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SN Applied Sciences

, 1:1671 | Cite as

Application of mathematical modeling value-at-risk (VaR) to optimize decision making in distribution networks

  • E. Khorshidi
  • V. R. GhezavatiEmail author
Research Article
  • 62 Downloads
Part of the following topical collections:
  1. 3. Engineering (general)

Abstract

Managers and capital masters of companies and factories try to adopt methods that maximize their profit and minimize their costs. A way for increasing profit is risk management. The risk management and the type of risk are defined in the literature of financial management. For measurement of the risk, many methods are defined which are all created in the recent century and it means the risk and risk management are rather new concepts. Among the newest tools for measuring the risk is value-at-risk (VaR), which was modified by Morgan (Riskmetrics technical document, Morgan Guaranty Trust Company, New York, 1996). VaR represents the maximum expected loss over a certain period of time and at a given confidence level. Many parametric, semi-parametric and nonparametric methods for VaR estimation have been developed. In this article using the mean–variance method one of the parametric techniques of VaR, has been tried to minimize the cost arisen due to locating the supply chain and minimize the maximum level of capital losses, to optimize decision making in the distribution network. The calculated model is tested with numerical examples by MATLAB and Lingo software, and this example supported the resulting model.

Keywords

Value at risk Supply chain Location-allocation Location Integer programming 

1 Introduction

Site selection is one of the most important issues in industrial engineering, which can lead to the reduction in costs and successfulness of industrial units. Site selection for service centers is known as a selection of a position for one or more service centers considering other centers and also existing limitations so that a particular goal would be optimized. Site selection problem is one of the network design problems which has been solved as one of the strategic decision making processes. In site selection problems, there are a set of nodes that certain demands and they must be fulfilled by a set of service centers through some transportation modes. The term site selection refers to modeling, formulation and solving those problems which are seeking the best location for the establishment of the service center and facilities [1]. The first scientific framework of this theory was introduced formally by Weber [2]. He was one of the theorists who theorized the site selection and cost minimization which constitute major transportation costs [3]. The site selection-allocation problem is also used as a combination of supply chain problems. A provided model which is a combination of the site-selection-allocation, path determination, and control problems developed by Kheybari et al. [4] is one of these problems.

Distribution refers to a stage in which a product is stored and transported from a supplier to a customer in a supply chain. For this purpose, raw materials, components and elements are transferred from a supplier to manufacturers and then final products are transferred to end customers. Performance of a distribution network should be measured through two aspects:
  1. 1.

    Customers’ demands must be met.

     
  2. 2.

    Cost of satisfying these needs.

     
Decisions regarded to the design of a supply chain network are classified as follow:
  1. 1.

    The role of facilities: What is the role of each facility? In addition, what is processed and run in each facility?

     
  2. 2.

    Location of facilities: Where should be the location of each facility?

     
  3. 3.

    Capacity allocation: How much capacity should be allocated to each facility?

     

Among the effective factors on the distribution networks, one can refer to taxes, tariffs, currency rate, demand risk etc. This research is going to consider the effect of the risk factors on the demand parameter. Site selection of the facilities was gradually proposed to supply chain which investigates the models of site selection in the supply chain [5]. Usually, in management of a supply chain, there are three levels of planning depending on time horizon: strategic, tactical and operational. Simchi-Levi et al. [6] stated that the strategic level deals with decisions which would have long-term impacts on the firm.

Sometimes, the meaning of application of supply chain network design is the same as strategic planning [7, 8]. Designing a bi-level and single period distribution network is also studied for managing a supply chain. This problem includes site selection for factories and distributing stockpiles. Moreover, the capacity of these facilities is determined from the possible selected choices, the related decisions on how to distribute products from a factory to distribution piles and also from stockpiles to customers [9].

Lejarza and Baldea [10] considered the issue of the supply chain in which the number and site of the distribution center are determined. Customers are facing random demands and each distribution center maintains a certain amount of guaranteed storage in order to achieve a certain service level for customers.

