Stochastic Inventory Management in a Shortage

The method of determining the parameters of stochastic inventory management system in conditions of economically based shortage

The main input parameter of the inventory management system is demand. In real conditions, demand often has a random pattern. In addition, it should be considered that replenishment of stocks (supply) also occurs with certain time fluctuations. Of course, it is possible to work with averaged values of demand and time of delivery, but in some cases, neglect of the stochastic nature of consumption and replenishment of inventory may result in incorrect calculations and, consequently, increase of logistic costs.

The main parameters of the inventory management system are the maximum desired stock, the order quantity, the size of safety stock, reorder point. However, we believe that an economically based shortage should also be present among the parameters of the inventory management system. To build an effective inventory management system in stochastic conditions, the algorithm is proposed (Fig. 1). Consider each of the stages of this algorithm in more detail.

Fig. 1. Algorithm of defining the parameters of the stochastic inventory management system in conditions of economically justified shortage

1. Removal of abnormal values.

A gross error or outliers is an error of a certain measurement included in a series of measurements, which for these conditions differs dramatically from other results of this series. In logistics, outliers can be results related to the emergence of conditions that do not have to repeat in the future – stop of production equipment, car accidents, theft of warehouse or transport supplies, supplier strike, etc. If the probability of such events in the future is extremely low, observations that are registered due to the above events must be removed from a series of data as gross errors, as they "differ dramatically from other results" (according to the definition). Possible criteria for checking the values for the presence of gross errors are shown in Table 1.

Table 1 Criteria for checking the statistical values for outliers

If an outlier is detected, the value must be excluded, and the numerical characteristics are recalculated. For subsequent calculations, the value of the sample mean and sample standard deviation is needed. The sample mean:

$\bar{x}=\frac{\sum_{i=1}^{N} x_{i} n_{i}}{N}=\sum_{i=1}^{n} x_{i} f_{i},$   (1)

where $x_{i}$ – the middle of the $i$-th interval; $n_{i}$ – the number of observations in the $i$-th interval; $N$ – thtal number of statistical observations; $n$ – the total number of intervals; $f_{i}$ – the frequency at which the values fall in the $i$-th interval.

Formulae for calculating dispersion and standard deviation are chosen depending on the number of observations (Table 2).

Table 2 Formulas for dispersion and standard deviation

2. Verification of the hypothesis concerning the normal distribution of demand and time of delivery.

The problem of identification of the distribution law describing statistical data is an extremely important stage of forecasting and planning.

A statistical hypothesis is any assumption concerning frequency function, probabilities (probability density function) or quantitative characteristics of numerical series.

To verify the hypothesis of the normal distribution of sales and delivery times, it is proposed to apply the Pearson chi-squared test. To do this, we must calculate the theoretical values of the probability of getting to each of the intervals. This is assisted by the Laplace transform (probability density function of the normal distribution).

The probability $p_{i}$ for the normal distribution of a random variable $X$ is calculated according to the formula:

$p_{i}=\varphi\left(x_{1}\right)-\varphi\left(x_{2}\right)=\varphi\left(\frac{x_{l i}-\bar{x}}{\sigma}\right)-\varphi\left(\frac{x_{r i}-\bar{x}}{\sigma}\right)$   (2)

where $x_{l i}$ and $x_{r i}$ – the lower and upper bounds of the intervals, respectively.

The values of the Laplace function for the positive values of the argument $x(0 \leq x \leq 5)$. For the values $x \text { > }5, \varphi(x)=0.5$ is taken. For negative $x$ values, the same table is used, given that the Laplace function is odd, i. e. $\varphi(-x)=-\varphi(x)$.

The value of the Pearson's criterion or Chi-squared criterion $\left(\chi^{2}\right)$ is calculated by the formula:

$\chi^{2}=\sum_{i=1}^{n} \frac{\left(f_{e_{i}}-f_{t_{i}}\right)^{2}}{f_{t_{i}}}$   (3)

where $n$ – the number of groups into which empirical data are divided; $f_{e i}$ – the frequency observed in the $i$-th group; $f_{t i}$ – theoretical frequency.

