5.9:	A situation where is is "possible to write programs that can
	access any I/O device without having to specify the device in
	advance".  That's the textbook's definition; I'd add that
	the specification of the device should not be implicit in the
	program's code, either.

5.13:	Several reasons (you only need to have given one):
	a) the CPU runs faster than the printer; you want to let
	jobs run to completion
	b) Sometimes, the program is CPU-bound, in which case if it
	had allocated the printer anything else that needs to print
	will block
	c) Most programs don't have permission to open the printer
	d) The print spooler can do things like accounting, printing
	banner pages, printing security labels, etc.

5.31:	Each character is effectively 10 bits, so you're printing
	5600 characters/second.  Each such character takes 100 useconds
	of CPU time, for a total of 560 milliseconds, or 56% of the CPU.

	But -- there's another way to look at it.  That assumes that
	there is perfect overlap of I/O with CPU time.  Assume that
	there's no overlap.  In that case, every 1/5600 of a second
	(178.6 usec), the system spends 100 usecs of CPU time before
	the next operation is started.  In that case, we don't achieve
	5600 characters a second, we achieve about 1/(.0001786+.0001)
	characters a second, or about 3589.  We spend .0001 seconds each
	time, for a total of .359 seconds of CPU per second, or 35.9%.

	Which answer is correct depends on how much overlap there actually
	is.  There is, in fact, some overlap, because once the I/O operation
	is started the system has to return to user level and block waiting
	for the I/O to finish, so 3589 characters/second is too low.  On
	the other hand, fielding the interrupt, activating the driver, and
	sending the next character to the device takes a noticeable fraction
	of the time for actually transmitting it, so we can't neglect it
	and we're not going to achieve 5600 characters/second.  The truth
	is somewhere in between (but I'll accept either of those answers).