The concept of risk is arisen from uncertainty for input parameters in a decision making process and shows the possible loss in the problem. Taymaz et al. [11] introduced the risk as an unwanted result for a decision making problem. Xu et al. [12] introduced the sustainability of risks in the logistic network design. Value at risk is a risk assessment and diagnosis that uses standard statistical techniques that are used every day in many tactical and operational problems. Value at risk indicates the maximum expected loss in a given planning horizon. In addition, there are other value-at-risk metrics that have been published independently and have been introduced as tools to maximize the portfolio of capital in order to maximize returns for a given level of risk [13]. VaR applied the confidence α in which the most values of VaR would not be exceeded to lose during the next h days [14, 15]. In the framework of parametric approaches, the first model is proposed by Morgan [16]. One major disadvantage of the mentioned model is the assumption of the financial yield normality while the experimental evidence indicates that the financial yield does not follow a normal distribution. The second disadvantage is that in order to estimate the fluctuation, the financial yield should be used; and the third one is that the parametric approach assumes independency of the yields. There are significant experimental evident on the normality of financial yield distribution [17, 18, 19, 20, 21]. The VaR also includes point estimation, but deviates from the median that requires some probability level details, and then provides a better expected point to occur in a better probability. VaR calculation method is divided into parametric and non-parametric approaches [22].

The reason that we considered risk in our model is uncertainty in the environment of decision space. In a decision making problem, there are some input parameters that fluctuates the results and they are not under decision maker control such as customers’ demand. Therefore, the decision maker should create the mathematical model in which the model is able to handle such uncertainty due to prevent future costs. Thus, we considered value at risk (VaR) as a new framework of controlling the risk in our problem.

In this research, the implementation of VaR regarded with the demands of customers for the distribution network site selection model under uncertainty is performed and the parametric method is used to solve this problem. In this paper, the parametric type of VaR (mean–variance) methods are applied to formulate the uncertainty.

2 Problem definition

In this research, the problem of the supply chain is considered regarding the demand under uncertainty with a normal distribution with minimization of costs such as fixed costs of locating the facility in a candidate site and costs of the services delivered to the customers by the facility as well as minimization of value at risk. The problem faced here includes a bi-level supply chain considering suppliers level and customers level. In different periods, the demand for different products is delivered probabilistically to the suppliers in most scenarios. The supplier is responsible for meeting the demands and he should minimize the costs of fulfilling and services to the customer.

VaR is a tool for controlling the risk in financial aspects. There are other tools such as robust optimization for controlling the risk but an expert tool for financial risk is VaR. Since in our paper, the total cost is based on the financial risk, therefore, we used VaR in this way. It should be noted this tool very recently is used in the optimization problems such as facility location. In the literature, we focused on the application of VaR in optimization problems due to its major impact in our model [23].

2.1 Problem assumptions

  1. 1.

    In site selection for a supply chain network, nodes are candidates for the locations of both customers and facilities.

     
  2. 2.

    Locating a facility in a candidate site has fixed and certain costs.

     
  3. 3.

    Only one system delivers services to each customer.

     
  4. 4.

    The amount of flow of each facility depends on its capacity.

     
  5. 5.

    The cost of delivering a service to a customer is fixed.

     
  6. 6.

    The amount of flow between a facility and each customer is determined by a standard normal distribution using variance and mean indexes.

     
  7. 7.

    The sum of demands is considered as the lost value which is not supplied considering the capacity of a facility and demands of a customer.

     

2.2 Proposed mathematical model

The site selection model of the supply chain is defined by implementing the VaR.

2.2.1 Indexes, parameters and decision variables

I

Set of customers’ sites; i = 1, 2,…, m

J

Set of candidate sites for facilities; j = 1, 2, …, n

h i

Average of demand for custome i

σ i

Amount of variance of demand for customer i

σ j

Total Amount of variance of demands assigned to candidate facility j

f j

Fixed cost of locating facility in candidate site j

b j

Total Amount of mean of demands assigned to candidate facility j

c ij

Cost of each transportation unit between customer i and candidate site j

p j

Amount of capacity for candidate facility j

Y ij

A binary variable that equals 1, if demand of customer i is supplied by candidate facility j, otherwise it is equal to zero

X j

A binary variable that equals 1, if a facility is located in site j otherwise it is equal to zero

VaR j

The highest expected loss in a given time horizon with the level of reliability at site j

