Optimization of Operations in Bus Company Service Workshops Using Queueing Theory

Abstract

Public transport companies are aware that the success of their operations largely depends on the proper sizing and optimization of their processes. Among the key activities are the maintenance and repair of the vehicle fleet. This paper presents the application of mathematical optimization methods from the field of operations research to improve the efficiency of service workshops for bus maintenance and repair. Based on an analysis of collected data using queueing theory, the authors assessed the current system performance and found that the queueing system still has spare capacity and could be downsized, which aligns with the company’s management goals. Specifically, the company plans to reduce the number of bus repair service stations (servers in a queueing system). The main question is whether the system will continue to function effectively after this reduction. Three specific downsizing solutions were proposed and evaluated using queueing theory methods: extending the daily operating hours of the workshops, reducing the number of arriving buses, and increasing the productivity of a service station (server). The results show that, under high system load, only those solutions that increase the productivity of individual service stations (servers) in the queueing system provide optimal outcomes. Other solutions merely result in longer queues and associated losses due to buses waiting for service, preventing them from performing their intended function and causing financial loss to the company.

1. Introduction

Various companies operating in different sectors of the economy are aware that the success of their business depends on the proper dimensioning of their activities. The correct dimensioning of operations (production) is influenced by several factors, such as: revenue, profit, costs, spatial limitations, quality and quantity of available workforce, monetary policies of the state, etc.

The role of operations research in this regard is primarily to assist in optimal decision-making for the establishment or reorganization of business operations within a company. In today’s world, the focus is primarily on finding the optimal balance between a company’s costs and profits. This problem is particularly relevant for large state or municipal companies, whose production capacities are still dimensioned for maximum revenue, rather than for maximum profit.

A similar situation applies to the bus company addressed in this article, which has overdimensioned service workshops. The bus company is owned by the municipality, and the company’s service workshops are located in a location that is highly attractive to other investors. For this reason, the company has decided to sell part of the workshops used for bus repairs. Since this decision is primarily of a business nature, the company’s management and transportation experts began to ask themselves what impact this business decision would have, as it would significantly affect the bus company’s operational processes. The main question arising from this is whether the capacity of the remaining service workshops will be sufficient to meet current needs. As a result, they commissioned a study, which was conducted by the authors of this research, and the results are presented here.

The objective of the research is to provide appropriate solutions for the operation of the service workshops in the newly created situation, based on the results of the mathematical method.

Many authors have already dealt with the mathematical modeling of bus maintenance systems, but they did not use queueing theory methods. Savsar M. developed a simulation model based on several processes occurring in a bus maintenance workshop. These processes include the arrival and decision process, assigning repair distributions and resource quantities, the repair process, counting the buses repaired, and the time in the system [1]. Hangany A. and Shafahi Y. addressed the problem of planning bus maintenance activities. They used mathematical programming and took into account the daily operational schedule for all buses, along with the available resources for maintenance. They then attempted to design daily schedules for inspections and maintenance of buses with minimal interruptions in bus operations [2]. Maze T.H. and Allen R.C., through a survey completed by 92 bus maintenance managers, developed a system for measuring the performance of service workshops [3]. Landowski B. developed a model for operation and maintenance processes in public transport, which he analyzed as a stochastic process and described as a homogeneous Markov process [4].

In the review of previous research and the scientific literature, no studies were found where queueing theory was used to optimize the operations of service workshops for bus repairs. This is likely due to the difficulty in obtaining data from companies, as the vehicle servicing and repair times, as well as the waiting times of vehicles in line before entering the workshops, are considered business secrets. These factors directly influence the business success of companies involved in vehicle maintenance. As a result, companies are reluctant to release this data for public use. The same applies to this study, as the authors are not allowed to disclose the name of the company or the city where the research was conducted.

The closest study related to the maintenance of buses using queueing theory was conducted by Young-neng Xu and others, who focused on the optimization of maintenance for urban rail transit vehicles [5]. Further research was conducted to optimize workshops where critical components were sent for repair and servicing [6]. Other studies on maintenance optimization include research by Han, Qing Tian, et al., who studied maintenance optimization for an aircraft carrier, where maintenance tasks need to be highly optimized due to significant spatial constraints on the ship [7]. Similarly, Bebic, D., et al. conducted a similar study on ship maintenance [8]. In a broader context, outside of vehicle maintenance, a study using queueing theory was conducted to assist with planning, managing, and controlling workforce allocation at different service levels in distributed multi-phase maintenance processes for software [9].

Queueing theory is also widely used in the field of transportation. In road traffic, delays and queue lengths on highways, particularly at toll stations, have been evaluated [10,11,12,13,14]. In the case of bus transportation, queueing theory has been applied to bus stops [15,16,17,18]. In recent years, numerous studies have focused on queue management at electric vehicle charging stations for buses [19,20,21,22] and cars [23,24,25,26,27]. These mathematical optimization methods are also used for traffic planning in cities [28], including the optimization of green traffic waves [29]. Queueing theory is also applied in the optimization of railway traffic [30], airports [31], such as for baggage screening queues [32], taxi traffic [33,34], and even cable car traffic [35]. Furthermore, queueing theory is applied in optimizing other areas, such as warehousing [36], manufacturing [37], and other activities where queues occur. These applications show that queueing theory can be a valuable tool in improving the efficiency of maintenance operations in various transport sectors, including buses.

