Computer composition and design work05 ——fifth verson

Computer Composition and Design Homework 5

3.20

[5] <§3.5> What decimal number does the bit pattern 0×0C000000 represent if it is a two’s complement integer? An unsigned integer?

1. The first is 0, If it is int Type then the description is a positive sign ,
   The calculation process is
   $$C(12)\times 16^6=201,326,592$$
2. unsigned int The calculation is the same as above

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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

Expressed as 2 Into the system for
$$0(operatorNumberposition)00011000(fingerCountposition)00000000000000000000000(SmallCountMinistrybranch)$$
It is not difficult to find that this number is positive , And the offset is 24 - 127 = -103
So the number is $1.0\times 2^{-103}$

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3.23

[10] <§3.5> Write down the binary representation of the decimal number
63.25 assuming the IEEE 754 single precision format

63.25 Binary representation as
$$111111.01=1.1111101\times 2^5$$
127+5 = 132 = 10000100(2)
The expression is as follows
$$01000010011111010000000000000000$$

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3.24

[10] <§3.5> Write down the binary representation of the decimal number
63.25 assuming the IEEE 754 double precision format.

1023+5 = 1028 = 10000000100(2)
$$0100000001001111101000000000000000000000000000000000000000000000$$

3.25

[10] <§3.5> Write down the binary representation of the decimal number
63.25 assuming it was stored using the single precision IBM format (base 16,
instead of base 2, with 7 bits of exponent).

$$63.25=111111.01(2)=3F.40(16)=0.3F40\times 16^2$$
Sign bit is 0, Exponential position $63+2=1000001(2)$
$$0100000111111010000000000000000$$
Look at the answer. It seems that the exponent is 64+2, I have doubts about this , Index 127 Used to indicate overflow ,0 Express 0, I think so 63 As an intermediate quantity

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3.26

[20] <§3.5> Write down the binary bit pattern to represent $-1.5625\times 10^{-1}$assuming a format similar to that employed by the DEC PDP-8 (the left most 12 bits are the exponent stored as a two’s complement number, and the rightmost 24 bits are the fraction stored as a two’s complement number). No hidden 1 is used. Comment on how the range and accuracy of this 36-bit pattern compares to the single and double precision IEEE 754 standards.

$$-1.5625\times 10^{-1}=-0.15625=-0.00101(2)=-0.101\times 2^{-2}$$

So the index $2userepaircodesurfaceinby000000000010,-2by111111111110$
The general expression is
$111111111110(12position)101000000000000000000000(24position）$

1. The index can be expressed as $-2048\sim 2047$
2. The base number can express $2^{-24}\sim 1-2^{-24}$

	DEC PDP-8	Single precision	Double precision
Minimum positive number	$2^{-24}\times 2^{-2048}$	$1\times 2^{-126}$	$1\times 2^{-1022}$
Maximum positive number	$(1-2^{-24})\times 2^{2047}$	$(1+1-2^{23})\times 2^{127}$	$(1+1-2^{52})\times 2^{1023}$

Should be DEC PDP-8 Can represent a wider range of

3.27

[20] <§3.5> IEEE 754-2008 contains a half precision that is only 16 bits
wide. Th e left most bit is still the sign bit, the exponent is 5 bits wide and has a bias
of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the
bit pattern to represent $-1.5625\times 10^{-1}$
assuming a version of this format, which
uses an excess-16 format to store the exponent. Comment on how the range and
accuracy of this 16-bit fl oating point format compares to the single precision IEEE
754 standard.

$$-1.5625\times 10^{-1}=-0.15625=-0.00101(2)=-1.01\times 2^{-3}$$

1. Symbol bit 1 A negative number
2. The expressible part of the index is 1-30, The offset for the 15, So the index is 12（01100) Express -3
3. The base part is marked with 0100 0000 00
   $$1011000100000000$$

3.29

[20] <§3.5> Calculate the sum of $2.6125\times 10^1$and $4.150390625\times 10^{-1}$by hand, assuming A and B are stored in the 16-bit half precision described in Exercise 3.27. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps.

$2.6125\times 10^1=26.125=1.1010001\times 2^4$

$4.150390625\times 10^{-1}=1.1010100111\times 2^{-2}$
Move to the left 6 Bits to align exponents

So there is （ Note that the half precision is only 11 Bits are used to save ）

1.1010001000

1.0000011010100111 / It should be noted that the beginning is 1, The rest needs to be left out

1.1010100010

Extra bits （ Yes 6 position ） More than significant digits （11） Half a bit , It should be carry

Last
$$1.1010100011(Into\; theposition)\times 2^4=26.546875$$

3.32

[20] <§3.9> Calculate $(3.984375\times 10^{-1}+3.4375\times 10^{-1})+1.771\times 10^3$by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit fl oating point format and in decimal.

   Keep a （Guard bit）、 Approximate position （Round bit） And viscous position （Sticky bit）.
   Keep a ： The lowest point after approximation 
   Approximate position ： The last of the reserved bits 
   Sticky bits ： All bits after the approximate bit are treated as one bit after or operation 

$3.984375\times 10^{-1}=1.1001100000\times 2^{-2}$

$3.4375\times 10^{-1}=1.0110000000\times 2^{-2}$

$1.771\times 10^3=1.1011101011\times 2^{10}$

(A)	1.1001100000	$\times 2^{-2}$
(B)+	1.0110000000	$\times 2^{-2}$
=	10.1111100000	$\times 2^{-2}$
=	1.0111110000	$\times 2^{-1}$

(A+B)	1.0111110000	$\times 2^{-1}$
=	0.000000000010111110000	$\times 2^{10}$
C	1.1011101011	$\times 2^{10}$（A+B The superfluous part is 101, exceed 0.5, Should carry ）
(A+B)+C=	1.1011101100	$\times 2^{10}$
=	1772

3.33

[20] <§3.9> Calculate $3.984375\times 10^{-1}+(3.4375\times 10^{-1}+1.771\times 10^3)$by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit fl oating point format and in decimal.

$3.984375\times 10^{-1}=1.1001100000\times 2^{-2}$

$3.4375\times 10^{-1}=1.0110000000\times 2^{-2}$

$1.771\times 10^3=1.1011101011\times 2^{10}$

(B)	0.0000000000010110000000	$\times 2^{10}$
+	1.1011101011	$\times 2^{10}$
=	1.1011101011	$\times 2^{10}$

(A)	0.0000000000011001100000	$\times 2^{10}$
+	1.1011101011	$\times 2^{10}$
=	1.1011101011	$\times 2^{10}$
=	1771

3.34

[10] <§3.9> Based on your answers to 3.32 and 3.33, does $(3.984375\times 10^{-1}+3.4375\times 10^{-1})+1.771\times 10^3=3.984375\times 10^{-1}+(3.4375\times 10^{-1}+1.771\times 10^3)$?

Is not the same , The first is 1772

The second is 1771

The exact solution should be 1771.742187