add Notes04.md

Blatt03/Blatt03.md

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+![pc1?](https://lsky.mhrooz.xyz/2024/11/28/2145179ee22cf.png)
+
+![gre1](https://lsky.mhrooz.xyz/2024/11/28/c787403eb7794.png)

Blatt03/Notes03.md

@@ -1018,4 +1018,75 @@ vi) Angenommen, der Lehrstuhl verlangt hohe Gebühren für den Transit Ihrer Nac
 
 vii) Konfigurieren Sie nun Router4 anstatt Router1 als BGP-Router Ihres AS, wobei die externen AS weiterhin an Router1 angeschlossen bleiben. Erläutern Sie anhand Ihrer Beobachtungen, inwiefern sich die beiden Szenarien unterscheiden. Theorie: Nennen Sie mindestens 2 Gründe, die für ein solches Szenario sprechen würden.
  现在将 router4 配置为您自治系统的 BGP 路由器，而外部自治系统仍然连接到 router1。根据您的观察，说明这两种场景有何不同。
- 理论：列出至少两个支持这种场景的理由。
+ 理论：列出至少两个支持这种场景的理由。
+
+
+
+## Appendix
+
+### Mathematical Explanation of the Fragmentation Process
+
+#### Definitions of Symbols
+
+- **StotalS_{\text{total}}**: Total size of the IP packet (in bytes), Stotal=9000S_{\text{total}} = 9000.
+- **SheaderS_{\text{header}}**: Size of the IP header (in bytes), Sheader=20S_{\text{header}} = 20.
+- **SMTUS_{\text{MTU}}**: Size of the MTU (in bytes), SMTU=1500S_{\text{MTU}} = 1500.
+- **Sdata_per_fragmentS_{\text{data\_per\_fragment}}**: Maximum data size per fragment (in bytes).
+- **NfragmentsN_{\text{fragments}}**: Total number of fragments.
+
+#### Calculation Steps
+
+1. **Maximum data size per fragment**
+
+   Sdata_per_fragment=SMTU−SheaderS_{\text{data\_per\_fragment}} = S_{\text{MTU}} - S_{\text{header}}
+
+   Substituting values:
+
+   Sdata_per_fragment=1500−20=1480 bytesS_{\text{data\_per\_fragment}} = 1500 - 20 = 1480 \text{ bytes}
+
+2. **Total size of the data section**
+
+   Sdata_total=Stotal−SheaderS_{\text{data\_total}} = S_{\text{total}} - S_{\text{header}}
+
+   Substituting values:
+
+   Sdata_total=9000−20=8980 bytesS_{\text{data\_total}} = 9000 - 20 = 8980 \text{ bytes}
+
+3. **Total number of fragments** Each fragment can carry Sdata_per_fragmentS_{\text{data\_per\_fragment}} bytes of data. The total number of fragments required is:
+
+   Nfragments=⌈Sdata_totalSdata_per_fragment⌉N_{\text{fragments}} = \lceil \frac{S_{\text{data\_total}}}{S_{\text{data\_per\_fragment}}} \rceil
+
+   Substituting values:
+
+   Nfragments=⌈89801480⌉=⌈6.068⌉=7N_{\text{fragments}} = \lceil \frac{8980}{1480} \rceil = \lceil 6.068 \rceil = 7
+
+4. **Data size of the last fragment** The total data size carried by the first Nfragments−1N_{\text{fragments}} - 1 fragments is:
+
+   Sdata_used=(Nfragments−1)⋅Sdata_per_fragmentS_{\text{data\_used}} = (N_{\text{fragments}} - 1) \cdot S_{\text{data\_per\_fragment}}
+
+   Substituting values:
+
+   Sdata_used=(7−1)⋅1480=8880 bytesS_{\text{data\_used}} = (7 - 1) \cdot 1480 = 8880 \text{ bytes}
+
+   The data size of the last fragment is:
+
+   Sdata_last=Sdata_total−Sdata_usedS_{\text{data\_last}} = S_{\text{data\_total}} - S_{\text{data\_used}}
+
+   Substituting values:
+
+   Sdata_last=8980−8880=100 bytesS_{\text{data\_last}} = 8980 - 8880 = 100 \text{ bytes}
+
+5. **Validation of total size** The total size of all fragments, including the header, should equal the original IP packet size:
+
+   Stotal=Sdata_total+SheaderS_{\text{total}} = S_{\text{data\_total}} + S_{\text{header}}
+
+   Verification:
+
+   8980+20=9000 bytes8980 + 20 = 9000 \text{ bytes}
+
+#### Summary
+
+- Total number of fragments: Nfragments=7N_{\text{fragments}} = 7.
+- Data size per fragment:
+  - The first Nfragments−1=6N_{\text{fragments}} - 1 = 6 fragments carry Sdata_per_fragment=1480S_{\text{data\_per\_fragment}} = 1480 bytes of data.
+  - The last fragment carries Sdata_last=100S_{\text{data\_last}} = 100 bytes of data.