By J. C. Burkill
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Note that, at time 50, process a has consumed only 10 ticks of execution, whereas it needed 12, and hence it has missed its first deadline. The second example is contained in Table 5. 775, which is below the bound, and hence this process set is guaranteed to meet all its deadlines. If a time-line for this set is drawn, all deadlines would be satisfied. Although cumbersome, time-lines can actually be used to test for schedulability. But how far must the line be drawn before one can conclude that the future holds no surprises?
More sophisticated graph reduction techniques can be used if sufficient semantic information is available. To give just a simple example of this, consider the following code: for I in 1.. 10 loop if Cond then -- basic block of cost 100 else 46 A. Burns and A. Wellings -- basic block of cost 10 end if; end loop; With no further information, the total ‘cost’ of this construct would be 10 × 100 + the cost of the loop construct itself, giving a total of, say, 1005 time units. It may, however, be possible to deduce (via static analysis of the code) that the condition Cond can only be true on at most three occasions.
Fortunately, the theorem can be reapplied to the reduced process set. Hence through successive reapplication, a complete priority ordering is obtained (if one exists). The following code in Ada implements the priority assignment algorithm; Set is an array of processes that is notionally ordered by priority; Set(N) being the highest priority, Set(1) being the lowest. The procedure Process_Test tests to see whether process K is feasible at that place in the array. The double loop works by first swapping processes into the lowest position until a feasible result is found, this process is then fixed at that position.
A Second Course in Analysis by J. C. Burkill