# Optimisation Models for Inventory Management with Limited Number of Stock Items

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Literature Review

## 3. Methods and Research Design

- (1)
- (2)

- (1)
- Deliveries are made at certain times at regular intervals—for example, once a week, monthly, etc.
- (2)
- The consumption of stocks is even.
- (3)
- High storage costs and, at the same time, high costs associated with stock shortages are observed.
- (4)
- In the case of a shortage of stock, needs cannot be met when the next supply is received.

#### 3.1. Staging of the Optimisation Model in the Presence of a Single Stock

_{i}and therefore costs for past deliveries (c

_{i}, i = 1, 2, …, m) already made known, which are carried out at m regular/fixed intervals (e.g., monthly).

_{1}, c

_{2}, …, c

_{m}}. The volumes of deliveries are also a one-dimensional array {x

_{1}, x

_{2}, …, x

_{m}}. A possible algorithm for finding the optimal solution is the following: The minimal element in the array is found {c

_{1}, c

_{2}, …, c

_{m}}. Its position in the array is stored in the variable “pos”. The optimal volume for stock delivery is x

_{pos}.

#### 3.2. Staging of the Optimisation Model in the Presence of More Than One Type of Stock and Where No Restrictions Are Imposed

_{1}, A

_{2}, …, A

_{n}, the volumes for each we will indicate with x

^{j}, j = 1, 2, …, n and the costs of each with c

^{j}, j = 1, 2, …, n.

_{i}

^{j}, i = 1, 2, …, m; j = 1, 2, …, n stocks of each product type and the corresponding c

_{i}

^{j}, i = 1, 2, …, m; j = 1, 2, …, n costs for each period are known for m periods, depending on the volume of stock of each product type for each of the periods.

^{j}

_{i0(j)}, j = 1, 2, …, n shall be determined by: i

_{0}(j) is such that c

^{j}

_{i0(j)}= min {c

_{1}

^{j}, c

_{2}

^{j}, …, c

_{m}

^{j}} for each j = 1, 2, …, n.

^{j}

_{i0(j)}is one of the intermediate values of x

_{i}

^{j}, i = 1, 2, …, m for each j = 1, 2, …, n, since in the case of small and large x

_{i}

^{j}, i = 1, 2, …, m the costs are higher.

^{j}

_{i0(j)}for each j = 1, 2, …, n.

_{3}

^{1}= 100, c

_{2}

^{2}= 110, c

_{6}

^{3}= 90 and c

_{8}

^{4}= 120, respectively. This means that if at the beginning of each month a stock of the products of the first 180 kg, the second 220 kg, the third 150 kg, and the fourth 210 kg is carried out, then it can be expected that the total costs will be minimal and will be about c

_{3}

^{1}+ c

_{2}

^{2}+ c

_{6}

^{3}+ c

_{8}

^{4}= 100 + 110 + 90 + 120 = 420 BGN per month.

_{i}

^{j}, i = 1, 2, …, m; j = 1, 2, …, n, except for non-negative conditions.

#### 3.3. Staging of the Optimisation Model in the Presence of More Than One Type of Stock and the Presence of Restrictions

_{3}

^{1}+ x

_{2}

^{2}+ x

_{6}

^{3}+ x

_{8}

^{4}= 180 + 220 + 150 + 210 = 760 kg, which exceeds the capacity of the warehouse and it is not possible to store these quantities of these stocks of the four products on a monthly basis. Of course, we can search Table 3 for other values close to the optimum values, so that their amount does not exceed 700 kg. However, it is difficult to apply this method to a large number of historical data and a large number of products.

_{j}, j = 1, 2, …, n in Table 1 and Table 2 by square regression, we obtain the dependency:

^{j}(x

^{j}) = a

^{j}+ b

^{j}x

^{j}+ d

^{j}(x

^{j})

^{2}, j = 1, 2, …, n

^{1}, x

^{2}, …, x

^{n}) = a

^{1}+ b

^{1}x

^{1}+ d

^{1}(x

^{1})

^{2}+ a

^{2}+ b

^{2}x

^{2}+ d

^{2}(x

^{2})

^{2}+ …

+ a

^{n}+ b

^{n}x

^{n}+ d

^{n}(x

^{n})

^{2}

^{1}+ x

^{2}+ … + x

^{n}≤ Q

^{j}≥ 0, j = 1, 2, …, n

_{j}≤ x

_{j}≤ V

_{j}, j = 1, 2, …, n

_{j}) and not exceed another specified level (V

_{j}).

c^{1} = 0.0075 (x^{1})^{2} − 2.4081 x^{1} + 308.58, | R^{2} = 0.8213, |

c^{2} = 0.0085 (x^{2})^{2} − 3.9563 x^{2} + 584.56, | R^{2} = 0.7413, |

c^{3} = 0.0125 (x^{3})^{2} − 4.0047 x^{3} + 417.6, | R^{2} = 0.9261, |

c^{4} = 0.0078 (x^{4})^{2} − 3.4187 x^{4} + 510.53, | R^{2} = 0.7711. |

^{1}, x

^{2}, x

^{3}, x

^{4}) = 0.0075 (x

^{1})

^{2}− 2.4081 x

^{1}+ 308.58 +

0.0085 (x

^{2})

^{2}− 3.9563 x

^{2}+ 584.56 +

0.0125 (x

^{3})

^{2}− 4.0047 x

^{3}+ 417.6 +

0.0078 (x

^{4})

