Exam OS1 (1DT044) / OSPP (1DT096), 2018-03-16
Mixed concepts
K - wait()
O - signal
G - SJF
  - Cache
  - Shell
H - Many-to-one
L - Operating system
B - Critical section
Q - Peterson's solution
C - Paging
S - Response time
A - System call
I - fork()
T - External fragmentation
D - Translation Lookaside-Buffer (a _cache_)
E - Context switch
N - pipe()
R - Throughput
P - Race condition
 - Deadlock avoidance
F - Time slicing / multitasking / Round-Robin
J - exec()
M - Message Passing (For *Shared Memory* conflicts need be avoided.)
 - File Allocation Table

Module 1
1.1 B
1.2 D
1.3 A
1.4 C
1.5 C

Module 2
2.1 D
2.2 Ready
    Running
    Terminated
    Waiting
2.3 a) fork()
    b) exec()
    d) exit()
    c) wait()
2.4 I/O completes (event)
    I/O request
    time slice expires (interrupt)
    fork a child
2.5 D
2.6 A

Module 3
3.1 C
3.2 D
3.3 
Avereage response time =
  0 + (2-1) + (5-4) + (11-8)     5
= --------------------------- = --- = 1,25
               4                 4

Avereage waiting time =
  (4-2 + 15-5) + (2-1 + 9-4 + 16-11) + (5-4) + (11-8 + 18-15)    30
= ----------------------------------------------------------- = ---- = 7,5
                                  4                               4
3.5 
Preemptive Shortest Job First:
---------------------------------------------------------------------------------
|   P1  |       P2      |     P3    |         P4        |            P1         |
|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20

Round-Robin, quantum = 3:
---------------------------------------------------------------------------------
|     P1    |     P2    |     P1    |     P3    | P2|     P4    |   P1  |   P4  |
|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20

Module 4
4.1 B
4.2 Mutual exclusion
    Hold and wait
    No preepmtion
    Circular wait
4.3 D
4.4 C
4.5 (First of all find totally available and then check at the end.)
Fill in the need matrix (1 pt):
   | A | B | C | D |
T0 | 2 | 4 | 6 | 0 |
T1 | 0 | 2 | 0 | 0 |
T2 | 2 | 0 | 0 | 1 |
T3 | 0 | 5 | 7 | 0 |
T4 | 0 | 0 | 0 | 0 |

Determine whether the state is safe by showing each step of the Banker’s algorithm 
using the available matrix and choice vector. (1 pt):
     |   Available   |
Step | A | B | C | D | Choice
  1  | 0 | 2 | 5 | 1 |   T4
  2  | 2 | 3 | 5 | 1 |   T1
  3  | 4 | 6 |11 | 1 |   T3
  4  | 4 | 6 |11 | 2 |   T0
  5  | 8 | 7 |11 | 2 |   T2
Total|12 |12 |14 | 3 | - OK!

-- OR --:
     |   Available   |
Step | A | B | C | D | Choice
  1  | 0 | 2 | 5 | 1 |   T1
  2  | 2 | 5 |11 | 1 |   T0
  3  | 6 | 6 |11 | 1 |   T2
  4  |10 |11 |14 | 2 |   T3
  5  |10 |11 |14 | 3 |   T4
Total|12 |12 |14 | 3 | - OK!

Determine whether the state is safe by showing each step of the Banker’s algorithm 
using the available matrix and choice vector. (1 pt)
 - Yes. It is safe to run.
 The vector of totally available is 
  <A,B,C,D> = <12,12,14,3>.

Module 5
5.1 C
5.2   4 GiB
    256 MiB
      4 KiB
5.3 A
5.4 B