IT Fundamentals: Binary Conversion, Memory Interleaving and Addressable Units

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

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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)

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[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