For the $\chi^{2}$ distribution, a table is created, where the critical value of the $\chi^{2}$ acceptance criterion for the selected significance level α and degrees of freedom is specified.

The hypothesis of the normal distribution is accepted when the actual $\chi^{2}$ value does not exceed the critical table one (Fig. 2).

Fig. 2. Illustration of deciding on a hypothesis about the normal distribution using the $\chi^{2}$ criterion graph (for degrees of freedom $k=2$)

Table values show the right border of the $\chi^{2}$ distribution graph, which means that it does not confirm the hypothesis with the significance level $\alpha$.

You can do without tabular values and use the function of the MS Excel 2010 CHISQ.INV.RT program (0.05; 2). As the first argument here is the significance level, as the second – the number of degrees of freedom. The value calculated by the MS Excel 2010 program is usually somewhat more exact than the tabular one.

To assess the correctness of the hypothesis concerning the belonging to the normal distribution, it is recommended to calculate an additional p-level or p-value (probability to obtain such or even greater value of the criterion in the equity of the null hypothesis).

3. Calculation of the optimal level of shortage (if the hypothesis for the normal distribution is accepted).

If the hypothesis of belonging of statistical data to the normal distribution is confirmed, it is possible to use the formula for the optimum level of shortage and tabular data for normal distribution.

The optimal level of shortage:

$S=\frac{c_{h}}{c_{h}+c_{d e f}}$   (4)

where $C_{h}$ – the cost of storing a unit of goods per 1 day; $C_{def}$ – losses from the shortage of a unit of goods per 1 day.

As for the shortage losses, there are certain problems with determining the value of these expenditures. There can be quite a lot of approaches. Consider a few of them:

1. if the client who wants to buy a product and is not able to do it because of the lack of goods in stock, purchases this product in another company. In this case, the shortage losses correspond to the value of lost profit from the sale of goods, which is in lack;
2. if the client who cannot buy a product because of its absence, is offered a discount with which he will be able to obtain a product if he waits for arriving of the new shipment. In this case, the shortage losses correspond to the value, which reduces the profit from the production unit through the discount;
3. if for satisfaction of the client and fast delivery of the missing goods, blitz-orders with an increased cost of delivery are done. Then the additional cost of unplanned supply is distributed between all units of the shipment, so this will be shortage losses;
4. one option can be a combination of others. For example, ordering the blitz-supply and simultaneously offering a discount, to make customer be waiting for the product and do not apply to the company-competitor. Then the shortage losses are summed up.

4. Calculation of the amount of safety stock (considering the optimal level of shortage).

If the demand (sales, need) and time of delivery (the time of order fulfillment) obey the normal distribution law, the calculation of the safety stock requires a table value Z (the number of entailing no deficit standard deviations).

The number of entailing no deficit standard deviations and service level are connected by the NORMSINV function of the MS Excel 2010 program. In order to find the number of entailing no deficit standard deviations at a certain service level, it is necessary to specify the =NORMSINV function (service level).

The probability of shortage (in this technique – the optimum level of shortage) is defined as

$\mathrm{d}=1-L$   (5)

where $L$ – service level

Considering the $Z$ values and parameters of the distribution of consumption volume and delivery time, the safety stock is calculated according to the formula:

$Z_{s}=z \cdot \sqrt{\bar{t} \cdot \sigma_{s}^{2}+\bar{S}^{2} \sigma_{t}^{2}}$   (6)

where $Z_{s}$ – the value of safety stock, un.; $z$ – the number of entailing no deficit standard deviations; $\bar{t}_{n}$ – average execution time of the order, days; $s_{s}$ – standard deviation of demand, un./day; $s_{t}$ – average consumption, un./day; st - standard deviation of delivery time, days.

