IT Fundamentals: Binary Conversion, Memory Interleaving and Addressable Units

This document covers topics such as binary conversion, memory interleaving, and addressable units in IT Fundamentals. It includes solved assignments and essays on these topics. The document also explains low-order and high-order interleaving with examples.

Running head: IT FUNDAMENTALS
IT FUNDAMENTALS
Name of the Student:
Name of the University:
Author Note:

1IT FUNDAMENTALS
Table of Contents
1a:-...................................................................................................................................................2
1b:-...................................................................................................................................................3
2:-.....................................................................................................................................................5
3a:-...................................................................................................................................................7
Low-Order Interleaving:-.............................................................................................................7
High-Order Interleaving:-............................................................................................................7
EXAMPLE:-................................................................................................................................9
3b:-...................................................................................................................................................8

2IT FUNDAMENTALS
1a:-
The required binary digits = 00111111010100000000000000000000
At first we find the exponent of this binary number and sign bit.
Sign bit is = 0
In an IEEE 754 floating point standard is a standard for manipulating floating quantities.
In a 32 bit binary number structure is (0-31Bit) 31st number means bit number 32 is the sign bit
30 to 23 bits are exponential and other 22 to 0 bit is called mantissa.
In this case 01111110 is the exponent and the exponent value E=126 (E=exponent)
0111 1110(2)
= 0 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20
= 0 + 64 + 32 + 16 + 8 + 4 + 2 + 0
= 64 + 32 + 16 + 8 + 4 + 2
=126(10)
Then we find the exponent value 126 and sign bit 0
So, we assume that, 1.10100000000000000000000 x 2E – 127
Then, 1.10100000000000000000000 x 2126 – 127 (the value of Exponent)
= 1.10100000000000000000000 x 2-1
= 1.625 x 2-1
=1.625/2
= 0.8125 (ANS)

3IT FUNDAMENTALS
[Then we convert the decimal value of 1.101
1 x 20 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3
= 1 + 0.5 + 0.125
= 1.625]
1b:-
Following number conversion:-
i. 0xAD9 into 3-base representation
(First convert this number in a base 10)
= 10*16ˆ2+13*16^1+9*16^0
= 10 *256+13*16 +9*1
= 2560+208+9
= 2777
After the conversion then we find
In this task we find 3 based representation
Then 2777/3 the answer is 925 and reminder is 2
= 925/3 then the answer is 308 and reminder is 1
= 308/3 then the answer is 102 and remainder is 2
= 102/3 then the answer is 34 and remainder is 0
= 34/3 then the answer is 11 and remainder is 1
= 11/3 then the answer is 3 and remainder is 2
= 3/3 then the answer is 1 and remainder is 0
= 1/3 then the answer is 0 (app.) and remainder is 1

4IT FUNDAMENTALS
Then the answer is 10210212 (ANS)
ii. 4518 into 2-base (binary) representation
First separate three number 4 5 and 1
Then find the octal value 4= 100, 5= 101, 1=001 then 451=100101001
4518 = (100101001)2 (ANS)
iii. 123.35 into octal representation (up to 3 octal points)
First convert this number in a decimal
So, assume that,
1 x 52 + 2 x 51 + 3 x 50 + 3 x 5-1
=1*25+2*5+3*1+ (3/5)
=25+10+3+0.6
=38.6
Then decimal to octal conversation is
38/8 the answer is 4 and the remainder is 6
4/8 the answer is 0 and the remainder is 4
Then we find the octal number
0.6*8=4.8
0.8*8=6.4 (4631)
0.4*8=3.2
0.2*8=1.6
………….

5IT FUNDAMENTALS
Then 0.6*8=4.8 (46)
0.8*8=6.4
So 46.4631 is the answer (ANS)
iv. 14.358 into decimal representation
So the decimal representation is
We determine,
1x 81 + 4 x 80 + 3 x 8-1 + 5 x 8-2
=1*8 +4*1+ (3/8) + (5/64)
= 8+4+0.375+0.078
= 12.453 (ANS)
2:-
MSG DB “1”
NMSG DB “0”
NUM DB 71H;
START: MOV AX,@DATA
MOV DS, AX
MOV AL, NUM
MOV BL, 02H;
MOV DX, 0000H;
MOV AH, 00H;
L1: DIV BL
CMP AH, 00H;
JNE NEXT
INC BH ;
NEXT:CMP BH,02H ;

6IT FUNDAMENTALS
JE FALSE ;
INC BL ;
MOV AX,0000H ;
MOV DX,0000H ;
MOV AL,NUM ;
CMP BL,NUM ;
JNE L1;
TRUE: LEA DX, MSG
MOV AH, 09H;
INT 21H
JMP EXIT
MOV AH, 09H;
INT 21H
EXIT:
MOV AH, 4CH
INT 21H
END START
INPUT: - 15
OUTPUT: - 0
Code Description:-

