## 4 Numerical Example
### Production Problem  
Consider a supply chain problem constituted of a production system, one principal stock, and two warehouses \((L=2)\) to satisfy a random demand. The manufacturing system is comprised of one machine producing one type of item during \(H=12\) months, with \(\Delta t=1\) month. For each demand of a product, the standard deviation is the same for all periods, \(\sigma d_{i}\)=2.2, \((\{i:1,2\})\), and it is assumed that initial level of inventory \(S(0)=10\). The average demand for warehouse 1 and warehouse 2 is: \(d_{1}(k)=d_{2}\ (k)=15\ \{k:0,\ \ldots,\ H-1\}\).  
* Lower and upper limits of production capacities: \(U_{min}=10\) and \(U_{max}=20\) and \(c_{p}=2\) mu, \(c_{h}=0.2\) mu, \(c_{si}=0.2\) mu/k, \(\{i:1,\ldots L=2\}\), \(w_{i}(0)=20\) with \(\{\text{i}:\ 1,\ 2\}\).
* Consumer satisfaction level is equal to 90% \(\theta_{i}=0.9(i=1,\ 2)\).
* Example 1:  
Tables 3-5 and Figure 6 show the optimum production and delivery planning and the lowest total cost for different values of the delivery time, \(\tau\). Meanwhile, Figure 6 illustrates the total cost (production, delivery, inventory) as a delivery time function, \(\tau\). So, the lowest total cost value corresponds to \(\tau^{*}=2\), with minimal cost equal to 13,772 mu.  
For the optimal planning of production, delivery quantities varied according to the delivery time, then impacted on total cost. Consequently, through the analysis of the variability of delivery time, we obtained the proper delivery time value that minimizes the total cost, which relates to \(\tau=1\), with minimal cost equal to \(24,885\) mu in our numerical Example 2. Moreover, compared with the first example, the higher value of the total cost relates to the higher value of the average demands. We can note that, as average demands increase, the process produces more. Then, we need to take more of the products in the process store and the purchase warehouses to satisfy the demand with a maximum service level. Consequently, the transported amount also increases with a faster delivery time.  
Example 3 (increased standard deviation):  
In this example, we try to increase the standard deviation to \(\sigma d_{i}=5\), (\(\{i:1,2\}\)) and use the same average demands presented in Example 2. We obtained the economic planning given by Tables 11-13.  
By increasing the standard deviation, we obtained the proper delivery time value, minimizing the total cost, which relates to \(\tau=1\), with minimal cost equal to \(26,955\ m\) in our numerical Example 3. Moreover, compared with the second example, the higher value of the total cost relates to the higher value of the standard deviation. From Figure 7, we can note that as standard deviation increases, the process produces more. Then, we need to take more of the products in the process store and the purchase warehouses to satisfy the demand with a maximum service level. Consequently, the transported amount also increases with a faster delivery time. However, at the same time, the increasing of the standard deviation will require more of the production system, and consequently, will impact on the degradation of the process as well as its strategy of maintenance and quality.