Bevin Shrestha

ID# 106003079

CSE 220

Homework #3 Part II

Due Date: November 05, 2008

PROBLEM A. Convert the following 2 signed integers from binary to Hexadecimal notation:
a.)11011110101011011011111011101111
Solution:
Given binary number is 1101 1110 1010 1101 1011 1110 1110 1111.
Lets group the four digits of binary number starting from left to right as follows:
^{(1111)2 =} _{1x 2}³ _{+ 1x 2}² _{+0x 2}¹ _{+1x 2}⁰ _{= (15)10} ^{= (F)}₁₆
⁽¹¹¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+1x 2}¹ _{+1x 0}⁰ _{= (14)10} ^{= (E)}₁₆
⁽¹¹¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+1x 2}¹ _{+0x 2}⁰ _{= (14)10} ^{= (E)}₁₆
⁽¹⁰¹¹⁾₂ ⁼ _{1x 2}³ _{+ 0x 2}² _{+1x 2}¹ _{+1x 2}⁰ _{= (11)10} ^{= (B)}₁₆
⁽¹¹⁰¹⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+0x 2}¹ _{+1x 2}⁰ _{= (13)10} ^{= (D)}₁₆
⁽¹⁰¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 0x 2}² _{+0x 2}¹ _{+0x 2}⁰ _{= (10)10} ^{= (A)}₁₆
⁽¹¹¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+1x 2}¹ _{+0x 2}⁰ _{= (14)10} ^{= (E)}₁₆
⁽¹¹⁰¹⁾ ₂⁼ _{1x 2}³ _{+ 1x 2}² _{+0x 2}¹ _{+1x 2}⁰ _{= (13)10} ^{= (D)}₁₆

^{Thus, (11011110101011011011111011101111)}₂ ^{= (deadbeef)₁₆}

^{b.)00001011101011011011111011101111
Solution:
Given binary number is 0000 1011 1010 1101 1011 1110 1110 1111.
Lets group the four digits of binary number starting from left to right as follows:
(1111)}₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+0x 2}¹ _{+1x 2}⁰ _{= (15)10} ^{= (F)}₁₆
⁽¹¹¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+1x 2}¹ _{+1x 0}⁰ _{= (14)10} ^{= (E)}₁₆
⁽¹¹¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+1x 2}¹ _{+0x 2}⁰ _{= (14)10} ^{= (E)}₁₆
⁽¹⁰¹¹⁾₂ ⁼ _{1x 2}³ _{+ 0x 2}² _{+1x 2}¹ _{+1x 2}⁰ _{= (11)10} ^{= (B)}₁₆
⁽¹¹⁰¹⁾₂ ⁼ _{1x 2}³ _{+ 1x 2}² _{+0x 2}¹ _{+1x 2}⁰ _{= (13)10} ^{= (D)}₁₆
⁽¹⁰¹⁰⁾₂ ⁼ _{1x 2}³ _{+ 0x 2}² _{+0x 2}¹ _{+0x 2}⁰ _{= (10)10} ^{= (A)}₁₆
⁽¹⁰¹¹⁾₂ ⁼ _{1x 2}³ _{+ 0x 2}² _{+1x 2}¹ _{+1x 2}⁰ _{= (14)10} ^{= (B)}₁₆
⁽⁰⁰⁰⁰⁾ ₂⁼ _{0x 2}³ _{+ 0x 2}² _{+0x 2}¹ _{+0x 2}⁰ _{= (13)10} ^{= (0)}₁₆

