I am trying to overcome a machine allocation problem with time horizon of 5 day. Production plan is hard to catch up, so my objective is to minimize total machines working time spent. Machines uses molds to produce and there are molds for each type of product. If a product is produced at the end of the day and if there will be production later day, total setup needed for that machine should be decreased by one. For this reason,
sets
i: mold type
j:jobs
k: days
parameters
x(i,k) ith mold production needed at day k
y(i,j) 1 if ith mold is compatible with jth machine
Decision variable
m(i,j,k) : 1 if ith mold processed in jth machine in day k 0 o/w
b(j,k) setup number of jth machine in day k
While computing the setup number for day 1, b(j,’1’), is simply equal to the sum of m(i,j,k).
For computing other days setup number I tried these but these made problem nonlinear and it takes months to solve.
b(j,'2')=e=sum(i,m(i,j,'2')) - sum(i,m(i,j,'2')*m(i,j,'1'))
By this way, if mold i is produced in both days, there will not be any setup made at second day. In order to restrain multiple setup reduction I put: sum(i,m(i,j,'2')*m(i,j,'1')) =l= 1
So, how can I decrease the setup number for a machine if it has used a mold a day before without making the problem nonlinear.
It is possible to linearize m(i,j,'2')*m(i,j,'1'):
Both(i,j) <= m(i,j,'2')
Both(i,j) <= m(i,j,'1')
Both(i,j) >= m(i,j,'2')+m(i,j,'1')-1
Both(i,j) is a binary variable
This transformation is done automatically by some solvers.
Note that there are alternative ways to model the start of a run, and often there are things to exploit (depending on the details).
Related
Detailed business problem:
I'm trying to solve a production scheduling business problem as below:
I have two plants producing FG A and B respectively.
Both the products consume the same Raw Material x
I need to create a 30 day production schedule looking at the Raw Material availability.
FG A and B can be produced if there is sufficient raw material available on the day.
After every 6 days of production the plant has to undergo maintenance and the production on that day will be zero.
Objective is to maximize the margin looking at the day level Raw material available and adhere to the production constraint (i.e. shutdown after every 6th day)
I need to build a linear programming to address the below problem:
Variable y: (binary)
variable z: cumulative of y
When z > 6 then y = 0. I also need to reset the cumulation of z after this point.
Desired output:
How can I build the statement to MILP constraint. Are there any techniques for solving this problem. Thank you.
I think you can model your maintenance differently. Just forbid any sequences of 7 ones for y. I.e.
y[t-6]+y[t-5]+y[t-4]+y[t-3]+y[t-2]+y[t-1]+y[t] <= 6 for t=1,..,T
This is easier than using your accumulator. Note that the beginning needs some attention: you can use historic data for this. I.e., at t=1, the values for t=0,-1,-2,.. are known.
Your accumulator approach is not inherently wrong. We often use it to model inventory. An inventory capacity is a restriction on how large the accumulated inventory can be.
I'm new to linear programming and trying to develop an ILP model around a problem I'm trying to solve.
My problem is analogous to a machine resource scheduling problem. I have a set of binary variables to represent paired-combinations of machines with a discrete-time grid. Job A takes 1 hour, Job B takes 1 hr and 15 minutes, so the time grid should be in 15 minute intervals. Therefore Job A would use 4 time units, and Job B would use 5 time units.
I'm having difficulty figuring out how to express a constraint such that when a job is assigned to a machine, the units it occupies are sequential in the time variable. Is there an example of how to model this constraint? I'm using PuLP if it helps.
Thanks!
You want to implement the constraint:
x(t-1) = 0 and x(t) = 1 ==> x(t)+...+x(t+n-1) = n
One way is:
x(t)+...+x(t+n-1) >= n*(x(t)-x(t-1))
Notes:
you need to repeat this constraint for each t.
A slightly better version is:
x(t+1)+...+x(t+n-1) >= (n-1)*(x(t)-x(t-1))
There is also a disaggregated version of this constraint that may help performance (depending on the solver: some solvers can do this disaggregation automatically).
Things can become interesting near the beginning and end of the planning period. E.g. machine started at t=-1.
Update:
A different approach is just to limit the "start" of a job to 1. I.e. allow only the combination
x(j,t-1) = 0 and x(j,t) = 1
for a given job j. This can be handled in a similar way:
start(j,t) >= x(j,t)-x(j,t-1)
sum(t, start(j,t)) <= 1
0 <= start(j,t) <= 1
H_ello lovely people,
my program is written as a scalable network framework. It consists of several components that run as individual programs. Additional instances of the individual components can be added dynamically.