2.2.2 Objective functions and limitations of the model

$${\text{Min Z}}_{1} = \mathop \sum \limits_{j \in J} VaR_{j}$$
(1)
$${\text{Min}}\,{\text{Z}}_{2} = \mathop \sum \limits_{j \in J} f_{j} X_{j} + \mathop \sum \limits_{j \in J} \mathop \sum \limits_{i \in I} h_{i} c_{ij} Y_{ij}$$
(2)
$$\begin{aligned} & {\text{s}} . {\text{t:}} \\ & \mathop \sum \limits_{j \in J} Y_{ij} = 1\quad \forall i \in I \\ \end{aligned}$$
(3)
$$\mathop \sum \limits_{i \in I} h_{i} Y_{ij} - p_{j} X_{j} \le 0\quad \forall j \in J$$
(4)
$$pr\left( {b_{j} - p_{j} \le VaR_{j} } \right) \ge \alpha \quad \forall j \in J$$
(5)
$$b_{j} = \mathop \sum \limits_{i \in I} h_{i} Y_{ij} \quad \forall j \in J$$
(6)
$$\sigma_{j} = \mathop \sum \limits_{i \in I} \sigma_{i} Y_{ij} \quad \forall j \in J$$
(7)
$$VaR_{j} \ge 0\quad \forall j \in J$$
(8)
$$X_{j} \in \left\{ {0,1} \right\}\quad \forall j \in J$$
(9)
$$Y_{ij} \in \left\{ {0,1} \right\}\quad \forall i \in I\;\;\forall j \in J$$
(10)

The first objective function minimizes the total amount of VaR for all activated facilities. The second objective function minimizes total costs including the cost of establishing facilities and also cost of transportation between facilities and customers.

