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Appointment Scheduling in ruby/rails

开发者 https://www.devze.com 2023-03-21 12:25 出处:网络
I am working on a call center software in rails and need to schedule appointments for agentsthat can handle calls for the customers. Having said the call center software, need to make sure that I sche

I am working on a call center software in rails and need to schedule appointments for agents that can handle calls for the customers. Having said the call center software, need to make sure that I schedule the appointments utilizing the entire agent's schedule as much as possible, leaving scope for least number of holes (where agent has no appointment).

Given an agent's schedule, for example 9:00AM to 5:30PM for a given day, with a lunch br开发者_如何学Pythoneak of 30 minutes between 1:00PM - 1:30PM, I need to schedule appointments of varying length in duration, some 60 minutes and some 90 minutes.

And if some reason, lunch break is leaving some holes in the schedule, I should be able to move lunch break 30minutes +/-, so instead of 1:00PM - 1:30PM it could be moved between 1:30PM - 2:00PM or 12:30PM - 1:00PM.

I started off creating lunch breaks as a kind of appointment which will provide the flexibility of moving the starts_at and finishes_at attributes. And the appointments being either 60 minutes or 90 minutes, which are multiple of 30 minutes and lunch also being 30 minutes, I started off splitting agents schedule into slots of 30 minutes each.

So for a given agent on a given day, looking at his schedule I instantiated an array of slots each with a duration of 30 minutes and starts_at and finishes_at attributes be like 9:00AM - 9:30AM, 9:30AM - 10:00AM, etc.

I need some help on looping through this array of appointment slots and pull either 2 consecutive slots or 3 consecutive slots depending upon 60 or 90 minute duration appointment, keeping in mind that I should be able to move the lunch +/- 30 minutes.

Any help is much appreciated.


Looking at your problem:

  1. Appointments are either 60 or 90 minutes long.
  2. Lunch can vary between a 90 minute interval 12:30-2:00

And we want to minimize the amount of minutes that have no appointments.

Now, you have a time interval to fill, which is 9:00AM to 5:30pm. Assuming appointments fall between 9:00-5:30, we can use a greedy algorithm for interval scheduling based on earliest finish time (source) with your additional constraint.

Basically the algorithm is as follows (in pseudo)

Let R be the set of all appointments
Let R11 be the set of appointments from R before 12:30 that are compatible with 12:30-1:00 and R12 be the set of appointments from R after 1:00 that are compatible with 12:30-1:00
Let R21 be the set of appointments from R before 1:00 that are compatible with 1:00-1:30 and R22 be the set of appointments from R after 1:30 that are compatible with 1:00-1:30
Let R31 be the set of appointments from R before 1:30 that are compatible with 1:30-2:00 and R32 be the set of appointments from R after 2:00 that are compatible with 1:30-2:00

Let R1Comb = findSet(R11) + 12:30-1:00 + findSet(R12)
Let R2Comb = findSet(R21) + 1:00-1:30 + findSet(R22)
Let R3Comb = findSet(R31) + 1:30-2:00 + findSet(R32)

Function findSet(R) 
    Let A be the time interval to fill    
    While R is not empty
        Choose a request r in R that has the smallest finishes_at
        Add r to A
        Remove all appointments in R that are not compatible with r
    EndWhile
    Return A
EndFunction

Return the R that has the smallest amount of holes in R1Comb, R2Comb, R3Comb

This algorithm makes use of a few concepts

  1. An appointment r1 is not compatible with r2 if they overlap.
  2. Because of #1, we know that Ri1/Ri2 for i=1,2,3 will not be conflicting with each other. Because if an appointment in Ri2 is not compatible with Ri1, then it is also not compatible with the lunch period, which is a contradiction because we took out all the non compatible appointments.
  3. Once we split the set of appointments, then it's a matter of solving 2 scheduling problems, which can be solved greedily.

An this algorithm is still O(n log n) because you are doing the greedy algorithm 6 times (a constant), and each greedy iteration is O(n log n), and the first few lines and the last line are all O(n).

People write theses on scheduling and it's not an easy problem. I suggest you looking at http://www.asap.cs.nott.ac.uk/watt/resources/university.html to get a better understanding.

Good luck :)

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