CS 537 Spring 2020, Project 2b: xv6 Scheduler

FAQ

Administrivia

Objectives

Overview

In this project, you’ll be implementing a simplified multi-level feedback queue (MLFQ) scheduler in xv6. Here is the overview of MLFQ scheduler in this assignment:

You will be implementing 2 system calls:

Reading:

xv6 scheduler code details

Most of the code for the scheduler is quite localized and can be found in kernel/proc.c, where you should first look at the routine scheduler(). It’s essentially looping forever and for each iteration, it looks for a runnable process across the ptable. If there are multiple runnable processes, it will select one according to some policy. The vanilla xv6 does no fancy things about the scheduler; it simply schedules processes for each iteration in a round-robin fashion. For example, if there are three processes A, B and C, then the pattern under the vanilla round-robin scheduler will be A B C A B C … , where each letter represents a process scheduled within a timer tick, which is essentially ~10ms, and you may assume that this timer tick is equivalent to a single iteration of the for loop in the scheduler() code. Why 10ms? This is based on the timer interrupt frequency setup in xv6 and you may find the code for it in kernel/timer.c. You can also find a code walkthrough in the discussion video.

Now to implement MLFQ, you need to schedule the process for some time-slice, which is some multiple of timer ticks. For example, if a process is on the highest priority level, which has a time-slice of 8 timer ticks, then you should schedule this process for ~80ms, or equivalently, for 8 iterations.

xv6 can perform a context-switch every time a timer interrupt occurs. For example, if there are 2 processes A and B that are running at the highest priority level (queue 3), and if the round-robin time slice for each process at level 3 (highest priority) is 8 timer ticks, then if process A is chosen to be scheduled before B, A should run for a complete time slice ( 8 timer ticks or ~80ms) before B can run. Note that even though process A runs for 8 timer ticks, every time a timer tick happens, process A will yield the CPU to the scheduler, and the scheduler will decide to run process A again (until it’s time slice is complete).

To make your life easier and our testing easier, you should run xv6 on only a single CPU . Make sure in your Makefile CPUS := 1.

MLFQ implementation details

Your MLFQ scheduler must follow these very precise rules:

  1. Four priority levels.
  2. Number the queues from 3 for highest priority queue down to 0 for lowest priority queue.
  3. Whenever the xv6 timer tick (10 ms) occurs, the highest priority ready process is scheduled to run.
  4. The highest priority ready process is scheduled to run whenever the previously running process exits, sleeps, or otherwise yields the CPU.
  5. The time-slice associated with priority queues are as follows:

    • Priority queue 3: 8 timer ticks (or ~80ms)
    • Priority queue 2: 16 timer ticks
    • Priority queue 1: 32 timer ticks
    • Priority queue 0 it executes the process until completion.
  6. If there are more than one processes on the same priority level, then you scheduler should schedule all the processes at that particular level in a round-robin fashion.
  7. The round-robin slices differs for each queue and is as follows:
    • Priority queue 3 slice : 1 timer ticks
    • Priority queue 2 slice : 2 timer ticks
    • Priority queue 1 slice : 4 timer ticks
    • Priority queue 0 slice : 64 timer ticks
  8. When a new process arrives, it should start at priority 3 (highest priority).
  9. At priorities 3, 2, and 1, after a process consumes its time-slice it should be downgraded one priority. At priority 0, the process should be executed to completion.
  10. If a process voluntarily relinquishes the CPU before its time-slice expires at a particular priority level, its time-slice should not be reset; the next time that process is scheduled, it will continue to use the remainder of its existing time-slice at that priority level.
  11. To overcome the problem of starvation, we will implement a mechanism for priority boost. If a process has waited 10x the time slice in its current priority level, it is raised to the next higher priority level at this time (unless it is already at priority level 3). For the queue number 0 (lowest priority) consider the maximum wait time to be 6400ms which equals to 640 timer ticks.
  12. To make the scheduling behavior more visible you will be implementing a system call that boosts a process priority by one level (unless it is already at priority level 3).

Create new system calls

You’ll need to create two system call for this project:

1. int getprocinfo(struct pstat *)

Because your MLFQ implementations are all in the kernel level, you need to extract useful information for each process by creating this system call so as to better test whether your implementation works as expected.

