Part II System Performance Modeling basic concepts operational

Part II System Performance Modeling: basic concepts, operational analysis (book, chap. 3) 1

Learning Objectives (1) �Introduce basic queuing concepts and notation. �Present communication-processing delay diagrams. �Discuss examples of service times and service demands. 2

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. 3

A Resource and its Queue Wi Si LINE Resource i Customers Si : service

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

Communication-Processing Delay Diagram Client C S 1 ccpu LAN Server S [r, m 1]

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

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

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. 9

A C/S System: solution • What is the minimum response time? Rr Dclient +

Calculating the Service Demands (1) • At the client – Dclient = Dccpu = 5 msec • At the network – Dnetwork = (m 1 + 7 m 2) / B • m 1 = m 2 = 1, 518 bytes • B = 10 Mbps – Dnetwork = 0. 0097 msec Ó 1998 Menascé & Almeida. All Rights Reserved. 11

Calculating the Server’s Service Demands (2) • Dserver = Dcpu + Ddisk • Ddisk = 10 * Sdisk = Avg. Seek + Avg. Latency + Transfer. Time = 0. 009 + 0. 0047 + 2, 048/20, 000 = 0. 0133 sec Ddisk = 10 * Sdisk = 0. 133 sec • Dcpu = 0. 010 sec • Dserver = 0. 143 sec Ó 1998 Menascé & Almeida. All Rights Reserved. 12

A Web Server and its Queues Sdisk 1 Vdisk 1 requests/sec Scpu Vcpu Xo

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

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

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

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? ULAN = XLAN * SLAN = 1, 000 * 0. 00015 = 0. 15 = 15% Ó 1998 Menascé & Almeida. All Rights Reserved. 18

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

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? Xserver = 7, 200 / 3, 600 = 2 tps Xdisk = Vdisk * Xserver = 4. 5 * 2 = 9 tps Udisk = Xdisk * Sdisk = 9 * 0. 02 = 0. 18 = 18% Ó 1998 Menascé & Almeida. All Rights Reserved. 20

Basic Performance Results Service Demand Law • The service demand Di is related to the system throughput and utilization by the following: Di = Vi * Si = (Xi/Xo)(Ui/Xi) = Ui / Xo Ó 1998 Menascé & Almeida. All Rights Reserved. 21

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? Ucpu = 90% Xserver = 30, 000 / (10*60) = 50 requests/sec Dcpu = Vcpu * Scpu = Ucpu / Xserver = 0. 90 / 50 = 0. 018 sec Ó 1998 Menascé & Almeida. All Rights Reserved. 22

Basic Performance Results Little’s Law N X R • 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 Ó 1998 Menascé & Almeida. All Rights Reserved. 23

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 (Nreq) was 3. • What was the average response time per NFS request at the server? “black box” = NFS server Xserver = 10, 800 / 1, 800 = 6 requests/sec Rreq = Nreq / Xserver = 3 / 6 = 0. 5 sec Ó 1998 Menascé & Almeida. All Rights Reserved. 24

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 Npackets = Rpacket * Xnetwork Npackets = 0. 05 * 512 = 25. 6 packets Ó 1998 Menascé & Almeida. All Rights Reserved. 25

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 = Xdisk = 20 requests/sec Srequest = 0. 00802 sec Udisk = Srequest * Xdisk = 0. 00802 * 20 = 16. 04% Ó 1998 Menascé & Almeida. All Rights Reserved. 26

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 27