Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Digital systems, linear algebra, introduction to formal proof / 5

Table of contents
- a 5
  - Solution a
- b 5
  - Solution b
- c 5
  - Solution c
- d 5
  - Solution d

a

Solution

For a normal half-precision floating-point number, write the stored fields as sign

s

, biased exponent

E

and ten-bit fraction

f

. The value is

(- 1)^{s} (1. f)_{2} 2^{E - 15}

for

1 \leq E \leq 30

. The leading one is implicit: the ten stored fraction bits give eleven significant binary digits including that one. The normalized binary representations and stored fields are

Value	Normalized binary representation	Sign, biased exponent, fraction
$1/2$	$1. 0_{2} 2^{- 1}$	`0 01110 0000000000`
$5/8$	$1.0 1_{2} 2^{- 1}$	`0 01110 0100000000`
$- 3/16$	$- 1. 1_{2} 2^{- 3}$	`1 01100 1000000000`
$15 \frac{1}{2}$	$1.111 1_{2} 2^{3}$	`0 10010 1111000000`

Thus the four 16-bit words are

0011100000000000, 0011100100000000, 1011001000000000, 0100101111000000.

Every conversion is exact: all four numbers have terminating binary representations within the available precision.

b

Solution

Both signs are positive, so perform addition of the aligned significands. The true exponents are

- 1

and

3

, a difference of four. Shift the smaller significand right four places, retaining its low bits:

\frac{5}{8} = 0.00010 1_{2} 2^{3}, 15 \frac{1}{2} = 1.11110 0_{2} 2^{3} .

Add at the common exponent:

1.11110 0_{2} + 0.00010 1_{2} 10.00000 1_{2} so 10.00000 1_{2} 2^{3} = 1.000000 1_{2} 2^{4} .

Renormalization raises the true exponent to

4

and the biased exponent to

E = 19 = 1001 1_{2}

. There are only seven fraction places to retain, so no floating-point rounding changes the result:

\frac{5}{8} + 15 \frac{1}{2} = 16 \frac{1}{8}, stored fields 010011 0000001000.

The complete word is 0100110000001000.

c

Solution

The smallest positive normal half-precision floating-point number uses

E = 1

and a zero fraction, while the largest uses

E = 30

and all ten fraction bits equal to one. Hence

x_{m i n, normal} = 1. 0_{2} 2^{1 - 15} = 2^{- 14}, x_{m a x, normal} = (2 - 2^{- 10}) 2^{15} = 65504.

The corresponding words are 0 00001 0000000000 and 0 11110 1111111111. The stored exponent

31

is reserved and cannot extend the finite normal range.

For a subnormal number,

E = 0

and the implicit leading bit becomes zero, while the scale remains

2^{1 - 15} = 2^{- 14}

. The least positive nonzero fraction is

0.000000000 1_{2} = 2^{- 10}

x_{m i n, subnormal} = 2^{- 10} 2^{- 14} = 2^{- 24}, word 000000 0000000001.

Using scale

2^{- 15}

here would be incorrect: the subnormal number convention uses the minimum normal scale, not the ordinary biased-exponent formula at

E = 0

.

d

Solution

For the unsigned interpretation of the binary string, the set positions have weights

2^{15}, 2^{13}, 2^{12}, 2^{11}, 2^{10}, 2^{8}, 2^{6}, 2^{5}, 2^{4}

. Thus

u = 2^{15} + 2^{13} + 2^{12} + 2^{11} + 2^{10} + 2^{8} + 2^{6} + 2^{5} + 2^{4} = 48496.

In 16-bit two's complement, the most significant bit is set, so subtract

2^{16}

from that unsigned interpretation:

z = 48496 - 65536 = - 17040.

For a half-precision floating-point number, split the fields at their actual boundaries rather than the printed groups of four:

101111 0101110000.

The sign is negative, the true exponent is

15 - 15 = 0

, and the fraction is

2^{- 2} + 2^{- 4} + 2^{- 5} + 2^{- 6} = 23/64

. It is a normal finite value, so

x = - (1 + 2^{- 2} + 2^{- 4} + 2^{- 5} + 2^{- 6}) 2^{0} = - \frac{87}{64} = - 1.359375.

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