Stochastic Inventory Management in a Shortage

Read this article. The study indicates that a stochastic inventory management system should be preceded by determining the economic feasibility of the shortage. How does using real-time statistics help determine purchasing?

Materials and methods of the study

Checking the algorithm on real data, sensitivity analysis of the model

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

Month number Sales volume, t
1 14
2 12
3 13
4 15
5 11
6 13
7 14
8 13
9 12
10 15
11 13
12 14

Table 4 Statistics of delivery time

Delivery number Time of delivery, day
1 5
2 6
3 5
4 7
5 5
6 4
7 5
8 6
9 5
10 5
11 6
12 6

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

Month number Sales volume, t Value of the left part of the condition for the anomalies of values \frac{\left|\bar{x}-x_{i}\right|}{S}
1 14 0.823045
2 12 1.234568
3 13 0.205761
4 15 1.851852
5 11 2.263374
6 13 0.205761
7 14 0.823045
8 13 0.205761
9 12 1.234568
10 15 1.851852
11 13 0.205761
12 14 0.823045

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

Confidence probability, p Number of values n=12
Romanovsky criterion Grubbs criterion
0.9 2.75 2.663
0.95 2.66 2.519
0.98 2.52 2.387
0.99 2.39 2.229

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

Interval number, i Lower border, \mathcal{X}_{l i} Upper border, \mathcal{X}_{r i} Number of values, \mathcal{n}_{i} Month number Frequency, f_{i}=n_{i} / n
1 11 11.8 1 5 0.083333
2 11.8 12.6 2 2, 9 0.166667
3 12.6 13.4 3 3, 6, 8, 11 0.333333
4 13.4 14.2 4 1, 7, 12 0.25
5 14.2 15.0 5 4, 10 0.166667

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

Interval number, і Number of observations, n{i} The first argument in the Laplace function, x_{1}=\left(x_{\mathrm{l}}-x\right) / \sigma The second argument in the Laplace function, x_{2}=\left(x_{\mathrm{r}}-x\right) / \sigma Laplace function first value \varphi\left(x_{1}\right) Laplace function second value \varphi\left(x_{2}\right) The probability of falling in the i-th interval, p_{i}=\varphi\left(x_{2}\right)-\varphi\left(x_{1}\right) Theoretical frequency, n \cdot p_{i} Pearson criterion components, K_{i}
1 1.00 –2.26 –1.44 –0.4881 –0.4251 0.06 0.76 0.08
2 2.00 –1.44 –0.62 –0.4251 –0.2324 0.19 2.31 0.04
3 4.00 –0.62 0.21 –0.2324 0.0832 0.32 3.79 0.01
4 3.00 0.21 1.03 0.0832 0.3485 0.27 3.18 0.01
5 2.00 1.03 1.85 0.3485 0.4678 0.12 1.43 0.23
Total 12.00 –3.09 1.03 –0.71 0.24 0.96 11.47 0.37

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

Parameter Formula for calculation Un. Scenario number

1 2 3 4
Optimal shortage level (5) % 0.3 1.8 6.4 1.4
Optimal safety stock level (6) un. 1.27 0.95 0.70 1.01
Optimal order quantity under deterministic conditions (7) un. 35.71 35.99 36.87 35.92
Optimal order quantity under stochastic conditions (11) un. 39.90 39.52 39.23 39.60
Optimum threshold level (12) un. 3.33 3.01 2.76 3.07

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.