3. Kernel Stack Desirable for security:
e.g., illegal parameters might be supplied

4. What’s on the kernel stack?
Upon entering kernel-mode:
task’s registers are saved on kernel stack
(e.g., address of task’s user-mode stack)
During execution of kernel functions:
Function parameters and return-addresses
Storage locations for ‘automatic’ variables

5. Practice Linux Xinu <--US--| and <--I--<--KS--|
8K per process (thread)
Separate from user stack
PCB contains pointer to the kernel stack
Xinu
<--I--<--KS--<--US--|
On top of user stack

6. PA0

7. PA0. 0-1 0. Step 2, `cs-status | head -1 | sed 's/://g'`
Step 6, cs-console, OR (control-spacebar)
.section data .section text .globl zfunction zfunction: pushl %ebp movl %esp, %ebp …. Leave ret Read In C, we count from 0

8. PA0. 2-3
Try “man end” and see what you can get Use “kprint” for output
Read “Print the address of the top of the run- time stack for whichever process you are currently in, right before and right after you get into the printos() function call.” carefully You can use in-line assembly Use ebp, esp

9. Others Course Workspace will also be afs based
Course Workspace will also be afs based
Moodle forum is up, please use it

10. Lecture 3 Scheduling

11. Workload Assumptions
1. Each job runs for the same amount of time. 2. All jobs arrive at the same time. 3. Once started, each job runs to completion. 4. All jobs only use the CPU (i.e., they perform no I/O). 5. The run-time of each job is known.

12. Tturnaround = Tcompletion − Tarrival
Scheduling Metrics
Performance: turnaround time
Tturnaround = Tcompletion − Tarrival
As Tarrival is now 0, Tturnaround = Tcompletion

13. First In, First Out Work well under our assumption
Relax “Each job runs for the same amount of time”
Convoy effect
120
100 20 40 60 80 20
110
Grocery, be nice
120
100 20 40 60 80

14. Shortest Job First SJF would be optimal
Relax “All jobs arrive at the same time.”
120
100 20 40 60 80 50
103.33
B/C
arrive
120
100 20 40 60 80 A B C

15. Shortest Time-to-Completion First
STCF is preemptive, aka PSJF
“Once started, each job runs to completion” relaxed
B/C
arrive
120
100 20 40 60 80 A B C A 50

16. Scheduling Metrics Performance: turnaround time
Tturnaround = Tcompletion − Tarrival
As Tarrival is now 0, Tturnaround = Tcompletion
Performance: response time
Tresponse = Tfirstrun − Tarrival

17. Turnaround time or response time
FIFO, SJF, or STCF
Round robin
120
100 20 40 60 80
R 40 8
120
100 20 40 60 80

18. Conflicting criteria Minimizing response time
requires more context switches for many processes
incur more scheduling overhead
decrease system throughput
Increase turnaround time
Scheduling algorithm depends on nature of system
Batch vs. interactive
Designing a generic AND efficient scheduler is difficult

19. Incorporating I/O Poor use of resources
Overlap allows better use of resources
CPU
Disk A A A A B A A A
120
100 20 40 60 80
CPU
Disk A B A B A B A B A A A
120
100 20 40 60 80

20. Multi-level feedback queue
Goal
Optimize turnaround time without priori knowledge
Optimize response time for interactive users Q6 Q5 Q4 Q3 Q2 Q1 A B
Rule 1: If Priority(A) > Priority(B)
A runs (B doesn’t).
Rule 2: If Priority(A) = Priority(B)
A & B run in RR.
1 and 2 are not enough C D

21. How to Change Priority
Rule 3: When a job enters the system, it is placed at the highest priority (the topmost queue).
Rule 4a: If a job uses up an entire time slice while running, its priority is reduced (i.e., it moves down one queue).
Rule 4b: If a job gives up the CPU before the time slice is up, it stays at the same priority level.

22. Examples Problems? 120 100 20 40 60 80 Q2 Q1 Q0 A B Q2 Q1 Q0 B B B B B
120
100 20 40 60 80 Q2 Q1 Q0 A B Q2 Q1 Q0 B B B B B B B B B B B B A A A A A A A A A A A A
120
100 20 40 60 80
Problems?

23. Priority Boost Q2 Q1 Q0 Q2 Q1 Q0 A A A A A A A
120
100 20 40 60 80
120
100 20 40 60 80
Rule 5: After some time period S, move all the jobs in the system to the topmost queue.

24. Gaming the scheduler Q2 Q1 Q0 Q2 Q1 Q0 A A A A A A A A A A A A A A A AB B B B B B B B B B B B B B B B B B B B B B B
120
100 20 40 60 80
120
100 20 40 60 80

25. Better Accounting
Rule 4a: If a job uses up an entire time slice while running, its priority is reduced (i.e., it moves down one queue).
Rule 4b: If a job gives up the CPU before the time slice is up, it stays at the same priority level.
Rule 4: Once a job uses up its time allotment at a given level (regardless of how many times it has given up the CPU), its priority is reduced (i.e., it moves down one queue).

26. Tuning MLFQ And Other Issues
How to parameterize?
The system administrator configures it
The users provides hints

27. Workload Assumptions
1. Each job runs for the same amount of time. 2. All jobs arrive at the same time. 3. Once started, each job runs to completion. 4. All jobs only use the CPU (i.e., they perform no I/O). 5. The run-time of each job is known.

28. MLFQ rules
Rule 1: If Priority(A) > Priority(B), A runs (B doesn’t).
Rule 2: If Priority(A) = Priority(B), A & B run in RR.
Rule 3: When a job enters the system, it is placed at the highest priority (the topmost queue).
Rule 4: Once a job uses up its time allotment at a given level (regardless of how many times it has given up the CPU), its priority is reduced (i.e., it moves down one queue).
Rule 5: After some time period S, move all the jobs in the system to the topmost queue.

29. Scheduling Metrics Performance: turnaround time
Tturnaround = Tcompletion − Tarrival
As Tarrival is now 0, Tturnaround = Tcompletion
Performance: response time
Tresponse = Tfirstrun − Tarrival
CPU utilization
Throughput
Fairness

30. A proportional-share or A fair-share scheduler
Each job obtain a certain percentage of CPU time.
Lottery scheduling tickets
to represent the share of a resource that a process should receive
If A 75 tickets, B 25 tickets, then 75% and 25% (probabilistically)
A B A A B A A A A A A B A B A A A A A A
higher priority => more tickets

31. Lottery Code
int counter = 0; Int winner = getrandom(0, totaltickets); node_t *current = head; while(current) { counter += current->tickets; if (counter > winner) break; current = current->next; } // current is the winner

32. Ticket currency
User A 100 (global currency) -> 500 (A’s currency) to A1 -> 50 (global currency) -> 500 (A’s currency) to A2 User B 100 (global currency) -> 10 (B’s currency) to B1 -> 100 (global currency)

33. More on Lottery Scheduling
Ticket transfer
Ticket inflation
Compensation ticket
How to assign tickets?
Why not Deterministic?

34. Next
Holiday next Monday
Work on PA0
Reading assigned later tonight