For the purposes of this research, a method from the field of operations research will be applied. First, the existing state of operations in the bus company’s service workshops will be analyzed, where data was obtained through the review of repair and maintenance records of the buses and through an interview with the workshop manager, who provided the necessary data for the calculations.

Next, queueing theory will be used to analyze the future state, where the operations in the workshops will be optimized, and various solutions will be proposed. For easier calculation using queueing theory, a tabular calculation was made in Microsoft Excel, where the necessary equations for the calculations were inserted.

This paper seeks to extend that body of knowledge by applying a classic multi-server queueing model to a bus company’s service workshops. Through this approach, we aim to provide insight into how such models can be used for practical business decisions regarding workshop downsizing while ensuring optimal operational efficiency.

Our focus is on a specific industrial application, namely the analysis of bus maintenance operations in service workshops. We used standard M/M/k models to assess the impact of reducing service station capacity. Nonetheless, we acknowledged the critical limitations of these models, such as the assumptions of Poisson arrivals and station homogeneity [15,16,20].

Future work will need to include real-time, timestamped data and advanced simulation methods to enable more accurate predictions and system optimization [24,26].

2. Materials and Methods

Based on the works of various authors who have addressed and presented queueing theory in their books [38,39,40,41,42], the authors of this article have studied queueing theory.

Queueing theory examines the quantitative properties of the organization of mass phenomena (processes). Massiveness and stochasticity are the fundamental properties studied by queueing theory. As such, this method is based on probability calculations and statistical analysis.

The foundations of queueing theory were established by the Scandinavian scientist Erlang, who studied the problems of telephone networks. This scientific discipline is now widely used in various fields, such as healthcare, manufacturing processes, banking, transportation, military activities, etc.

The basic structure of a queueing system consists of:

-

Arrival of users,

-

Queue of waiting users,

-

Service,

-

Departure of users.

Queueing theory determines the relationship between the properties of arrivals, waiting for service, service, customer behavior in the system, and user departures. The efficiency of a queueing system, depending on the characteristics of the problem and the goal of the problem, is expressed through various parameters, such as: the average number of users arriving per unit of time; the average number of users that can be served per unit of time; the average number of users departing from the system; the probability that users will be served immediately upon arrival; the average waiting time for service; the average number of users waiting for service; the average time users spend in the system; the ratio of the average number of served users to the average number of arriving users (relative throughput of the system); the average number of busy servers; the average utilization rate of working time.

The basic input data from which the analysis is derived are:

λ—arrival rate (intensity) of arrivals,

μ—service rate (intensity) per server,

k—number of servers.

The arrival rate and the service rate are the fundamental properties of a queueing system. They determine the organization of the system. It is sufficient for only one parameter to be stochastic.

Depending on the size and characteristics of the queueing system, we distinguish between:

-

Single-server queueing system,

-

Multi-server queueing system,

-

Closed queueing system,

-

Multi-server closed queueing system.

In a closed queueing system, the arrival rate is dependent on the system’s state. The system consists of a finite number of units, which can cease to operate at any given moment. For a unit that has stopped functioning to be reactivated, service is required. A characteristic of this system is the limited number of arrival producers.

In most cases, the arrival rate follows the Poisson distribution law of arrival probabilities.

$$\mathrm{P}\mathrm{n}(\mathrm{t})={\displaystyle \frac{{(\lambda \cdot \mathrm{t})}^{\mathrm{n}}}{\mathrm{n}!}}\cdot {\mathrm{e}}^{-\lambda \mathrm{t}}$$

(1)

Pn(t)—the probability of n arrivals in a time interval t,

λ—the average number of arrivals expected in that time interval (arrival rate or intensity),

For the event flow to be treated as Poisson, the following conditions must be met:

• The probability Pn(t) depends on the length of the time interval and the number of Arrivals
• The probability Pn(t) is independent of the number of arrivals before the time interval t,
• In a sufficiently small time interval, two or more arrivals cannot occur simultaneously.

Therefore, it holds that:

$$\underset{\mathrm{t}\to 0}{\lim }{\displaystyle \frac{{\displaystyle \sum _{\mathrm{i}=2}^{\mathrm{n}}\mathrm{P}\mathrm{i}(\mathrm{t})}}{\mathrm{t}}}=0$$

(2)

In the case where the time interval is one time unit (t = 1), the Poisson distribution takes the following form:

$$\mathrm{P}\mathrm{n}={\displaystyle \frac{{\lambda }^{\mathrm{n}}}{\mathrm{n}!}}\cdot {\mathrm{e}}^{-\lambda }$$

(3)

From this, it follows that the probability of no arrivals occurring in the time interval t₀ is:

$$\mathrm{P}\mathrm{o}={\mathrm{e}}^{-\lambda {\mathrm{t}}_0}$$

(4)

This probability is equal to the probability that the interval between two consecutive arrivals will be greater than or equal to t₀, so the distribution function of the time interval between two consecutive arrivals can be written as:

$$\begin{gathered} \mathrm{F}(\mathrm{t})=0, \mathrm{t}<0\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em} \\ \mathrm{F}(\mathrm{t})=1-{\mathrm{e}}^{-\lambda \mathrm{t}}, \mathrm{t}\ge 0 \end{gathered}$$

(5)

or distribution law:

$$\begin{gathered} \mathrm{f}(\mathrm{t})=0, \mathrm{t}<0\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em} \\ \mathrm{f}(\mathrm{t})=1-{\mathrm{e}}^{-\lambda \mathrm{t}}, \mathrm{t}\ge 0 \end{gathered}$$

