(a) Calculate the efficiency of each layer. i. HTTP ii. TCP iii. IP iv. Ethernet (b) In practice, the data is much larger than 1 KB. Assume that the data is 6 KB and the payload of Ethernet is only 1500 bytes including its header. Recalculate the efficiency of each layer again. Note that Ethernet has to break large packets into 1500 bytes, hence, more overhead. i. HTTP ii. TCP iii. IP iv. Ethernet v. Ethernet vi. Ethernet vii. Ethernet In effect we are transmitting 6000 bytes but with 4 (128 + 20 + 20 + 18) overhead which results in 6000/4(1500 + 128 + 20 + 20 + 18) = 0.84 (c) Explain which layer is End-to-End (E2E) connection and which layer is Point-to-Point (P2P) connection and why? i. HTTP ii. TCP iii. IP iv. Ethernet (d) Explain which layer is connection-oriented and which layer is connection-less-oriented and why?. 5 pts i. HTTP ii. TCP iii. IP iv. Ethernet Show the detail of your work.

1. Consider the layers in TCP/IP reference model (Figure 1.22) of your text Assume that the user data is 1 KB, Sender Data HTTP Application Application HTTP SMTP RTP DNS Data HTTP Transport Transport Data HTTP Transport Packet Header Network Transport TCP UDP CRC Data HTTP Transport Packet Header Frame Header Data Link Layers Protocols 100101010110101100101010110101001010101001011 Physical Internet IP ICMP Link DSL SONET 802.11 Ethernet GRANDMETRIC WAICHE HTTP header is 128 bytes, TCP header is 20 bytes, IP header is 20 bytes, and Ethernet header is 18 bytes. Assume that the maximum payload for Ethernet is 1500 bytes excluding the header