Principles of Computer Organization Second Edition (Tang shuofei) Chapter 6 after class exercise answers 6.27 second sub question steps and answers,

Principles of Computer Organization Second Edition (Tang shuofei) Chapter 6 after class exercise answers 6.27 second sub question steps and answers,

（2）[2-3×(13/16)]-[2-4×(-5/8)]\x0b
x= 2-3×(13/16)= 2-011×0.110 100\x0b
y= 2-4×(-5/8)=2-100×(-0.101000)\x0b
[x] Order complement and tail complement = 11101; 00.110100-x0b [y] order complement and tail complement = 11100; 11.011000-x0b
1) For order: x0B [&; e] complement = [ex] complement + [- ey] complement, x0B = 11101 + 00100 = 00001
If [&; e] complement > 0, ey should be aligned to ex, then:
[ey] complement + 1 = 11100 + 00001 = 11101 / x0B [&ᦇ 61508; e] complement + [- 1] complement = 00001 + 11111 = 0 / x0B
So far, ey = ex
\X0B [y] complement = 11101; 11.101100 (x0B 2)
Mantissa operation: x0B [MX] complement + [- my] complement = 0.110 100
+ 0 0 .0 1 0 1 0 0 \x0b
--------------------------------
0 1 .0 0 1 0 0 0
\Results normalization: right rule [X-Y] complement = 11101; 01.001 000
=11,110；00.100 100
\Rounding: no rounding is required
\If there is no overflow, then: X-Y = 2-010 × (0.100 100) \ \ x0B = 2-2 × (9 / 16)