1. جلسه دهم شبکه های کامپیوتری به نــــــــــــام خدا

2. Link Utilization Error Free Operation Stop and Wait: Sliding Window (W=window Size=2 n -1): 2

3. Link Utilization Probability of frame error = p Stop and Wait Utilization Probability of a frame requiring exactly k transmission = p k-1 (1-p) Expected Number of Transmission for a frame (N r ): Utilization should be divided by N r : Selective Reject Utilization Exact same idea as stop and wait 3

4. Link Utilization Go back N Utilization Each frame error requires re-transmission of L packets where L>=1 Total number of frames that should be transmitted if the original frame must be transmitted k times=f(k)=1+(k-1)L 4

5. 5

6. Multiple Access Control Sublayer 6 shared wire or medium

7. Multiple Access Links 7 Two types of “links”: point-to-point – PPP for dial-up access – point-to-point link between Ethernet switch and host broadcast (shared wire or medium) – old-fashioned Ethernet – 802.11 wireless LAN shared wire (e.g., cabled Ethernet) shared RF (e.g., 802.11 WiFi) humans at a party (shared air, acoustical) shared RF (satellite)

8. 8 Point-to-point networks Point-to-point networks are those in which, when a message is sent from one computer to another, it usually has to be sent via other computers in the network.

9. 9 Broadcast networks Shared channel Broadcast networks have a single communication channel that is shared by all the machines on the network.

10. Channel Allocation 10 Channel Allocation problem: Single channel has to be shared among multiple users Solutions: Static allocation: user gets a channel for the whole “conversation” Dynamic allocation: channel capacity allocated for a user is not fixed; depends on usage

11. Static Channel Allocation 11 Using FDM or TDM to divide channel capacity among N users Inefficient bandwidth use: some users may be idle Users may be denied access even when other users do not use there allocated capacity Long time service delay T

12. Static Channel Allocation 12 Delay in a single M/M/1 Queue Channel can transfer C bits/sec Frames arrive randomly with an average rate of λ frames/sec and the inter-arrival times are exponentially distributed Frame length values has an exponential distribution with an average length of 1/µ bits. Mean delay of frames in the queue is given by: Example: C=100 Mbps, 1/µ=10000 bits, λ =5000 frames/sec, T=???? sec Delay if we divide the channel into N sections Assume we divide the same stream into N smaller streams and serve each of them in a separate queue There will be N channels with a capacity of C/N bits/sec for each There will be N streams with an average rate of λ /N frames/sec each Mean delay of each queue: Example: If we divide the above queue into 10 sections, =2

13. Dynamic Channel Allocation 13 In data communication, the peak to mean ratio of traffic can be very high. In many applications, users require the channel in random times Static allocation will be a waste of channel resources It is best to assign the channel based on user demand Different protocols for coordination: ALOHA : pure and slotted CSMA : carrier sense multiple access protocols Collision free protocols Limited contention protocols Wireless LAN protocols

14. Dynamic Channel Allocation 14 Assumptions for analysis of dynamic channel allocation algorithms: Station model: N independent stations generating frames within interval Δ t with probability = λ Δ t where λ is frame rate (frames/second) Single communication channel Collision occurs when 2 frames are transmitted simultaneously Time is continuous, or slotted (discrete intervals) Stations may have the hardware to notice: Collision of frames The presence of another signal on the channel (carrier sense)

15. 15 Pure Aloha Station transmits as soon as it has a frame If a collision is detected, it waits for a random time interval and then retransmits

16. ALOHA Efficiency 16 Notes and assumptions: Large number of stations Frames of equal length Transmit time for frame on the channel is normalized to unity Stations can sense channel collision and re-try transmission if there is a collision

17. ALOHA Efficiency 17 Analysis: All stations together, generate an average of S new frames per frame time If S>1, every frame suffers a collision, so for reasonable operation, 0<S<1 A station does not send a new frame until the old frame has been transmitted successfully. So, S can be seen as a measure of throughput as well. Probability of k attempts per frame time by all stations, old and new combined is Poisson with mean G per frame time. Low load => rare collision => S~G Increase the load => More collisions => G>S Throughput is then the offered load, G, times the probability of no collision, P 0 => S=G*P 0

18. ALOHA Efficiency 18 Probability of k frames generated per frame-time (t) is (Poisson distribution) Probability that there is no frame generated in a given interval t is No collision if no other frame generated in two frame-times. Mean number of frames generated during 2t sec is 2G so: ALOHA Efficiency: S=Ge -2G