(6)

The mean value of the time interval between two arrivals is:

$$\alpha ={\displaystyle \underset{0}{\overset{\infty }{\int }}\mathrm{t}\lambda {\mathrm{e}}^{-\lambda \mathrm{t}}}\mathrm{d}\mathrm{t}={\displaystyle \frac{1}{\lambda }}$$

(7)

and the variance:

$${\sigma }_{\mathrm{t}}^2={\displaystyle \underset{0}{\overset{\infty }{\int }}(\mathrm{t}-\frac{1}{\lambda }}{)}^2\lambda {\mathrm{e}}^{-\lambda \mathrm{t}}\mathrm{d}\mathrm{t}={\displaystyle \frac{1}{{\lambda }^2}}$$

(8)

In most queuing systems, the goal is to reduce the service time, which is most accurately described by the exponential distribution with the following probability distribution function:

$$\mathrm{F}(\mathrm{t})=1-{\mathrm{e}}^{-\mu \mathrm{t}}$$

(9)

or distribution law:

$$\mathrm{f}(\mathrm{t})=\mu {\mathrm{e}}^{-\mu \mathrm{t}}$$

(10)

In multi-server queuing systems, multiple customers can be served simultaneously. Figure 1 illustrates such a system. If the system is capable of serving up to k customers at the same time, it is referred to as a k-server queuing system.

An important parameter in queueing systems is the utilization factor. In multi-server queueing systems, we distinguish between:

• Utilization factor per server:

$$\rho ={\displaystyle \frac{\lambda }{\mu }}$$

(11)

• Utilization factor of the system:

$${\rho }^{*}={\displaystyle \frac{\lambda }{\mathrm{k}\cdot \mu }}$$

(12)

If ρ* < 1, the system is able to serve the customers, although some congestion may still occur. However, if ρ* > 1, the queue grows without bound and congestion increases indefinitely.

Other important performance indicators of a multi-server queuing system:

• The probability that there are no users in the system:

$${\mathrm{p}}_0={({\displaystyle \sum _{\mathrm{n}=0}^{\mathrm{k}}\frac{{\rho }^{\mathrm{n}}}{\mathrm{n}!}}+{\displaystyle \frac{{\rho }^{\mathrm{k}}}{\mathrm{k}!}}\cdot {\displaystyle \frac{{\rho }^{*}}{1-{\rho }^{*}}})}^{-1}$$

(13)

• Probability that there are n users in the system:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{n}}={\displaystyle \frac{\rho }{\mathrm{n}}}\cdot {\mathrm{p}}_{\mathrm{n}-1}, \mathrm{n}<\mathrm{k} \\ \hspace{1em}{\mathrm{p}}_{\mathrm{n}}={\rho }^{*}\cdot {\mathrm{p}}_{\mathrm{n}-1}, \mathrm{n}\ge \mathrm{k} \end{gathered}$$

(14)

• Average number of users being served:

$$\mathrm{Q}={\displaystyle \frac{{\rho }^{\mathrm{k}}}{\mathrm{k}!}}\cdot {\displaystyle \frac{{\rho }^{*}}{{(1-{\rho }^{*})}^2}}\cdot {\mathrm{P}}_0$$

(15)

• Average number of users waiting for service:

$$\mathrm{S}={\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{k}-1}\mathrm{n}{\mathrm{p}}_{\mathrm{n}}}+\mathrm{k}\cdot (1-{\displaystyle \sum _{\mathrm{n}=0}^{\mathrm{k}-1}{\mathrm{p}}_{\mathrm{n}}})$$

(16)

• Average number of users in the system (waiting + being served):

$$\mathrm{T}=\mathrm{Q}+\mathrm{S}$$

(17)

• Average waiting time of a user before service:

$${\mathrm{W}}^{*}={\displaystyle \frac{\mathrm{Q}}{\lambda }}$$

(18)

• Average time a user spends in the system (waiting + being served):

$$\mathrm{W}={\mathrm{W}}^{*}+{\displaystyle \frac{1}{\mu }}$$

(19)

3. Results—Solving a Real-World Case of Queueing Theory in the Bus Company’s Maintenance Workshops

Since the authors of this article promised the management of the company that the name of the company and the city, the workshop where the research was conducted, will not be disclosed in the publication of the research, the data provided below pertains to the maintenance workshop of a bus company with around 220 buses. The data was obtained from the head of the maintenance workshop.

3.1. Problem Description

The bus company’s service workshops have 60 workstations in larger facilities. These workstations are mainly service stations (servers) and lifts for vehicle repairs. On average, there are two mechanics per service station. The workshops are open for an average of 12 h per working day. In these service workshops, 220 domestic buses are serviced, including 57 urban buses and 163 suburban and intercity buses. All buses belong to the company that owns the workshops (domestic buses).

For maintenance, servicing, and repairs, an urban bus requires 1350 working hours per year, while a suburban or intercity bus requires 950 working hours. The average number of bus visits per day is 65. The workshops were originally designed to maintain, service, and repair 500 buses, so their current capacity is too large. Currently, these workshops carry out 60,000 working hours for external clients, but this is insufficient. The profit loss for the company if a bus is out of service for one day is approximately 580 EUR.

