2.23: To implement 'down', disable interrupts and check the variable
you're decrementing.  If >0, decrement it, enable interrupts, and
continue.  If =0, block and schedule another process, enabling
interrupts.

To implement 'up', disable interrupts and increment the
variable. If a process was sleeping on this variable, wake it.  Then
re-enable interrupts.

The essence of the answer is that testing and changing the
variable has to be done with interrupts masked.

2.39: The algorithm that has optimal average turnaround is shortest
job first. We therefore schedule them as 3, 5, 6, 9, with X
inserted in the appropriate place.

2.42: The simplest solution is to give the boost only if the process
was actually blocked waiting for user input.  Other solutions
include rate-limiting the priority boost or requiring some output
before each input line -- output on such terminals was slow, no
faster than 10-13 characters/second.  Other answers are possible as
well.

3.15: A deadlock is not possible.  A process with two of the
resources can run to completion.  A process with one resource needs
a second before it can run.  Assume each process grabs one.  Both
will try for another.  There are three, so there's one resource
free;one of the requests will succeed.  The other process will wait
until the first is done.