In this section, we solve the multiobjective MINLP model presented in

Section 3 with the interactive approach in

Section 4. In addition, we investigate the model by minimizing only the costs. Therefore, in the numerical experiments, tools for solving single- and multiobjective problems are needed. However, multiobjective problems are solved with MITSPA utilizing scalarization. Via the scalarization, multiple objectives are reformulated as a single-objective problem and by solving this problem, a Pareto optimal solution for the original problem can be obtained. Thus, we only need a solver capable of solving single-objective MINLP problems. For this purpose, we use the method employing the branch-and-bound idea called SCIP (Solving Constraint Integer Programs) [

45,

46,

47] in GAMS [

48] with SoPlex and Ipopt. The relative tolerance is set as 0.000001 and and the maximum time limit was set to 86,400 s.

#### 5.1. Cost Minimization Results

We begin the numerical experiments by investigating the single-objective case minimizing the total costs i.e., using (

8) as the only objective. The affects of the hiatus on the costs are also considered. The data used is described in

Appendix A except the parameters related to the costs which are omitted due to their commercial nature. However, the repeatable single-objective example with an artificial cost data are given in

Appendix B. In total, the problem has 887 linear and 61 nonlinear constraints. Furthermore, we have 963 continuous, 77 binary, and 41 integer variables. Note that the scenario without the hiatus has 39 constraints less than the base scenario. The CPU time of the single-objective base scenario having the hiatus is 579 s. For the scenario without the hiatus, the algorithm stopped after 86,439 s reaching the time limit with the 0.0005% relative gap between the dual and the primal bound.

In

Figure 4, the solutions for both scenarios with and without the hiatus are presented.

Figure 4a illustrates the values of the objectives (

1)–(

8) in the interval from 0 to 1 such that 0 represents the value of the ideal vector for the corresponding objective in the base scenario and 1 represents the nadir value in the base scenario. The solution for the base scenario is given in the red line and the solution without the hiatus is given in the blue line.

Figure 4b,c shows the schedules obtained for the base scenario and the scenario without the hiatus, respectively. From these schedules, we see how many canisters are planned to be encapsulated at each period. The blue bars in

Figure 4b,c refer to canisters of OL1-2, the orange refers to LO1-2, and purple refers to OL3.

As we see, when the costs are minimized, the model gains advantage from the hiatus and the solution without the hiatus has around 1.7% higher costs than the solution of the base scenario. More significant differences of the solutions can be seen from the other objective values. For example, the scenario without the hiatus needs around 5.9% more storage time than the base scenario and, even if the both scenarios have the same ending time, the scenario without the hiatus has the encapsulation facility in operation 20 years more than the base scenario. However, the central tunnel is 11.6% longer in the base scenario than in the scenario without the hiatus.

From the schedules obtained, we see that the first three periods, or in other words 15 years, looks similar in the both scenarios. After that, one period has a minor difference and then again the next three periods are similar. Then, the rest of OL1-2 is disposed of in two periods in the base scenario and in five periods in the case without the hiatus. Finally, OL3 is disposed of and the base scenario needs one period less than the case without the hiatus, but the disposal ends at the same time in the both cases. The similar beginning in the schedules indicates that the decision about whether to have the hiatus or not does not significantly affect the disposal pace before the rest of OL1-2 will be disposed of when the costs are minimized. However, for OL1-2, the maximum canister power is higher in the base scenario than in the scenario without the hiatus. Moreover, the spacing between the disposal tunnels is larger while the canister spacing is the same being minimal.

#### 5.2. Multiobjective Optimization Results

In the following, we solve the multiobjective model presented in

Section 3 with MITSPA. First, we focus on the basic scenario with the hiatus. The data used for the parameters are given in

Appendix A other than the cost related parameters which are excluded due to their commercial nature. Compared with the single-objective case, besides having eight objectives instead of one, we need some additional constraints and variables to rewrite the scalarized problem (

44) as in [

37]. The weighting coefficients

${\mathit{\lambda}}^{\phantom{\rule{-0.166667em}{0ex}}U}$ and

${\mathit{\lambda}}^{\phantom{\rule{-0.166667em}{0ex}}A}$ are selected as is suggested in [

39] such that

when

${\mathit{f}}^{nad}-{\mathit{f}}^{R}>0$ and

${\mathit{f}}^{R}-{\mathit{f}}^{\mathit{id}}>0$. Otherwise, the coefficient

$1/({\mathit{f}}^{nad}-{\mathit{f}}^{id})$ is used. Furthermore, we set the coefficient

$\rho =0.0001$.

The decision maker selects the ideal vector

${\mathit{f}}^{id}$ as the first reference point

${\mathit{f}}_{1}^{R}$. Here, we present three different solutions in Step 3 of MITSPA. In

Figure 5, the results for the first reference point are illustrated.

Figure 5a shows the values of the objectives in the interval from 0 to 1. This interval is the same as in

Section 5.1 such that 0 represents the ideal vector of the base scenario and 1 represents the nadir value. The solutions are referred to with the reference point and the value of the parameter

q; for instance, the solution r1q1 is a solution obtained with the reference point 1 when

$q=1$.

Figure 5b–d presents the disposal schedules for the solutions r1q1, r1q3, and r1q8, respectively. In those figures, blue refers to OL1-2, orange to LO1-2, and purple to OL3.

Based on the objective function values illustrated in

Figure 5a, the length of the central tunnel and the total costs seem to be conflicting objectives. In addition, the smaller number of canisters yields to the longer central tunnel. Nevertheless, this sounds reasonable since, with fewer canisters, the power of the canister may be larger, which leads to the larger spacings between canisters. This in its turn lengthens the central tunnel. Furthermore, the lengths of the disposal and central tunnels are also conflicting objectives. Indeed, if the canister heat powers are equal, we may increase the canister spacing or the disposal tunnel spacing to control the temperature on the surface of canisters in the disposal facility. The increase in the canister spacing will increase the disposal and central tunnel lengths. However, the increase in the disposal tunnel spacing will increase only the central tunnel length, but the increase is typically much larger. The schedules in

Figure 5b–d propose the same ending time and the place for the hiatus.