^{Thus, (00001011101011011011111011101111)}₂ ^{= (0badbeef)₁₆}

^{PROBLEM B. Convert the following 2 signed real numbers from single floating point notation to hexadecimal.
a.) -0.0008162677986547351
The integral part is 0}₁₀ ^{= 0}₂^{.
The fractional part is:
0.0008162677986547351 x 2 = 0.0016325355973094702
0.0016325355973094702 x 2 = 0.0032650711946189404
0.0032650711946189404 x 2 = 0.0065301423892378808
0.0065301423892378808 x 2 = 0.0130602847784757616
0.0130602847784757616 x 2 = 0.0261205695569515232
0.0261205695569515232 x 2 = 0.0522411391139030464
0.0522411391139030464 x 2 = 0.1044822782278060928
0.1044822782278060928 x 2 = 0.2089645564556121586
0.2089645564556121586 x 2 = 0.4179291129112243712
0.4179291129112243712 x 2 = 0.8358582258224487424
0.8358582258224487424 x 2 = 1.6717164516448974848
0.6717164516448974848 x 2 = 1.3434329032897949696
0.3434329032897949696 x 2 = 0.6868658065795899392
0.6868658065795899392 x 2 = 1.3737316131591798784
0.3737316131591798784 x 2 = 0.7474632263183597568
0.7474632263183597568 x 2 = 1.4949264526367195136
0.4949264526367195136 x 2 = 0.9898529052734390272
0.9898529052734390272 x 2 = 1.9797058105468780544
0.9797058105468780544 x 2 = 1.9594116210937561088
0.9594116210937561088 x 2 = 1.9188232421875122176
0.9188232421875122176 x 2 = 1.8376464843750244352
0.8376464843750244352 x 2 = 1.6752929687500488704
0.6752929687500488704 x 2 = 1.3505859375000977408
0.3505859375000977408 x 2 = 0.7011718750001954816
0.7011718750001954816 x 2 = 1.4023437500003909632
0.4023437500003909632 x 2 = 0.8046875000007819264
0.8046875000007819264 x 2 = 1.6093750000015638528
0.6093750000015638528 x 2 = 1.2187500000031277056
0.2187500000031277056 x 2 = 0.4375000000062554122
0.4375000000062554122 x 2 = 0.8750000000125108224
0.8750000000125108224 x 2 = 1.7500000000250216448
0.7500000000250216448 x 2 = 1.5000000000500432896
0.5000000000500432896 x 2 = 1.0000000001000865792}

^{Thus, taking the integral part from the above obtained result, we get
(0.0008162677986547351)}₁₀
^{= (0.000000000011010101111110101100111)}₂
^{= 1.1010101111110101100111 x 10^-11}

_{Sign = 1}

_{Exponent = -11 + 127 = (116) = (01110100)2}

^{Mantissa = (10101011111101011001110)}₂

^{So, (-0.0008162677986547351)}₁₀
^{= [1]_sign [0111 0100]}_exponent ^{[101 0101 1111 1010 1100 1110]}_mantissa
^{= (BA55FACE)}₁₆

^{b.) -9651108864}
9651108864 – 2^33 = 1061174272

1061174272 – 2^29 = 524303360

524303360 – 2^28 = 255867904

255867904 – 2^27 = 121650176

121650176 – 2^26 = 54541312

54541312 – 2^25 = 20986880

20986880 – 2^24 = 4209664

4209664 – 2^22 = 15360

15360 – 2^13 = 7160

7160 – 2^12 = 3072

3072 – 2^11 = 1024

1024 – 2^10 = 0

The binary representation would be 1000111111010000000011110000000000

Let’s, normalize the number...

1000111111010000000011110000000000 = 1.000111111010000000011110000000000 x 2³³

Now we represent the individual parts...

Sign = 1 because the number is negative

Exponent = 33 + 127 = 160

Fraction = 1.000111111010000000011110000000000

Now we make our 32 bit representation...

1 10100000 00011111101000000001111 (That makes all 32 bits, the rest of the fraction is truncated).

1101 0000 0000 1111 1101 0000 0000 1111

= (d00fd00f)₁₆

Problem C: Convert the following 2 signed real numbers from single floating point notation to hexadecimal. Show all work:

2.796762340086568 X 10^-29
-5.056837726336028 x 10^-10

a.) 2.796762340086568 X 10^-29

We get the following after some calculations:

1 .00011011101000000001101 x 2^-95

Now let’s represent the individual parts...

