Ó 1998 Menascé & Almeida. All Rights Reserved.1 Part II System Performance Modeling: basic concepts, operational analysis (book, chap. 3)

Ó 1998 Menascé & Almeida. All Rights Reserved.2 Learning Objectives (1) Introduce basic queuing concepts and notation. Present communication-processing delay diagrams. Discuss examples of service times and service demands.

Ó 1998 Menascé & Almeida. All Rights Reserved.3 Learning Objectives (2) Discuss operational analysis: utilization law forced flow law service demand law Little’s law Present examples that show the use of operational analysis to model Web performance

Ó 1998 Menascé & Almeida. All Rights Reserved.4 A Resource and its Queue Resource i LINE Customers WiWi SiSi S i : service time W i : waiting time resource: CPU, disk, network, etc.

Ó 1998 Menascé & Almeida. All Rights Reserved.5 Queuing Basic Concepts Total time spent by a request during the j th visit to a resource i: –Service time (S i j ): period of time a request is receiving service from resource i, such as CPU or disk. –Waiting time (W i j ): the time spent by a request waiting access to resource i

Ó 1998 Menascé & Almeida. All Rights Reserved.6 Communication-Processing Delay Diagram RrRr S 1 ccpu S 2 ccpu Client CLANServer S W 1 scpu S 1 scpu W 1 sio W 2 scpu S 2 scpu S 1 sio [r,m1] [r,m2]

Ó 1998 Menascé & Almeida. All Rights Reserved.7 Basic Queuing Concepts Service Demand (D i ) is the sum of all service times for a request at resource i D scpu = S 1 scpu + S 2 scpu Queuing Time (Q i ) is the sum of all waiting times for a request at resource i Q scpu = W 1 scpu + W 2 scpu

Ó 1998 Menascé & Almeida. All Rights Reserved.8 Basic Queuing Concepts Residence Time (R’ i ) at resource i is the sum of service demand plus queuing time. R’ i = Q i + D i Response time (R r ) of a request r is the sum of that request’s residence time at all resources. R server = R’ cpu + R’ disk

Ó 1998 Menascé & Almeida. All Rights Reserved.9 A C/S System: example Consider that a transaction t in a C/S system uses 5 msec of CPU at the browser, 10 msec of CPU at the server, and reads ten 2048-byte blocks from the server’s disk. The average seek time at the disk is 9 msec, the average latency is 4.17 msec and the transfer rate is 20 MB/sec. Consider that the client and server are connected by a 10 Mbps Ethernet and that a request going from the client to the server takes a full packet (1,518 bytes) and the reply requires 7 packets.

Ó 1998 Menascé & Almeida. All Rights Reserved.10 A C/S System: solution What is the minimum response time? R r  D client + D network + D server we are ignoring all waiting times.

Ó 1998 Menascé & Almeida. All Rights Reserved.11 Calculating the Service Demands (1) At the client –D client = D ccpu = 5 msec At the network –D network = (m 1 + 7m 2 ) / B m 1 = m 2 = 1,518 bytes B = 10 Mbps –D network = msec

Ó 1998 Menascé & Almeida. All Rights Reserved.12 Calculating the Server’s Service Demands (2) D server = D cpu + D disk D disk = 10 * S disk S disk = AvgSeek + AvgLatency + TransferTime = ,048/20,000,000 = sec D disk = 10 * S disk = sec D cpu = sec D server = sec

Ó 1998 Menascé & Almeida. All Rights Reserved.13 A C/S System: solution What is the minimum response time? R r  D client + D network + D server R r  = sec.

Ó 1998 Menascé & Almeida. All Rights Reserved.14 A Web Server and its Queues Disk 2 Disk 1 CPU requests/sec XoXo S cpu V cpu S disk2 V disk2 S disk1 V disk1 Web Server

Ó 1998 Menascé & Almeida. All Rights Reserved.15 A Web Server and its Queues: parameters and notation (1) V i : average number of visits to queue i by a request; S i: average service time of a request at queue i per visit to the resource; i average arrival rate of requests to queue i D i service demand of a request at queue i, D i = V i x S i

Ó 1998 Menascé & Almeida. All Rights Reserved.16 A Web Server and its Queues: parameters and notation (2) N i: average number of requests at queue i, waiting or receiving service from the resource X i: average throughput of queue i, i.e. average number of requests that complete from queue i per unit of time X o: average system throughput, defined as the number of requests that complete per unit of time.

