Problem 1:

Correct Gantt diagram:

arrival: a     b c
time   : 0123456789012345678901234567890123456789
queue 0: aaaa  bbbbccc  cc
queue 1:     aa       aa  bbbbbbbbaaaaaaaa
queue 2:                                  bbaaaaa

turnaround:
a: 40
b: 29
c: 9

wait:
a: 19
b: 15
c: 2


-----------------------------------------------------
wrong understanding of the Round Robing Policy (added for info about computing turnaround and wait)
non-renewable budget per queue. This is not in line with
a Round Robin policy for the two first queues. It is shown here just to give
the resulting turnaround and wait times.

arrival: a     b c
time   : 0123456789012345678901234567890123456789
queue 0: aaaa  bbbbccc  c
queue 1:     aa       aa bbbbbbbbaaaac
queue 2:                              bbaaaaaaaaa

turnaround:
a: 40
b: 24
c: 20

wait:
a: 19
b: 10
c: 13
----------------------------------------------------



Mechanism:

A good explanation will involve: timer, interrupt, service routine,
context switch.

Starvation is possible:

If Pa and Pb repeated infinitely bursts of 3 and waits of 2 while Pc
had a burst of 5, then Pc would starve.


Problem 2:

1. See course. Size. Permissions/ACL. Timestamps (e.g., last modification).

2. Logical blocks are 2^11B. With 32 bits pointers one can address
2^32 logical blocks, amounting to a maximum of 2^32 x 2^11 B = 2^43 B = 8TiB.

3. Disk is 1TiB=2^40B. Hence, 2^40/2^11 logical blocks, which amounts to 2^29 blocks,
with 1 bit per block, we get 2^26B for the bitmap.


4. 12 x 2^11B + (1 x (2^11/2^2) x 2^11) + (1 x (2^11/2^2)^2 x 2^11) +
(1 x (2^11/2^2)^3 x 2^11) ~ 2^38B

5. Random access. Overhead. Corruption.

6. External fragmentation. Access time. Overhead.

Problem 3:

1. external fragmentation.

2. a. check paging in course.

2. b.
0x113 -> 0xC513(142423 oct = 50451 dec), invalid
0x8F0 -> 0x45F0(42760 oct = 17904 dec, 0xBFF0(137760 oct = 49136 dec)

2. c.

0x2A --> 0x200(1000 oct = 512 dec) 0xC00(6000 oct = 3072 dec)
0x0A --> 0xF00(7400 oct = 3840 dec) 0x500(2400 oct = 1280 dec)
0xC0 --> 0x500(2400 oct = 1280 dec) 0xD00(6400 oct = 3328 dec)
0xC4 --> 0x900(4400 oct = 2304 dec) 0x700(3400 oct = 1792 dec)


Problem 4

1. left as an exercise. Be aware of the setuid bit.

2. Modifying and running an executable owned by root with setuid on is
powerful.