1. Introduction

2. Literature Review

3. Research Methodology

3.1. Modeling Ideas

The modeling problem central to integrating Unmanned Aerial Vehicles (UAVs) with ground vehicles for medical supply delivery aims to address intrinsic deficiencies in healthcare delivery mechanisms.

Given a transport vehicle equipped with a set of isomorphic UAVs $\mathrm{U}=\{{\mathrm{U}}_1,{\mathrm{U}}_2,...{\mathrm{U}}_{\mathrm{k}}\}$, K is the number of UAVs. The task that needs to be executed is to distribute emergency supplies from the starting point warehouse (set as ${\mathrm{b}}_0$) to the set of customers with given material needs, represented as C = {1, 2... n}, and return to the destination warehouse (set as ${\mathrm{b}}_{\mathrm{n}+1}$) after completing the distribution task. The demand for each material demand customer is expressed in g, with vehicles traveling at an average speed of ${\mathrm{V}}^{\mathrm{T}}$ and UAVs flying at an average speed of ${\mathrm{V}}^{\mathrm{D}}$. In our problem setting, the customers of material demand points are divided into two main parts: the UAV delivery customer set and the vehicle delivery customer set. Establish a path optimization model for this type of dual subject delivery problem based on meeting corresponding constraint conditions, optimize the total completion time and customer satisfaction objectives, and seek the optimal integrated delivery path.

Figure 1. Delivery sketch map.

Figure 1. Delivery sketch map.

3.1.1. Objective Functions

Our article has two main objectives: one is to minimize delivery time and maximize customer satisfaction.

Objective 1: Minimizing the Delivery Time

The driving plan of a vehicle is actually an orderly arrangement of customers in the vehicle delivery customer set ${\mathrm{C}}^{\mathrm{T}}$, that is, the vehicle delivery route is marked as $\{{\mathrm{b}}_0,{\mathrm{b}}_1...{\mathrm{b}}_{{\mathrm{n}}^{\mathrm{r}}}\}$. To ensure that vehicles complete emergency supplies distribution and return to the warehouse in the shortest possible time, the optimal sequence for vehicle delivery to customers is determined. We introduce a variable ${\mathrm{x}}_{\mathrm{i}\mathrm{j}}$,where i, j ∈ ${\mathrm{C}}^{\mathrm{T}}$, indicating whether the vehicle is traveling from material demand point $\mathrm{i}$ to point$\mathrm{j}$. If so, then ${\mathrm{x}}_{\mathrm{i}\mathrm{j}}=1$, otherwise it is 0. The time taken for the vehicle to travel from point $\mathrm{i}$to point $\mathrm{j}$ is:

$${\mathrm{t}}_{\mathcal{i}\mathcal{j}}^{\mathrm{T}}={\displaystyle \frac{{\mathrm{L}}_{\mathcal{i}\mathcal{j}}^{\mathrm{T}}}{{\mathrm{V}}^{\mathrm{T}}}}$$

(1)

The time for vehicle delivery to customers can be expressed as:

$${\mathrm{t}}^{\mathrm{T}}=\sum _{(\mathcal{i},\mathcal{j}\in {\mathrm{C}}^{\mathrm{T}})}{\mathrm{t}}_{\mathcal{i}\mathcal{j}}^{\mathrm{T}}{\mathrm{x}}_{\mathcal{i}\mathcal{j}}$$

(2)

In this issue, if the customer’s demand for UAV delivery involves parallel delivery, the completion time of delivery is the longest time for the UAV to return to the truck among all UAV delivery paths. Due to the use of homogeneous UAVs, the payload of the UAV is set to${\mathrm{Q}}_{\mathrm{k}}^{\mathrm{D}}$ $(\mathrm{k}=\mathrm{1,2}...\mathrm{K})$. Considering the battery capacity of the UAV, the maximum distance for a single flight of the UAV is L, and the Euclidean distance between customer points for material needs is ${\mathrm{L}}_{\mathcal{i}\mathcal{\text{'}}\mathcal{j}}^{\mathrm{D}}$, and

$${\mathrm{L}}_{\mathcal{i}\mathcal{\text{'}}\mathcal{j}}^{\mathrm{D}}=\sqrt{{({\mathrm{A}}_{\mathcal{i}\mathcal{\text{'}}}-{\mathrm{A}}_{\mathcal{j}\mathcal{\text{'}}})}^2+{({\mathrm{B}}_{\mathcal{i}\mathcal{\text{'}}}-{\mathrm{B}}_{\mathcal{j}\mathcal{\text{'}}})}^2}(\mathcal{i},\mathcal{j}\in {\mathrm{C}}^{\mathrm{D}})$$