Set constraint (3) indicates that each customer must be assigned to one facility in order to be served. Set constraint (4) ensures that each customer can be assigned to a facility once the facility is activated. In addition, each facility can cover total demands up to its capacity. Set constraint (5) which is given in a non-linear form indicates that the probability of losing demand up to the amount of VAR for each facility must be at least a. Considering the fact that a non-linear term can lead to a remarkable increase in the solution time, then this limitation is first rewritten and then converted to a linear model using a linear integer model. Set constraints (6) and (7) determine the total amount of mean and variance of demands assigned to candidate facility j, respectively. The stated probability in set constraint (5) can be converted to a standard normal function probability considering that α is the reliability value, then one can extract the accumulative distribution function value from the table of standard normal function which is equal to \(Z_{\alpha }\). These descriptions are indicated in the following calculations:
$$\begin{aligned} & pr\left( {b_{j} - p_{j} \le VaR_{j} } \right) \ge \alpha \to pr\left( {Z \le \frac{{VaR_{j} + p_{j} - b_{j} }}{{\sqrt {\sigma_{j} } }}} \right) \\ & \quad \ge \alpha \to \frac{{VaR_{j} + p_{j} - b_{j} }}{{\sqrt {\sigma_{j} } }} \ge Z_{\alpha } \to VaR_{j} + p_{j} - b_{j} \ge Z_{\alpha } *\sqrt {\sigma_{j} } \\ \end{aligned}$$
In order to increase the speed of solving the problem, one can make the limitation as follows by changing the variable \(\sqrt {\sigma_{j} } = w_{j}\):
$$VaR_{j} + p_{j} - b_{j} \ge Z_{\alpha } w_{j} \quad \forall j \in J$$
(11)
Now, set constraint (5) is substituted by constraint (11). Considering the integrals, this limitation is non-linear. Therefore, the fragmentation model can be used to make it linear. Since the limit of variable \(\sigma_{j}\) considering the problem parameters can be determined, in order to increase the speed of problem solving one can approximate \(w_{j}\). Assume that the maximum amount of \(w_{j}\) be a. To do so, the interval [0, a] is divided into L parts. The more the L is, the more accurate the approximation would be achieved. The limitations 12–30 are added to the model in order to make it linear. In this paper, we assume that the maximum value for wj be 120. Thus, interval [0, 120] are separated into 12 sub-intervals where break points are: 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110 and 120. Therefore, the mathematical model in order to approximate linearization is as follows:
$$\begin{aligned} w_{j} & = \lambda_{j0} \sqrt 0 + \lambda_{j1} \sqrt {10} + \lambda_{j2} \sqrt {20} + \lambda_{j3} \sqrt {30} + \lambda_{j4} \sqrt {40} \\ & \quad + \,\lambda_{j5} \sqrt {50} + \lambda_{j6} \sqrt {60} + \lambda_{j7} \sqrt {70} + \lambda_{j8} \sqrt {80} + \lambda_{j9} \sqrt {90} \\ & \quad + \,\lambda_{j10} \sqrt {100} + \lambda_{j11} \sqrt {110} + \lambda_{j12} \sqrt {120} \quad \forall j \in J \\ \end{aligned}$$
(12)
$$\begin{aligned} \sigma_{j} & = \lambda_{j0} 0 + \lambda_{j1} 10 + \lambda_{j2} 20 + \lambda_{j3} 30 + \lambda_{j4} 40 \\ & \quad + \,\lambda_{j5} 50 + \lambda_{j6} 60 + \lambda_{j7} 70 + \lambda_{j8} 80 + \lambda_{j9} 90 \\ & \quad + \,\lambda_{j10} 100 + \lambda_{j11} 110 + \lambda_{j12} 120\quad \forall j \in J \\ \end{aligned}$$
(13)
$$\mathop \sum \limits_{l = 0}^{12} \lambda_{jl} = 1\quad \forall j \in J$$
(14)
$$\lambda_{jl} \ge 0\quad \forall j \in J,\;\;l = 0,1, \ldots ,12$$
(15)
$$\lambda_{j0} \le \delta_{j0} \quad \forall j \in J$$
(16)
$$\lambda_{j1} \le \delta_{j0} + \delta_{j1} \quad \forall j \in J$$
(17)
$$\lambda_{j2} \le \delta_{j1} + \delta_{j2} \quad \forall j \in J$$
(18)
$$\lambda_{j3} \le \delta_{j2} + \delta_{j3} \quad \forall j \in J$$
(19)
$$\lambda_{j4} \le \delta_{j3} + \delta_{j4} \quad \forall j \in J$$
(20)
$$\lambda_{j5} \le \delta_{j4} + \delta_{j5} \quad \forall j \in J$$
(21)
$$\lambda_{j6} \le \delta_{j5} + \delta_{j6} \quad \forall j \in J$$
(22)
$$\lambda_{j7} \le \delta_{j6} + \delta_{j7} \quad \forall j \in J$$
(23)
$$\lambda_{j8} \le \delta_{j7} + \delta_{j8} \quad \forall j \in J$$
(24)
$$\lambda_{j9} \le \delta_{j8} + \delta_{j9} \quad \forall j \in J$$
(25)
$$\lambda_{j10} \le \delta_{j9} + \delta_{j10} \quad \forall j \in J$$
(26)
$$\lambda_{j11} \le \delta_{j10} + \delta_{j11} \quad \forall j \in J$$
(27)
$$\lambda_{j12} \le \delta_{j11} \quad \forall j \in J$$
(28)
$$\mathop \sum \limits_{l = 1}^{12} \delta_{jl} = 1\quad \forall j \in J$$
(29)
$$\delta_{jl} \in \left\{ {0,1} \right\}\quad \forall j \in J,\;\; l = 0,1, \ldots ,$$
(30)

Above procedure says that in order to compute linear approximation of \(w_{j}\), firstly, It should be clear that the point is like \(w_{j}\), which lies between two consecutive extremes of defined distances.

Then, the value of \(\sqrt {w_{j} }\) is computed exactly according to the same convex linear combination which was used for \(w_{j}\) (for more information, we can refer to Ghezavati et. al. [24]).

2.3 Implementation of genetic algorithm for the proposed model

In this paper, we use Lingo solver for solving the proposed model for small-sized examples and we aim to use a genetic algorithm for solving large-sized examples. For this purpose, we need to code the genetic algorithm in a language programming. In this way, we choose MATLAB software to code the genetic algorithm.