^{2}− 3.4187 x

^{4}+ 510.53

^{1}+ x

^{2}+ x

^{3}+ x

^{4}≤ 700,

x

^{1}≥ 0, x

^{2}≥ 0, x

^{3}≥ 0, x

^{4}≥ 0

## 4. Results and Discussion

^{1}= 139.46, x

^{2}= 214.12, x

^{3}= 147.54, x

^{4}= 198.88 and min: Z = 483.74

^{1}= 139 kg., x

^{2}= 214 kg., x

^{3}= 148 kg., x

^{4}= 199 kg., which is actually the optimal integer solution to the task. This means that supplies of 139 kg must be made from the first type of stock, supplies of 214 kg must be made from the second type of stock, the third type of stock must be supplied at a rate of 148 kg, and the fourth type of stock must be supplied at a rate of 199 kg. This value is naturally higher than the minimum of the function, provided that there are no limits on the number of supplies of each type of stock, it is the optimal solution in case of restrictions.

## 5. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## References

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j i | x^{1} | x^{2} | … | x^{j} | … | x^{n} |
---|---|---|---|---|---|---|

1 | x_{1}^{1} | x_{1}^{2} | … | x_{1}^{j} | … | x_{1}^{n} |

2 | x_{2}^{1} | x_{2}^{2} | … | x_{2}^{j} | … | x_{2}^{n} |

… | … | … | … | … | … | … |

i | x_{i}^{1} | x_{i}^{2} | … | x_{i}^{j} | … | x_{i}^{n} |

… | … | … | … | … | … | … |

m | x_{m}^{1} | x_{m}^{2} | … | x_{m}^{j} | … | x_{m}^{n} |

j i | c^{1} | c^{2} | … | c^{j} | … | c^{n} |
---|---|---|---|---|---|---|

1 | c_{1}^{1} | c_{1}^{2} | … | c_{1}^{j} | … | c_{1}^{n} |

2 | c_{2}^{1} | c_{2}^{2} | … | c_{2}^{j} | … | c_{2}^{n} |

… | … | … | … | … | … | … |

i | c_{i}^{1} | c_{i}^{2} | … | c_{i}^{j} | … | c_{i}^{n} |

… | … | … | … | … | … | … |

m | c_{m}^{1} | c_{m}^{2} | … | c_{m}^{j} | … | c_{m}^{n} |

Product | A^{1} | A^{2} | A^{3} | A^{4} |
---|---|---|---|---|

x^{i}Month | x^{1} | x^{2} | x^{3} | x^{4} |

1 | 200 | 300 | 180 | 250 |

2 | 120 | 220 | 210 | 230 |

3 | 180 | 250 | 160 | 235 |

4 | 130 | 240 | 140 | 260 |

5 | 240 | 190 | 110 | 320 |

6 | 260 | 210 | 150 | 180 |

7 | 210 | 280 | 170 | 200 |

8 | 150 | 310 | 200 | 210 |

9 | 110 | 245 | 215 | 170 |

10 | 165 | 315 | 165 | 150 |

11 | 170 | 230 | 130 | 205 |

12 | 160 | 180 | 120 | 215 |

Costs Month | c^{1} | c^{2} | c^{3} | c^{4} |
---|---|---|---|---|

1 | 130 | 180 | 110 | 160 |

2 | 140 | 110 | 125 | 145 |

3 | 100 | 130 | 95 | 150 |

4 | 125 | 120 | 105 | 170 |

5 | 170 | 150 | 130 | 210 |

6 | 180 | 115 | 90 | 155 |

7 | 135 | 160 | 100 | 130 |

8 | 130 | 170 | 120 | 120 |

9 | 120 | 135 | 135 | 165 |

10 | 110 | 165 | 98 | 180 |

11 | 105 | 117 | 108 | 125 |

12 | 115 | 155 | 118 | 122 |

A_{1} | A_{2} | A_{3} | A_{4} | ||||
---|---|---|---|---|---|---|---|

x^{1} | c^{1} | x^{2} | c^{2} | x^{3} | c^{3} | x^{4} | c^{4} |

110 | 120 | 180 | 155 | 110 | 130 | 150 | 180 |

120 | 140 | 190 | 150 | 120 | 118 | 170 | 165 |

130 | 125 | 210 | 115 | 130 | 108 | 180 | 155 |

150 | 130 | 220 | 110 | 140 | 105 | 200 | 130 |

160 | 115 | 230 | 117 | 150 | 90 | 205 | 125 |

165 | 110 | 240 | 120 | 160 | 95 | 210 | 120 |

170 | 105 | 245 | 135 | 165 | 98 | 215 | 122 |

180 | 100 | 250 | 130 | 170 | 100 | 230 | 145 |

200 | 130 | 280 | 160 | 180 | 110 | 235 | 150 |

210 | 135 | 300 | 180 | 200 | 120 | 250 | 160 |

240 | 170 | 310 | 170 | 210 | 125 | 260 | 170 |

260 | 180 | 315 | 165 | 215 | 135 | 320 | 210 |

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Vasilev, J.; Milkova, T. Optimisation Models for Inventory Management with Limited Number of Stock Items. *Logistics* **2022**, *6*, 54.
https://doi.org/10.3390/logistics6030054

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Vasilev J, Milkova T. Optimisation Models for Inventory Management with Limited Number of Stock Items. *Logistics*. 2022; 6(3):54.
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Vasilev, Julian, and Tanka Milkova. 2022. "Optimisation Models for Inventory Management with Limited Number of Stock Items" *Logistics* 6, no. 3: 54.
https://doi.org/10.3390/logistics6030054