Maintenance and repairs are currently carried out at 60 service stations distributed across three buildings. The first building has 25 service stations, the second has 14, and the third has 21 service stations. Each service station has its own entrance and exit to the building, allowing buses to enter and leave independently of one another. Between these buildings, there is a large parking area where buses can wait for service (Figure 2).

Since the company plans to sell Building 3, which contains 21 service stations, it will reduce the number of service stations from 60 to 39.

3.2. Analysis of the Current Situation

From Figure 2, we can see that when buses enter the system and wait in the parking lot, they can choose any of the 60 service stations, as each has its own entrance to the building. Buses can leave the system independently of one another. This ensures the highest efficiency of the maintenance workshops.

In this case, it is a multi-server queuing system, whose model [38] is shown in Figure 3. Buses (users) enter the system at a single point, and this queuing system has one common queue from which all 60 servers (service stations) are fed. Then, buses exit the system through a single exit.

Since we did not obtain data on the exact times when the buses arrived at the service, we used a Poisson distribution to model the bus arrivals in our study.

The current situation will be analyzed using a multi-server queuing system, which has 60 service stations (servers) as shown in Figure 3. Data was obtained from the head of the technical workshops.

• Number of service stations (servers): k = 60
• Number of workers per service station: d = 2
• Number of company buses: m = 220
• Number of city buses: m₁ = 57
• Number of suburban or intercity buses: m₂ = 163
• Number of working hours required for the maintenance, servicing, and repair of a city bus: h_d1 = 1350 h/year
• Number of working hours required for the maintenance, servicing, and repair of a suburban or intercity bus: h_d2 = 950 h/year
• Average number of visits for company buses: λ_d = 65 visits/day
• Workshop working hours: t_d = 12 h/day
• Number of working hours performed for external clients: h_dz = 60,000 h/year
• Number of working days per year: l = 252 days
• Profit loss if a bus is out of service: j = 580 EUR/day

Calculation of input parameters for the multi-server queuing system:

• Average number of working hours per bus:

$${\mathrm{h}}_{\mathrm{d}}={\displaystyle \frac{{\mathrm{h}}_{\mathrm{d}1}\cdot {\mathrm{m}}_1+{\mathrm{h}}_{\mathrm{d}2}\cdot {\mathrm{m}}_2}{\mathrm{m}}}={\displaystyle \frac{57\cdot 1350+163\cdot 950}{220}}=1053.6 \mathrm{h}/\mathrm{year}$$

(20)

• Average number of hours a bus spends on maintenance, servicing, and repair in the workshops:

$$\mathrm{h}={\displaystyle \frac{{\mathrm{h}}_{\mathrm{d}}}{\mathrm{d}}}={\displaystyle \frac{1053.6}{2}}=525.8 \mathrm{h}/\mathrm{year}$$

(21)

• Number of workshop hours occupied by 220 in-house buses:

$${\mathrm{h}}_{220}=\mathrm{h}\cdot \mathrm{d}=526.8\cdot 220=115896 \mathrm{h}/\mathrm{year}$$

(22)

• Service time per one bus visit:

$${\mathrm{t}}_{\mathrm{a}}={\displaystyle \frac{{\mathrm{h}}_{220}}{{\lambda }_{\mathrm{d}}\cdot \mathrm{l}}}={\displaystyle \frac{115896}{65\cdot 252}}=7.075 \mathrm{h}$$

(23)

• Number of working hours completed in the workshops in one year:

$${\mathrm{h}}_{\mathrm{d}\mathrm{l}}={\mathrm{m}}_1\cdot {\mathrm{h}}_{\mathrm{d}1}+{\mathrm{m}}_2\cdot {\mathrm{h}}_{\mathrm{d}2}+{\mathrm{h}}_{\mathrm{d}\mathrm{z}}=57\cdot 1350+163\cdot 950+60000=291800 \mathrm{h}/\mathrm{year}$$

(24)

• Arrival rate (for internal and external buses):

$$\lambda ={\displaystyle \frac{{\mathrm{h}}_{\mathrm{d}\mathrm{l}}}{{\mathrm{t}}_{\mathrm{a}}\cdot \mathrm{l}\cdot \mathrm{d}}}={\displaystyle \frac{291800}{7.075\cdot 252\cdot 2}}=81.83 \mathrm{arrivals} \mathrm{per} \mathrm{day}$$

(25)

• Service rate (service intensity) per service station (server):

$$\mu ={\displaystyle \frac{{\mathrm{t}}_{\mathrm{d}}}{{\mathrm{t}}_{\mathrm{a}}}}={\displaystyle \frac{12}{7.075}}=1.7 \mathrm{visits}/\mathrm{day}$$

(26)

The basic parameters (λ, μ, k) were entered into a computer subprogram for calculating a multi-server queuing system, developed in Microsoft Excel.

Results:

• Service factor per service station: ρ = 48.16
• System service factor: ρ* = 0.802
• Average number of buses waiting for service: Q = 0.268
• The average number of buses currently being serviced: S = 48.135
• Average number of buses in the system: T = 48.403
• Average waiting time for service: W* = 2 min, 22 s
• Average time a customer (bus) spends in the system (waiting + service): W = 7 h, 5 min, 48 s
• Utilization of working time:

$$\eta ={\displaystyle \frac{\mathrm{S}}{\mathrm{k}}}={\displaystyle \frac{48.135}{60}}=0.802=80.2\%$$

(27)

• If one bus operates on average 16 h a day, the loss of profit for one bus per hour is: 580/16 = 36.25 EUR. Based on this, the loss of profit can be calculated if there are on average S = 48.1353 buses in the service workshops with an average W = 7.058 h.

$$\mathrm{J}={\displaystyle \frac{\mathrm{j}}{16}}\cdot \mathrm{S}\cdot \mathrm{W}={\displaystyle \frac{580}{16}}\cdot 48.1535\cdot 7.058=\text{12,315} \mathrm{EUR}$$

(28)

3.3. Analysis of the Future State with Possible Solutions

The future state refers to the situation when the building with 21 service stations will be sold. In this case, the queueing system will have 60 − 21 = 39 service stations (servers). All other data remains the same as in Section 3.2.

