Note for prog assignment:

>>  Example task set:
>>>
>>> T     A  C  D
>>> T1   0   3  12
>>> T2   0   4  15
>>> T3   2   5   9
>>> T4   5   2   8
>>>
>>  so, I can calculate
>> U1(0) = 3/12-0 --> 1/4
>> U2(0) = 3+4/15-0= 7/15  --> max speed
>> In this case,* I can pick up 7/15(0.467) as max speed,so T1 is going to
>> execute before executing of T2
>> (Because T1's deadline is earlier than T2)
>> So, T1 execute for 3/(7/15)=45/7sec(6.42sec) [0, 6.423] speed 0.467
>>
>> However, T3 arrive at 2, so I need to take acceptance test all remain
>> task(T1, T2, T3)
>>
>> U1(2)=RC(S1,2)/(D1-2) -->( 6.42-2 )/ (12-2) = 0.442
>> U2(2) = (RC(S1, 2) + RC(S2, 2)) / (D2-2) --> (4.42 + 4) / (15-2) =
>> 0.6477
>> -->T2's speed is 1, not 0.467 right?
>
> No. The original execution time of T1 is 3. How come it is (6.42 - 2) =
> 4.42 > 3 at t = 2 ?? You need to change it back to the one under the max
> speed of 1.0. That is 3*((6.42-2)/6.42)<-----------the remaining cmp time
> for T1 at t=2.


(calculating the remaining cmp time, the time should be measured as under
the speed of 1.0).