7IT FUNDAMENTALS
MSG DB “1” is the Message database 1 . User just press a key to continue 0.
NMSG DB “0” is the Message database 0 .NUM DB 71H it is the algorithm of number
Entering. In this task the number is 71H. The binary number of this enter number is 01110001.
START: MOV AX,@DATA is the 16 bit data register which is moved in AX.
MOV DS, AX is the Move the data register in DS. MOV AL, NUM is to Copy a number into
the register. MOV BL, 02H; means The Separating starts from 2, Hence BH is equate to
02H. MOV DX, 0000H; it means To evade Division overflow mistake and that is DX
MOV AH,00H ; it is To evade Division overflow mistake. L1: DIV BL is the Loop checking
for prime number. CMP AH, 00H; Remains is equated with 00H to AH
JNE NEXT is the Conditional Jump for the next. INC BH; BH is
incremented if the Amount is separable by BL value. NEXT: CMP BH, 02H; compare the
value If BH > 02H so it is not prime .JE FALSE; is the Jump in equal if statement is not
a Prime No. INC BL ; to Increase BL value
MOV AX,0000H To evade the Division of overflow mistake
MOV DX,0000H ; Then To avoid Divide overflow mistake
MOV AL,NUM ; To Move the Defaulting no to AL
CMP BL,NUM ; To Run the loop until BL matches Value
JNE L1; It means the Jump to patterned again with incremented value of BL number
Then the important part to show the enter number is a Prime No or not
TRUE: LEA DX, MSG means that if the statement is true then LEA assign in the DX. MOV
AH, 09H; is Moved to the number and Used to print a string. INT 21H is the one integer
value. JMP EXIT is to jump the exit value and show this number prime or not. FALSE: LEA
DX, NMSG if the prime condition is not match then address the in bit message 0.
MOV AH, 09H; and then Print a string value. INT 21H to Show this integer value.
EXIT: Exit this program. MOV AH, 4CH if the number is prime, so it is move
INT 21H Input Integer. END START End this program and continue to the loop.
3a:-
The technique which is used for increasing the throughput is named as Memory
Interleaving. The basic requirement states about the memory system that gets divided into
independent based banks and thus it suggests it can answer read or write based requirements and
those are independent in real.
High Order and Low Order is the types of memory interleaving.

8IT FUNDAMENTALS
Low-Order Interleaving:-
The LOI implies that the small memory bits are used for the purpose of Section
identification in the memory location. LOI usually spreads the memory bits in the address and
supplies this in the memory location. The High Order Bits that are displaced in the address bar.
This address bar are stored in the LOI Segment. This consequences in horizontal formatting
sequenced in the Section.
But in this LOI section always depends on the structures of address like address 0 to 3 is comes
under the 0 to 3 sequence then no 4 address is not comes under the no 4 section . It is cyclic
process and no 4 address bits again comes under the section 0. Then address 4 to 7 again comes
under the section 0 to 3.
High-Order Interleaving:-
For the purpose of HOI, high range bits are used in the Section organizations. This is a
natural procedure which is found within the AA26-A27 bus lines. Those bus lines are sequenced
in 4 memory Sections in a sequence of 2to 4 decoder. The collection of those address bar are
then collected in the memory location. The HOI is the procedure of transfer of address from one
address bar to another address bar. According to the Section address those bits are stored on the
memory. The high order bits are sequenced or maintain the sequence in a consecutive manner
and are stored within the same Section. Whenever the high order beats reach the boundary order
line, the stored data does not follows the sequenced structure. The sequence beat that maintains
the sequence consecutively are known as the High order interleaving.
In an example the address of 0 to 64 bits are comes under the section 0 then 64 bits to 128
bits are categorized in section 1 same as 128 to 192 bits and 192 to 256 bits are comes into the
section 2 and 3 respectively.

9IT FUNDAMENTALS
EXAMPLE:-
Assume module is 8 (addresses)
Memory bit 32.
So,
HIGH ORDER INTERLEAVING
SECTION 0 0 1 2 3
SECTION 1 4 5 6 7
SECTION 2 8 9 10 11
SECTION 3 12 13 14 15
SECTION 4 16 17 18 19
LOW ORDER INTERLEAVING p q r s
SECTION 0 0 8 16 24
SECTION 1 1 9 17 25
SECTION 2 2 10 18 26
SECTION 3 3 11 19 27
SECTION 4 4 12 20 28
SECTION 5 5 13 21 29
SECTION 6 6 14 22 30
SECTION 7 7 15 23 31

10IT FUNDAMENTALS
SECTION 7 28 29 30 31
3b:-
When the value of K is taken 4, the interleaving way results to be 16.
Let X be the interleaving method,
X= 2^k
In this solution k =4 so,
So we easily find the value of X which is the (24= 16)
For every arrangement, the k secondary location bits choose its own section.
According to the problem, the number of bits becomes = (32*4k=128k)
Then we say that (16*8= 128)
Which implies, 128k= 27K
Then, the Addressable unit = (2^7*2^10)
=2^ (10+7)
=2^17
Addressable unit = 2^17
Therefore, seventeen bits are required for allocation of each location
According to the calculation, the Memory banks will be 8= (23)
So, each Section memory = Bits/memory bank
= 128/8=16
[Therefore the calculated Section memory is 16
which implies, 16= 24

11IT FUNDAMENTALS
So 4 bits offset Section
The bit number is (17-4=13 bits)]
Therefore, for a HOI, the address memory organization can be represented as:
Chip is 0100 chip no is 4
And offset chip (13) is 0000000001101
Therefore, For LOI, the address memory organization can be represented as:
Offset chip is 0001000010100
And number of chips is 1101.