The question I have concerns a hobby project that I'm working on to help out my wife with her work.
I realize that the problem I'm facing is quite similar to what's been answered on SO in numerous threads. However, I can't seem to find any thread that would address one little specific difference that I need to count in.
Here's a detailed description of the problem:
A teacher has numerous students that she teaches. Each student needs to have two lessons every week at least two days apart (i.e. Monday + Wednesday, Tuesday + Thursday etc.). The lessons are individual meaning there's only one student and the teacher in a class at a time. The teacher needs to have exactly one lunch break every day. The lunch break can be any time between two classes - the only condition is: a day cannot start or end in a lunch break. Each student provides their availability during a week (a list of possible time slots they can attend the class).
So far, this looks like a "regular" scheduling problem that there's bunch of material available online on.
But here's the catch: students are in different age meaning their lesson takes either 30 or 45 minutes. Younger students' lesson takes 30 minutes and olders' - 45 minutes. The goal of the algorithm is to find the optimal weekly schedule for the teacher. By optimal I mean a schedule where the teacher needs to spend least amount of time in school - all lessons back to back with no need for the teacher to wait for another student.
My first attempt was to come with all possible permutations of students' classes, lunch breaks and week days but for a 3 student schedule I came up with 13! permutations (that's roughly around 6 bln permutations). Here's why 13:
3 students - 2 classes each
4 lunch breaks (a 4 day week to keep the numbers small)
3 day breaks (a "night" so to speak)
The above gave me 13 elements to try in different permutations. However, after calculating the factorial I gave that up - 6 bln permutations will not be tried in any reasonable time on a "regular" machine. And that's just counting 3 students, the real-life data will be closer to 9 or 10. Obviously, the nonsense permutations (2 "nights" in a row, 2 classes for the same student on the same day and so on) would be thrown right away but still...
That was when I realized I'm not going to do this inventing the wheel by myself.
I did some research and came across Hopcroft-Karp algorithm. However I don't think it can be applied in a mixed timespan slots scenario (30/45 minute lessons). Am I wrong? Then I found some info on genetic algorithms. However, again I am not sure if they can take the mixed timespan condition into account?
I would greatly appreciate any pointers as to which directions my research should go.
Related
I have this problem:
You need to develop a meal regime based on the data entered by the user. We have a database with the meals and their prices (meals also have a mark whether they are breakfast, cause on breakfast we eat smth different from lunch and dinner most often). The input receives the amount of money (Currency is not important) and the number of days. At the output, we must get a meal regime for a given number of days. Conditions:
Final price does not differ from the given one by more than 3%.
meals mustn't repeat more than once every 5 days.
I found this not effective solution: We are looking for an average price per day = amount of money / number of days. Then, until we reach the given number of days, we iterate throught each breakfast, then lunch and dinner (3 for loops, 2 are nested) and if price is not too different, then we end the search and add this day to the result list. So the design now looks like this:
while(daysCounter < days){
for(){
for(){
for(){
}
}
It looks scary, although there is not a lot of data (number of meals is about 150). There are thoughts that it is possible to find a more effective solution. Also i think about dynamic programming, but so far there are no ideas how to implement it.
Dynamic programming won't work because a necessary part of your state is the meals from the last 5 days. And the number of possibilities for that are astronomical.
However it is likely that there are many solutions, not just a few. And that it is easy to find a working solution by being greedy. Also that an existing solution can be improved fairly easily.
So I'd solve it like this. Replace days with an array of meals. Take the total you want to spend. Divide it up among your meals, in proportion to the median price of the options for that meal. (Breakfast is generally cheaper than dinner.) And now add up that per meal cost to get running totals.
And now for each meal choose the meal you have not had in the last 5 days that brings the running total of what has been spent as close as possible to the ideal total. Choose all of your meals, one at a time.
This is the greedy approach. Normally, but not always, it will come fairly close to the target.
And now, to a maximum of n tries or until you have the target within 3%, pick a random meal, consider all options that are not eaten within the last or next 5 days, and randomly pick one (if any such options exist) that brings the overall amount spent closer to the target.
(For a meal plan that is likely to vary more over a long period, I'd suggest trying simulated annealing. That will produce more interesting results, but it will also take longer.)
There is a carwash that can only service 1 customer at a time. The goal of the car wash is to have as many happy customers as possible by making them wait the least amount of time in the queue. If the customers can be serviced under 15 minutes they are joyful, under an hour they are happy, between 1 hour to 3 hours neutral and 3 hours to 8 hours angry. (The goal is to minimize angry people and maximize happy people). The only caveat to this problem is that each car takes a different amount of time to wash and service so we cannot always serve on first come first serve basis given the goal we have to maximize customer utility. So it may look like this:
Customer Line :
Customer1) Task:6 Minutes (1st arrival)
Customer2) Task:3 Minutes (2nd arrival)
Customer3) Task:9 Minutes (3rd)
Customer4) Task:4 Minutes (4th)
Service Line:
Serve Customer 2, Serve Customer 1, Serve Customer 4, Serve Customer 3.
