Q1 (a)
(b)
(c)
(d)
(e)
(f)
A (96 Mb)
B(34 Mb)
C(18 Mb)
A leaves
D(48 Mb)
E(8Mb)
96
96 34
128 128 128 128
96 34 18
128 128 128 128
128 128 128 128
48 34 18
128 128 128 128
34 18
0,61
F Rate A (96 Mb)
96
B(34 Mb) 16 32 64 128
34 96
C(18 Mb) 16 32 64 128
18 34 96
A leaves 16 32 64 128
18 34
48 34 18 8
128 128 128 128
D(48 Mb) 16 32 64 128
18 34 48
F Rate
The blocks that are not allocated are not considered in the fragmentation rate
128 128 128 128 0,70
F Rate E(8Mb) 8 18 34 48
16 32 64 128 0,35
F Rate
The blocks that are not allocated are not considered in the fragmentation rate
16 32 64 128
With best fit and inequal partitionning, the FRate is best Due to the block size constraints, best fit has no real impact here 0,39
Q2
t0
first fit P1 H P4 P5 H P2 H P3 P6
32 16 64 8 4 18 22 6 30
(a)
(b)
(c )
(d)
P7(8)
P2 leaves
P3 leaves
P8(22)
P1 P7 H P4 P5 H P2 H P3 P6
32 8 8 64 8 4 18 22 6 30
P1 P7 H P4 P5 H P3 P6
32 8 8 64 8 44 6 30
P1 P7 H P4 P5 H P6
32 8 8 64 8 50 30
P1 P7 H P4 P5 P8 H P6
32 8 8 64 8 22 28 30
F Rate t0
next fit P1 H P4 P5 H P2 H P3 P6
P7(8) 32 16 64 8 4 18 22 6 30
P1 H P4 P5 H P2 P7 H P3 P6
P2 leaves 32 16 64 8 4 18 8 14 6 30
P1 H P4 P5 H P7 H P3 P6
32 16 64 8 22 8 14 6 30
P3 leaves P1 H P4 P5 H P7 H P6
32 16 64 8 22 8 20 30
0,22
P8(22) P1 H P4 P5 P8 P7 H P6
32 16 64 8 22 8 20 30
surprisely, first fit performs best F Rate
0,44
Q3 c. a. 5 levels (256, 128, 64, 32 and 16 Mb) b. A, B, D, F and H on the fourth level (32 Mb) C, E, G, I, J on the fifth level (16 Mb) d, e.recursive searches are supposed to be in the [0 4] interval (tree deep) A 3 t0 A B C D E F G H I J B 0 256 1 C 2 128 1 1 1 1 1 D 0 64 1 1 1 1 E 0 32 2 2 3 3 3 5 4 6 6 6 F 2 16 2 2 2 2 4 4 4 4 G 1 256 256 256 256 256 256 256 256 256 256 256 H 1 I 0 J 1 max is 3 and obtained for A min is 0 and obtained for B, D, E, I f. A B C D E F G H I J Free
Allocated 18 28 9 24 11 19 4 29 7 13 F Rate
Q4 64 Kb = 2^16 then d = 16 4 Gb = 2^32 then m = 32 d = 16, m = 32 then p = 16 and we have 2^16 = 65536 frames / pages 1024 entries * 64 kb = 2^10 pages, with a page size of 2^16 we have 2^10 * 2^16 = 2 ^26 = 64 Mb In the best case 2^32 / 2^26 = 2^6 = 16, thus 1/16 of the memory can be loaded at a maximum in the TLB a.
A(140) = 3 pages B(96) = 2 pages C(220) = 4 pages D(205) = 4 pages E(550) = 9 pages
Hole 14 4 7 8 5 13 12 3 9 3 16 0,82
Buddy 32 32 16 32 16 32 16 32 16 16 The buddies that are not allocated are not considered for the frame rate
b.
(a) (b) A loaded B loaded 0 A0 A0 1 A1 A1 2 A2 A2 3 B0 4 B1 5 6 7 8 9 10 11 12
c.
(c ) C loaded A0 A1 A2 B0 B1 C0 C1 C2 C3
Page table - D 0 1 2 3
d. A C D
Process 140 220 205
(d) (e ) (f) B leaves D loaded A leaves A0 A0 A1 A1 A2 A2 D0 D0 D1 D1 C0 C0 C0 C1 C1 C1 C2 C2 C2 C3 C3 C3 D2 D2 D3 D3
Page table - E 0 1 2 3 4 5 6 7 8
3 4 9 10
Pages 192 256 256
Holes 52 36 51
At F Rate = 1- 52 / (52+36+51) = 0,62
e.
8F0F0 corresponds to p = 8 and d = F0F0 then it becomes CF0F0 with frame = C (i.e. 12)
0 1 2 5 6 7 8 11 12
(g) (h) C leaves E loaded E0 E1 E2 D0 D0 D1 D1 E3 E4 E5 E6 D2 D2 D3 D3 E7 E8
