Question

Problems 1. [40 pointsl Consider a three layer protocol in which Layer 3 encapsulates Layer 2 and Layer 2 encapsulates Layer 1. Assume minimalist headers with fixed length packets. Assume the following characteristics of the layers: Layer 1, 6 octet address length, 512 octet payload; Layer 2, 4 octet address length , 256 octet payload; Layer 3, 8 octet address length, 1024 octet payload. Note that in the minimalist header arrangement no error detection or correction will be used; however, there must be a scheme (that you must devise) to allow a multipacket datagram at each separate layer. You may assume that there is some sort of routing or other address translation protocol that wl identify which addresses are to be used. Assume that the data communications channel in use do form a data communications network. (Hint: do recall what is needed for a data communications network as contrasted with an arbitrary graph.) 1.1. For each of the three layers, separately calculated, how many items (nodes) can be addressed? Do not simply show an answer, but ex plain your reasoning (hint: combinatorics) 1.2. We have discussed functors as a formal, theoretical description be- tween layers of a network. In terms of these three layers, illustrate the functors between the layers, and explain how functors address the differences in topology and relevant information content at the different layers 1.3. Assume that the datagram has 5 layer 1 packets. How does this datagram encapsulate in layer 2? 1.4. Taking the above result, or a made-up one of your choosing if you cannot calculate such a result, how does this data from layer 2 e capsulating layer 1 data then encapsulate in layer 3?

Answer 1

11 Lets us calculate the number of nodes that can be addressed: Layer 1:-6 octet address length 6 8 address length = 48 bits [Since 1 octet contains 8]. So we see that for Layer 1 the Address bits are 48 in number. Hence No. of different nodes that can be calculated 248 The Range of Addresses identified will be [O to (248 - 1)]. Layer 2-4 octet address length 4*8 address length 32 bits [Since 1 octet contains 8]. So we see that for Layer 2 the Address bits are 32 in number. Hence No. of different nodes that can be calculated 232. The Range of Addresses identified will be [O to (232- 1)] Layer 3:-8 octet address length = 8 8 address length = 64 bits [Since 1 octet contains 8] . So we see that for Layer 3 the Address bits are 64 in number. Hence No. of different nodes that can be calculated 264 The Range of Addresses identified will be [O to (264 - 1)] 1.2 When the Datagram packets travel from one network to other they need to do so via Access Points situated at the boundaries of the connecting netwroks. These access points are also known as functors which functions by partitioning a given datagram packet say from Layer 1 into 2 fragments (IC-Interface Control Information and SDU = Servicve Data Unit). In ICI bits are added to check the headers of the packet when it travels via networks which contains information like Header Seq Header Size# and other relevant control Information required. In SDU bits are added to check for error detectiona and correction so that the data packets doesnt get corrupted when it travels via interconnecting networks. So, in our case we have functors between Layer 1 and Layer 2 and also in between Layer 2 and Layer 3 to guarantee corrupt free datagram packet movement via networks. Below is the attached image for how functors work between network layers:

e ocle: Adciess Legt 512 crtel ay oad The Data packet when accessed vla Service polnt Service Access Points i Functors LAYER 1 ICI Interface Control Information) SDU (Service Data Unit octe: Adcress Lengtr 256 ccel pay oad The Data packet when accessed vla Service polnt LAYER 2 iČl (Interface Control Information) SDUくService Data Unit) 1024 oc:e: pay sad δ ccel Acdiess Length LAYER 3 1.3 Lets Assume our Network Model of 5 layers as:- Layer 1 - Physical Layer Layer 2- Data Link Layer Layer 3 - Network Layer Layer 4 - Transport Layer Layer 5 - Application Layer So, in out layer 2 it performs the following:- Provides node-to-node data transfer (between two directly connected nodes) also handles error detection and correction from the physical layer, if any Two sublayers exist here as well - the Media Access Control (MAC) layer and the Logical Link Control (LLC) layer This way in a 5-Layer Network Model, Layer 2 can be encapsulated Please let me know in case of any clarifications required. Thanks!