Download lab2 source, unpack and examine the source for weensyos2 using the following commands:
% tar xvzf weensyos2.tgz % ls weensyos2 COPYRIGHT lib.c schedos-3.c schedos-kern.h x86.h GNUmakefile lib.h schedos-4.c schedos-loader.c x86struct.h answers.txt mergedep.pl schedos-app.h schedos-symbols.ld x86sync.h bootstart.S mkbootdisk.c schedos-boot.c schedos-x86.c conf schedos-1.c schedos-int.S schedos.h elf.h schedos-2.c schedos-kern.c types.h %
Change into the weensyos2 directory and run gmake run-schedos.
This will build and run the single operating system you'll use in
WeensyOS 2, the "scheduler OS" or SchedOS. As before, this will start up
Bochs, but not the emulated computer. To start the emulated computer, type
"c" at the <bochs:1>
prompt.
After a moment you should see a window like this:
The SchedOS consists of a kernel and four user processes. The processes
are extremely simple: the schedos-1
process prints 320 red
"1
"s, the schedos-2
process prints 320 green
"2
"s, and so forth. Each process yields control to the kernel
after each character, so that the kernel can choose another process to run.
After printing all 320 characters, each process exits.
The four processes
coordinate their printing with a shared variable, cursorpos
,
located at memory address 0x190000. Note that just like Minilab 1, these mini-processes share
data/code segment with each other so the modification of cursorpos
in one process will be reflected in another process.
The kernel initializes
cursorpos
to point at address 0xB8000, the start of CGA
console memory. Processes write their characters into
*cursorpos
, and then increment cursorpos
to the
next position.
Read and understand the SchedOS process code.
Specifically, read and understand schedos-1.c
.
Read and understand the comments in
schedos-app.h
. The basic feeling should be familiar to you
from WeensyOS 1.
The kernel's job is very simple. At boot time, it initializes the hardware,
initializes a process descriptor for each process, and runs the first
process. At that point it loses control of the machine until a system call
or interrupt occurs. System calls and interrupts restart the kernel by
effectively calling the trap
function. Note that this simple
kernel has no persistent stack: every time a system call occurs, the
kernel stack starts over again from the very top, and any previous stack
information is thrown away. Thus, all persistent kernel data must be
stored in global variables.
procdescriptor_t
defined in
schedos-kern.h
. This is a lot like the process descriptor
structure from WeensyOS 1.schedos-kern.c
.start
function from schedos-kern.c
, which
initializes the kernel.trap
function from schedos-kern.c
, which
handles all interrupts and system call traps. This is like the interrupt()
function from WeensyOS 1.SchedOS supports two system calls, sys_yield
and
sys_exit
. The sys_yield
call yields control to
another process, and sys_exit
exits the current process,
marking it as nonrunnable. The kernel implementations of these system
calls (in trap()
) both call the schedule
function. This function is SchedOS's scheduler: it chooses a process from
the current set of runnable processes, then runs it. In the first part of
this problem set, you'll focus on this function, and SchedOS's scheduling
algorithms.
Read and understand the schedule
function
from schedos-kern.c
.
Exercise 1. What is the name of the scheduling
algorithm schedule()
currently implements? (What is
scheduling_algorithm 0
?)
Exercise 2. Add code to schedule()
so that scheduling_algorithm 1
implements strict priority
scheduling. Your implementation should give schedos-1
higher
priority than schedos-2
, which has higher priority than
schedos-3
, which has higher priority than
schedos-4
. Thus, process IDs correspond to priority levels
(assuming that numeric priority levels are defined as usual, where smaller
priority levels indicate higher priority). You will also
need to change schedos-1.c
so that the schedos
processes actually exit via sys_exit()
, instead of just
yielding forever. Test your code.
Exercise 3. Calculate the average turnaround
time and average wait time across all four jobs for
scheduling_algorithm
s 0 and 1. Assume that printing 1
character takes 1 millisecond and that everything else, including a context
switch, is free.
Exercise 4A. Add another scheduling algorithm,
scheduling_algorithm 2
, that acts like
scheduling_algorithm 1
except that priority levels are defined
by a separate p_priority
field of the process descriptor.
