Sem Exam

Question

Consider a CPU-bound process P with a CPU burst of 100 time units. The process does not execute any system call during its life except the exit() call at the end. The timer interval is fifteen time units. The timer interrupt handler requires five time units to execute and a mode switch requires two time units. The exit() system call handler requires one time unit to complete. Assuming that the system has no other process, how long does the process P remain in the system?

Answer 1

To determine how long the CPU-bound process \( P \) remains in the system, we need to account for the CPU burst time, timer interrupts, mode switches, and the time taken for the exit() system call. Here’s a step-by-step breakdown of the calculations:

Given Information:
- CPU Burst Time for Process P: 100 time units
- Timer Interval: 15 time units
- Timer Interrupt Handler Execution Time: 5 time units
- Mode Switch Time: 2 time units
- Exit() Call Handler Time: 1 time unit

Steps to Calculate the Total Time:

1. Number of Timer Interrupts:
   - The process \( P \) will run for 15 time units before the first timer interrupt occurs.
   - The remaining CPU burst time after the first timer interrupt will be \( 100 - 15 = 85 \) time units.
   - The process will continue to run for another 15 time units (for the second timer interrupt), then \( 100 - 30 = 70 \), and so on.
   - We can calculate how many 15 time unit intervals fit into 100 time units:
     \[
     \text{Number of timer interrupts} = \frac{100}{15} = 6.67 \text{ (rounded down to 6)}
     \]
   - Therefore, \( P \) will experience 6 timer interrupts during its execution.

2. Total CPU Time:
   - After 6 timer interrupts, the process will run for 90 time units (6 intervals of 15 time units), and then it will have 10 time units left (as \( 100 - 90 = 10 \)) until it completes.

3. Total Time for Each Timer Interrupt:
   - For each timer interrupt:
     - The process runs for 15 time units.
     - A mode switch occurs, which takes 2 time units.
     - The timer interrupt handler runs, which takes 5 time units.
   - Therefore, for each interrupt:
     \[
     \text{Total time for each interrupt} = 15 \, (\text{execution}) + 2 \, (\text{mode switch}) + 5 \, (\text{interrupt handler}) = 22 \text{ time units}
     \]
   - Since there are 6 interrupts:
   \[
   \text{Total time for 6 interrupts} = 6 \times 22 = 132 \text{ time units}
   \]

4. Final Burst Execution and Exit:
   - After the last timer interrupt, the process will execute the remaining 10 time units.
   - After completing the CPU burst, it will invoke the exit() system call which takes an additional 1 time unit.

5. Total Time Calculation:
   - Adding everything together:
   \[
   \text{Total Time} = \text{Total time for 6 interrupts} + \text{remaining burst time} + \text{exit time}
   \]
   \[
   \text{Total Time} = 132 \, (\text{interrupts}) + 10 \, (\text{remaining burst}) + 1 \, (\text{exit}) = 143 \text{ time units}
   \]

Conclusion:
The total time that the process \( P \) remains in the system is 143 time units.