(3)

The distance from the UAV takeoff point to the material demand point is ${\mathrm{L}}_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}\mathrm{\text{'}}}({\mathrm{i}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}},{\mathrm{j}}^{\mathrm{\text{'}}}\in {\mathrm{C}}^{\mathrm{D}})$, and the nth flight route (i.e. the number of UAV flights) is defined using the path set ${\mathrm{P}}_{\mathrm{n}}$. The element ${\mathrm{P}}_{{\mathrm{n}}_{\mathrm{i}}}$ in the route represents the order of customer i being delivered by the UAV in route n. If the m-th customer of the UAV ${\mathrm{U}}_{\mathrm{k}}$ is set to ${\mathrm{k}}_{\mathrm{m}}$ (${\mathrm{k}}_{\mathrm{m}}$=0, which means that the m-th customer has not been delivered, and ${\mathrm{P}}_{{\mathrm{n}}_{{\mathrm{i}}^{\mathrm{*}}}}$ represents the launch and recovery points of the UAV, then the completion time of one operation flight for the UAV delivery group ${\mathrm{G}}_{{\mathrm{N}}_1}$ is:

$${\mathrm{t}}_{{\mathrm{G}}_{\mathrm{i}}}^{\mathrm{D}}=\mathrm{m}\mathrm{a}\mathrm{x}\left({\displaystyle \frac{{\mathrm{x}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}}\left(\sum _{\mathrm{i}\mathrm{\text{'}}=1,\mathrm{j}\mathrm{\text{'}}={\mathrm{k}}_{\mathrm{m}}}{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{\mathrm{i}}},{\mathrm{n}}_{\mathrm{j}}}^{\mathrm{D}}+{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{{\mathrm{i}}^{\mathrm{*}}}},{\mathrm{n}}_{\mathrm{i}}}^{\mathrm{D}}+{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{\mathrm{j}}},{\mathrm{n}}_{{\mathrm{i}}^{\mathrm{*}}}}^{\mathrm{D}}\right)}{{\mathrm{V}}^{\mathrm{D}}}}\right),\mathrm{n}\in \mathrm{1,2}....\mathrm{n}$$

(4)

The completion time for UAV delivery to discrete customers is:

$${\mathrm{t}}_{{\mathrm{j}}^{\mathrm{*}}}=2{\mathrm{f}}_{{\mathrm{j}}^{\mathrm{*}}}.{\displaystyle \frac{{\mathrm{L}}_{{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{*}}}}{{\mathrm{V}}^{\mathrm{D}}}}$$

(5)

So the total time to complete all grouping and discrete delivery customers is:

$${\mathrm{t}}^{\mathrm{D}}=\sum _{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{t}}_{{\mathrm{G}}_{\mathrm{i}}}^{\mathrm{D}}+\sum _{{\mathrm{j}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{D}}\cap {\mathrm{j}}^{\mathrm{*}}\notin \mathrm{G}}{\mathrm{t}}_{{\mathrm{j}}^{\mathrm{*}}}$$

(6)

The main objective of healthcare material distribution is to deliver emergency materials to customers in demand faster. The equivalent form of the optimization objective function is as follows:

$${\mathrm{F}}_1=\mathrm{m}\mathrm{i}\mathrm{n}({\mathrm{t}}^{\mathrm{T}}+{\mathrm{t}}^{\mathrm{D}})$$

(7)

Objective 2: Maximizing Customer Satisfaction

When a public event occurs, the requirements of the delivered customers, especially their satisfaction with the delivery time, also need to be taken seriously.

