Number of processes OS switches context from A to B main use fork, and the child process call execvp() Can compiler do something bad by adding privileged instructions?

Lecture 3 Scheduling

How to develop scheduling policy What are the key assumptions? What metrics are important? What basic approaches have been used in the earliest of computer systems?

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.

Scheduling Metrics Performance: turnaround time T turnaround = T completion − T arrival As T arrival is now 0, T turnaround = T completion

First In, First Out Work well under our assumption Relax “Each job runs for the same amount of time” Convoy effect

Shortest Job First SJF would be optimal Relax “All jobs arrive at the same time.” ABC B/C arrive

Shortest Time-to-Completion First STCF is preemptive, aka PSJF “Once started, each job runs to completion” relaxed ABC B/C arrive A

Scheduling Metrics Performance: turnaround time T turnaround = T completion − T arrival As T arrival is now 0, T turnaround = T completion Performance: response time T response = T firstrun − T arrival

Turnaround time or response time FIFO, SJF, or STCF Round robin

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

Incorporating I/O Poor use of resources Overlap allows better use of resources CPU Disk A A A A A A A B CPU Disk A A A A A A A B BB B

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.

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 C D Rule 1: If Priority(A) > Priority(B) A runs (B doesn’t). Rule 2: If Priority(A) = Priority(B) A & B run in RR.

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.

Example Q2 Q1 Q0 A A A B B A

Example with I/O Q2 Q1 Q0 B A B A B A B A B A B A B A B A B A B A B A B A Problems: Starvation Program can game the scheduler Program may change its behavior over time

Priority Boost Rule 5: After some time period S, move all the jobs in the system to the topmost queue Q2 Q1 Q0 A A A Q2 Q1 Q0 A A A A

Gaming the scheduler Q2 Q1 Q Q2 Q1 Q0 B A B A B A B A B A B A B A B A B A B A B A B AA B A AA BB AA BB AA BB AA BB AA BB

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).

Tuning MLFQ And Other Issues How to parameterize? The system administrator configures it Default values available: on Solaris, there are 60 queues time-slice 20 milliseconds (highest) to 100s milliseconds (lowest) priorities boosted around every 1 second or so. The users provides hints: command-line utility nice

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.

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.

Scheduling Metrics Performance: turnaround time T turnaround = T completion − T arrival As T arrival is now 0, T turnaround = T completion Performance: response time T response = T firstrun − T arrival CPU utilization Throughput Fairness

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

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

Lottery Fairness Study

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

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

Stride Scheduling: a deterministic fair-share scheduler Deterministic but requires global state What if a new job enters in the middle

Scheduling Workload assumption Metrics MLFQ Lottery Scheduling and stride scheduling

Next Work on PA0 Reading: chapter 12-16

PA0

PA Step 2, `cs-status | head -1 | sed 's/://g'` Step 6, cs-console, OR (control-spacebar) 1..section.data.section.text.globl zfunction zfunction: pushl %ebp movl %esp, %ebp …. Leave ret Read In C, we count from 0http://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax

PA0. 2,3, and 5 2.Try “man end” and see what you can get Use “kprintf” for output 3.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 5.syscallsummary_start(): should clear all numbers syscallsummary_stop(): should keep all numbers

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