*Model 3*

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 *c<sup>s</sup>* = 25EUR are known and the price increases during the period [0, *<sup>T</sup>*] with <sup>ρ</sup>*<sup>p</sup>* = 0.000786, then the growth pattern will be *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 1.000786*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>p</sup>* = 0.786).

When using the EOQ model without considering the price increase, 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:

$$\text{TC}\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 + 20 \ast 1.000786^{40} \ast 25 \ast 40) \ast 1.001^{360} + \dots + (400 + 20 \ast 1.000786^{360} \ast 25 \ast 40) \ast 1.001^{40} = 292146 \text{ EUR}$$

When applying model 3, the time between deliveries is found by Formula (26):

$$t\_{\rm so} = \frac{40}{1.000786^{100} \sqrt{1 - 0.786}} = 80 \text{ days}$$

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

$$\begin{aligned} TC\_0 &= (400 + 20 \ast 25 \ast 80) \ast 1.001^{400} + (400 + 20 \ast 1.000786^{80} \ast 25 \ast 80) \ast \\ 1.001^{320} + \cdots &+ (400 + 20 \ast 1.000786^{320} \ast 25 \ast 80) \ast 1.001^{80} = 290915 \text{ EUR} \end{aligned}$$

The savings will be as follows:

$$
\Delta TC = 292146 - 290915 = 1331 \text{ EUR}
$$

If the price decreases during the period [0, *T*] with ρ*<sup>p</sup>* = −0.003, the growth pattern has the form *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 0.997*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>p</sup>* = −3).

With the EOQ model, excluding price reductions, 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:

$$T\mathbb{C}\_{w} = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 + 20 \ast 0.997^{40} \ast 25 \ast 40) \ast 1.001^{360} + \cdots$$

$$\cdots + (400 + 20 \ast 0.997^{360} \ast 25 \ast 40) \ast 1.001^{40} = 165954 \text{ EUR}$$

When applying model 3, the time between deliveries is found by Formula (26):

$$t\_{so} = \frac{40}{0.997^{100} \sqrt{1+3}} = 25 \text{ days}$$

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

$$T\mathbb{C}\_{0} = (400 + 20 \ast 25 \ast 25) \ast 1.001^{400} + (400 + 20 \ast 0.997^{25} \ast 25 \ast 25) \ast 1.001^{375} + \cdots$$

$$\cdots + (400 + 20 \ast 0.997^{375} \ast 25 \ast 25) \ast 1.001^{25} = 164244 \text{ EUR}$$

The savings will be as follows:

$$
\Delta TC = 165954 - 164244 = 1730 \text{ EUR}
$$