Also implement a system call that lets processes set their own priority
level. If more than one process has the same priority level, your
scheduler should alternate among them.
Exercise 4B. Add another scheduling algorithm,
scheduling_algorithm 3
, that implements proportional-share
scheduling. In proportional-share scheduling, each process is allocated an
amount of CPU time proportional to its share. For example, say
schedos-1
has share 1 and schedos-4
has share 4.
Under proportional-share scheduling, schedos-4
will run 4
times as often as schedos-1
(at least until it exits); so we
might expect to see output like "441444414444144...
". (Note
that this is a form of priority scheduling, but the priority levels are
defined differently: larger shares indicate higher priority.)
In this section of the problem set, you'll investigate synchronization issues. But synchronization isn't interesting without concurrency, and right now, our operating system is cooperatively multithreaded: each application decides when to give up control. We introduce concurrency by turning on clock interrupts and introducing preemptive multithreading. When a clock interrupt happens, the CPU will stop the currently-running process -- no matter where it is -- and transfer control to the kernel. This indicates that the current process's time quantum has expired, so the kernel will switch to another process. However, note that clock interrupts will never affect the kernel: this simple kernel runs with interrupts entirely disabled. Interrupts can only happen in user level processes.
Change scheduling_algorithm
back to 0.
Then change the interrupt_controller_init(0)
call in
schedos-kern.c
to interrupt_controller_init(1)
.
This turns on clock interrupts.
After running gmake run-schedos
, you should
see a window like this:
Note: It is better to do this portion of the lab using bochs. Qemu's interrupt handling is too fast; you may not be able to observe the same interrupt effects on qemu.
Exercise 5. Explain what has happened. Why does
this output look different from the output without clock interrupts? Be
specific (talk about particular lines of schedos-1.c
).
But we're not done! Let's cause clock interrupts to happen a little bit more frequently.
The HZ
constant in
schedos-kern.h
equals the number of times per second that the
clock interrupts the processor. It is set to 100 by default, meaning the
clock interrupts the processor once every 10 milliseconds. Set this
constant to 1000, so that the clock interrupts the processor every
milliscond.
After running gmake run-schedos
, you should
see a window like this:
Note that the output has less than 320*4 characters! Clearly there is a
race condition somewhere! (Your particular output may differ. You may
actually see 320*4 characters with occasional gaps, which
demonstrates a related race condition. If you still see 320*4 characters
with no gaps, try raising HZ
to 2000 or 3000 -- but not much
higher, since Bochs starts acting weird.)
Exercise 6. Explain what has happened. Why does
this output look different from the earlier two? Be specific (talk about
particular lines of schedos-1.c
, and why the higher clock rate
made a difference). Where is the race condition?
Exercise 7. Implement a synchronization mechanism that fixes this race condition. Your code should always print out 320 * 4 characters with no spaces. (The ordering of characters may vary, however, due to clock interrupts.)
There are lots of ways to implement the synchronization mechanism; here are a couple.
x86sync.h
directly.x86sync.h
to build a lock
data type, then use lock_acquire
and lock_release
operations. Note that all four processes must share the same lock! It
does no good to implement a different lock object per process.lock_acquire
and
lock_release
operations.However, you must not turn off clock interrupts. That would be cheating. Some hints:
x86sync.h
atomic
operations to work.cursorpos
points to a 16-bit integer, so the C
statement cursorpos++;
actually increments the address stored
in cursorpos
by 2 bytes, not one.obj/schedos-[1-4].sym
files, which tells you
where each symbol is located. Note that cursorpos
has the
same address in each process.This completes the problem set.
You will electronically hand in code and a small writeup containing answers to the numbered exercises and extra-credit exercises.
Answer the numbered exercises 1, 3, 5, 6 by editing the file in the weensyos2/ directory named
answers.txt
. Only ASCII text files are accepted. Please do not
hand in Microsoft word documents nor PDF files.
For coding exercises, it's OK for answers.txt
to just refer to
your code (as long as you comment your code).
Run the command make tarball
to create a file named
weensyos2-yourusername.tar.gz
. This tarfile contains the answers.txt file.
Submit the weensyos2-yourusername.tar.gz
file here.