The primary population is matrix \(Y_{ij}\); i.e. chromosomes are elements of matrix \(Y_{ij}\). The primary population is the same as the size of the population from random matrices \(Y_{ij}\). For example, if the population involves 200 individuals, the random matrix \(Y_{ij}\) will be the matrix to meet the limitations. The cross-over used for solving the model is a uni-point cross-section in a way that we have for example 2 parents namely \(Y_{ij}\) and \(Y_{ij}\). \(Y_{1}\) and \(Y_{2}\) are two I*J matrices. A random number is selected between 1 and I and using row r, 2 new children are made; i.e. matrixes \(Y_{1}\) and \(Y_{2}\) are substituted in rows 1 − r. Doing so maintains the limitation of the problem in the new children:
$$\mathop \sum \limits_{j} Y_{ij} = 1 \quad \forall i \in I$$
In order to solve the model, a multi-point mutation is used. In this kind of mutation, some genes are selected randomly in the chromosomes and their values would be changed. In this problem, a row is selected randomly and then new values are allocated to this row based on limitation \(\sum\nolimits_{j} {Y_{ij} = 1}\). In fact, it tries to make a situation in which the limitations of the problem would not be contradicted in the new child. Therefore, for row R the new value allocation would be as follows:
$$\mathop \sum \limits_{j} Y_{rj} = 1$$

The steps of the proposed genetic algorithm are described as follows:

Step 1 Initialize the input parameters of the algorithm including: the probability of the crossover and mutation, population size, the maximum number of iterations.

Step 2 The required number of initial solutions are generated randomly.

Step 3 Feasibility of the initial solutions is checked.

Step 4 The required solutions (called parents) are selected for crossover operation according to the crossover probability.

Step 5 The crossover operation is performed on parents to generate new solutions (called offspring).

Step 6 Feasibility of the offspring solutions is checked.

Step 7 The required solutions (called parents) are selected for mutation operation according to the mutation probability.

Step 8 The mutation operation is performed on parents to generate new neighborhood solutions.

Step 9 Feasibility of the neighborhood solutions is checked.

Step 10 Create the pool of solutions including the current solutions, offspring solutions, and neighborhood solutions.

Step 11 The fitness function for all solutions in the pool is computed according to the objective function in the Lp-metric method.

Step 12 According to the population size, the required number of solutions are selected according to the roulette cycle.

Step 13 Check the stopping criterion. If it is satisfied to stop and report the best solution, otherwise, go to step 4.

2.4 Implementation of Lp-metric approach

Since the model function has two objectives, the problem is solved at first with the first objective function and solution \(Z_{1}^{*}\) is obtained. Then it is solved with the second objective function to give solution \(Z_{2}^{*}\). Moreover, in order to consider both objective functions simultaneously, the Lp-metric technique is used [25]:
$$minz = \left[ {w_{1} \frac{{\left( {z_{1} - z_{1}^{*} } \right)^{p} }}{{z_{1}^{*} }} + w_{2} \frac{{\left( {z_{2} - z_{2}^{*} } \right)^{p} }}{{z_{2}^{*} }}} \right]^{1/p}$$
where \(w_{1} = 0.5\), \(w_{2}\) = 5 and \(p = 1.\)

The system by which code writing and implementing is done is involves 4 GB RAM, CPU: Core i5, 230 GHz. Once the proposed model is transferred to a single objective function, it is solved by Lingo 8 software for small sized examples and also it is solved by the proposed genetic algorithm for large sized examples that are coded by MATLAB 2011.

3 Numerical example

In order to show the effect of VaR on the mathematical model, a problem with 10 customers and 8 facility sites from 5 data sets and different input sets are considered. Later, a problem with 20 customers sites, 15 facilities sites and 5 data sets and different input sets, in addition to a problem with 50 customers sites and 40 facilities sites, and finally a problem with 70 customers site and 50 facilities sites are considered. Furthermore, it is attempted to consider the cost of each transportation unit between location j and customer i, fixed cost of locating the facility for site j (f), mean customer’s demand (h) and capacity of facility j (p) as being different. Therefore, in all problems \(\sigma_{i}\) is assumed to be the normal distribution with a mean of 12 and variance of 2. Other parameters are also considered as \(\alpha = 95\%\) and a = 120 and L = 12.