5. Calculation of the amount of optimal order quantity considering the possibility of shortage.

The model of inventory management with the possibility of shortage provides the calculation of the optimal order quantity by the formula

$Q_{\text {opt } t}=\sqrt{\frac{2 D C_{s}}{C_{h}}} \cdot \sqrt{\frac{C_{h}+C_{d e f}}{C_{d e f}}}$   (7)

where $D$ – demand for goods during a certain period of time, un.; $C_{s}$ – the cost of order delivery, which does not depend on the size of the order, UAH/un.; $C_{h}$ – the cost of storing a unit of goods during a certain period of time, UAH/un.; $C_{def}$ – the cost of shortage of a unit of goods during a certain period of time, UAH/un.

6. Calculation of the interval between orders.

To calculate the optimal interval between orders, first the number of deliveries per year should be determined:

$K=D / Q_{o p t^{-}}$   (8)

The interval between orders is calculated by dividing the number of days in the period by the number of deliveries for that period:

$I=T / K$   (9)

7. Calculation of the optimal order quantity in stochastic conditions.

The order quantity can be calculated as follows:

$Q_{i}^{\prime}=S \cdot\left(t_{i}+\bar{t}\right)+Z \cdot \sqrt{\left(t_{i}+\bar{t}\right) \cdot \sigma_{s}^{2}+\bar{S}^{2} \sigma_{t}^{2}}-Z_{T i}-Z_{t i}$   (10)

where $Q_{i}^{\prime}$ – the quantity of the order $i$, un.; $t_{i}$ – time interval between orders, days; $Z_{ti}$ – the level of the current inventory when issuing an order $i$, un; $Z_{ti}$ – the quantity of the stocks on the way not received at the moment of issuing an order $i$, un.

If the inventory level at the time of the order is zero and there are no stocks on the way, the order quantity can be obtained by the formula:

$Q_{i}^{\prime}=S \cdot\left(t_{i}+\bar{t}\right)+Z \cdot \sqrt{\left(t_{i}+\bar{t}\right) \cdot \sigma_{s}^{2}+\bar{S}^{2} \sigma_{t}^{2}}$   (11)

8. Calculation of the threshold level of the inventory management system.

The threshold level of inventory determines the level at which it is necessary to make the next order for replenishment of stocks. In stochastic conditions of consumption and receiving orders, the threshold level is determined by the formula:

$Q_{R O P}=\overline{S t}+Z \cdot \sqrt{\bar{t} \sigma_{s}^{2}+\bar{S}^{2} \sigma_{t}^{2}}=\overline{S t}+Z_{S}$   (12)

Consider the application of the above methods on a practical example. Let there be sales statistics for 12 months of the year, as well as statistics of delivery time of goods (Tables 3, 4).

Table 3 Statistics of sales volume

Table 4 Statistics of delivery time

1. Extraction of abnormal values from the statistics.

Since the number of our statistical values is 12, we can use the criteria of Romanovsky or Grubbs. The formulas view to apply these criteria (Table 1) is the same, differing only in tabular data, which are compared with the left part of the condition (inequality).

Because the number of values presented in the statistics is less than $30(n=5 \text { < }30)$, to calculate the standard deviation, use the "corrected" formula (Table 2).

Calculated values of the sample mean (1) and sample corrected standard deviation (Table 2): $\bar{x}=13.2 \text { un., } \sigma=0.972 \text { un. }$

Now you can calculate the values of the left part of the condition for the anomalies of values (Table 5).

Compare the obtained values of the left part of the condition for the abnormal values with the table ones (Table 6).