From the first iteration, the decision maker selects the solution r1q8 as the current solution **f**_{1} since it is a good compromise between the solutions presented. It minimizes the additional water pools, and the total costs are still at the reasonable level. The next reference point ${\mathit{f}}_{2}^{R}$ adapts the solution r1q8 expect that the number of canisters is selected to be minimal and the ending and the operation time are increased. These changes aim to minimize the canisters needed while keeping the costs reasonable and trying to avoid the additional water pools.

The results of the second iteration are illustrated in

Figure 6. From

Figure 6a, we see that the solutions r2q5 and r2q7 are quite similar, and the solution r2q4 is clearly a different kind. Both solutions r2q5 and r2q7 need less canisters than the current solution

**f**_{1}, but one additional pool is needed and the storage time is increased. On the other hand, the solution r2q4 lengthens the operation time as was suggested by the reference point

${\mathit{f}}_{2}^{R}$, but also the number of canisters is increased. However, the central tunnel is shorter than in the current solution. In schedules given in

Figure 6b–d, we see that, in the solution r2q4 in

Figure 6b, the hiatus is one period shorter than in other schedules.

The decision maker selects the solution r2q7 as the current solution

**f**_{2}. This is selected since the number of canister is the lowest and also the costs are low. Furthermore, the schedule in

Figure 6d looks most feasible from the practical point of view due to the calm beginning. The reference point

${\mathit{f}}_{3}^{R}$ adjusts this solution such that the central and disposal tunnels are set to the minimum and the storage, ending, and operation time are deteriorated.

The solutions of the third iteration all have longer storage, ending, and operation time than the current solution

**f**_{2} as is seen in

Figure 6a. Additionally, the total costs are increased. Furthermore, the total length of the central and disposal tunnels are shorter in all the solutions, and no additional water pools are needed. In this iteration, the solutions have less variety. Indeed, the differences of the solutions r3q3 and r3q4 are hard to see from

Figure 6a. However, the solution r3q3 has a bit longer storage time, few extra canisters, longer disposal tunnel, and it is a slightly more expensive than the solution r3q4. However, the central tunnel in the solution r3q3 is shorter than in the solution r3q4. The schedules in

Figure 7b–d show that the biggest differences between the solutions can be seen in the second part of OL1-2. However, the solution r3q3 has a few extra canisters in the last period of LO1-2 compared to the solution r3q4. By considering all the schedules presented, the two-shift work is utilized only when OL3 is encapsulated.

The decision maker selects the solution r3q4 as the final solution

${\mathit{f}}^{\ast}={\mathit{f}}_{3}$. Both solutions r3q3 and r3q4 would be good alternatives, but the solution r3q4 is selected due to the smaller number of canisters and the shorter disposal tunnel. During the interactive process, we learned that the model suggests schedules with heavy starting which will decrease in later periods. The behavior of this kind is very hard to employ in practice, and, thus, we investigate what happens, if the reference point is the solution r3q4, or, in other words

${\mathit{f}}_{4}^{R}={\mathit{f}}_{3}$, and we add the following constraint:

This constraint guarantees that, at every period, the hiatus is not applied and the number of canisters to be encapsulated needs to be at least as many as in the previous period.

Here, we observe only one solution, namely r4q5. As we see from

Figure 8a, the additional constraint increases the number of the canisters needed together with the length of the disposal tunnel and the costs. On the other hand, the storage, ending, and operation times get shorter. From the schedule point of view in

Figure 8b, the encapsulation before the hiatus is very even except the last period when the two-shift work is needed. Compared to the solution r3q4, the solution r4q5 has a longer hiatus taking 25 years, and the disposal ends one period earlier.

All the results obtained during the interactive process suggest a hiatus between 15 to 25 years. This indicates that the model gains advantage from the hiatus since only a 5-year hiatus is forced. Unfortunately, the hiatus has its own risks. Therefore, we test how well we can achieve the objective values of the solution

**f**^{*} with the scenario without the hiatus. Thus, we set

${\mathit{f}}_{5}^{R}={\mathit{f}}_{4}^{R}$ and the result is shown in

Figure 9.

By omitting the hiatus, the objective values for the operation time and the central tunnel increase as well as the number of canisters, the disposal tunnel, and the costs but only 1.2%. However, the objective values for the ending and storage time decrease. The schedule of the solution r5q1 in

Figure 9b is similar to the solution r3q4 for the first nine periods. After that, the minimal amount of canisters are encapsulated until the last period when as many canisters as possible are encapsulated by using the two-shift work, and the final disposal ends one period sooner than in the solution r3q4. Again, the similar beginning in the schedules of the base scenario and the scenario without hiatus can be observed as in the single-objective case minimizing costs. However, in this case, the differences in the maximum canister power and the disposal tunnel spacing are less significant.

The CPU times of solving each scalarized problem (

44) with the augmentation term in the base scenario reported here varies from 645 s to 6008 s while the average CPU time is 3101 s and the median CPU time is 1898 s. The CPU time of the solution r4q5 is 7265 s and of the solution r5q1 3943 s making both alternative scenarios more time-consuming than the average problem of the base scenario. In general, it seems that the scalarized problems with the smaller value of the parameter

q (i.e., closer to the

${L}_{\infty}$ metric) are computationally harder than the scalarized problems with the higher value of

q (i.e., closer to the

${L}_{1}$ metric). Similar behavior is also observed in [

37].