Created input data for solving the model by “MATLAB” and “Lingo” are shown in Table 1.
Table 1

Input data

Data

I

J

c

f

h

P

Sets and inputs

1

10

8

[5–15]

[10–20]

[20–25]

[20–40]

2

10

8

[5–15]

[15–50]

[100–250]

[150–300]

3

10

8

[5]

[10]

[10–25]

[15–35]

4

10

8

[5]

[10]

[10–25]

[20–50]

5

10

8

[5]

[10–20]

[20–25]

[20–40]

6

20

15

[5–15]

[10–20]

[20–25]

[20–40]

7

20

15

[5–15]

[10–20]

[20–25]

[20–50]

8

20

15

[5–15]

[10–20]

[20–25]

[40–50]

9

20

15

[5–15]

[10–20]

[20–30]

[20–60]

10

20

15

[5–15]

[10–20]

[20–30]

[20–50]

11

50

40

[5–15]

[10–20]

[10–30]

[20–60]

12

70

60

[5–15]

[10–20]

[20–30]

[20–50]

Results of input data from Table 1 for solving model considering just the first objective function is shown in Table 2. These results have gotten by “MATLAB” and “lingo”. In Fig. 1, the solution run time for the proposed genetic algorithm and Lingo solver is illustrated. As can be seen, due to the Np-hard type of the mathematical model, the solution run time for Lingo has an exponential pattern and therefore, by increasing the size of the model actually Lingo is unable to solve the model. In this way, the solution run time of the genetic algorithm by MATLAB software has a linear form and this makes possible to solve the large-sized examples by this algorithm.
Table 2

Comparison between results of MATLAB and LINGO

MATLAB

LINGO

Data

Result

Time (s)

Result

Time (s)

In the first function

1

9.057

137

9.041503

945

2

9.6325

88

9.6288

38

3

56.5394

110

56.5213

192

4

3.1173

156

3.1088

199

5

34.3619

62

34.3438

465

6

92.221

151

92.4037

6480

7

49.0423

248

49.041

7800

8

36.7346

257

36.724

8640

9

1.4371

142

1.4213

7560

10

99.9337

196

99.9124

5200

11

39.144

355

221.617

34 h

12

51.6566

450

Infeasible

Fig. 1

Comparison of solution runs time of genetic algorithm and Lingo

Results of input data from Table 1 for solving model considering just second objective function are shown in Table 3. These results have gotten by “MATLAB” and “Lingo”.
Table 3

Comparison between results of MATLAB and LINGO

MATLAB

LINGO

Data

Result

Time (s)

Result

Time (s)

In second function

1

1286

92

1286

1

2

10640

68

10,640

1

3

985

95

985

1

4

875

91

875

1

5

1297

53

1297

1

6

2504

71

2504

1

7

2628

46

2628

1

8

2509

48

2509

1

9

2737

62

2737

1

10

2856

58

2856

1

11

4615

150

4615

43

12

6663

310

6663

190

Results of input data from Table 1 for solving model considering both of objective functions are shown in Table 4. These results have gotten by “MATLAB” and “lingo”.
Table 4

Comparison between results of MATLAB and LINGO

MATLAB

LINGO

Data

Result

Time (s)

Result

Time (s)

After implementing LP-metric

1

0.1844

115

0.184307

1983

2

0.0918

12

0.0918

106

3

0.0355

65

0.0332

1690

4

0.04

134

0.04

394

5

0.01045

118

1034

186

6

0.2015

137

0.2011

7095

7

0.2159

243

0.2153

9300

8

0.2758

218

0.2742

9660

9

1.174

155

1.1622

9480

10

0.1459

130

0.145

6900

11

0.4516

328

Not computed

12

0.3498

1537

Infeasible

Now testing will be done to show the effect of increasing each input on the value of objective functions. The first 5 numerical examples of Table 4 are used in the test. Considering the importance of the objective function results, it is decided to ignore the calculated time. It should be noted that each test has done by Lingo.