As we can see, all the values of the left part of the condition concerning anomalies (Table 5) are smaller than any value of the Romanovsky criterion (Table 6). This means that with a probability of at least 99%, all statistics are not abnormal or gross values. According to the Grubbs criterion (Table 6), the last value of the criterion that corresponds to 99 % of non-membership with the abnormal value, is violated for the sales volume in the month No. 5 (2.26 > 2.229). If we accept the need for 99 % of the absence of abnormal values in the statistics, the sales number of the 5th month must be removed from statistics. The sample mean and standard deviation must be recalculated. We believe that the confidence probability of 0.98 will be enough to test the anomalies of the statistical series. This would mean that no statistics values have been removed as abnormal.

Table 5 Calculation of the values of the left side of the condition for values abnormality

Table 6 Critical values of the Romanovsky and Grubbs criteria $\left(\beta_{q} \text { and } v_{p}\right)$

2. Verification of the hypothesis concerning the normal distribution law.

Let us calculate the required number of groups by the formula $k=1+3.322 \cdot \lg (12)=4.6$. As the rounding of the integers in the calculations is carried out in a greater direction, $k=5$.

Find the intervals of sales volumes for each of 5 periods. For this purpose, in the monthly statistics (Table 3) we find the largest and smallest value: $x_{\min }=11, x_{\max }=15$.

If you calculate with the MS Excel 2010 program, you can use the MAX and MIN functions.

The width of each of the five intervals will be 0.8 units $(w=0.8)$. Thus, any value of demand between 11 and 11.8 units is included in the 1st interval. According to Table 3, it is the value of the month No. 5.

For the 2nd interval, the left border begins with all numbers that are greater than the right border of the 1st interval, that is, after 11.8.

The right border of the 2nd interval is calculated similarly: $x_{r 2}=x_{I 2}+w=11.8+0.8=12.6$.

Similarly, the boundaries of other intervals are determined.

Then the frequency of hitting values in each of the five intervals must be calculated (the last column of Table 7).

Table 7 Specifying the number of observations in each interval and frequency

The procedure for calculating the Pearson criterion components is given in Table 8. Let us explain how to get columns 5 and 6. We find the value of $\varphi\left(x_{1}\right)=\varphi(-2.26)$. Since $\varphi(-x)=-\varphi(x)$ we find the values in the table corresponding to the value of $x{1}$ modulo: $\varphi(2.26)=0.4881$. So, the value of $\varphi(-2.26)=-\varphi(2.26)=-0.4881$. Similarly, we find the value of the Laplace function for all other arguments $x{1}$ and $x{2}$.

All the components of the $\chi^{2}$ formula are calculated in Table 8, column 9. The total value of the $\chi^{2}$ criterion is 0.37. Because the number of intervals is $n=5$, the number degrees of freedom equals $k=5–3=2$.

The calculated value of $\chi^{2}$ equals 0.37, tabular is 6.0, and $0.37 \text { < } 6.0$, so the hypothesis of the normal distribution is accepted.

The test of the hypothesis for the normal distribution of delivery time is carried out similarly. Here are just the results. Distribution parameters of delivery time: sample mean – 4.67 days, sample corrected standard deviation – 1.03 days. The calculated value of $\chi^{2}$ equals 1.84, table – 6.0. Consequently, $1.84 \text { < } 6.0$, and the hypothesis of the normal distribution is accepted.

3. Calculation of the optimal shortage level.

Set the data about the inventory management system. Let the product cost is $100  \mathrm{UAH}$, and the storage cost is approximately 50 % of its cost per year, therefore, the annual cost of the storage of a unit of goods will be $100 \cdot 0.5=50 \mathrm{UAH}$, a day – $50 / 365=0.137 \mathrm{UAH}$.