In this way, no one waited in line for more than 15 minutes before being served. I know I should use some form of priority queue to implement this but I honestly know how should I give priority to different customers. I cannot give priority to customers with the least amount of work needed since they may be the last customer to have arrived for example (others would be very angry) and I cannot just compare based on time since the first guy might have a task that takes the whole day.So how would I compare these customers to each other with the goal of maximizing happiness?
Cheers
This is similar to ordering calls to a call center. It becomes more interesting when you have gold and silver customers. Anyway:
Every car has readyTime (the earliest time it can be washed, so it's arrival time - which might be different for each car) and a dueTime (the moment the customer becomes angry, so 3 hours after readyTime).
Then use a constraint solver (like OptaPlanner) to determine the order of the cars (*). The order of the cars (which is a genuine planning variable) implies the startWashingTime of each car (which is a shadow variable), because in your example, if customer 1 is ordered after customer 2 and if we start at 08:00, we can deduce that customer 1's startWashingTime is 08:03.
Then the endWashingTime is startWashingTime + washingDuration.
Then it's just a matter of adding 2 constraints and let the solver solve() it:
endWashingTime must be lower than dueTime, this is a hard constraint. This is to have no angry customers.
endWashingTime must be lower than startTime plus 15 minutes, this is a soft constraint. This is to maximize happy customers.
(*) This problem is NP-complete or NP-hard because you can relax it to a knapsack problem. In practice this means: you can't write an algorithm for it that scales out and finds the optimal solution in reasonable time. But a constraint solver can get you close.
I asked few days ago, a question about how to transform a University Class Scheduling Problem into a Boolean Satisfiability Problem.
(Class Scheduling to Boolean satisfiability [Polynomial-time reduction])
I got an answer by #Amit who was very elegant and easy to code.
Basically, his answer was like this : instead of considering courses, he considered time-intervals.
So for the i-th course, he just indicted all the possible intervals for this course. And we obtain a solution when there is at least 1 true-interval for every course and when no interval overlap an other.
This methods works very well when we consider only courses and nothing else. I generalize it by encoding the room inside the interval.
for example, instead of [8-10] to say that a course can happen between 8am and 10am.
I used [0.00801 - 0.01001] to say that a course can happen between 8am and 10am in the room 1.
I'm sure that you are currently wandering "why use double ?" well, because here come my problem :
To continue to generalize this method, I encode also the n° of the teacher in this interval.
I used [1.00801 - 1.01001] to say that a course can happen between 8am and 10am in the room 1 and be taught by the teacher n°1.
Here is what I got for now :
like this [1.008XX - 1.010XX] can happen in the same time as [2.008YY - 2.010YY], which is true, if the teacher 1 is teaching in the room X between 8am and 10am, the teacher 2 can teach also in Y between 8am and 10am, if and only if the room is available.
The problem is : with this method I cannot assure that XX and YY will be different and that YY will be available, because [1.008XX - 1.010XX] don't overlap [2.008XX - 2.010XX], so for now, the solver consider this possible.
And I still don't have any clue on how to assure this, by using this interval-method...
I need a way to encode {Interval, room and teacher-id} in order that :
a teacher cannot be in 2 places in the same interval.
there cannot be 2 teachers in the same room for the same interval.
there is a least 1 interval true by course.
Thanks in advance for your help,
Best regards !
Follow up question: Class Scheduling to Boolean satisfiability [Polynomial-time reduction] Final Part
This answer is extension of Part 1's answer, and uses the same notations when possible.
Ok, assume each interval is assigned to one teacher (if more than one teacher can take the interval, just have multiple instances of it, with different teachers per instance), so to indicate teacher t teaches in a classroom p at time x to y, we can use the old variable that this class is given - V_{i,j} - for the class and interval.
For each teacher t , and for each pair of intervals c=(x1,y1), d=(x2,y2) in classes (a,b) the teacher might participate in, add the clause:
Q_{t,i,j} = Not(V_ac) OR Not(V_bd) OR Smaller(y1,x2) OR Smaller(y2,x1)
Intuitively, the above clause guarantees a teacher cannot be in the same time in two places - no intervals overlap that the same teacher is assigned to them.
By chaining each pair (i,j) for each teacher t with AND to the original formula, it satisfies your first constraint - a teacher cannot be in 2 places in the same interval. - since each teacher cannot be in two places at the same time.