3.1 First example

In the first test, 5 units will be added to the value of “c” in each example and the result has been shown in the following table. Input data for this test are shown in Table 5.
Table 5

Model inputs by increase “c”

Data

I

J

f

h

p

1

10

8

[10–20]

[20–25]

[20–40]

2

10

8

[15–50]

[100–250]

[150–300]

3

10

8

[10]

[10–25]

[15–35]

4

10

8

[10]

[10–25]

[20–50]

5

10

8

[10–20]

[20–25]

[20–40]

We use input data of Table 5 to perform this test and achieve effect modality of increase “c” on the objective functions, results by “MATLAB” and “Lingo” are illustrated in Table 6.
Table 6

Result of increase “c”

Data

New first objective function

Old first objective function

New second objective function

Old second objective function

New general objective function

Old general objective function

1

9.0415

9.057

2276

1286

0.121

0.1844

2

9.6288

9.6325

18,685

10,640

0.0522

0.0918

3

56.5213

56.5394

1960

985

0.18293

0.0355

4

3.1088

3.1173

1740

875

0.0201

0.04

5

34.3619

34.3619

2238

1297

0.0138

0.01045

Considering Table 6, it is obvious that by increasing the value of “c”, the first objective function does not change significantly but the second objective function increases while the second objective function has a descending trend due to a decrease in the total costs. Figure 2 shows the comparison between the old and new values of the second objective function. In addition, with respect to the Lp-metric method, it does not have a special effect on the general objective function.
Fig. 2

The comparison between the old and new values of the second objective function

3.2 Second example

In this test, the value of parameter “f” has increased for 2 units. Input data for this test are shown in Table 7.
Table 7

Model inputs by increase “f”

Data

I

J

c

h

P

1

10

8

[5–15]

[20–25]

[20–40]

2

10

8

[5–15]

[100–250]

[150–300]

3

10

8

[5]

[10–25]

[15–35]

4

10

8

[5]

[10–25]

[20–50]

5

10

8

[5]

[20–25]

[20–40]

We use input data of Table 7 to perform this test and achieve effect modality of increase “f” on the objective functions, results by “MATLAB” and “Lingo” are illustrated in Table 8.
Table 8

Result of increase “f”

Data

New first objective function

Old first objective function

New second objective function

Old second objective function

New general objective function

Old general objective function

1

9.0415

9.057

1296

1286

0.1852

0.1844

2

9.6288

9.6325

10,652

10,640

0.0919

0.0918

3

56.5121

56.5394

987

985

0.0325

0.0355

4

3.1088

3.1173

877

875

0.0478

0.04

5

34.2438

34.3619

1307

1297

0.1038

0.01045

As can be seen, the increasing the value of parameter “f” does not show a distinguished effect on the first objective function and causes just an insignificant reduction. However, its effect on the second objective function is obvious and increases the cost, but it is still not as much effective as parameter “c”.

3.3 Third example

In this section, the effect of increasing 2 units for parameter “h” on the objective functions will be checked. Input data for this test are shown in Table 9.
Table 9

Model inputs by increase “h”

Data

I

J

c

f

p

1

10

8

[5–15]

[10–20]

[20–40]

2

10

8

[5–15]

[15–50]

[150–300]

3

10

8

[5]

[10]

[15–35]

4

10

8

[5]

[10]

[20–50]

5

10

8

[5]

[10–20]

[20–40]

We use input data of Table 9 to perform this test and achieve effect modality of increasing parameter “h” on the objective functions, results by “MATLAB” and “Lingo” are reported in Table 10.
Table 10

Result of increase “h”

Data

New first objective function

Old first objective function

New second objective function

Old second objective function

New general objective function

Old general objective function

1

12.0415

9.057

1391

1286

0.1866

0.1844

2

11.6288

9.6325

10,770

10,640

0.0948

0.0918

3

75.8962

56.5394

1085

985

0.0360

0.0355

4

17.6843

3.1173

975

875

0.0416

0.04

5

49.6171

34.3619

1403

1297

0.1066

0.01045

According to the obtained results, increasing “h” has a significant impact on increasing the first and second objective functions and therefore increases the general objective function.

Figure 3 shows the comparison between the values for the first objective function. As can be seen, by increasing parameter “h” in the model, the value of the first objective increases. This is pattern is because of the fact that by increasing the holding cost, the model tries to minimize the number of products in the warehouse and this leads to increasing the value at risk for satisfying the demands. So, the level of VaR increases in this regard. However, as can be seen from Fig. 4, in this analysis the model tries to minimize holding cost and this leads to decrease total costs of the model and therefore, the total cost has a descending trend.
Fig. 3

Comparison between the values for the first objective function by increasing parameter “h”

Fig. 4

Comparison between the values for the second objective function by increasing parameter “h”

