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lecture5-4up(peronal planning-squre-in-out)

lecture5-4up(peronal planning-squre-in-out) - ISyE...

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ISyE 323 Lecture #5 Prof. Jeff Linderoth September 17, 2009 ISyE Lecture #5 Outline Outline Production Planning Example ((not) Section 3.4) General Algebraic Modeling Personnel Scheduling Yummy – A Blending Problem ISyE Lecture #5 2 ISyE Lecture #5 Personnel Scheduling Outline Production Planning Example ((not) Section 3.4) Background Model Steps Final Model General Algebraic Modeling Personnel Scheduling Yummy – A Blending Problem ISyE Lecture #5 3 ISyE Lecture #5 Personnel Scheduling Problem Statement Union Airways needs to schedule customer service agents at check-in desks and gates. A 24-hour day is divided into 5 (overlapping) 8-hour shifts: 1. 6:00 AM – 2:00 PM 2. 8:00 AM – 4:00 PM 3. 12:00 PM – 8:00 PM 4. 4:00 PM – 12:00 AM 5. 10:00 PM – 6:00 AM Agents’ pay is different for different shifts The airline knows how many workers are required for each portion of the day. The goal is to find the schedule of workers to cover the requirements at minimum cost. ISyE Lecture #5 4
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ISyE Lecture #5 Personnel Scheduling Shifts and Requirements The requirements for each time period, the shifts that cover them, and the pay for each shift, are given by: Shift Minimum Number of Time Period 1 2 3 4 5 Agents Needed 6 AM–8 AM 48 8 AM–10 AM 79 10 AM–12 PM 65 12 PM–2 PM 87 2 PM –4 PM 64 4 PM –6 PM 73 6 PM –8 PM 82 8 PM –10 PM 43 10 PM–12 AM 52 12 AM–6 AM 15 Cost/day/agent $170 $160 $175 $180 $195 ISyE Lecture #5 5 ISyE Lecture #5 Personnel Scheduling Decision Variables and Objective Function Let x j = the number of agents assigned to shift j , j = 1 , . . . , 5 Is divisibility assumption satisfied? The objective function is given by Z = 170 x 1 + 160 x 2 + 175 x 3 + 180 x 4 + 195 x 5 . We want to minimize Z . ISyE Lecture #5 6 ISyE Lecture #5 Personnel Scheduling Functional Constraints The number of agents on duty during each time period must be greater than or equal to the required number. For example, from 2 PM–4 PM, we need at least 64 workers. Since shifts 2 and 3 both cover the 2 PM–4 PM time slot, we know x 2 + x 3 64 NB: these constraints come from the data table just like in previous examples if we treat the ’s as 1’s ISyE Lecture #5 7 ISyE Lecture #5 Personnel Scheduling The LP minimize Z = 170 x 1 + 160 x 2 + 175 x 3 + 180 x 4 + 195 x 5 subject to x 1 48 (6–8 AM) x 1 + x 2 79 (8–10 AM) x 1 + x 2 65 (10 AM–12 PM) x 1 + x 2 + x 3 87 (12 PM–2 PM) x 2 + x 3 64 (2 PM–4 PM) x 3 + x 4 73 (4 PM–6 PM) x 3 + x 4 82 (6 PM–8 PM) x 4 43 (8 PM–10 PM) x 4 + x 5 52 (10 PM–12 AM) x 5 15 (12 AM–6 AM) x j 0 j = 1 , . . . , 5 Some of these constraints are redundant —which are they? ISyE Lecture #5 8
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ISyE Lecture #5 Personnel Scheduling Optimal Solution The optimal solution to this LP is ( x 1 , x 2 , x 3 , x 4 , x 5 ) = (48 , 31 , 39 , 43 , 15) .
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