Your second constraint there cannot be 2 teachers in the same room for the same interval. is also satisfied by the fact that there cannot be two classes that overlap the time and class.
The 3rd constraint there is a least 1 interval true by course. is satisfied by the F1 clause, since you have to choose at least one interval (with one teacher assigned) for each course.
Here is my scenario,
I run a Massage Place which offers various type of massages. Say 30 min Massage, 45 min massage, 1 hour massage, etc. I have 50 rooms, 100 employees and 30 pieces of equipment.When a customer books a massage appointment, the appointment requires 1 room, 1 employee and 1 piece of equipment to be available.
What is a good algorithm to find available resources for 10 guests for a given day
Resources:
Room – 50
Staff – 100
Equipment – 30
Business Hours : 9AM - 6PM
Staff Hours: 9AM- 6PM
No of guests: 10
Services
5 Guests- (1 hour massages)
3 Guests - (45mins massages)
2 Guests - (1 hour massage).
They are coming around the same time. Assume there are no other appointment on that day
What is the best way to get ::
Top 10 result - Fastest search which meets all conditions gets the top 10 result set. Top ten is defined by earliest available time. 9 – 11AM is best result set. 9 – 5pm is not that good.
Exhaustive search (Find all combinations) - all sets – Every possible combination
First available met (Only return the first match) – stop after one of the conditions have been met
I would appreciate your help.
Thanks
Nick
First, it seems the number of employees, rooms, and equipment are irrelevant. It seems like you only care about which of those is the lowest number. That is your inventory. So in your case, inventory = 30.
Next, it sounds like you can service all 10 people at the same time within the first hour of business. In fact, you can service 30 people at the same time.
So, no algorithm is necessary to figure that out, it's a static solution. If you take #Mario The Spoon's advice and weight the different duration massages with their corresponding profits, then you can start optimizing when you have more than 30 customers at a time.
Looks like you are trying to solve a problem for which there are quite specialized software applications. If your problem is small enough, you could try to do a brute force approach using some looping and backtracking, but as soon as the problem becomes too big, it will take too much time to iterate through all possibilities.
If the problem starts to get big, look for more specialized software. Things to look for are "constraint based optimization" and "constraint programming".
E.g. the ECLIPSe tool is an open-source constraint programming environment. You can find some examples on http://eclipseclp.org/examples/index.html. One nice example you can find there is the SEND+MORE=MONEY problem. In this problem you have the following equation:
S E N D
+ M O R E
-----------
= M O N E Y
Replace every letter by a digit so that the sum is correct.
This also illustrates that although you can solve this brute-force, there are more intelligent ways to solve this (see http://eclipseclp.org/examples/sendmore.pl.txt).
Just an idea to find a solution:
You might want to try to solve it with a constraint satisfaction problem (CSP) algorithm. That's what some people do if they have to solve timetable problems in general (e.g. room reservation at the University).
There are several tricks to improve CSP performance like forward checking, building a DAG and then do a topological sort and so on...
Just let me know, if you need more information about CSP :)
This question already has answers here:
Best Fit Scheduling Algorithm
(4 answers)
Closed 9 years ago.
I have the following problem to solve, perhaps you could give me some ideas regarding the problem - solving approach:
There are:
8 classrooms
16 teachers
201 students
149 parents
241 appointments (most of the parents need to see multiple teachers, either because the have more than one child or because one child is being taught by two or more teachers)
2 days.
For each day:
7 classrooms are available for 20 hours per day.
1 classroom is available for 10 hours per day.
Each teacher occupies one classroom
Each appointment lasts for one hour
Further constrains:
- For each parent, all appointments must be sequential (having at maximum 1 hour pause)
- Each parent should have to visit the school one day only.
- For each teacher, all appointments within the day must be sequential (having at maximum 2 hours pause)
- Out of the 16 teachers, 3 can only be present one of the two days.
I'm trying to find an approach to produce the appointments schedule, obviously without having to calculate all possible variations until all the requirements are met. Any ideas?
You need to define your constraints a bit more; i.e., what is the relationship between students and teachers? What is the relationship between students and parents? Do parents have to have individual appointments, or are the parents of a single student allowed to meet with a single teacher together?
I'd approach this with an initial (in testing) naive approach; just pick your highest constrained resource (looks like the teachers that can only be present for one of the two days), schedule those using the first available resources, and then continue through the set of resources, scheduling them using the first available resources that match their constraints and see if you can schedule the entire set. If not, you need to find your limiting resource(s), and apply some heuristics to your matching to find the best way to optimize those limited resources.
It's a bit of a tricky problem; have fun!
Take a look at the curriculum course track of ITC2007 and it's implementation in Drools Planner or unitime. Both of them use meta-heuristics such as tabu search and simulated annealing.