• Arrival rate (for internal and external buses): λ = 81.83 visits/day
• Service rate (service intensity) per service station: μ = 1.7 visits/day
• Number of service stations in the queuing system: k = 39
• Calculation of the system service factor:

$${\rho }^{*}={\displaystyle \frac{\lambda }{\mu \cdot \mathrm{k}}}={\displaystyle \frac{81.83}{1.7\cdot 39}}=1.234$$

(29)

Since the system’s service factor ρ* > 1, the accumulation of waiting users for service will increase without limit. This means that the 39 service stations (servers) of the queuing system will not be sufficient for the current load of the service workshop.

For the queuing system to be properly operable and the problem solvable, the service factor of the system must be less than 1 (ρ* < 1). From Equation (29), it is evident that this can be achieved in two ways:

• Increase μ by increasing the number of working hours per day,
• Reduce λ by not accounting for the 60,000 working hours performed for external clients. The workshops will service and repair only domestic buses.

The problem can also be solved by increasing the productivity of the service station (server). This can be performed in two ways:

• Increase the productivity of the worker. This reduces the annual number of standard working hours required for repairs, servicing, and maintenance of the bus (which is not recommended as it may reduce the safety of the completed work),
• Combine individual operations in servicing or repairing buses at each service station. This allows multiple workers to perform different tasks simultaneously at each service station, thereby reducing the time the bus spends in the service workshops.

3.3.1. Increase in the Number of Working Hours per Day

The possibility for the system to be solvable increases with the number of working hours per day. Instead of 12 h, 16 h will be worked (in two shifts).

Data:

k = 39; hdz = 60,000 h/year

d = 2; l = 252 days/year

m = 220; h_d = 1053.6 h/year

m₁ = 75; h = 525.8 h/year

m₂ = 163; h220 = 115,896 h/year

h_d1 = 1350 h/year; t_a = 7.075 h

h_d2 = 950 h/year; h_dl = 291,800 h/year

λ_d = 65 visits/day; t_d = 16 h/day (The data that has changed:)

• Arrival rate (for internal and external buses): λ = 81.83 visits/day
• Service rate (service intensity) per service station:

$$\mu ={\displaystyle \frac{{\mathrm{t}}_{\mathrm{d}}}{{\mathrm{t}}_{\mathrm{a}}}}={\displaystyle \frac{16}{7.075}}=2.26 \mathrm{visits}/\mathrm{day} (\mathrm{The}\mathrm{data}\mathrm{that}\mathrm{has}\mathrm{changed})$$

(30)

• Number of service stations (servers) in the queuing system: k = 39

The basic data (λ, μ, k) has been re-entered into a computer program for calculating a multi-server queuing system created in Microsoft Excel. The tabular calculation display is shown in

Table 1. Output from the computer program (in EXCEL) for calculating the queuing system.

