Let IC be the instruction count, then the total number of cycles on the unmodified machine is
(0.48*1.0 + 0.33*1.4 + 0.17*1.7 + 0.02*1.2)*IC = 1.255*IC
If 25% of the data transfer instructions are changed, then the total number of cycles on the modified machine is
((0.48 - 0.33*0.25)*1.0 + 0.33*1.4 + 0.17*1.7 + 0.02*1.2)*IC = 1.1725*IC
If the cycle time of the unmodified machine is C, then the total time on the unmodified machine is 1.255*IC*C, the total time on the modified machine is 1.1725*IC*1.1*C = 1.28975*IC*C. The unmodified machine is
1 - 1.28975/1.255 = 2.77% faster.
The problem is that lw sign extends the offset A_lower. So,
A_upper_adjusted = A_upper + bit 15 of A (the "sign" bit of A_lower)
Let A and B be the numbers (unsigned) in $t3 and $t4 respectively. If there is a carry out, then A+B >= 2^n. So, if we do
addu $t2, $t3, $t4
The number in $t2 is A+B-2^n. Since both A and B are less than 2^n, we conclude that $t2<$t3 and $t2<$t4. If there is no carry out, we have $t2>=$t3 and $t2>=$t4. Thus, we have
addu $t2, $t3, $t4
sltu $t2, $t2, $t3
alternatively, 1's complement is the largest (unsigned) number which doesn't cause a carry out when it is added to the original number
nor $t2, $t3, $t3 #1's complement
sltu $t2, $t2, $t4
Fig.4.4 on page 222 shows that if overflow and OldSet = 1, then A >= 0 and B < 0 which means A < B is false; if overflow and OldSet = 0, then A < 0 and B >= 0 which means A < B is true. Thus we have the following truth table:
OldSet overflow NewSet
1 0 1
0 0 0
0 1 1
1 1 0
That is
NewSet = OldSet XOR overflow