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 :)
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.)
I was given a task of putting students into groups (to prepare a coding camp), but with several constraints. Though I've finished the task by hand, I'd like to know is there already exist some algorithms for tasks like this, or how can I design such an algorithm.
Background: 40 students in total, with these attributes:
gender: F/M
grade: Year 1/2
school: School 1/School 2/...
early assessment result: Rank from 1 to 40
Constraints: All of them needs to be satisfied.
Exactly 4 people per group
Each group needs to have at least a girl
Each group needs to have at least a Year 2 student
4 group members needs to come from 4 different schools
Each group needs to have at least a student who ranked top 10 in early assessment
What I'm expecting:
The Best: An existing algorithm/program for these kind of problems
Or, An algorithm for this specific problem
Or at least, Some ideas of creating an algorithm for this specific problem
My thoughts:
Since I've successed in making groups by hand, I know that such a solution indeed exists for my current dataset. But if I need an algorithm to find a solution for me, it should first try to check whether a solution even exists, by check if the number of girl / Year 2 students is greater than 10 (with pigeonhole principle), and some other conditions. And obviously, Constraint 5 is the easiest, and can provide a base solution for the rest. However, I still can not find a systematic way of doing it. Perhaps bruteforce and randomization can help? I'm not sure.
And sorry, since the data is confidential, I can not post it.
Update: After consulting a friend, here is a possible method:
First put the top 1 to 10 into 10 different groups.
Then iterate through groups. If the only person in the group is a boy/girl, try to add a girl/boy from a different school.
Then the problem size is reduced from 2^40 to 2^20, making bruthforce a viable solution.
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.
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.
EDIT: Just to make sure someone is not breaking their head on the problem... I am not looking for the best optimal algorithm. Some heuristic that makes sense is fine.
I made a previous attempt at formulating this and realized I did not do a great job at it so I removed that question. I have taken another shot at formulating my problem. Please feel free to provide any constructive criticism that can help me improve this.
Input:
N people
k announcements that I can make
Distance that my voice can be heard (say 5 meters) i.e. I may decide to announce or not depending on the number of people within these 5 meters
Goal:
Maximize the total number of people who have heard my k announcements and (optionally) minimize the time in which I can finish announcing all k announcements
Constraints:
Once a person hears my announcement, he is be removed from the total i.e. if he had heard my first announcement, I do not count him even if he hears my second announcement
I can see the same person as well as the same set of people within my proximity
Example:
Let us consider 10 people numbered from 1 to 10 and the following pattern of arrival:
Time slot 1: 1 (payoff = 1)
Time slot 2: 2 3 4 5 (payoff = 4)
Time slot 3: 5 6 7 8 (payoff = 4 if no announcement was made previously in time slot 2, 3 if an announcement was made in time slot 2)
Time slot 4: 9 10 (payoff = 2)
and I am given 2 announcements to make. Now if I were an oracle, I would choose time slots 2 and time slots 3 because then 7 people would have heard (because 5 already heard my announcement in Time slot 2, I do not consider him anymore). I am looking for an online algorithm that will help me make these decisions on whether or not to make an announcement and if so based on what factors. Does anyone have any ideas on what algorithms can be used to solve this or a simpler version of this problem?
There should be an approach relying upon a max-flow algorithm. In essence, you're trying to push the maximum amount of messages from start->end. Though it would be multidimensional, you could have a super-sink, which connects to each value of t, then have each value of t connect to the people you can reach at this time and then have a super-sink. This way, you simply have to compute a max-flow (with the added constraint of no more than k shouts, which should be solvable with a bit of dynamic programming). It's a terrifically dirty way to solve it, but it should get the job done deterministically and without the use of heuristics.
I don't know that there is really a way to solve this or an algorithm to do it the way you have formulated it.
It seems like basically you are trying to reach the maximum number of people with exactly 2 announcements. But without knowing any information about the groups of people in advance, you can't really make any kind of intelligent decision about whether or not to use your first announcement. Your second one at least has the benefit of knowing when not to be used (i.e. if the group has no new members then you can know its not worth wasting the announcement). But it still has basically the same problem.
The only real way to solve this is to use knowledge about the type of data or the desired outcome to make guesses. If you know that groups average 100 people with a standard deviation of 10, then you could just refuse to announce if less than 90 people are present. Or, if you know you need to reach at least 100 people with two announcements, you could choose never to announce to less than 50 at once. Obviously those approaches risk never announcing at all if the actual data does not meet what you would expect. But that's always going to be a risk, since you could get 1 person in the first group and then 0 in all of the rest, no matter what you do.
Or, you could try more clearly defining the problem, I have a hard time figuring out how to relate this to computers.
Lets start my trying to solve the simplest possible variant of the problem: Lets assume N people and K timeslots, but only one possible announcement. Lets also assume that each person will only ever stay for one timeslot and that each person who hasn't yet shown up has an equally probable chance of showing up at any future timeslot.
Given these simplifications, at each timeslot you look at the payoff of announcing at the current timeslot and compare to the chance of a future timeslot having a higher payoff, eg, lets assume 4 people 3 timeslots:
Timeslot 1: Person 1 shows up, so you know you could get a payoff of 1 by announcing, but then you have 3 people to show up in 2 remaining timeslots, so at least one of those timeslots is guaranteed to have 2 people, so don't announce..
So at each timeslot, you can calculate the chance that a later timeslot will have a higher payoff than the current by treating the remaining (N) people and (K) timeslots as being N independent random numbers each from 1..k, and calculate the chance of at least one value k being hit more than or equal to the current-payoff times. (Similar to the Birthday problem, but for more than 1 collision) and then you need to decide hwo much to discount based on expected variances. (bird in the hand, etc)
Generalization of this solution to the original problem is left as an exercise for the reader.