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UNB/Year 4/Semester 1/CS3418/10-6-2023.md
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Lecture Topic: Scheduling
Scheduling Criteria:
- CPU Utilization
- Throughput
- Turnaround Time
- Waiting Time: Time spent in ready queue
- Response Time
Multiprocessing Scheduler Design Choices:
- One Queue for all processes
- A queue per process
CPU: Homogeneous or heterogenous
| CPU 0 | CPU 1 |
| ----- | ----- |
| Cache | Cache |
| ISA | ISA |
Affinity:
Soft - May be scheduled on multiple CPUs
Hard - Has to be scheduled on a designated CPU
Load Balancing: Keeping jobs evenly distributed
- Job pushing: Jobs may be pushed to another CPU
- Job stealing: Unused CPU may steal jobs from the queue
CPU alternates between computation and input output
Load stalls minimization
A load stall is when the CPU is idle when memory access is being performed
Instruction reordering
The CPU may perform operations out of order to better align with CPU computation and input output cycles, to minimize load stalls
Hyper threading:
CPU bursts are most frequently short
There are different kinds of algorithms that solve a few kinds of problems:
- First Come First Serve: When job arrives it immediately gets worked on
- Tie Breaker: When two jobs arrive at the same time
So, P0 comes first, but since we have a tie between P1 and P2. In the first come first serve case:
| Process | Arrival | Duration | Wait Time | Response Time | Turnaround |
| ------- | ------- | -------- | --------- | ------------- | ---------- |
| P0 | 0 | 3 | 0 | 0 | 3 |
| P1 | 1 | 20 | 2 | 2 | 22 |
| P2 | 1 | 5 | 22 | 22 | 27 |
So the average wait time is
(0 + 2 + 22)/3 = 8
If there are different arrival times:
| Process | Arrival | Duration | Wait | Turnaround |
| ------- | ------- | -------- | ---- | ---------- |
| P0 | 0 | 3 | 0 | 3 |
| P1 | 2 | 20 | 6 | 26 |
| P2 | 1 | 5 | 2 | 7 |
In this case the average wait time is
(0 + 6 + 2)/3 = 2.666666...
Much better.
We can reorder jobs to reduce the wait time for jobs
The turnaround time also improves (do the calculation in your head dummy)
The throughput remains the same however
Kinds of algorithms:
Shortest Job first
| Process | Arrival | Duration | Wait Time | Response Time | Turnaround |
| ------- | ------- | -------- | --------- | ------------- | ---------- |
| P0 | 0 | 9 | | | |
| P1 | 1 | 15 | | | |
| P2 | 1 | 7 | | | |
The Gantt Chart
| | P0 | | P2 | | P3 | |
| --- | --- | --- | --- | --- | --- | --- |
| 0 | | 9 | | 16 | | 31 |
Lecture Topic: Scheduling
Scheduling Criteria:
- CPU Utilization
- Throughput
- Turnaround Time
- Waiting Time: Time spent in ready queue
- Response Time
Multiprocessing Scheduler Design Choices:
- One Queue for all processes
- A queue per process
CPU: Homogeneous or heterogenous
| CPU 0 | CPU 1 |
| ----- | ----- |
| Cache | Cache |
| ISA | ISA |
Affinity:
Soft - May be scheduled on multiple CPUs
Hard - Has to be scheduled on a designated CPU
Load Balancing: Keeping jobs evenly distributed
- Job pushing: Jobs may be pushed to another CPU
- Job stealing: Unused CPU may steal jobs from the queue
CPU alternates between computation and input output
Load stalls minimization
A load stall is when the CPU is idle when memory access is being performed
Instruction reordering
The CPU may perform operations out of order to better align with CPU computation and input output cycles, to minimize load stalls
Hyper threading:
CPU bursts are most frequently short
There are different kinds of algorithms that solve a few kinds of problems:
- First Come First Serve: When job arrives it immediately gets worked on
- Tie Breaker: When two jobs arrive at the same time
So, P0 comes first, but since we have a tie between P1 and P2. In the first come first serve case:
| Process | Arrival | Duration | Wait Time | Response Time | Turnaround |
| ------- | ------- | -------- | --------- | ------------- | ---------- |
| P0 | 0 | 3 | 0 | 0 | 3 |
| P1 | 1 | 20 | 2 | 2 | 22 |
| P2 | 1 | 5 | 22 | 22 | 27 |
So the average wait time is
(0 + 2 + 22)/3 = 8
If there are different arrival times:
| Process | Arrival | Duration | Wait | Turnaround |
| ------- | ------- | -------- | ---- | ---------- |
| P0 | 0 | 3 | 0 | 3 |
| P1 | 2 | 20 | 6 | 26 |
| P2 | 1 | 5 | 2 | 7 |
In this case the average wait time is
(0 + 6 + 2)/3 = 2.666666...
Much better.
We can reorder jobs to reduce the wait time for jobs
The turnaround time also improves (do the calculation in your head dummy)
The throughput remains the same however
Kinds of algorithms:
Shortest Job first
| Process | Arrival | Duration | Wait Time | Response Time | Turnaround |
| ------- | ------- | -------- | --------- | ------------- | ---------- |
| P0 | 0 | 9 | | | |
| P1 | 1 | 15 | | | |
| P2 | 1 | 7 | | | |
The Gantt Chart
| | P0 | | P2 | | P3 | |
| --- | --- | --- | --- | --- | --- | --- |
| 0 | | 9 | | 16 | | 31 |
The problem with this kind of algorithm is determining the duration of jobs