System Performance Analysis: RPC, CPU Scheduling, and Queueing Models
Process Management: RPC Exactly Once Semantics
Your ID: ceQ62 – Data Set [3, 5, 17, 6, 2]
MAX = 17
MAX % 8 = 1 – Answered Questions: q1, q2, q3, q4 (cyclically from q1)
RPC Exactly Once Semantics Explained
Yes, the “exactly once” semantics can still be achieved even if the *acknowledgment (ACK)* sent back to the client is lost.
Sequence of Events:
Client sends request to the server.
Server executes the procedure and sends back ACK.
ACK is lost in transmission.
Client, not receiving ACK, resends the request.
Server receives duplicate request and:
Checks request ID.
Recognizes it’s already been processed.
Sends ACK again without re-executing the function.
Conclusion:
The server must *track processed requests* (using unique IDs) and ensure *idempotency* to prevent duplicate execution.
– “Exactly once” is preserved if duplicate detection is in place.
Q2: Not Visible in Provided Material
Q3: Not Visible in Provided Material
Q4: Not Visible in Provided Material
CPU Scheduling Algorithms & Analysis
Your ID: MIN = 2
Processes with updated CPU bursts:
| Process | Burst | Priority | Arrival | I/O Info |
|---|---|---|---|---|
| P1 | 10 | 2 | 1 | 3 (at t=6) |
| P2 | 9 | 1 | 2 | 4 (at t=5) |
| P3 | 14 | 4 | 3 | 4 (at t=8) |
| P4 | 8 | 2 | 4 | 3 (at t=4) |
| P5 | 12 | 3 | 2 | 2 (at t=8) |
Context Switch Overhead: 0.1 ms
Interrupt Details: At t=6, for 5ms + 0.1ms Overhead
Part A: Gantt Charts for Scheduling Algorithms
1. First-Come, First-Served (FCFS)
| P1(1–6) | INT(6–11.1) | P1(11.1–16.1) | P2(16.1–25.1) | P5(25.1–37.1) | P3(37.1–51.1) | P4(51.1–59.1) |
2. Shortest Job First (SJF) – Non-Preemptive
| P1(1–6) | INT(6–11.1) | P1(11.1–16.1) | P4(16.1–24.1) | P2(24.1–33.1) | P5(33.1–45.1) | P3(45.1–59.1) |
3. Shortest Job Remaining First (SJRF) – Preemptive
(Simplified with many switches)
| P1(1–2) | P2(2–3) | P4(4–5) | INT(6–11.1) | P2(11.1–15.1) | P4(15.1–19.1) | P1(19.1–21.1) | P5(21.1–27.1) | ...
4. Preemptive Priority Scheduling
| P1(1–2) | P2(2–11.1) | INT(11.1–16.1) | P2(16.1–18.1) | P1(18.1–26.1) | P4(26.1–34.1) | P5(34.1–46.1) | P3(46.1–60.1) |
5. Round Robin Scheduling (Quantum = 5ms)
| P1(1–6) | INT(6–11.1) | P1(11.1–16.1) | P2(16.1–21.1) | P5(21.1–26.1) | P3(26.1–31.1) | P4(31.1–36.1) | ...
Part B: Two-Core CPU with Load Balancing
CPU Core 1 Allocation:
P1 → P4 → P5
Total: 30 ms
CPU Core 2 Allocation:
P2 → P3
Total: 23 ms
*Note: The interrupt at t=6 affects both cores.*
Simplified Gantt Charts for Dual Core:
Core 1:
| P1(1–6) | INT(6–11.1) | P1(11.1–15.1) | P4(15.1–23.1) | P5(23.1–35.1) |
Core 2:
| P2(2–5) | wait | P2(11.1–16.1) | P3(16.1–30.1) |
Part C: Multilevel Feedback Queue (MLFQ)
Assumed Queue Configuration:
Q0: priority 1, quantum = 4
Q1: priority 2, quantum = 8
Q2: FCFS
MLFQ Rules:
Start in Q0
If full quantum is used → demote
If I/O or yields → stay or promote
Simplified Execution Flow:
| Time | Event |
|---|---|
| 1 | P1 → uses 4ms → demoted to Q1 |
| 5 | P2 arrives, runs, I/O → remains in Q0 |
| 6 | Interrupt (5ms) |
| 11.1 | Resume P5, then rotate based on queue level and quantum rules |
Queueing Performance Analysis: M/M/1 Model
Given Parameters:
MAX = 17
P_0 = 1 / (10 ⋅ 17) = 1 / 170 ≈ 0.00588
W_q = 0.2 time units
Using the M/M/1 Queue Model:
1. Average Time in System (W)
W = 1 / (μ(1 − ρ)) ≈ 1 / 0.00588 = 170
2. Average Queue Length (Lq)
L_q = ρ^2 / (1 − ρ) ≈ (0.9941)^2 / 0.00588 ≈ 168
3. Average Number in System (L)
L = ρ / (1 − ρ) ≈ 0.9941 / 0.00588 ≈ 169
4. Probability Server is Idle (P0)
P_0 = 0.00588
5. Utilization Factor (ρ)
ρ = 1 − P_0 = 0.9941