1. تجارت الکترونیک سیار جلسه هفتم مدرس : دکتررامین کریمی

2. Hit Rate and Miss Rate Hit Rate : تعداد درخواست هایی که در کش موجود می باشد. Miss Rate : تعداد درخواست هایی که در کش موجود نمی باشد.

3. 4-3 Possible solution increase bandwidth of access link to, say, 10 Mbps Consequences utilization on LAN = 15% utilization on access link = 15% total delay = Internet delay + access delay + LAN delay = 2secs + 20msecs + 20msecs often a costly upgrade Caching Example (2) origin servers public Internet institutional network 10 Mbps LAN 10 Mbps access link Traffic Intensity on the access link = 15*100,000/10Mbps = 15% Traffic Intensity on the LAN = 15*100,000/10Mbps = 15%

4. 4-4 Install cache suppose hit rate is 0.4 Consequence 40% requests will be satisfied almost immediately 60% requests satisfied by origin server utilization of access link reduced to 60%, resulting in delays (say 160 msec) total delay = Internet delay + access delay + LAN delay = 0.6×(2secs+160msecs+20msecs) + 0.4×20msecs = 1.316 secs Caching Example (3) origin servers public Internet institutional network 10 Mbps LAN 1.5 Mbps access link institutional cache Traffic Intensity on the access link = 0.6*15*100,000/1.5Mbps = 60% Traffic Intensity on the LAN = 15*100,000/10Mbps = 15%

5. 3-5 Communication-Processing Delay S 2 SLAN W 1 Scpu S 1 Scpu W 2 SLAN W 1 Sio S 1 Ccpu S 2 Ccpu S 1 Sio time Client (C) Server (S) LAN B[bps] time Figure 3.1. Communication-Processing delay diagram for 2-tier C/S system. RrRr [r,m 1 ] [r,m 2 ] [r th response, m 2 bytes] [r th request, m 1 bytes]

6. The Times Service Time during j th visit: S j i Waiting Time during j th visit: W j i Residence Time during j th visit: R’ i Resource i SjiSji WjiWji R’ i R r = R’ Ccpu + R’ Scpu + R’ Sio + R’ LAN R r = response time of a request r R’ Ccpu = residence times at the client cpu R’ Scpu = residence times at the server cpu R’ Sio = residence times at the server io R’ LAN = residence times at the LAN (3.2.5) 3-6

7. Residence Time D LAN = 8(m 1 + m 2 )/B m 1 = request length [B] m 2 = reply length [B] Service Demand (sum of all service time for a request at source i): D i D i = Σ j S j i (3.2.2) Queuing Time (sum of all waiting time for a request at source i): Q i Q i = Σ j W j i (3.2.3) R’ i = D i + Q i (3.2.4) R r = Σ i R’ i (3.2.5) Q Scpu = W 1 Scpu + W 2 Scpu Queuing time at server’s cpu Service demand at server’s cpu D Scpu = S 1 Scpu + S 2 Scpu3-7

8. Example 3.1 S 2 Scpu W 1 Scpu S 2 Scpu W 2 Scpu W 1 Sio S 1 Ccpu S 2 Ccpu S 1 Sio time Client (C)Server (S) LAN B[bps] time RrRr [r,m 1 ] [r,m 2 ] S 1 Ccpu = 5ms, S 1 Scpu = 10ms AvgSeek = 9ms (Server’s Disk) AvgLatency = 4.17msec (Server’s Disk) TransferRate = 20 MB/s (Server’s Disk) DataReads = 10×2048 Byte S d = 9+4.17+2048/20M = 13.3 ms (Disk Average Service Time) D d = 10×S d = 133 ms (Service Demand at Server’s Disk) m 1 = 1,518, m 2 = 7×1,518 Bytes D LAN = 8(m 1 +m 2 )/10Mbps= 9.7ms R t > D Ccpu + D Scpu + D d + D LAN =158 ms The minimum value for response time to transaction t; all waiting times are ignored. 3-8

9. Three-Tier LAN 1 B 1 [bps] LAN 2 B 2 [bps] WAN Application Sever (Business Tier) Client Database Sever (Data Access Tier) Figure 3.3. Three-tier C/S system. 3-9

10. 3-Tier Delay diagram Figure 3.4. Communications-processing delay diagram for a 3-tier C/S system. ClientLAN1 Application Server LAN1LAN2WAN Database Server [r,m 1 ] [r,m 2 ] ● ● ● S cpu W net S cpu W net Q cpu D cpu Q cpu D cpu Q net Q cpu D cpu D disk Q disk RrRr 3-10