ρ=	48.13529
ρ*=	0.802254
		n	n!	ρn	(ρn)/n!	Pn	n.Pn
		0	1	1	1	1.29152× 10−16	0
		1	1	1	1	1.29152 × 10−16	0
		2	1	36.2079646	36.2079646	4.67632 × 10−15	4.67632 × 10−15
		3	2	1311.016701	655.5083503	8.46599 × 10−14	1.6932 × 10−13
		4	6	47,469.24629	7911.541048	1.02179 × 10−12	3.06536 × 10−12
		5	24	1,718,764.789	71,615.19955	9.24922 × 10−12	3.69969 × 10−11
		6	120	62,232,974.65	518,608.1221	6.69791 × 10−11	3.34895 × 10−10
		7	720	2,253,329,343	3,129,624.088	4.04196 × 10−10	2.42518 × 10−9
		8	5040	81,588,469,092	16,188,188.31	2.09073 × 10−9	1.46351 × 10−8
		9	40,320	2.95415 × 1012	73,267,668.67	9.46263 × 10−9	7.57011 × 10−8
		10	362,880	1.06964 × 1014	294,763,683.7	3.80692 × 10−8	3.42623 × 10−7
		11	3,628,800	3.87294 × 1015	1,067,279,303	1.37841 × 107	1.37841 × 10−6
		12	39,916,800	1.40231 × 1017	3,513,091,928	4.53721 × 10−7	4.99093 × 10−6
		13	479,001,600	5.07749× 1018	10,600,159,015	1.36903 × 10−6	1.64283 × 10−5
		14	6,227,020,800	1.83846× 1020	29,523,860,185	3.81305 × 10−6	4.95697 × 10−5
		15	8.7178 × 1010	6.65668 × 1021	76,357,063,177	9.86163 × 10−6	0.000138063
		16	1.3077 × 1012	2.41025 × 1023	1.84316 × 1011	2.38046 × 10−5	0.00035707
		17	2.0923 × 1013	8.72702 × 1024	4.17106 × 1011	5.38699 × 10−5	0.000861918
		18	3.5569 × 1014	3.15988 × 1026	8.88385 × 1011	0.000114736	0.001950518
		19	6.4024 × 1015	1.14413 × 1028	1.78703 × 1012	0.000230798	0.00415437
		20	1.2165 × 1017	4.14265 × 1029	3.40552 × 1012	0.000439828	0.008356738
		21	2.4329 × 1018	1.49997 × 1031	6.16535 × 1012	0.000796264	0.015925289
		22	5.1091 × 1019	5.43108 × 1032	1.06302 × 1013	0.00137291	0.028831115
		23	1.124 × 1021	1.96648 × 1034	1.74954 × 1013	0.002259558	0.049710285
		24	2.5852 × 1022	7.12024 × 1035	2.75423 × 1013	0.003557131	0.081814011
		25	6.2045 × 1023	2.57809 × 1037	4.15521 × 1013	0.00536652	0.12879647
		26	1.5511 × 1025	9.33475 × 1038	6.01807 × 1013	0.00777243	0.194310751
		27	4.0329 × 1026	3.37992 × 1040	8.38085 × 1013	0.010823995	0.281423872
		28	1.0889 × 1028	1.2238 × 1042	1.1239 × 1014	0.014515364	0.39191483
		29	3.0489 × 1029	4.43114 × 1043	1.45336 × 1014	0.018770421	0.525571789
		30	8.8418 × 1030	1.60442 × 1045	1.8146 × 1014	0.023435819	0.67963874
		31	2.6525 × 1032	5.80929 × 1046	2.1901 × 1014	0.028285443	0.848563291
		32	8.2228 × 1033	2.10343 × 1048	2.55803 × 1014	0.033037365	1.02415832
		33	2.6313 × 1035	7.61608 × 1049	2.89441 × 1014	0.037381742	1.196215749
		34	8.6833 × 1036	2.75763 × 1051	3.17578 × 1014	0.041015661	1.353516797
		35	2.9523 × 1038	9.98481 × 1052	3.38201 × 1014	0.043679223	1.485093584
		36	1.0333 × 1040	3.6153 × 1054	3.49874 × 1014	0.045186736	1.581535762
		37	3.7199 × 1041	1.30903 × 1056	3.51895 × 1014	0.045447771	1.636119739
		38	1.3764 × 1043	4.73971 × 1057	3.44362 × 1014	0.044474899	1.645571267
		39	5.2302 × 1044	1.71615 × 1059	3.28122 × 1014	0.042377515	1.610345573
		39	2.0398 × 1046	6.21384 × 1060	3.04632 × 1014	0.039343681	1.534403565
		∑	2.094 × 1046	6.39033 × 1060	3.79228 × 1015	0.489779171	16.30935228
Q=	7.1269197
S=	36.207964
T=	43.334884
W*	0.0870942
W=	0.5295720

.

Since the results W and W* are related to the workday, which has 16 working hours, the correct result is as follows:

$$\begin{gathered} {\mathrm{W}}^{*}=0.08709\cdot 16=1.39 \mathrm{h}=1 \mathrm{h}, 23 \min \mathrm{and} 31 \mathrm{s} \\ \mathrm{W}=0.52957\cdot 16=8.47 \mathrm{h}=8 \mathrm{h}, 28 \min \mathrm{and} 23 \mathrm{s} \end{gathered}$$

Utilization of working time:

$$\eta ={\displaystyle \frac{\mathrm{S}}{\mathrm{k}}}={\displaystyle \frac{36.2079}{39}}=0.928=92.8\%$$

(31)

The loss of profit, if the system contains an average of S = 36.2079 buses and an average of W = 8.47 h (the calculation is performed using Equation (28)), is:

$$\mathrm{J}=11117.18 \mathrm{EUR}$$

3.3.2. Workshops Serve Only Domestic Buses

In this case, the arrival rate is reduced, and we have λ = λ_d per day. The 60,000 working hours for external customers (for buses that are not owned by the company) are not taken into account in the calculation.

Using the equations from Section 3.2, the following data is obtained:

k = 39; h_dz = 0 h/year

d = 2; l = 252 days/year

m = 220; h_d = 1053.6 h/year

m₁ = 75; h = 525.8 h/year

m₂ = 163; h₂₂₀ = 115896 h/year

h_d1 = 1350 h/year; t_a = 7.075 h

h_d2 = 950 h/year; h_dl = 231,800 h/year

λ = λ_d = 65 visits/day; t_d = 12 h/year

μ = 1.7 visits/day

The basic data (λ, μ, k) have been inserted into a computer program for calculating the multi-server queuing system, and the following results were obtained:

Q = 42.9704681

S = 38.2352941

T = 81.2057622

W* = 0.66108412

W = 1.24931942

Since the results W and W* are related to a working day of 12 h, the correct result is as follows:

$$\begin{gathered} {\mathrm{W}}^{*}=0.66108\cdot 12=7.93 \mathrm{h}=7 \mathrm{h} \mathrm{and} 56 \min \\ \mathrm{W}=1.2493\cdot 12=8.47=14.991 \mathrm{h}=14 \mathrm{h} \mathrm{and} 59 \min \end{gathered}$$

Utilization of working time:

$$\eta ={\displaystyle \frac{\mathrm{S}}{\mathrm{k}}}={\displaystyle \frac{38.2057}{39}}=0.98=98\%$$

(32)

The loss of profit, if the system contains an average of S = 38.2057 buses and an average W = 14.991 h (calculated using Equation (28)), is as follows:

$$\mathrm{J}=20761.88 \mathrm{EUR}$$

3.3.3. Increasing the Productivity of the Service Station (Server)

If the number of standard hours required for the repair of a bus hd (number of working hours per year) is reduced, the arrival rate λ also decreases. According to Equation (30), this makes the queuing system stable again (resulting in a finite queue).

