6-1 The famous Y. S. Chang Restaurant is open 24 hours a day. Waiters and busboys report for duty at 3am, 7am, 11am, 3pm, 7pm, or 11pm, and each works an 8-hour shift. The following table shows the minimum number of workers needed during the six periods into which the day is divided. Chang’s scheduling problem is to determine how many waiters and busboys should report for work at the start of each time period to minimize the total staff required for one day’s operation. (Hint: Let Xi equal the number of waiters and busboys beginning work in time period i, where i = 1, 2, 3, 4, 5, 6.) If Chang’s were able to reduce the number of required waiters and busboys by 1 during some period, during which period should they make the reduction?

Period | Time | Number of Waiters and Busboys Required | 1 | 3am-7am | 3 | 2 | 7am-11am | 12 | 3 | 11am-3pm | 17 | 4 | 3pm-7pm | 11 | 5 | 7pm-11pm | 12 | 6 | 11pm-3am | 4 |

In the QM,
Let X1 = Number of waiters and busboys at period 1
Let X2 = Number of waiters and busboys at period 2
Let X3 = Number of waiters and busboys at period 3
Let X4 = Number of waiters and busboys at period 4
Let X5 = Number of waiters and busboys at period 5
Let X6 = Number of waiters and busboys at period 6

Period 1 = 3 Period 2 = 14 Period 3 = 3 Period 4 = 8 Period 5 = 4 Period 6 = 0

Period 1, 3 and 5 drop from total 32 workers to 31 each if any of those 3 periods required minimum is reduce by 1. The period I choose (if have to choose 1 only) to make the reduction is period 1. When period 1 reduces by 1, the total number of workers required would drop from 32 to 31. Also, logically it makes sense to have lesser number of workers during the graveyard shift, as the number of customers should be lesser. And also it would save on company transport cost as going to work at 3am means no public transport and therefore