Network Protocol Layers PDF

Transport Layer HEADER Note: Each layer introduces or presents services PC where the OS supports at the form of protocols which should be the TCP/IP Stack...

Transport Layer HEADER Note: Each layer introduces or presents services PC where the OS supports at the form of protocols which should be the TCP/IP Stack included in the header. NO Application Layer Message (which is going to be split) devices NO Transport Layer Network Data 1 Data 2 Data N devices HEADER Segment 1 Segment 2 Segment N MAC HEADER Segment 1 Segment 2 Segment N Routers Network Layer LLC Trailer Packet 1 Packet 2 Packet N Data link Layer  Two sublayers Bridges and Packet 1 Packet 2 Packet N (Media access control (MAC) and Switches Logical link control (LLC)) Frame 1 Frame 2 Frame N Raw bits 1 (0’s and 1’s) Raw bits 2 (0’s and 1’s) Raw bits 3 (0’s and 1’s) Repeaters Physical Layer Digital signal 1 Digital signal 2 Digital signal 3 and Hubs Some MAC header items: Source MAC Address Repeaters regenerate the digital signal Hubs are multiple ports repeaters and Destination MAC Address excluding the noise (two ports) Internetworking Link (channel, Media, Line) carries digital signal devices Some Network header items: Source IP Address and Destination IP Address Internetworking devices: Switches, repeaters, hubs, routers, bridges 1. Every PC has a network interface card (NIC). 2. Every NIC has a unique MAC address which is Owned by IP version 4: 32 bits set by manufacturer (48 bits). Local ISP IP version 6: 128 bits Interface A refers to Interface B refers to PSTU network JU network Router 1. Routers work at IP layer (LAYER 3) in which IP addresses exist. 2. Switches work at Data link layer (LAYER Aggregate Switch 2) in which MAC addresses exist. Router understands only IP addresses (Computer Center) Switch understands only MAC addresses School 1 (Main School 2 School 3 School 24 Switch) (Main Switch) (Main Switch) (Main Switch) EE (Main ME (Main IE (Main CPE (Main Switch) Switch) Switch) Switch) CPE 1 Switch CPE 2 Switch CPE 3 Switch CPE 4 Switch CPE 5 Switch PC1 PC2 PC20 PC1 PC2 PC20 PC1 PC2 PC20 PC1 PC2 PC20  Switching Table Port # MAC Addr Switch (Aware of MAC Hub (Aware of nothing) addresses) PORT 1 PORT 2 PORT 3 PORT 4 PORT 1 PORT 2 PORT 3 PORT 4 PC 1 PC 2 PC 3 PC 4 PC 1 PC 2 PC 3 PC 4 Hub regenerates the receiving signal and then broadcast it to all PC1 wants to send other ports. something to PC4 Unicast: One-to-one Multicast: One-to-many Broadcast: One-to-all Network Performance: Delay and throughput Delay consists of four types, namely, propagation delay, transmission delay, processing delay, as well as queueing delay. Propagation delay refers to the time elapsed from the moment any bit is transmitted till the moment it is received by the destination. Keep in mind that the propagation speed relies on the channel type. Dis tan ce ( m ) T prop  Velocity (Pr opagation speed , m / s ) Transmission delay refers to the time elapsed from the moment the first bit of the message is transmitted till the moment the last bit is transmitted. Message Length (bits) Ttrans  Rb ( Source data rate, bits per second (bps) ) Processing and queueing delays are non-deterministic delays as there are no specific formulas exist  Source (A) Switch Destination (B) A wants to send a message to B Timing Diagram A B S TTrans TProp End-to-End delay (TD) TTrans TProc TProp TTrans  Question: Suppose there is a 100-Mbps point-to-point link between the earth and the moon. The distance from the moon to the earth is approximately 385,000 km, and data travels over the link at the speed of light (3 × 108 m/s). A camera on the moon takes pictures on the earth and saves them in digital format to disk. If the Mission Control on earth wants to download the most current image of size 5 × 106 bytes. What will be the amount of time that will elapse between the moment when the request goes out and the moment when transfer is completed? Earth MOON 3.85 x108 ( m ) TProp T prop   