 Solution

ID: 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/d/solution

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.

2^{16}

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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