The time satisfaction is represented by a number between 0 and 1. 0 represents the time when the vehicle or UAV reaches the customer’s demand point as very dissatisfied, while 1 represents the time when the vehicle or UAV reaches the customer’s demand point as very satisfied. Assuming that the latest expected time point for vehicles or UAVs to arrive at the material demand customer ${\mathrm{i}}_0$ is ${\mathrm{t}}_{{\mathrm{i}}_0}^h$, and the maximum acceptance time range for the material demand customer is $[0,{\mathrm{T}}_{{\mathrm{i}}_0}^h]$, the time satisfaction is shown in Figure 2.

The time satisfaction equation can be expressed as follows:

$${\mathrm{S}}_1=\left\{\begin{array}{c}1\\ {\displaystyle \frac{{\mathrm{t}}_{{\mathrm{i}}_0}^h-{\mathrm{t}}_{{\mathrm{i}}_0}}{{\mathrm{t}}_{{\mathrm{i}}_0}^h-{\mathrm{T}}_{{\mathrm{i}}_0}^h}}\\ 0\end{array},{{\mathrm{t}}_{{\mathrm{i}}_0}^h\le {\mathrm{t}}_{{\mathrm{i}}_0}\le \mathrm{T}}_{{\mathrm{i}}_0}^h\right.$$

(8)

From the perspective of distribution material quantity, satisfaction is positively correlated with the satisfaction rate of distribution material quantity. Using 0 to indicate that the number of materials delivered by UAVs or vehicles to customers in need completely does not meet the demand, and the demand customers are very dissatisfied. Using 1 to indicate that the number of materials delivered to customers in need completely meets the demand, and the demand customers are very satisfied. The satisfaction chart regarding the satisfaction rate of material delivery volume is as follows Figure 3:

The relationship expression is as follows:

$${\mathrm{S}}_2=\left\{\begin{array}{c}0\\ {\displaystyle \frac{{\mathrm{u}}_{{\mathrm{i}}_0}-{\mathrm{u}}_h}{{\mathrm{u}}_{\mathrm{l}}-{\mathrm{u}}_{{\mathrm{i}}_0}}}\\ 1\end{array},{\mathrm{u}}_{\mathrm{l}}\le {\mathrm{u}}_{{\mathrm{i}}_0}\le {\mathrm{u}}_h\right.$$

(9)

The optimal objective function for customer satisfaction obtained by combining the time satisfaction sector and the quantity satisfaction is:

$${\mathrm{F}}_2=\mathrm{m}\mathrm{a}\mathrm{x}(\mathsf{\alpha }{\mathrm{S}}_1+\mathsf{\beta }{\mathrm{S}}_2)$$

(10)

$${\mathrm{s}.\mathrm{t} \mathrm{g}}_{\mathrm{i}}^{\mathrm{D}}\mathrm{o}\mathrm{r}{\mathrm{g}}_{\mathrm{i}}^{\mathrm{T}}\le {\mathrm{g}}_{\mathrm{i}}$$

(11)

$${{\mathrm{t}}_{{\mathrm{i}}_0}\le \mathrm{T}}_{{\mathrm{i}}_0}^h$$

(12)

The demand and time are equally important, here α Related to β Taking the mean of 0.5, the constraint (27) indicates that the number of materials delivered by vehicles or UAVs to the material demand point cannot exceed the customer’s demand, and the constraint (28) indicates that the time for delivery to the material demand point cannot exceed the maximum tolerance time.

3.1.2. Constraints

The outlined constraints are in place to ensure the material supply process is efficient and effective in a integrated delivery system between vehicles and UAVs. These constraints dictate the behavior of the vehicles and UAVs, optimizing the delivery process and enhancing overall system efficiency.

$$\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}}{\mathrm{x}}_{0\mathrm{i}}=\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}}{\mathrm{x}}_{\mathrm{i}0}=1$$

(13)

$$\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}\cup \left\{0\right\},\mathrm{j}\in {\mathrm{C}}^{\mathrm{T}}\cup \{\mathrm{n}+1\}}{\mathrm{x}}_{\mathrm{i}\mathrm{j}}=\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}\cup \left\{0\right\},\mathrm{j}\in {\mathrm{C}}^{\mathrm{T}}\cup \{\mathrm{n}+1\}}{\mathrm{x}}_{\mathrm{i}\mathrm{j}}$$