The components initially register with IP and Port at a central unit. This manager periodically sends to the components where other components can be found. But not only that, each component is assigned a weight / probability / chance of how often it should be addressed compared to the others.
As an example: 1Master, Component A, B, C
All Components registered at Master, Master sends to A: [B(127.0.0.1:8080, 3); C(127.0.0.1:8081. 5)]
A runs in a loop and calculates the communication partner over and over again from this data.
So, A should request B and C in a 3 to 5 ratio. How many requests each one ultimately gets depends on the running performance. This is about the ratio.
Of course, the numbers 3 and 5 come periodically and change dynamically. And it's not about 3 components but potentially hundreds.
My idea was:
Add 3 and 5. Calculate a random number between 1 and 8. If it is greater than 3, take C else B ....
But I think that's not a clean solution. Probably computationally intensive in every loop. In addition, management and data structures are expensive. In addition, I think that a random number from the STL is not balanced enough.
Can someone perhaps give me a hint, how I implemented this cleanly or does someone have experiences with it or an idea?
Thank you in every case;)
I have an idea for you:
Why not try it with cummulative probabilities?
1.) Generate a uniformly distributed random number.
2.) Iterate through your list until the cumulative probability of the visited element is greater than the random number.
Look at this (Java code but will also work in C++), (your hint that you use C++ was very good!!!)
double p = Math.random();
double cumulativeProbability = 0.0;
for (Item item : items) {
cumulativeProbability += item.probability();
if (p <= cumulativeProbability) {
return item;
}
}
I have difficulty formulating the constraints correctly. Dumbed down version of the problem:
There are 12 time units, 3 products, demand d_{i,t} for the product i at time $t$ is known in advance and the resources r_{i,t} (all 8, product i uses the non-i resources) necessary for product i at time t are known as well. We need to minimize holding costs h_i by deciding how much of product i we need to produce at time t, this is called x_{i,t}. Starting inventory for every product is 6. To help I introduce stock level s_{i,t}. This equates to the following formulation:
I got this working using the Excel solver but I need to do this in AIMMS. The Stock variable s is giving me issues, I can't get it to work using an if statement to condition on if t=1 nor would I know how to split it in two constraints, given the first iteration of the second constraint would need to reference the first constraint.
You can specify index domain in your constraint attributes as follows:
(t,i) | t > 1
if t > 1 statement should work if the set of time instances is a subset of integers. If not - you should use ord(t) > 1, i.e.
if ord(t) > 1
then
Your_Constraint
endif
I'm currently implementing a histogram that will show a very large scale data using Qt and I have some doubts about which data structure(s) I should be using for my problem. I will be displaying the amount of queries received from users of an application and the way I should display is as follows -in a single application that will show different histograms upon clicking different "show me this data etc." buttons-
1) Display the histogram of total queries per every month -4 months of data here, I
kept four variables and incremented them as I caught queries belonging to those months
in the CSV file-
2) Display the histogram of total queries per every single day in a selected month -I was thinking of using 4 QVectors to represent the months for this one, incrementing every element of the vector (day), as I come by that specific day -e.g. the vector represents the month of August and whenever I come across a data with 2011-08-XY , I will increment the (XY + 1)th element of that vector by 1- my second alternative is to use 4 QLinkedList's for the sake of better complexity but I'm not sure if the ways I've come up with are efficient enough and I'm willing to listen to any other idea.
3) Here's where things get a bit complicated. Display the histogram of total queries per every hour in a selected day and month. The data represented is multiplied in a vast manner and I don't know which data structure -or combination of structures- I should use to implement this one. A list of lists perhaps?
Any ideas on my problems at 2) and 3) would be helpful, Thanks in advance.
Actually, it shouldn't be too unmanageable to always do queries per hour. Assuming that the number of queries per hour is never greater than the maximum int value, that's only 24 ints per day = 32 bits or 64 depending on your machine. Assuming 32 bits, then you could get up to 28 years worth of data per MB.
There's no need to transfer the month/year - your program can work that out. Just assign hour 0 to the earliest point in your data, which you keep as a constant, then work out the date based on hours passed since then.
This avoids having to have a list of lists or anything fancy - just use an array where each address contains the number of hours since hour 0, and the number of queries for that hour.
Why don't you simply use a classical database?
When you start asking these kind of question I think it is a good time to consider a more robust structure.There are multiple data structures implemented inside any DB, optimized either for different access type. You should considerate at least lookup, insertion, deletion, range queries. There is no structure which is better than the others in all costs, so there is always a trade-off.
Qt has some database classes you can use. I never used the Qt SQL library, but I think you should give it a shot. Fortunately, there is a Qt SQL programming guide at the end of the page linked.