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. 首位为0，如果是int类型那么说明是正号，
   计算过程为
   $$C(12)\times 16^6=201,326,592$$
2. unsigned int的计算同上

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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

表示为2进制为
$$0(符号位)00011000(指数位)00000000000000000000000(小数部分)$$
不难发现这个数是正数，且偏移量为24 - 127 = -103
所以该数为 $1.0\times 2^{-103}$

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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

63.25二进制表示为
$$111111.01=1.1111101\times 2^5$$
127+5 = 132 = 10000100(2)
于是表示如下
$$01000010011111010000000000000000$$

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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).

$$63.25=111111.01(2)=3F.40(16)=0.3F40\times 16^2$$
标志位为0，指数位为$64+2=1000001(2)$
$$0100000111111010000000000000000$$
看答案貌似写指数位是64+2 这个IBM偏移量固定为64

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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.

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

所以指数$2用补码表示为000000000010，-2为111111111110$
总的表示为
$111111111110(12位)101000000000000000000000(24位）$

1. 指数可以表达的数字为$-2048\sim 2047$
2. 底数可表达$2^{-24}\sim 1-2^{-24}$

	DEC PDP-8	单精度	双精度
最小正数	$2^{-24}\times 2^{-2048}$	$1\times 2^{-126}$	$1\times 2^{-1022}$
最大正数	$(1-2^{-24})\times 2^{2047}$	$(1+1-2^{23})\times 2^{127}$	$(1+1-2^{52})\times 2^{1023}$

应该为DEC PDP-8能表示的范围更大

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. 符号位为1表示负数
2. 指数可表达的部分为1-30，偏移量为15，所以指数用12（01100)表示-3
3. 底数部分以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.

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

$4.150390625\times 10^{-1}=1.1010100111\times 2^{-2}$
向左移动6位以对齐指数

于是有（注意半精度只有11位用于保存）

1.1010001000

1.0000011010100111 /需要注意开头是1，剩下的部分需要舍去

1.1010100010

额外的位（有6位）超过有效位（11）位的一半，应该是进位

最后
$$1.1010100011(进位)\times 2^4=26.546875$$

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.

  保留位（Guard bit）、近似位（Round bit）和粘滞位（Sticky bit）。
  保留位：近似后的最低位
  近似位：保留位的后一位
  粘滞位：近似位后的所有位进行或运算后视作一位

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

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

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

(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多余的部分是101，超过0.5，应进位）
(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 $(3.984375\times 10^{-1}+3.4375\times 10^{-1})+1.771\times 10^3=3.984375\times 10^{-1}+(3.4375\times 10^{-1}+1.771\times 10^3)$?

并不相同，第一个为1772

第二个是1771

精确的解应该为1771.742187