1. First, (wm, wl) = (.573, .248); then wmvl/vm = .290; so the answer is (.572, .958). This in fact is the correct result to six decimals.
2. The answer is not affected, since the normalization routine truncates to eight places and can never look at this particular byte position. (Scaling to the left occurs at most once during normalization, since the inputs are normalized.)
3. Overflow obviously cannot occur at line 09, since we are adding two-byte quantities, or at line 22, since we are adding four-byte quantities. In line 30 we are computing the sum of three four-byte quantities, so this cannot overflow. Finally, in line 32, overflow is impossible because the product fufv must be less than unity.
4. Insert ‘
JOV OFLO; ENT1 0 ...