3.4 Forth example

In this analysis, 2 units are added to parameter “p” and the outputs are gained again. Input data for this test are shown in Table 11 and then the outputs are reported.
Table 11

Model inputs by increasing “p”

Data

I

J

c

f

h

1

10

8

[5–15]

[10–20]

[20–25]

2

10

8

[5–15]

[15–50]

[100–250]

3

10

8

[5]

[10]

[10–25]

4

10

8

[5]

[10]

[10–25]

5

10

8

[5]

[10–20]

[20–25]

We use input data of Table 11 to perform this test and achieve effect modality of increasing parameter “p” on the objective functions, results by “MATLAB” and “Lingo” are reported in Table 12.
Table 12

Result of increase “p”

Data

New first objective function

Old first objective function

New second objective function

Old second objective function

New general objective function

Old general objective function

1

1.6284

9.057

1286

1286

0.1955

0.1844

2

7.6288

9.6325

10,640

10,640

0.1859

0.0918

3

43.3129

56.5394

985

985

0.0549

0.0355

4

0.6886

3.1173

875

875

0.04

0.04

5

26.3811

34.3619

1297

1297

0.1182

0.01045

Above results show that increasing of parameter “p” does not influence the second objective function but considerably reduces the first objective function with an insignificant effect on the general objective function.

4 Results and discussion

It should be noted that increasing the capacity of facility j leads to a reduction of VaR in the first objective function. Considering that during solving the model by software, the objective function is simulated individually. Also, it can be observed that in the second objective function the values obtained from Lingo and MATLAB are the same, but the run time in Lingo for small dimensions is 1 s and as the dimensions increase, this time increases as well. Meanwhile, change of the value of parameters and problem inputs affects the amount of cost value (second objective function) more than the first one.

In addition, it can be seen that any increase in the value of fixed costs (f) leads to a small decrease in the first objective function of the proposed model. Once the value of capacities of the facilities increased in the model, the first objective function reduces more than previous satiations, however, such an event cannot change the global objective function. According to the definition of VaR in previous sections, the value of the objective function improves by the length of the interval of demand once the technique of VaR is applied.

Table 13 illustrates the consequence of changing each parameter individually on first, second, and global objective functions.
Table 13

Sensitivity analysis for the value of parameters on the VaR and costs

Parameter

Value at risk

Total cost

General

c

Reduce

Exceeded

Indeterminate

f

Reduce

Exceeded

Indeterminate

h

Exceeded

Exceeded

Increase

p

Reduce

Indifferent

Increase

5 Conclusion

In this study, a framework for evaluating and managing the investment in facility location problem regarded with site selection by applying a general site selection problems framework with respect to risk concepts is presented using the VaR approach. The best advantage of VaR is the creation of a structured approach to making precise risk decisions. The process of achieving the right VaR depends on its numerical value. In this article, the decision maker tries to optimize the value of VaR in the site selection problem and the costs associated with fixed costs where the manager would be able to forecast the total risks which can occur in the future.

It was also worked in this research to create a certain and linear method by implementing the VaR and linearization of the model by fragmentation linear approximation method. Also, by considering the fact that the obtained mathematical model is bi-objective, each objective function was solved separately by lingo and MATLAB using Lp-metric and the obtained results are compared. Meanwhile, it is proved that applying the VaR, one can minimize the costs and because of the profit by predicting the maximum loss. In this paper, demand was considered uncertain and it followed a normal distribution function. In this condition, even if the demand was considered as a risk factor and the amount of customer’s demand was not determined, one can manage the risk and control the maximum loss using the criteria of VaR measurement to be able to minimize the total costs.

For future researches, authors can consider the main parameters regarded to VaR as a fuzzy numbers since tactical financial parameters are almost uncertain. In addition, implementing the proposed method for designing distribution networks and reverse logistics is necessary. Furthermore, developing for VaR concept using the fuzzy theory can be a suitable research in this area. Finally, applying multi objective metaheuristic methods such as multi objective particle swarm optimization (MOPSO) method will be another interesting improvement.

Notes

Compliance with ethical standards

Conflict of interest

The author declares that they have no conflict of interest.

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Copyright information

© Springer Nature Switzerland AG 2019

Authors and Affiliations

  1. 1.School of Industrial EngineeringIslamic Azad University, South Tehran BranchTehranIran

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