*Model 2*

If at the beginning of the period [0, *T*] uncontrollable parameters of the logistic process such as T = 400 days, r = 0.001, µ = 25 units/day, and *p* = 20 EUR/unit are known and the cost of delivery increases during the period [0, *<sup>T</sup>*] with <sup>ρ</sup> <sup>=</sup> 0.00075, then the growth pattern will be *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 1.0023*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = 2.3).

When using the EOQ model, excluding the increase in delivery costs, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\text{TC}\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 1.0022^{40} + 20 \ast 25 \ast 40 \ast 1.001^{400})$$

$$1.001^{360} + \dots + (400 \ast 1.0022^{360} + 20 \ast 25 \ast 40) \ast 1.001^{40} = 258365 \text{ EUR}$$

When applying model 2, the time between deliveries is found by Formula (20):

$$t\_{so} = 1.0023^{100} \ast 40 = 50 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\begin{aligned} T\mathbb{C}\_{\mathcal{O}} &= (400 + 20 \ast 25 \ast 50) \ast 1.001^{400} + (400 \ast 1.0022^{50} + 20 \ast 25 \ast 50) \ast 1.001^{50} \\ 1.001^{350} + \cdots &+ (400 \ast 1.0022^{320} + 20 \ast 25 \ast 50) \ast 1.001^{50} = 258068 \text{ EUR} \end{aligned}$$

The savings will be as follows:

$$
\Delta TC = 258365 - 258068 = 297 \text{EUR}
$$

If the cost of delivery decreases during the period [0, *T*] with ρ*<sup>c</sup>* = −0.0018, the growth pattern has the form *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 0.9982*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = −1.8).

When using the EOQ model, excluding delivery cost reduction, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\begin{aligned} T\mathbb{C}\_{w} &= (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 0.9982^{40} + 20 \ast 25 \ast 40) \ast 1.001^{360} + \cdots + (400 \ast 0.9982^{360} + 20 \ast 25 \ast 40) \ast 1.001^{40} = 254627 \text{ EUR} \end{aligned}$$

When applying model 2, the time between deliveries is found by Formula (20):

$$t\_{so} = 0.9982^{100} \ast 40 = 33 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\begin{aligned} TC\_o &= (400 + 20 \ast 25 \ast 33) \ast 1.001^{400} + (400 \ast 0.9982^{33} + 20 \ast 25 \ast 33) \\ \ast 1.001^{367} + \cdots &+ (400 \ast 0.9982^{367} + 20 \ast 25 \ast 33) \ast 1.001^{33} = 254513 \text{ EUR} \end{aligned}$$

The savings will be as follows:

$$
\Delta T \text{C} = 254627 - 254513 = 114 \,\text{EUR}.
$$
