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Exact(5)

Proof We first prove that ⋃ j = j k z R k, j has not only exactly k late orders but also the maximum total processing time among all schedules for k = 0, 1, …, t − 1.

This, along with (iii) in Lemma 1, p l 0 = max { p i | J i ∈ J ∖ ⋃ j = j 0 z R 0, j }, and ⋃ j = j 0 z R 0, j being a set of early orders with the maximum total processing time among all schedules, implies that ⋃ j = j 0 z R 1, j does not only have exactly a late order but also the maximum total processing time among all schedules.

Similarly, we see that ⋃ j = m z R 0, j is a set of early orders with the maximum total processing time among all schedules for m = z − 1, …, j 0, where the orders in R 0, j are delivered at time T j.

Further, by Step 4(a) and Step 4(b) of Algorithm NF, we see that ⋃ j = m z R 1, j does not only exactly have a late order but also the maximum total processing time among all schedules for m = j 0 − 1, …, j 1.

By the definition of N 0, z and Step 4(a) and Step 4(b) of Algorithm NF, we see that R 0, z is a set of early orders delivered at time T z with the maximum total processing time among all schedules.

Similar(55)

Canonical analysis revealed that the response surface of daily finished vessel quantity and total processing time were both saddle surfaces that contained neither global maximum nor global minimum.

(a) Total processing time.

The total processing time was 1872 s (~31 min).

Hence, total processing time in cheesemaking is decreased.

The results indicate that the maximum total expected indirect loss is much higher than the maximum total expected direct loss.

Moreover, user can transmit data with maximum total power.

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