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Labor Shift Scheduling Problem Labor Shift Scheduling Problem

Labor Shift Scheduling Problem - PowerPoint Presentation

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Labor Shift Scheduling Problem - PPT Presentation

Unable to hire new police officers because of budget limitations the Gotham City Police commissioner is trying to utilize the force better The minimum requirements for police patrols for weekdays are noted below ID: 487650

police problem sept maximize problem police maximize sept computer officers metacreation time intelligence enters leaves artificial x3subject sample number

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Slide1

Labor Shift Scheduling ProblemUnable to hire new police officers because of budget limitations, the Gotham City Police commissioner is trying to utilize the force better. The minimum requirements for police patrols for weekdays are noted below:

Time PeriodMinMidnight - 4:00 am64:00 am - 8:00 am48:00 am - noon14noon - 4:00 pm84:00 pm - 8:00 pm128:00 pm - Midnight16

The police officers can start their shifts at the starting time of any of the above time periods. However they have to patrol for 8 hours.

What is the minimum number of officers required to satisfy the requirements?Express this an LP problem.

1Slide2

June7/07 Raeside cartoon lampoons

VPD, Times Colonist. 2Slide3

I have copies of the first part of the text for those who requested one (plus 2 extras). Hopefully our textbook will arrive soon.Assignment #1 has been posted: due Sept. 20 (or Sept. 24 with 10% late penalty)

Programming Project 1 has been posted:due Sept. 27 (or Oct. 1 with a 10% late penalty).3Slide4

Musical Metacreation, by Philippe Pasquier

Friday, September 14, 12:00pm, ECS 660While artificial intelligence aims to make "machines do things that would require intelligence if done by humans", metacreation is a new field devoted to endow machines with creative behavior. The presentation will propose an overview of past and current works on computational creativty conducted at the Metacreation, Agents and Multi-Agent Systems (MAMAS) laboratory. We will introduce and motivate problems, we will describe systems that address these, and empirical evaluation studies that evaluate our solutions. While doing so, we will highlight a number of open problems in artificial intelligence, computer music and generative art.4Slide5

Interested in meeting other people in your faculty from all different stages in their academic careers? Come join other eager students to make connections and discuss the involvement of women in Computer Science and Engineering.

Or just come for cookies and tea! :Where: ECS 660When: 2:30pm on Friday, September 14Who: Everyone welcome!5Slide6

Standard form (chapters 1-7):Maximize

cT xsubject toAx ≤ band x ≥ 0What can we do if our problem is not already in standard form?6Slide7

7Another sample problem:

Maximize 5 x1 + 5 x2 + 3 x3subject to x1 + 3 x2 + x3 ≤ 3 -x1 + 3 x3 ≤ 22 x1 - x2 + 2 x3 ≤ 4 x1 + 3 x2 - x3 ≤ 2

x1, x2, x3 ≥ 0How will we type in this problem as input?Slide8

8

Another sample problem:Maximize 5 x1 + 5 x2 + 3 x3subject to x1 + 3 x2 + x3 ≤ 3 -x1 + 3 x3 ≤ 22 x1 - x2

+ 2 x3 ≤ 4 x1 + 3 x2 - x3 ≤ 2

x1, x2

, x

3

≥ 0

3 4

5

5 3

3 1 3

1

2

-

1

0 3

4

2 -

1

2

2

2 3 -

1Slide9

9The initial dictionary:

X4 = 3.00- 1.00 X1 - 3.00 X2 - 1.00 X3 X5 = 2.00+ 1.00 X1 + 0.00 X2 - 3.00 X3 X6 = 4.00- 2.00 X1 + 1.00 X2 - 2.00 X3 X7 = 2.00- 2.00 X1 - 3.00 X2 + 1.00 X3 ------------------------------------------z = -0.00+ 5.00 X1 + 5.00 X2 + 3.00 X3 How should we represent this in a tableau/array for our computer program?Slide10

10

X1 enters. X7 leaves. z = -0.000000 After 1 pivot:X4 = 2.00- 1.50 X2 - 1.50 X3 + 0.50 X7 X5 = 3.00- 1.50 X2 - 2.50 X3 - 0.50 X7 X6 = 2.00+ 4.00 X2 - 3.00 X3 + 1.00 X7 X1 = 1.00- 1.50 X2 + 0.50 X3 - 0.50 X7 ------------------------------------------z = 5.00- 2.50 X2 + 5.50 X3 - 2.50 X7 Slide11

11

X3 enters. X6 leaves. z = 5.000000 After 2 pivots:X4 = 1.00- 3.50 X2 + 0.50 X6 + 0.00 X7 X5 = 1.33- 4.83 X2 + 0.83 X6 - 1.33 X7 X3 = 0.67+ 1.33 X2 - 0.33 X6 + 0.33 X7 X1 = 1.33- 0.83 X2 - 0.17 X6 - 0.33 X7 ------------------------------------------z = 8.67+ 4.83 X2 - 1.83 X6 - 0.67 X7 Slide12

12

X2 enters. X5 leaves. z = 8.666667 After 3 pivots:X4 = 0.03+ 0.72 X5 - 0.10 X6 + 0.97 X7 X2 = 0.28- 0.21 X5 + 0.17 X6 - 0.28 X7 X3 = 1.03- 0.28 X5 - 0.10 X6 - 0.03 X7 X1 = 1.10+ 0.17 X5 - 0.31 X6 - 0.10 X7 -------------------------------------------z = 10.00- 1.00 X5 - 1.00 X6 - 2.00 X7 How could we argue that this must be the optimal solution?