The queuing system is analyzed by varying the number of working hours needed for repair, maintenance, and servicing of a single bus per year. This causes a change in the basic input parameters of the queuing system, which are then successively entered into the subprogram for calculating the multi-server queuing system. The equations from Section 3.2 are also used. The results are presented in

Table 2. Results of a multi-server queuing system if the productivity of the workstation is increased- part 1.

hd	h	h220	ta	hdl	λ	μ	ρ*	Cp (%)	Ct (%)
500	250	55,000	3.3578	170,000	100.455	3.5738	0.7207	110.72	52.5437
520	260	57,200	3.4921	174,400	99.0909	3.4364	0.7394	102.615	50.6454
540	270	59,400	3.6264	178,800	97.8283	3.3091	0.758	95.1111	48.7472
560	280	61,600	3.7607	183,200	96.6558	3.1909	0.7767	88.1429	46.8489
580	290	63,800	3.895	187,600	95.5643	3.0809	0.7953	81.6552	44.9506
600	300	66,000	4.0293	192,000	94.5455	2.9782	0.814	75.6	43.0524
620	310	68,200	4.1636	196,400	93.5924	2.8821	0.8327	69.9355	41.1541
640	320	70,400	4.2979	200,800	92.6989	2.792	0.8513	64.625	39.2559
660	330	72,600	4.4322	205,200	91.8595	2.7074	0.87	59.6364	37.3576
680	340	74,800	4.5665	209,600	91.0695	2.6278	0.8886	54.9412	35.4594
700	350	77,000	4.7009	214,000	90.3247	2.5527	0.9073	50.5143	33.5611
720	360	79,200	4.8352	218,400	89.6212	2.4818	0.9259	46.3333	31.6629
740	370	81,400	4.9695	222,800	88.9558	2.4147	0.9446	42.3784	29.7646
760	380	83,600	5.1038	227,200	88.3254	2.3512	0.9632	38.6316	27.8664
780	390	85,800	5.2381	231,600	87.7273	2.2909	0.9819	35.0769	25.9681
785	393	86,350	5.2717	232,700	87.5825	2.2763	0.9866	34.2166	25.4935
790	395	86,900	5.3053	233,800	87.4396	2.2619	0.9912	33.3671	25.019
795	398	87,450	5.3388	234,900	87.2985	2.2477	0.9959	32.5283	24.5444
799	400	87,890	5.3657	235,780	87.1868	2.2364	0.9996	31.8648	24.1648
800	400	88,000	5.3724	236,000	87.1591	2.2336	1.0005	31.7	24.0699
1054	527	101,159	7.0755	291,792	81.8254	1.696	1.2371	0	0

h_d (h/year)—average number of working hours per bus; h (h/year)—average number of hours a bus spends on maintenance, servicing, and repair; in the workshops; h₂₂₀ (h/year)—number of workshop hours occupied by 220 in-house buses; t_a (h)—service time per one bus visit; h_dl (h/year)—number of working hours completed in the workshops in one year; λ (arrivals/day)—arrival rate (intensity of arrival flow); μ (visits/day)—service rate (intensity of service per service station (server); ρ*—the system service factor; C_p (%)—percentage increase in workstation productivity; C_t (%)—percentage reduction in service time.

and

Table 3. Results of a multi-server queuing system if the productivity of the workstation is increased- part 2.

hd	hd1	hd2	Q	S	T	W*	W	h	J
500	640.6606	450.8352	0.0893	28.1084	28.1978	0.0106	3.368	0.7207	3471.19
520	666.287	468.8686	0.1374	28.8385	28.9739	0.0166	3.5087	0.7393	3715.293
540	691.9134	486.9021	0.2078	29.584	29.7721	0.0254	3.6519	0.785	3973.445
560	717.5399	504.9355	0.3096	30.291	30.6007	0.0384	3.7991	0.7766	4248.655
580	743.1663	522.9689	0.4563	31.0192	31.4756	0.0573	3.9532	0.7953	4546.366
600	768.7927	541.0023	0.666	31.7468	32.4129	0.0845	4.1139	0.814	4873.141
620	794.4191	559.0357	0.9657	32.4736	33.4394	0.1238	4.2874	0.8328	5239.491
640	820.0456	577.0691	1.373	33.2015	34.5989	0.1808	4.4788	0.8513	5663.221
660	845.672	595.1025	2.025	33.929	35.954	0.2643	4.6963	0.8699	6171.386
680	871.2984	613.1359	2.9566	34.6561	37.6128	0.3895	4.9561	0.8886	6812.587
700	896.9248	631.1693	4.3892	35.3839	39.7732	0.5831	5.284	0.9072	7680.474
720	922.5513	649.2027	6.7132	36.1113	42.8245	0.8988	5.734	0.9259	8974.025
740	948.1777	667.2361	10.8515	36.8392	47.6907	1.4638	6.4334	0.9445	11212.62
760	973.8041	685.2696	19.6036	37.5676	57.1713	2.6633	7.7673	0.9632	16228.69
780	999.4305	703.303	47.1572	38.2937	85.451	6.4505	11.689	0.9818	36501.66
785	1005.837	707.8113	66.222	38.475	104.698	9.0732	14.345	0.9865	54887.78
790	1012.244	712.3197	105.59	38.6575	144.248	14.491	19.796	0.9912	104358.3
795	1018.65	716.828	235.082	38.8398	273.922	32.315	37.654	0.9958	376938.6
799	1023.776	720.4347	2650.63	38.9853	2689.61	364.82	370.19	0.9996	36386766
800	1025.057	721.3364
1054	1350	950