Sign = 0 because the number is positive

Exponent = -95 + 127 = 32

Fraction = 1 .00011011101000000001101

Now let’s make our 32 bit representation...

0 00100000 00011011101000000001101

0001 0000 0000 1101 1101 0000 0000 1101

= (100dd00d)₁₆

b.) -5.056837726336028 x 10^-10

We get the following after some calculations:

1 .00010110000000001010101 x 2^-31

Now let’s represent the individual parts...

Sign = 1 because the number is negative

Exponent = -31 + 127 = 96

Fraction = 1 .00010110000000001010101

Now lets make our 32 bit representation...

1 01100000 00010110000000001010101

1011 0000 0000 1011 0000 0000 0101 0101

= (b00b0055) ₁₆

^{PROBLEM D. Convert the following 2 numbers from hexadecimal to single floating point notation.}

^{a.)0xfeedface}

^{Solution:
First of all, lets convert this number into a binary number.
(f)}₁₆ ^{= (15)}₁₀ ^{= (1111)}₂
^(e)₁₆ ^{= (14)}₁₀ ^{= (1110)}₂
^(e)₁₆ ^{= (14)}₁₀ ^{= (1110)}₂
^(d)₁₆ ^{= (13)}₁₀ ^{= (1111)}₂
^(f)₁₆ ^{= (15)}₁₀ ^{= (1111)}₂
^(a)₁₆ ^{= (10)}₁₀ ^{= (1010)}₂
^(c)₁₆ ^{= (12)}₁₀ ^{= (1100)}₂
^(e)₁₆ ^{= (14)}₁₀ ^{= (1110)}₂

^{Thus, [1]_sign [11111101]}_exponent ^{[1101111111110101100]}_mantissa

^{sign = negative
exponent - bias = 253 - 127 = 126
1 + fraction = 1 + (}₂^-1 _{+ 2}^-2 _{+ 2}^-4_{+ 2}^-5_{+ 2}^-6 _{+ 2}^-7 _{+ 2}^-8 _{+ 2}^-9 _{+ 2}^-10 _{+ 2}^-11 _{+ 2}^-12 _{+ 2}^-13 _{+ 2}^-15 _{+ 2}^-16_{) = 1.874923706
Thus, Answer = -1.874923706 x 10}¹²⁶

_{b.)0x0badfood}

_{Solution:
First of all, lets convert this number into a binary number.
(0)16} ^{= (0)}₁₀ ^{= (0000)}₂
^(b)₁₆ ^{= (11)}₁₀ ^{= (1011)}₂
^(a)₁₆ ^{= (10)}₁₀ ^{= (1010)}₂
^(d)₁₆ ^{= (13)}₁₀ ^{= (1101)}₂
^(f)₁₆ ^{= (15)}₁₀ ^{= (1111)}₂
⁽⁰⁾₁₆ ^{= (0)}₁₀ ^{= (0000)}₂
⁽⁰⁾₁₆ ^{= (0)}₁₀ ^{= (0000)}₂
^(d)₁₆ ^{= (15)}₁₀ ^{= (1111)}₂

^{Thus, [0]_sign [00010111]}_exponent ^{[01011011111000000001111]}_mantissa

^{sign = positive
exponent - bias = 23 - 127 = -104
1 + fraction = 1 + (}₂^-2 _{+ 2}^-4 _{+ 2}^-5_{+ 2}^-7_{+ 2}^-8 _{+ 2}^-9 _{+ 2}^-10 _{+ 2}^-11 _{+ 2}^-20 _{+ 2}^-21 _{+ 2}^-22 _{+ 2}^-23_{) = 1.358888506
Thus, Answer = 1.358888506 x 10}^-104

_{PROBLEM E. Add the following 2 signed hexadecimal integers by first converting them to binary.}

