# Trustworthiness of Situational Awareness: Significance and Quantification
## 5 Numerical Results
### Case 2: Low-Trust Information Source
In the following, we calculate the score of the metrics that affect the trustworthiness of the link for this case. _Confidentiality:_ Since the plaintext information (without encryption) sent from the MS traverses the wired link between the CHN and the GN, the confidentiality score for \(l_{2}\) or \(l_{CHN-GN}\) can be assigned as 0. _Integrity:_ Since no HMAC of the message is transmitted along with the message on the link \(l_{2}\) in this case, the integrity score for \(l_{2}\) can be ascribed as 0. _Availability:_ Assuming that the link can carry the messages and is not overloaded with data traffic nor is affected by any denial-of-service attack, an availability score of 1 can be attributed to \(l_{2}\). _Quality:_ Assuming a bit error rate of \(48\times 10^{-5}\)[20], the quality score for \(l_{2}\) can be calculated as \(1-0.00048=0.99952\). Since, in our model, there are four metrics that influence the trustability of a link, we assign a weight of 0.25 (i.e., \(1/4=0.25\)) to each metric. Using Equation (4), the trustworthiness score of the link \(l_{2}\) or \(l_{GCS-GN}\) can be determined as \(\mathcal{T}_{l_{2}}=(0.25)(0)+(0.25)(0)+(0.25)(1)+(0.25)(0.99952)=0.49988\). **Link \(l_{3}\) or \(l_{GN-RN}\):** The trustworthiness of the wired link \(l_{3}\) or \(l_{GN-RN}\) between the GN and the RN in the core network can be calculated similarly to the wired link \(l_{2}\), and can be given as \(\mathcal{T}_{l_{3}}=(0.25)(0)+(0.25)(0)+(0.25)(1)+(0.25)(0.99952)=0.49988\). **Link \(l_{4}\) or \(l_{RN-DRN}\):** The trustworthiness of the wired link \(l_{4}\) or \(l_{RN-DRN}\) between the RN and the DFN can be calculated similar to the wired link \(l_{2}\), and can be given as \(\mathcal{T}_{l_{4}}=(0.25)(0)+(0.25)(0)+(0.25)(1)+(0.25)(0.99952)=0.49988\). **Link \(l_{5}\) or \(l_{DRN-CBN}\):** The trustworthiness of the wired link \(l_{5}\) or \(l_{DRN-CBN}\) between the DFN and the CBN can be calculated similarly to the wired link \(l_{2}\), and can be given as \(\mathcal{T}_{l_{5}}=(0.25)(0)+(0.25)(0)+(0.25)(1)+(0.25)(0.99952)=0.49988\). Now that we have calculated the trustworthiness of the MS as well as all the nodes and links in the path between the MS and the CBN for this case, we can calculate the trustworthiness of SA information provided by the MS at the CBN using Equation (2) as:  
\[\mathcal{T}_{COP}^{MS} = \mathcal{T}_{MS}\cdot\mathcal{T}_{CHN}\cdot\mathcal{T}_{GN} \cdot\mathcal{T}_{RN}\cdot\mathcal{T}_{DFN}\cdot\] \[\mathcal{T}_{l_{1}}\cdot\mathcal{T}_{l_{2}}\cdot\mathcal{T}_{l_{ 3}}\cdot\mathcal{T}_{l_{4}}\cdot\mathcal{T}_{l_{5}}\] \[= (0.5625)(0.625)(0.625)(0.625)(0.625)\times\] \[(0.49875)(0.49988)(0.49988)(0.49988)(0.49988)(0.49988)\] \[= 0.00267\]  
We clarify that this example can have different cases; for example, the CHN node could encrypt the information and could also calculate the message hash before sending it to the GN, and thus would result in a different trustworthiness score of the MS at the CBN. However, we present this case to illustrate how a trustworthiness score from an SA source appears at the CBN node. The trustworthiness score at the CBN node depends not only on the security capabilities of the SA information source but also on the devices and links along the path from the SA information source to the CBN.