Question

3.21 [10] <§3.5> If the bit pattern 0×0C000000 is placed into the Instruction Register, what MIPS instruction will be executed?

3.22 [10] <§3.5> What decimal number does the bit pattern 0×0C000000 represent if it is a floating point number? Use the IEEE 754 standard.

Given the following 32-bit binary sequences representing single precision IEEE

754 floating point numbers:

a = 0100 0000 1101 1000 0000 0000 0000 0000

b = 1011 1110 1110 0000 0000 0000 0000 0000

Perform the following arithmetic and show the final addition and multiplication results in both normalized binary format and IEEE 754 single-precision format. Show your steps (Do not convert a and b to decimal base, compute the addition and multiplication, then convert the results back to normalized binary and single precision).

a)     a + b

b)     a × b

Answer 1

3.21)
0x0C000000
Converting 0C000000 to binary
0 => 0000
C => 1100
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
So, in binary 0C000000 is 00001100000000000000000000000000

=>  00001100000000000000000000000000
=>  000011 00000000000000000000000000
=>  I-type instruction
000011 is jal
=>  jal 0
Answer: jal 0

3.22)
in binary 0C000000 is 00001100000000000000000000000000
0 00011000 00000000000000000000000
sign bit is 0(+ve)
exp bits are 00011000
Converting 00011000 to decimal
   00011000
   => 0x2^7+0x2^6+0x2^5+1x2^4+1x2^3+0x2^2+0x2^1+0x2^0
   => 0x128+0x64+0x32+1x16+1x8+0x4+0x2+0x1
   => 0+0+0+16+8+0+0+0
   => 24
in decimal it is 24
so, exponent/bias is 24-127 = -103
frac bits are

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.0 * 2^-103
Answer: 2^-103

1 * 2^-103 in decimal is 9.860761315262648e-32
so, 00001100000000000000000000000000 in IEEE-754 single precision format is 9.860761315262648e-32
Answer: 2^-103 or 9.860761315262648e-32

3.23)
a = 0 10000001 10110000000000000000000
sign bit is 0(+ve)
exp bits are 10000001
Converting 10000001 to decimal
   10000001
   => 1x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+1x2^0
   => 1x128+0x64+0x32+0x16+0x8+0x4+0x2+1x1
   => 128+0+0+0+0+0+0+1
   => 129
in decimal it is 129
so, exponent/bias is 129-127 = 2
frac bits are 1011

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.1011 * 2^2

b = 1 01111101 11000000000000000000000
sign bit is 1(-ve)
exp bits are 01111101
Converting 01111101 to decimal
   01111101
   => 0x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+0x2^1+1x2^0
   => 0x128+1x64+1x32+1x16+1x8+1x4+0x2+1x1
   => 0+64+32+16+8+4+0+1
   => 125
in decimal it is 125
so, exponent/bias is 125-127 = -2
frac bits are 11

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.11 * 2^-2

a)
a + b
= 1.1011 * 2^2 - 1.11 * 2^-2
= 1.1011 * 2^2 - 0.000111 * 2^2
= 1.100101 * 2^2

single precision:
--------------------
sign bit is 0(+ve)
exponent bits are (127+2=129) => 10000001
   Divide 129 successively by 2 until the quotient is 0
      > 129/2 = 64, remainder is 1
      > 64/2 = 32, remainder is 0
      > 32/2 = 16, remainder is 0
      > 16/2 = 8, remainder is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10000001
   So, 129 of decimal is 10000001 in binary
frac/significant bits are 10010100000000000000000

so, a+b in single-precision format is 0 10000001 10010100000000000000000

Answer:
----------
normalized binary format:           1.100101 * 2^2
IEEE 754 single-precision format:   0 10000001 10010100000000000000000

b)
a x b
= 1.1011 * 2^2 x -1.11 * 2^-2
= 1.1011 x -1.11
= -10.111101
= -1.0111101 * 2^1

single precision:
--------------------
sign bit is 1(-ve)
exponent bits are (127+1=128) => 10000000
   Divide 128 successively by 2 until the quotient is 0
      > 128/2 = 64, remainder is 0
      > 64/2 = 32, remainder is 0
      > 32/2 = 16, remainder is 0
      > 16/2 = 8, remainder is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10000000
   So, 128 of decimal is 10000000 in binary
frac/significant bits are 01111010000000000000000

so, a x b in single-precision format is 1 10000000 01111010000000000000000

normalized binary format:           -1.0111101 * 2^1
IEEE 754 single-precision format:   1 10000000 01111010000000000000000