1.28 sec End-to-End 8 delay (TD) 3x10 m / s TTrans TProp 40x10 6 (bits) Ttrans  6  0.4 100 x10 bps Assuming that the request is very short message (i.e., of length equals one bit) TD =2x1.28+0.4=2.96 Question: Hosts A and B are each connected to a switch S via 10-Mbps links as shown in the figure below. The propagation delay on each link is 20 μs. S is a store-and-forward device; it begins retransmitting a received packet 35 μs after it finished receiving it. Calculate the total time required to transmit 10,000 bits from A to B (a) As a single packet; (b) As two 5,000-bit packets sent one right after the other. 10-Mbps link 10-Mbps link Source (A) Switch Destination (B) TProp = 20 μs TProp = 20 μs Message length: 10,000 bits A store-and-forward switch; it begins retransmitting a received packet 35 μs after it finished receiving it (Tproc) a) TD = 2075μs A B S TTrans TProp =20μs End-to-End delay (TD) TTrans = 10000/10x10^6bps=1000μs TProc = 35 μs TProp TTrans  A B S TTrans TProp =20μs End-to-End delay (TD) TTrans = 5000/10x10^6bps=500μs TProc = 35 μs TProp =20μs TTrans = 5000/10x10^6bps=500μs TTrans = 5000/10x10^6bps=500μs b) TD = 1500+75=1575μs Throughput Is the rate at which the data is reliably delivered to the destination (bps). Goal: Is to maximize the throughput of the link (channel). Summery: Channel capacity (bps) Message Length (bits) Throughput  End  to  end delay (sec) Link efficiency (unitless): Ttrans (sec) Link efficiency  End  to  end delay (sec) Bandwidth of the link: Is the range of frequencies that can go through the channel without degradation (W in Hz). Bandwidth of the signal: Is the range of frequencies that are available in the signal (B in Hz). Shannon’s theorem (Channel capacity, C, which is the maximum data rate that can be supported by the channel): Signal power Rb  C  W log 2 1  SNR  SNR  Noise power Question: If we want to send data at a rate of 8000 bps through a channel with bandwidth of 1000 Hz. (a) What is the minimum SNR (signal to noise ratio) required? Answer: (a) 8000 = 1000 log2(1+ S/N), SNR = 256 –1 = 255. How bits “0” and “1” are mapped into a digital signal? Encoding scheme Mapping Three encoding schemes: NRZ 1: Level High, 0: Level Low 1. Non- return to zero (NRZ) NRZ-I 1: There is a transition of the current state 2. NRZ- Inverted (NRZ-I) 0: No transition 3. Manchester Manchester 1: 0: Note: Both NRZ-I and Manchester are synchronized with system clock.  Question: Show the NRZ, Manchester, and NRZI encoding for the bit sequence shown in the figure below. Assume that the NRZI signal starts out as “low”. 1100101010101100 Bits 1 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 CLK NRZ NRZ-I MAN CHES TER Multiplexing Frequency division multiplexing (FDM, analog transmission) Time division multiplexing (TDM, digital transmission) FDM Terms and conditions: 1. Bandwidth of the medium (channel) exceeds the bandwidth of the signal 2. Each signal is modulated into a different carrier frequency 3. The carrier frequencies are separated so signals do not overlap. TDM Conditions: 1. Data rate of the medium exceeds the data rate of the signal (digital) to be transmitted. Digital transmission has several advantages over analog transmission: 1. Analog circuits require amplifiers, and each amplifier adds distortion and noise to the signal. In contrast, digital amplifiers regenerate an exact signal, eliminating cumulative errors. In other words, repeaters take out cumulative problems in transmission. Thus, can transmit longer distances. 2. Easy to apply encryption to digital data 3. Better integration if all signals are in one form. Can integrate voice, video and digital data. 4. Easier to multiplex large channel capacities with digital 5. Higher bandwidth 6. Lower error rate Link Categories Simplex: where signals travel in only one direction Half duplex: where signals travel in both directions but only one direction at a time. Full duplex: where signals travels in both directions and at the same time.