I. DEC-TO-HEX-TO-BINARY
0 = 0 = 0000 0000
1 = 1 = 0000 0001
2 = 2 = 0000 0010
3 = 3 = 0000 0011
4 = 4 = 0000 0100
5 = 5 = 0000 0101
6 = 6 = 0000 0110
7 = 7 = 0000 0111
8 = 8 = 0000 1000
9 = 9 = 0000 1001
10 = A = 0000 1010
11 = B = 0000 1011
12 = C = 0000 1100
13 = D = 0000 1101
14 = E = 0000 1110
15 = F = 0000 1111
TECHNIQUE TO BINARY PREDICTION
1 + 0 ---> 1
0 + 1 ---> 1
1 + 1 ---> 0
Example:
A) You are given the first (2) binaries, namely,
0000 0011 and 0000 0100. What is the next binary?
0000 0011
+ 0000 0100
--------------------
0000 0101
Tip: You don't even have to calculate once you know the pattern.
Study the 5 proceeding binaries:
0000 0110
0000 0111
0000 1000
0000 1001
0000 1010
Notice how the position of 1 changes:
101 --> 110 --> 111 --> 1000 --> 1001 --> 1010 --> 1011
Here's the pattern rule:
00 - none occupied
01 - occupy the first position
10 - occupy the second position
11 - occupy all positions
100 - none occupied
101 - occupy the first position
110 - occupy the second position
111 - occupy all positions
... and so on.
TECHNIQUE TO HEXADECIMAL PREDICTION
It begins from 0 to FF.
Pattern:
0 - 0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
1 - 10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
...and so on.
Techniques can simplify things, rids you of unnecessary stress, and increases your productivity by optimizing your time. ^^,)
1 = 1 = 0000 0001
2 = 2 = 0000 0010
3 = 3 = 0000 0011
4 = 4 = 0000 0100
5 = 5 = 0000 0101
6 = 6 = 0000 0110
7 = 7 = 0000 0111
8 = 8 = 0000 1000
9 = 9 = 0000 1001
10 = A = 0000 1010
11 = B = 0000 1011
12 = C = 0000 1100
13 = D = 0000 1101
14 = E = 0000 1110
15 = F = 0000 1111
TECHNIQUE TO BINARY PREDICTION
1 + 0 ---> 1
0 + 1 ---> 1
1 + 1 ---> 0
Example:
A) You are given the first (2) binaries, namely,
0000 0011 and 0000 0100. What is the next binary?
0000 0011
+ 0000 0100
--------------------
0000 0101
Tip: You don't even have to calculate once you know the pattern.
Study the 5 proceeding binaries:
0000 0110
0000 0111
0000 1000
0000 1001
0000 1010
Notice how the position of 1 changes:
101 --> 110 --> 111 --> 1000 --> 1001 --> 1010 --> 1011
Here's the pattern rule:
00 - none occupied
01 - occupy the first position
10 - occupy the second position
11 - occupy all positions
100 - none occupied
101 - occupy the first position
110 - occupy the second position
111 - occupy all positions
... and so on.
TECHNIQUE TO HEXADECIMAL PREDICTION
It begins from 0 to FF.
Pattern:
0 - 0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
1 - 10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
...and so on.
Techniques can simplify things, rids you of unnecessary stress, and increases your productivity by optimizing your time. ^^,)
II. EFFECTIVE ACCESS TIME
SAMPLE PROBLEM FROM CMPS 369:
A demand paging system provides a TLB (10 ns access time), cache memory (40 ns access time), and main memory (100 ns access time, including the cache "miss"). The page table is found in the TLB 40% of the time, and not in the TLB but in the cache 20% of the time. A process memory location is found in the cache 60% of the time. Calculate the EAT.
[CPU] ---> TLB ---> cache ---> RAM
Given:
There are 6 possible scenarios.
PAGE TABLE DATA
1. TLB cache [TLB: 10 ns; CACHE: 40 ns] ---> 0.60( 10 ns + 40 ns) ---> 30
2. TLB memory [TLB: 10 ns; RAM: 100 ns] ---> 0.40(10 ns + 100 ns) ---> 44
3. cache cache [CACHE: 10 ns + 40 ns; CACHE: 40 ns] ---> 0.60(10 ns + 40 ns + 40 ns) ---> 54
4. cache memory [CACHE: 10 ns + 40 ns; RAM: 100 ns] ---> 0.40(10 ns + 40 ns + 100 ns) ---> 60
5. memory cache [RAM: 10 ns + 100 ns; CACHE: 40 ns] ---> 0.60(10 ns + 100 ns + 40 ns) ---> 90
6. memory memory [RAM: 10 ns + 100 ns; RAM: 100 ns] ---> 0.40(10 ns + 100 ns + 100 ns) ---> 84
Next Step:
TLB ---> 40% ---> 0.40(30 + 44) ---> 29.6
CACHE ---> 20% ---> 0.20(54 + 60) ---> 22.8
RAM ---> 40% (Difference of 100% - [TLB + CACHE]) ---> 0.40(90 + 84) ---> 69.6
Final Step:
EAT = 29.6 + 22.8 + 69.6 = 122
A demand paging system provides a TLB (10 ns access time), cache memory (40 ns access time), and main memory (100 ns access time, including the cache "miss"). The page table is found in the TLB 40% of the time, and not in the TLB but in the cache 20% of the time. A process memory location is found in the cache 60% of the time. Calculate the EAT.
[CPU] ---> TLB ---> cache ---> RAM
Given:
There are 6 possible scenarios.
PAGE TABLE DATA
1. TLB cache [TLB: 10 ns; CACHE: 40 ns] ---> 0.60( 10 ns + 40 ns) ---> 30
2. TLB memory [TLB: 10 ns; RAM: 100 ns] ---> 0.40(10 ns + 100 ns) ---> 44
3. cache cache [CACHE: 10 ns + 40 ns; CACHE: 40 ns] ---> 0.60(10 ns + 40 ns + 40 ns) ---> 54
4. cache memory [CACHE: 10 ns + 40 ns; RAM: 100 ns] ---> 0.40(10 ns + 40 ns + 100 ns) ---> 60
5. memory cache [RAM: 10 ns + 100 ns; CACHE: 40 ns] ---> 0.60(10 ns + 100 ns + 40 ns) ---> 90
6. memory memory [RAM: 10 ns + 100 ns; RAM: 100 ns] ---> 0.40(10 ns + 100 ns + 100 ns) ---> 84
Next Step:
TLB ---> 40% ---> 0.40(30 + 44) ---> 29.6
CACHE ---> 20% ---> 0.20(54 + 60) ---> 22.8
RAM ---> 40% (Difference of 100% - [TLB + CACHE]) ---> 0.40(90 + 84) ---> 69.6
Final Step:
EAT = 29.6 + 22.8 + 69.6 = 122
III. ITSD Club 2014-2015 Video
<soon>
IV. Countdown Until Capstone Due
12 Days