To be more specific, this system call returns 0 on success and -1 on failure. If success, some basic information about each process: its process ID, how many timer ticks have elapsed while running in each level, which queue it is currently placed on (3, 2, 1, or 0), and its current procstate (e.g., SLEEPING, RUNNABLE, or RUNNING) will be filled in the pstat structure as defined

struct pstat {
  int inuse[NPROC]; // whether this slot of the process table is in use (1 or 0)
  int pid[NPROC];   // PID of each process
  int priority[NPROC];  // current priority level of each process (0-3)
  enum procstate state[NPROC];  // current state (e.g., SLEEPING or RUNNABLE) of each process
  int ticks[NPROC][4];  // number of ticks each process has accumulated at each of 4 priorities
  int wait_ticks[NPROC][4]; // number of ticks each process has waited before being scheduled
};

The file can be seen pstat.h. Do not change the names of the fields in pstat.h.

Action items to successfully add pstat.h:

2. int boostproc(void)

This system call simply increases the priority of current process one level unless it is already in queue number 3.

[EDITED] for simplicity this system call returns 0 unconditionally.

In reality existence of this system call makes the system susceptible to attack, where the process increases its priority right after it is demoted. But for the scope of this assignment it’ll make it easier for you to test if things are working correctly!

Tips

Code

We suggest that you start from the source code of xv6 at ~cs537-1/xv6-sp20 , instead of your own code from p1b as bugs may propagate and affect this project.

> cp -r ~cs537-1/xv6-sp20 ./

Testing

Testing is critical. Testing your code to make sure it works is crucial. Writing testing scripts for xv6 is a good exercise in itself, however, it is a bit challenging. As you may have noticed from p1b, all the tests for xv6 are essentially user programs that execute at the user level.

Basic Test

You can test your MLFQ scheduler by writing workloads and instrumenting it with getprocinfo systemcall. Following is an example of a user program which spins for a user input iterations. You can download this example here.

#include "types.h"
#include "stat.h"
#include "user.h"
#include "pstat.h"

int
main(int argc, char *argv[])
{
    struct pstat st;

    if(argc != 2){
        printf(1, "usage: mytest counter");
        exit();
    }

    int i, x, l, j;
    int mypid = getpid();

    for(i = 1; i < atoi(argv[1]); i++){
        x = x + i;
    }

    getprocinfo(&st);
    for (j = 0; j < NPROC; j++) {
        if (st.inuse[j] && st.pid[j] >= 3 && st.pid[j] == mypid) {
            for (l = 3; l >= 0; l--) {
                printf(1, "level:%d \t ticks-used:%d\n", l, st.ticks[j][l]);
            }
        }
    }
    
    exit();
    return 0;
}

If you run ticks-test 10000000 the expected output is something like below:

level:3      ticks-used:8
level:2      ticks-used:16
level:1      ticks-used:32
level:0      ticks-used:160

The ticks used on the last level will be somewhat unpredictable and it may vary. However, on most machines, we should be able to see that the ticks used at levels 3, 2, and 1 as 8, 16 and 32 respectively. As there are no other programs are running, there won’t be any boost for this program after it reaches bottom queue (numbered 0) (It won’t wait for 500 ticks). If you invoke with small counter value such as 10 (ticks-test 10), then the output should be like this:

level:3      ticks-used:1
level:2      ticks-used:0
level:1      ticks-used:0
level:0      ticks-used:0

Write Your Own Tests

These tests are by no means exhaustive. It is important (and a good practice) to write your own test programs. Try different tests which will exercise the functionality of the scheduler by using fork(), sleep(n), and other methods.

Submitting Your Implementation

  1. Please create a subdirectory ~cs537-1/handin/LOGIN/p2b/src
mkdir -p ~cs537-1/handin/LOGIN/p2b/src 

To submit your solution, copy all of the xv6 files and directories with your changes into ~cs537-1/handin/<cs-login>/p2b/src. One way to do this is to navigate to your solution’s working directory and execute the following command:

cp -r . ~cs537-1/handin/<cs-login>/p2b/src/

Consider the following when you submit your project: