 So let's start. I guess all of you are familiar with the Trezang salesman problem. So we are given a list of cities and the distances between them. And our aim is to find a closed route that visits each city exactly once with the minimum cost. Now we have something more. We have Trezang salesman problem with time in those. And this time we also have some time in those for each city. And this means that you need to visit each city within this time in those. That is determined. And this problem has also practical applications. As you see here, for instance, vehicle routing, this is vehicle routing problem with time in those. And here you have more than one vehicle traveling through the cities that again you have to obey this time in those constraint. Let me demonstrate this through an example. Here we are given a graph with four nodes. And on the right hand side, you see the early start time and late start time for each city. So this is a time in those. And this route you see right now, which starts at zero goes to three, sorry, zero, three, one, two. It has cost of five. Now let's trace this. At time equals zero, we are at node zero. Then at time equals two, we are at node three. And then if you check node two, sorry, if you check node three, the early start time is four. This means you need to wait. So we wait until t equals four. And then at time equals five, we are at node one. But now let's check the late start time for node one. So it is four, but not t is five. So this thing is this route is not a feasible one. Now let's check another route. I don't know why it is transitioning so fast, but okay. Now let's check this route. This has cost six. We will start from, again, zero, one, three, two. So at t equals zero, we are again at zero. Now at t equals three, we are at node one. Let's check. We are within the time in the frame. Now we go to node three. It's t equals four. And this is still okay. We go to node two. Now early start time is 12. So we need to wait. But that's fine. We can wait. And we arrive to node zero at t equals 13. For node zero, we don't assume an early start or late start time. And so this is a feasible route. Let me mention one more thing. Here our aim is to minimize the total cost. In this case, it was six. It's not to minimize the total time. This is another problem called makespan problem. It's closely related, but it's a different problem. And in this work, our aim is to provide unconstrained binary problem formulation for this problem, so that you can solve it using, for instance, quantum algorithm like quantum annealing on d-bay. And let me also mention that this problem is MP hard, like TSB. So you have seen this since Monday. I won't go into details again. But our aim is to come up with an ising model for the problem, so that we can use d-bay and quantum annealing to solve the problem on d-bay machines. So for this, as I mentioned, we need to come up with ising model, which uses spin variables, which are minus one or one. But often it is more convenient to work with binary variables instead. And this leads us to Cuba formulation. Here, xis are binary variables. They are either zeros or ones. And qid's are some real coefficients. This is quadratic in the sense that we have at most quadratic terms. So we don't have terms like xi, xj, xk. And it's unconstrained in the sense that we are minimizing some function, but we don't have any constraints. And once you can express your problem in this form, it's really easy to obtain the equivalent ising model through this transformation. And then you are done. You can use quantum annealing to find the solution to your problem. Now, to formulate the Cuba formulation for TSB with time in those, we need to consider three things. First of all, we need to have a route that visits each city exactly once. And secondly, this route should obey the timing of constraints. And thirdly, we need to minimize the total cost of the route. The state of art before our work was due to Papalistas. They had a Cuba formulation, but the problem was in their formulation, it was possible to obtain solution which consisted of subtours. But we need a closed single tour. Furthermore, they had an assumption that early start time is equal to zero for all cities. And they did not allow any waiting times. So what we did was we came up with three different formulations. They are different in their nature, as I will explain now. Sorry. In the first one, the main binary variables are of the form X, U, V, I, which is equal to one. If nos U and V are at consecutive positions, I minus one and I in the tour. And this results in a model of O, N cubed plus and delta variables. Here N is the number of cities, and delta is the largest number appearing in the latest start time. The second one, we have the main binary variable X, V, I. And this is one if nos V is visited at I step in the tour. We call this node-based model. And this is how the Cuba for TSP is usually formulated. However, if you use this binary variable, we end up with a higher order model, not a quadratic one. So it has better number of variables, as you can see, but it's higher order. And thirdly, we have another model here where the main binary variable is X, U, V. This is developed from an existing integer linear program, which existed in the literature. And this has N squared plus and delta number of variables. Now I will try to explain the first formulation briefly, how we came up with the formulation. First of all, we need to have the root constraint. And in the first term, you can see that this is forcing that only one of the variables of the form X, 0, V, 1 is equal to 1. Otherwise, we have a penalty term. So I mentioned that we will have an unconstrained formulation. So what we do to incorporate the constraint is to use the penalty method here so that whenever the summation is not equal to 1, we will have a here positive penalty. And this will increase the objective function, and this is something not desired. So the first one is for T equals 1. And the second one ensures that exactly one edge is visited at each time point. And similarly this term here makes sure that the two ends at 0 and exactly one edge is visited also at time N plus 1. Furthermore, to ensure that we have a closed single cycle, we make sure that each city is left exactly once. And also, we need to make sure that if we are traversing some edge UV at time I minus at time I, then we are visiting some edge VW at I plus 1. So here V is common. And this makes sure that we have a single closed route, not sub-tours. And now we have the time in those constraints. Here we define AI. This is not a variable, but I will explain it in the next slide. This denotes the arrival time to the I visited city. We have an integer variable omega I, and this denotes the waiting time at the I visited city. And overall, we call this the service time. So you can imagine this as you are going to some city, you wait there and you give service at that city. And we need to make sure that this is greater than or equal to the early start time of the I visited city. And this is the term on the left. And also we need to make sure that the arrival time is less than or equal to the latest start time of the city visited in the I location. And this is the term on the right-hand side. And for this AI, we came up with some recursive relation. And so this is saying arrival time to the city visited at I position is equal to the arrival time to the city visited at I minus 1 position plus waiting time there, plus the cost of moving between them. And if we express this explicitly, we obtain the following. The first one is trivial because it is just the cost of the edge between no zero and the first visited city. And then arrival time to the I city is expressed as a total waiting time until reaching city visited at position I plus cost of all the edges in between. So we will replace this AI. You saw in the previous slide with this expression. Involving binary and integer variables. So we come up with this time of constraints. This is for I equals 1. But again, we need to use a penalty method because we cannot have constraints in Cuba. But first we need to convert inequalities into equalities. And we do this by using slack variables, XI. They are integer variables. And so this inequality on the left appears as this term on the right-hand side in the objective function. And this is denote binary encoding because those omega variables and XI variables, they are integers. But they need to be converted into binary variables first. And similarly, we convert each inequality into some penalty term on the right-hand side. And finally, this is the cost of the route. This is basically the, so the first one is just the cost of traveling from city zero to the first visited city. Then we have the cost of all the edges in between. And finally we have the last visited edge cost. And this is overall how our final hematonic looks like. It involves the route constraint term, time window term, and the cost term. Here P1, P2, P3, they are penalty coefficients. Indeed, you can take P3 to P1. And then you should adjust P1 and P2 so that they are not large enough to override the cost. But they are not small enough so that they get violated. And very briefly to mention, for the last formulation I showed you in the previous slide, there exists an integer linear program, which means an integer linear program has a linear object with linear inequalities. What we did was to, we realized that some of the inequalities were redundant and we removed them. And then we also made some simplification for the variables. And then we also used the penalty method to obtain the cuba formulation and this resulted in a cuba formulation with n squared plus n squared delta variables. To compare the number of variable requirements by each model, we tried some real life instances with large number of cities. And in this graph you can see this. And as you can see for 200 cities, the number of required variables reached 10 to 6 for the age-based model. So from the synthetics, this was expected. And you can see that node-based is the best, but however, it's not quadratic. So that's the problem with that. Although our aim was not to have some experimental results, we tried running some instances on D-Way. We created some random instances with 3, 4, and 5 cities. And here you can see the number of required physical variables after the embedding. We used D-Way's advantage with chimerotopology. And since our instances were small in the number of cities, you can see that age-based required less number of physical variables compared to ILP, both physical and logical. And to select the penalty values, we used simulated annealing. So we kind of did a grid search. We picked some P1, P2 pairs, and we checked the highest probability of returning optimal solution in those cases. So those numbers are denoting probability. And then for each instance, we selected the P1, P2 pair which has the highest probability of acceptance. And here are some experimental results from D-Way machine. So those are energy histograms. Let me explain what I mean by optimal feasible and feasible. Optimal means that the obtained binary assignment results in a route which corresponds to the optimal solution. But for instance, slack variables may be assigned in the wrong way. So we are not interested in that. But it's enough if it is encoding an optimal route. And this goes for the feasible as well. The results show that ILP formulation performed better than the age-based one. Because in the age-based model, we could not find any optimal solution, for instance, for instances with 5 cities. So although this looks like a really small instance with 5 cities, we were not able to get any optimal solution. And here, this red line, it denotes real ground state energy. And you can see that most of the time, the lowest energy we could get and the ground state energy, there was a gap. The annealing time was 15 microseconds here. We tried to increase it, but it did not help. We also tried different chain strengths, and we picked the best one among those. And finally, let me again mention that besides this slack variable assignment, there is also another thing in the age-based formulation. The same optimal route could be obtained by assigning different weighting times. Because imagine that you go to a city. Even though you don't need to wait, you can wait there. Our formulation was allowing this so that there were many different bit assignments, which correspond to the same route. We thought that this would help while finding the optimal solution, but it ended up that ILP was performing better. And I can add that the choice of the penalty values is really challenging, because we needed to tune both parameters at the same time. And finally, I think that constraint-hybrid solver recently provided by D-Wave can be a better alternative for such problems, where you have lots of constraints. And the reason is, in that solver, you don't need to use this penalty method. You can directly list your constraints. And we recently tried this, indeed, on a real-world problem. And it's really give comparable results compared to classical solvers. So I believe this can be a better alternative for solving such constraint problems. And thanks for listening. I will be happy to answer if you have any questions. From online, you can just turn on... This is from the previous session. Yes, there is a question here. What were the other solvers you compared to the hybrid solver CQM? It was not for this problem. We solved vehicle optimization problem. So we did not try hybrid solver for TSB. And for this vehicle optimization problem, we tried solving it with Ruby, CBC, versus hybrid solver of D-Wave, constraint solver. And this work is also available in ARCA. Question from here. Can you... Let's open the chat. I think there is maybe another question. Thanks. No, OK. Thank you. Then I think we have no other questions. So let's thank the speaker again.