t b a b a

The makespan of schedule S1 is equal to the largest completion time over all machines:
C A1
max = 8.
(ii) Schedule S2:
[J1 ↦→ M2, J2 ↦→ M1, J3 ↦→ M3, J4 ↦→ M4, J5 ↦→ M3, J6 ↦→ M2, J7 ↦→ M1, J8 ↦→ M2]
The schedule vector is:
| S2〉 =| 010011〉⊗ | 000010〉⊗ | 100101〉⊗ | 110100〉
⊗ | 100011〉⊗ | 010111〉⊗ | 000101〉⊗ | 011010〉)
The completion times for schedule S2 for all machines are:
C A2 (M1) = 2 + 5 = 7, C A2 (M2) = 3 + 7 + 10 = 20, C A2 (M3) = 5 + 3 = 8, C A2 (M4) = 4.
with makespan C A2
max = 20.
The schedule vector can also be in a superposition of some basic states, for example, the
schedule vector could be in an equal superposition of the two schedules, S1 and S2:
| S〉 = 1/ √ 2(| S1〉+ | S2〉)
= 1/ √ 2(| 000001〉⊗ | 010001〉⊗ | 000110〉⊗ | 110100〉
⊗ | 100011〉⊗ | 010111〉⊗ | 100011〉⊗ | 000001〉
+ | 010011〉⊗ | 000010〉⊗ | 100101〉⊗ | 110100〉
⊗ | 100011〉⊗ | 010111〉⊗ | 000101〉⊗ | 011010〉).
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