## 3 Solution Algorithms
### Determining Rescue Route and Rescue Sweptly
#### 3.2.2 Initialization  
In the process of initial population generation, the genes of sub-chromosomes EM need to meet constraints (10) and (11). In order to improve the quality of the initial population, rescue centers are sorted from the perspective of two objective functions.  
Let \(o_{ir}^{1}\) denote the sequence number of the rescue center \(r\) sorted in descending order according to the minimum expected travel time of the rescue route for rescue resource \(i\). Correspondingly, let \(o_{ir}^{2}\) represent the sequence number of the rescue center \(r\) sorted in ascending order according to the reliability of the rescue route for rescue resource \(i\). Generally, the higher the amount of resources provided by the top rescue centers, the  
Figure 2: A simple example of a chromosome.  
better the quality of chromosomes for this object. Let us set the comprehensive ranking value of rescue centers for rescue resource \(i\) as \(o_{ir}\) and the weight of random sorting as \(\omega_{i}\), respectively. Then, comprehensive ranking can be defined as follows.  
\[o_{ir}=\omega_{i}\cdot o_{ir}^{1}+(1-\omega_{i})\cdot o_{ir}^{2} \tag{20}\]  
For sub-chromosome k, the first stage of this algorithm has obtained the set of feasible routes between each rescue center and the accident location, and at this stage, we only need to select an appropriate route.  
The process of generating a chromosome based on random sorting is as follows:  
* Step 1: Parameter setting: set \(R\), \(I\), \(k_{rs}\), \(o_{ir}^{1}\), \(o_{ir}^{2}\), \(ED_{i}\) and \(EN_{ir}\). Let \(g=1\) and \(\omega_{i}=0\).
* Step 2: For \(g=1\) to _popsize_ (the number of chromosomes), repeat Step 3 to Step 5.
* Step 3: For \(i=1\) to \(I\), calculate the comprehensive ranking value \(o_{ir}\) and sort the rescue centers according to \(o_{ir}\) in descending order.
* Step 4: For \(r=1\) to \(R\), arrange the supply of rescue resource \(i\) in the order of \(o_{ir}\): let the supply of the first rescue center be \(EM_{ir}=\min\{EN_{ir},ED_{i}\}\), \(ED_{i}=ED_{i}-\min\{EN_{ir},ED_{i}\}\) and calculate the supply of the subsequent rescue center \(EM_{ir}=\min\{EN_{ir},ED_{i}\}\). If \(ED_{i}=0\), set \(\omega_{i}=\omega_{i}+1/(popsize-1)\) and proceed to Step 3.
* Step 5: For \(r=1\) to \(R\), generate a random integer number \(k_{r}\in[1,k_{rs}]\) as a gene of sub-chromosome k.  
For example, if the demand for the rescue resources \(ED_{i}\) (\(I=3\)) is (50, 40, 40), the rescue capacity \(EN_{ir}\) and the comprehensive ranking value \(o_{ir}\) of the rescue centers (\(r=4\)) are shown in Figure 3, and the sub-chromosome of the supply quantity for rescue resources will be generated according to the above chromosome generation method. For resource 1, according to the comprehensive ranking value \(o_{1r}\), the first place is rescue center 1, and its supply quantity can be determined as min [50, 20] = 20. The second place is rescue center 2, and it can be determined that the supply quantity is min [50, 51, 52, 20, 10] = 10. Similarly, the third place is rescue center 4, and its supply quantity is min [50, 51, 52, 20, 10, 30] = 20. The demand of resource 1 has been met, and we can obtain a feasible sub-chromosome of the supply quantity by repeating this operation, as shown in Figure 3.  
Obviously, the above chromosome generation method satisfies constraints (10) and (11).  
Figure 3: Illustration of sub-chromosome EM generation.