_{a.)
0x10000fa5
0x0af50af5}

_{Solution:
First of all, lets convert this hexadecimal number 0x10000fa5 into its equivalent binary number.
(1)16} ^{= (0001)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
^{(f)₁₆ = (1111)}₂
^(a)₁₆ ^{= (1010)}₂
⁽⁵⁾₁₆ ^{= (0101)}₂

^{Thus, 0x10000fa5 = (00010000000000000000111110100101)}₂

^{Similarly, lets convert this hexadecimal number 0x0af50af5 into its equivalent binary number.
(0)}₁₆ ^{= (0000)}₂
^{(a)₁₆ = (1010)}₂
^(f)₁₆ ^{= (1111)}₂
⁽⁵⁾₁₆ ^{= (0101)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
^(a)₁₆ ^{= (1010)}₂
^(f)₁₆ ^{= (1111)}₂
⁽⁵⁾₁₆ ^{= (0101)}₂

^{Thus, 0x0af50af5 = (00001010111101010000101011110101)}₂

^{Now, let’s add (00010000000000000000111110100101)}₂ ^{and (00001010111101010000101011110101)}₂^{.
That is,
(00010000000000000000111110100101)}₂ ^{+ (00001010111101010000101011110101)}₂ ^{= (11010111101010001101010011010)}₂

^{PROBLEM F. Add the following 2 signed hexadecimal integers by first converting them to binary.}

^{a.)
0xfaded0af
0xfa1afe15}

^{Solution:
First of all, let’s convert this hexadecimal number 0xfaded0af into its equivalent binary number.
(f)} ₁₆ ^{= (1111)}₂
^(a) ₁₆ ^{= (1010)}₂
^(d) ₁₆ ^{= (1101)}₂
^(e) ₁₆ ^{= (1101)}₂
^(d) ₁₆ ^{= (1101)}₂
⁽⁰⁾₁₆ ^{= (0000)}₂
^(a) ₁₆ ^{= (1010)}₂
^(f) ₁₆ ^{= (1111)}₂

^{Thus, 0xfaded0af = (11111010110111011101000010101111)}₂

^{Similarly, let’s convert this hexadecimal number 0xfa1afe15 into its equivalent binary number.
(f)₁₆ = (1111)}₂
^(a)₁₆ ^{= (1010)}₂
⁽¹⁾₁₆ ^{= (0001)}₂
^(a)₁₆ ^{= (1010)}₂
^(f)₁₆ ^{= (1111)}₂
^(e)₁₆ ^{= (1110)}₂
⁽¹⁾₁₆ ^{= (0001)}₂
⁽⁵⁾₁₆ ^{= (0101)}₂

^{Thus, 0xfa1afe15 = (11111010000110101111111000010101)}₂

^{Now, let’s add (11111010110111011101000010101111)}₂ ^{and (11111010000110101111111000010101)}₂^{.
That is,
(11111010110111011101000010101111)}₂ ^{+ (111110100001101011111110000101011)}₂ ^{= (10111011110001001111001100110011011010)}₂

^{PROBLEM G. Add the following 2 signed, single floating point numbers by first converting them to binary.
0x0000acdc
0x00abacab}

^Solution:
Translate to binary first...

0x 0 0 0 0 a c d c

0000 0000 0000 0000 1010 1100 1101 1100

0x 0 0 a b a c a b

0000 0000 1010 1011 1010 1100 1010 1011

Now let’s translate each of them to single floating point numbers...

0 00000000 00000001010110011011100

Sign = 0, so it is positive

Exponent = 0-127 = -127

Fraction = 1.00000001010110011011100

0 00000001 01010111010110010101011

Sign = 0, so it is positive

Exponent = 1-127 = -126

Fraction = 1.01010111010110010101011

Now, let’s add:

0.10000000101011001101110 x 2^-126 + 1.01010111010110010101011 x 2^-126 = 1.11011000000001100011001 x 2^126

Therefore the complete 32-bit representation of the added value is...

0 00000001 11011000000001100011001

Which can also be written as...

0000 0000 1110 1100 0000 0011 0001 1001

= (00ec0319) ₁₆