Ó 1998 Menascé & Almeida. All Rights Reserved.17 Basic Performance Results Utilization Law The utilization (U i ) of resource i is the fraction of time that the resource is busy. U i = X i * S i = i * S i

Ó 1998 Menascé & Almeida. All Rights Reserved.18 Utilization Law: example A network segment transmits 1,000 packets/sec. Each packet has an average transmission time equal to 0.15 msec. What is the utilization of the LAN segment? U LAN = X LAN * S LAN = 1,000 * = 0.15 = 15%

Ó 1998 Menascé & Almeida. All Rights Reserved.19 Basic Performance Results Forced Flow Law By definition of the average number of visits V i, each completing request has to pass V i times, on the average, by queue i. So, if X o requests complete per unit of time, V i *X o requests will visit queue i. X i = V i * X o

Ó 1998 Menascé & Almeida. All Rights Reserved.20 Forced Flow Law: example Database transactions perform an average of 4.5 I/O operations on the database server. During a one-hour monitoring period, 7,200 transactions were executed. What is the average throughput of the disk? If each I/O takes 20 msec on the average, what is the disk utilization? X server = 7,200 / 3,600 = 2 tps X disk = V disk * X server = 4.5 * 2 = 9 tps U disk = X disk * S disk = 9 * 0.02 = 0.18 = 18%

Ó 1998 Menascé & Almeida. All Rights Reserved.21 Basic Performance Results Service Demand Law The service demand D i is related to the system throughput and utilization by the following: D i = V i * S i = (X i /X o )(U i /X i ) = U i / X o

Ó 1998 Menascé & Almeida. All Rights Reserved.22 Service Demand Law: example A Web server running on top of a Unix system was monitored for 10 minutes. It was observed that the CPU was 90% busy during the monitoring period. The number of HTTP requests counted in the log was 30,000. What is the CPU service demand of an HTTP request? U cpu = 90% X server = 30,000 / (10*60) = 50 requests/sec D cpu = V cpu * S cpu = U cpu / X server = 0.90 / 50 = sec

Ó 1998 Menascé & Almeida. All Rights Reserved.23 Basic Performance Results Little’s Law The average number of customers in a “black box” is equal to average time each customer spends in the “box” times the throughput of the “box”. N = R * X X R N

Ó 1998 Menascé & Almeida. All Rights Reserved.24 Little’s Law: example (1) An NFS server was monitored during 30 min and the number of I/O operations performed during this period was found to be 10,800. The average number of active requests (N req ) was 3. What was the average response time per NFS request at the server? “black box” = NFS server X server = 10,800 / 1,800 = 6 requests/sec R req = N req / X server = 3 / 6 = 0.5 sec

Ó 1998 Menascé & Almeida. All Rights Reserved.25 Little’s Law: example (2) The average delay experienced by a packet when traversing a network segment is 50 msec. The average number of packets that cross the network per second is 512 packets/sec (network throughput). What is the average number of packets in transit in the network? “black box” = network segment N packets = R packet * X network N packets = 0.05 * 512 = 25.6 packets

Ó 1998 Menascé & Almeida. All Rights Reserved.26 Little’s Law: example (3) The disk of a Web server receives requests at a rate of 20 requests/sec. The average disk service time, considering both random and sequential requests, is 8.02 msec. What is the average disk utilization? “black box” = disk disk = X disk = 20 requests/sec S request = sec U disk = S request * X disk = * 20 = 16.04%

Ó 1998 Menascé & Almeida. All Rights Reserved.27 Part II: Summary Basic Concept of Queuing Theory and Operational Analysis terminology and notation service time and service demand waiting time and queuing time Basic Performance Results and Examples utilization law forced flow law service demand law Little’s Law