(14)

$$\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}\cup \left\{0\right\},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{x}}_{\mathrm{i}\mathrm{j}\mathrm{\text{'}}}=0$$

(15)

$$\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}}{\mathrm{g}}_{\mathrm{i}}^{\mathrm{T}}+\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{g}}_{\mathrm{i}}^{\mathrm{D}}\le {\mathrm{Q}}^{\mathrm{T}}$$

(16)

$$\sum _{\mathrm{i},\mathrm{j}\in {\mathrm{C}}^{\mathrm{T}}}{\mathrm{x}}_{\mathrm{i}\mathrm{j}}+\sum _{\mathrm{i}\mathrm{\text{'}},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{x}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}}=1$$

(17)

$$\sum _{\mathrm{i}\mathrm{\text{'}},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{x}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}}=\sum _{\mathrm{i}\mathrm{\text{'}},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{x}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}}$$

(18)

$$\sum _{\mathrm{i}\mathrm{\text{'}}=1,\mathrm{j}\mathrm{\text{'}}={\mathrm{n}}_{\mathrm{m}},\mathrm{n}\in {\mathrm{P}}_{\mathrm{n}}^{\mathrm{D}}}{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{\mathrm{i}}},{\mathrm{n}}_{\mathrm{j}}}^{\mathrm{D}}+{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{{\mathrm{i}}^{\mathrm{*}}}},{\mathrm{n}}_{\mathrm{i}}}^{\mathrm{D}}+{\mathrm{L}}_{{\mathrm{P}}_{{\mathrm{n}}_{\mathrm{j}}},{\mathrm{n}}_{{\mathrm{i}}^{\mathrm{*}}}}^{\mathrm{D}}<\mathrm{L}$$

(19)

$$h_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{p}}=1,{\mathrm{i}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}}\cap {\mathrm{i}}^{\mathrm{*}}\ne 0,\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}$$

(20)

$$h_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{p}}+{\mathrm{x}}_{\mathrm{j}\mathrm{\text{'}}\mathrm{k}\mathrm{\text{'}}}\ge 1,{\mathrm{i}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}},\mathrm{k}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}$$

(21)

$$\sum _{\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{h}_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{p}}=1,\forall {\mathrm{i}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}}$$

(22)

$$\sum _{\mathrm{i}\in {\mathrm{C}}^{\mathrm{T}}}{\mathrm{x}}_{\mathrm{i}\mathrm{j}}=h_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{p}},{\mathrm{j}=\mathrm{j}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}}$$

(23)

$$\sum _{\mathrm{i}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}}{\mathrm{g}}_{\mathrm{i}\mathrm{\text{'}}}^{\mathrm{D}}\le {\mathrm{Q}}^{\mathrm{D}}$$

(24)

$${\mathsf{\alpha }}_{\mathrm{i}}-{\mathsf{\alpha }}_{\mathrm{j}}+1\le (\mathrm{n}+2)+2(1-{\mathrm{x}}_{\mathrm{i}\mathrm{j}})\forall \mathcal{i}\in \{0....\mathrm{n}\},\forall \mathrm{j}\in \{1....\mathrm{n}+1\},1\le {\mathsf{\alpha }}_{\mathrm{i}}\le \mathrm{n}+2$$

(25)

$${\mathsf{\alpha }}_{\mathrm{i}}-{\mathsf{\alpha }}_{\mathrm{j}}+1\le \mathrm{n}(1-{\mathrm{x}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}})\forall \mathrm{i}\mathrm{\text{'}},\mathrm{j}\mathrm{\text{'}}\in {\mathrm{C}}^{\mathrm{D}}1\le {\mathsf{\alpha }}_{\mathrm{i}}\le \mathrm{n}$$

(26)

$${\mathrm{w}}_{\mathrm{i}\mathrm{\text{'}}{\mathrm{i}}^{\mathrm{*}}\mathrm{j}\mathrm{\text{'}}}^{\mathrm{u}}=0$$

