will be 7.
Case 1:
One thing is very clear that counting semaphore variable value reduce from 10 to 7 means it will be a 3P or Down operation.
10 – 3 = 7 so, after 3P operation it will become 7. Option B correct
Case 2:
If the counting semaphore variable reduces from 10 to 7 means it will be a V or Up operation along with a P or Down operation or vice-versa but in such a way that it reduces by 3.
It will be possible by only option D. (13P and 10V)
First 13P operation: 10 – 13 = -3, then 10V operation: -3 + 10 = 7
So, Correct answer: C. b, d