h_d (h/year)—average number of working hours per bus; h_d1 (work h/year)—average number of working hours for repairs, servicing, and maintenance of a city bus; h_d2 (work h/year)—average number of working hours for repairs, servicing, and maintenance of a suburban or intercity bus; Q—average number of buses waiting for service; S—average number of buses being serviced; T—average number of buses in the system (waiting + in service); W* (h)—average waiting time of a bus before service; W (h)—average time a bus spends in the system (waiting + service); J (EUR)—loss in profit if T buses spend W time in the queuing system.

.

Data:

• Number of service stations (servers): k = 39
• Number of workers per service station: d = 2
• Number of company buses: m = 220
• Number of city buses: m₁ = 57
• Number of suburban or intercity buses: m₂ = 163
• Workshop working hours: t_d = 12 h/day
• Number of working hours performed for external clients: h_dz = 60,000 h/year
• Number of working days per year: l = 252 days
• Profit loss if a bus is out of service: j = 580 EUR/day

In the case where the work in the maintenance workshops is reorganized in such a way that individual maintenance operations are combined at a single service station and three workers (previously two) work simultaneously at that station, the productivity of the service station increases by C_p = 50%. As seen in Table 2, in this case, a single visit of a bus to the maintenance workshop is reduced to t_a = 4.7 h, and the waiting time for service is reduced to W* = 0.5831. Consequently, the loss of profit due to bus downtime in the maintenance workshops is only EUR 7680.

The diagram in Figure 4 shows how the duration of a single bus visit and the total time a bus spends in the system change depending on the average number of working hours per year per bus. The data is taken from Table 2 and Table 3.

If the average number of working hours for repairs, servicing, and maintenance of a bus in service workshops (h_d) is reduced from 790 to 620 h per year, the duration of a single bus visit t_a linearly decreases from 5.3 to 4.03 h. Consequently, the total time a bus spends in the service workshops W decreases from 19.8 to 4.1 h. However, this reduction is no longer linear. In this case, t_a is reduced by 24%, while W is reduced by as much as 79%. As a result, the cost due to the bus being out of service is also reduced by 79%, amounting to only EUR 5239.49 (Table 2 and Table 3).

4. Discussion

The optimization of operations in the analyzed service workshops of a bus company is justified in terms of reducing production space (service workshops), as this increases the efficiency of space utilization with the same volume of work. If the company decides to eliminate 21 of the existing 60 service stations, the current system would no longer be able to maintain the same number of buses as before. Therefore, three potential solutions were proposed to enable the company to continue maintaining its fleet using the remaining 39 service stations.

The first solution, which involves increasing the daily working time from 12 to 16 h, would result in an average of 36.2 buses being present in the system for 8.47 h. This would lead to approximately EUR 11,117 in lost profit due to buses being out of operation (unable to transport passengers during this time). The average waiting time for service would be 1 h and 23 min.

The second solution keeps the working day at 12 h, as before, but limits service exclusively to the company’s own fleet (internal buses), discontinuing services for external clients and forfeiting the corresponding revenue. In this case, about 38 buses would be in the system for an average of 14.9 h, resulting in EUR 20,760 in losses due to idle buses. This outcome is less favorable, as the average waiting time for service increases to 7.56 h.

The third solution involves increasing the productivity of each service station by 50%. Instead of two workers per station, an additional worker is assigned so that three workers operate simultaneously. Tasks at the service station are reorganized by merging individual service operations. As a result, the total time a bus spends in the system is reduced to 4.7 h, and losses amount to only EUR 7680, as only 32.4 buses are on average unavailable for passenger transport during servicing. The average waiting time for service in this case is just 0.89 h.

The optimal solution is the third one, as it does not reduce the volume of work but significantly lowers production costs. Additionally, the losses due to buses being out of service are the smallest, and most importantly, the waiting time for service is drastically reduced.

Based on the analysis of the three proposed solutions for organizing work in the service workshops, the following additional conclusions can be made:

• A high utilization of working hours does not necessarily lead to the best solution, as high utilization also increases waiting times for service (see Table 2 and Table 3). From this perspective, the first solution (increasing the number of working hours per day) is better than the second (servicing only internal buses).
• The second solution, in which the company forfeits 60,000 working hours per year due to workshop downsizing (cost reduction), is not optimal. Why should workshops not also serve external clients if there is demand?
• If the value of the system service factor ρ* is close to 1, even a small reduction in service time (ta) significantly reduces the waiting time for a bus to receive service. Therefore, it is more reasonable to optimize the service station workflow.
• In the case of merging service operations at individual service stations, the throughput of the queuing system increases, and as a result, profit losses due to buses being out of service decrease.