(27)

$${\mathrm{t}}_{\mathrm{j}}^{\mathrm{T}}={\mathrm{t}}_{\mathrm{i}}^{\mathrm{T}}+{\mathrm{t}}_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{T}}+{\mathrm{f}}_{{\mathrm{j}}^{\mathrm{*}}}{\mathrm{t}}_{{\mathrm{j}}^{\mathrm{*}}}^{\mathrm{D}}+h_{{\mathrm{i}}^{\mathrm{*}}\mathrm{j}}^{\mathrm{p}},{\mathrm{t}}_{{\mathrm{G}}_{\mathrm{i}}}^{\mathrm{D}},{\mathrm{i}=\mathrm{i}}^{\mathrm{*}}\in {\mathrm{C}}^{\mathrm{T}}$$

(28)

$${\mathrm{t}}_{\mathrm{j}\mathrm{\text{'}}}^{\mathrm{D}}={\mathrm{t}}_{\mathrm{i}\mathrm{\text{'}}}^{\mathrm{D}}+{\mathrm{t}}_{\mathrm{i}\mathrm{\text{'}}\mathrm{j}\mathrm{\text{'}}}^{\mathrm{D}}$$

(29)

4. Results and Discussions

5. Improved Artificial Bee Colony

5.1. Results Analysis

This section uses the same example and uses the improved bee colony algorithm to optimize the paths for vehicle and UAV delivery to customers, as shown in Figures 17, 19, 22, and 24. The indicators for solving vehicle paths using the improved bee colony algorithm and the classical artificial bee colony algorithm are compared, as shown in Table 4. At the same time, provide a comparison chart of the iterative optimization curves of the algorithm, as shown in Figures 18, 19, 21, 23, and 24.

Figure 17. Improved Bee Colony Algorithm for Path Optimization

Figure 17. Improved Bee Colony Algorithm for Path Optimization

Figure 18. Comparison of iterative optimization curves for solving vehicle paths

Figure 18. Comparison of iterative optimization curves for solving vehicle paths

Figure 19. Solving path optimization for the class1 of UAV delivery customers using the improved the bee colony algorithm

Figure 19. Solving path optimization for the class1 of UAV delivery customers using the improved the bee colony algorithm

Figure 20. Comparison chart of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the class 1 of UAV customer

Figure 20. Comparison chart of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the class 1 of UAV customer

Figure 21. Improved bee colony algorithm to solve the path optimization for the class 2 of unmanned aerial vehicle delivery customers

Figure 21. Improved bee colony algorithm to solve the path optimization for the class 2 of unmanned aerial vehicle delivery customers

Figure 22. Comparison of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the class 2 of UAV customers

Figure 22. Comparison of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the class 2 of UAV customers

Figure 23. Improved bee colony algorithm to solve the path optimization for the class 3 of unmanned aerial vehicle delivery customers

Figure 23. Improved bee colony algorithm to solve the path optimization for the class 3 of unmanned aerial vehicle delivery customers

Figure 24. Comparison chart of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the UAV customers of class 3

Figure 24. Comparison chart of iterative optimization curves between improved bee colony algorithm and artificial bee colony algorithm for solving the UAV customers of class 3

Based on the above figures, it is evident that as the number of iterations increases, the flight time decreases gradually. This result proves that the enhanced algorithm proposed in this article effectively optimizes the path and reduces the time for vehicle-UAV integrated delivery. Moreover, our model is cost-effective, making it suitable for real-time applications.

The results of improving the bee colony algorithm compared to artificial bee colony algorithm for solving vehicles and UAVs are shown in Table 5:

By comparing and analyzing the results in Table 5.2, the improved bee colony algorithm using simulated annealing strategy reduces the number of optimization iterations by 47% compared to the artificial bee colony algorithm, and also reduces the running time by 45%. This indicates that the improved bee colony algorithm can obtain the optimal solution with fewer iterations and shorter running time. The total path time obtained by solving also decreased by 14.7%, indicating that the improved bee colony algorithm can effectively jump out of the local optimal solution in the mid-term, reflecting the superiority of the improved bee colony algorithm.

6. Conclusions and Discussions

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