The goods are sold at the price of $150 \text { UAH }$, the profit from the sale is $50   \mathrm{UAH}$. Calculate the optimal parameters of the inventory management system by the following scenarios:

1. If in the absence of goods, the client refuses to expect the supply and goes to the competitor, the shortage losses are $C_{\text {def }}=50 \mathrm{UAH}$.
2. In the second version, we can offer the client a discount of 5 % of the goods price in case he agrees to wait for the next delivery, i. e. the sale price will be $150∙0.95= =142.5 UAH$ and profit will decrease by $50–42.5=7.5 \mathrm{UAH}$. Thus, in this option $C_{\text {def }} =7.5 \mathrm{UAH}$.
3. In the case of a decision on extra delivery, the cost of goods is increased by $2  \mathrm{UAH}$. Then the profit from the sale of goods is reduced by $2  \mathrm{UAH}$, i. e. $C_{\text {def }} =2 \mathrm{UAH}$.
4. In this option, the client is offered to wait for the next delivery and given a 5 % discount. Thus, there is a combination of options 2 and 3. The shortage losses will be $7.5+2= =9.5 \mathrm{UAH}$.

The value of the optimal shortage level for the above four scenarios is shown in Table 9.

It can be seen that the worst option is the client's leave to a competitor company, in this case, the acceptable shortage level is only 0.3 %. The maximum shortage level is valid in case of extra supply – 6.4 %. However, this case is very sensitive to the unit cost of the blitz-delivery, that is, additional costs per unit of goods delivered.

Table 8 Calculation of theoretical frequencies and Pearson criterion

Table 9 Results of calculating the optimal parameters of the inventory management system with stochastic demand and delivery time

4. Calculation of the safety stock level (considering the optimal shortage level).

Table values of non-deficiency standard deviations $Z$ for the four values of the optimal shortage level are equal to 2.76; 2.06; 1.52; 2.2.

When the MS Excel 2010 NORMSINV function is used to obtain the tabular values, we set the given service level in unit fractions as an argument.

To calculate the safety stock value, we recall the statistical parameters of demand and delivery time:

1. for demand: sample mean – 13.2 units per month, i. e. 0.44 units per day, corrected standard deviation – 0.0324 units per day;
2. for delivery time: sample mean – 4.67 days, corrected standard deviation is 1.03 days.

The value of the optimum safety stock level for the above four scenarios is shown in Table 9. Safety stock quantity differs significantly for different scenarios – from 0.7 to 1.27 units (almost twice).

5. Calculation of the optimal order quantity considering the possibility of shortage.

Let us present the initial data included in the formula:

– annual demand – $159 \text { un. }$;
– delivery costs – $200 \mathrm{UAH} ;$;
– costs for storing one unit of goods during a year – $50 \mathrm{UAH} ;$;
– losses from shortage of one unit of goods during a year:
1 scenario – $50 \cdot 365=18,250$; 2 scenario – $7.5 \cdot 365=2737.5$; 3 scenario – $2 \cdot 365=730$; 4 scenario – $9.5 \cdot 365=3467.5 \text { UAH }$.

The value of the optimal order quantity for the above four scenarios can be seen in Table 9. The values are not very different for different shortage losses. This is caused by the known statement on the stability of the Wilson model. For the following calculations, it will be enough to take a certain average value of the optimal order quantity – 36 units.

6. Calculation of the interval between orders.

To calculate the optimal interval between the orders, the number of deliveries per year must be determined by the formula (8): $K=4.42$.

According to (9), the interval between orders is calculated as dividing 365 days in the period by the amount of supply during this period (4.42): $I=82.6 \text { days. }$

7. Calculation of the optimal order quantity in the stochastic conditions.

The results of previous calculations are used to determine the optimal order quantity in the stochastic conditions (Table 9).

It can be seen that the values of the optimal order quantity considering the stochastic consumption and delivery frequency differ slightly depending on the number of non-deficient standard deviation (2.76, 2.06, 1.52, 2.2) and varies from 39.23 to 39.90 un. Thus, we can claim that the shortage losses weakly affect the order quantity both for differential and stochastic parameters.

8. Calculation of the threshold level (ROP – Reorder Point) of the inventory management system.

Let us calculate the value of the threshold level, at which you should make an order for replenishment of stocks (Table 9). It is seen that the value of the threshold level varies depending on the amount of safety stock from 2.76 to 3.33 un.