Solutions and grading by Elena Oranskaya 
(eoranska@cs.princeton.edu, office hours MW 12:30-1:30 or by appointment)

Problem 1.
The halt instruction can only be excuted when the 'jump' instruction in location F6 is taken. 
That only happens when the contents of register 0 equal the contents of register 3, namely the
binary value 5. So the program loops, incrementing the value in register 0 with every iteration
until it reaches 5.

Problem 2.
10	11FF	load R1 with contents of memory location FF
12	1200	load R2 with contents of memory location 00
14	5312	add content of R1 and R2 and save result in R3
16	2000	load R0 with value 0
18	B320	jump to address 20 if contents of R3 equal content of R0 (else continue)
1A	3100	store contents of R1 into memory location 00
1C	32FF	store contents of R2 into memory location FF
1E	B024	jump to address 24 - unconditionally
20	3000	store value 0 into memory location 00
22	30FF	store value 0 into memory location FF
24	C000	halt

The solution above is one possible correct answer. There exist a few variations, which will 
also produce the desired result.

Problem 3.
A person types 30 words per minute, each of length 5 characters, which amounts to 
5 * 30 = 150 characters per minute.

1 minute = 60 seconds and therefore the typing rate also equals
150 / 60 = 2.5 characters per second.

The time that elapses between the typing of two characters is
1/ 2.5 = 0.4 seconds.

Since the machine can execute 1000000 instructions per second, it executes
0.4 * 1000000 = 400000 instructions between the typing of two consecutive
characters.

Problem 4.
Complications may arise if processes intend to alter the file. Simultaneous alteration will
result in the inconsistency of data, which means that changes made by once process may be
overwritten by the other and vice versa. In certain cases errors will result even if only
one of the processes is writing to the file while the other is reading. So a universal 
solution would be to only grant simultaneous read access. In special cases when an interprocess
communication protocol has been established between the processes they may get write access.

Problem 5.
Access time = 0.017 + 0.008 = 0.025 seconds
Time slice = 0.05 seconds, so the process spends
0.025/0.05 = 50% of its time slice waiting for a read opration to complete.

The machine executes 1000000 instructions per second, so during the waiting period it can
execute 0.025 * 1000000 = 25000 instrucitons.

Problem 6.
Greater throughput would be achieved if one of the processes was I/O bound and the other -
compute bound. The reason is that as we saw in Problem 5, I/O operations take a long time, 
during which the compute bound process can execute and get a chunk of its computation 
accomplished. Furthermore, the modern I/O systems have DMA channels, which allow them to 
transfer data in and out of memory without interrupting the CPU. So the two processes demand 
fewer of the same resources thus minimizing the system delays. 

If we had two I/O bound processes they may compete for the same peripherals, and would have to 
be run intermittenly, increasing the time spent in doing interrupt handling
in addition to greater usage of
the slower I/O devices.

Problem 7 (extra credit).
00	2000	load R0 with value 0
02	2201	load R2 with value 1
04	11FE	load R1 with contents of memory location FE
06	8312	AND the contents of registers R1 and R2 and save the reslut in R3
08	B304	jump address 04 (reload memory location FE) if the result of the AND 
		was 0 - i.e. printer not ready
0A	35FF	store the contents of R5 to memory location FF - i.e. print!
0C	C000	halt