Notes on System performance. (Check these references. Jeff Buzen: Fundamental laws of computer system performance, ACM SIGMETRICS.P. Denning and J. Buzen: The Operational Analysis of Queuing Network Models, ACM Computing Surveys, September 1978.http://www.cs.washington.edu/homes/lazowska/qsp/Images/Chap_03.pdf )By a system, in this case, we are referring to those that could be modeled as queuing systems. A modern computer system is one such example, a service center with multiple servers is another kind, etc. In this case, the customers desiring and receiving service from systems are called clients, customers, jobs, task, requests, interactions, etc. (depending on the context).Consider a dynamic system S. It is dynamic because its states (system states) are changing in time. Even if the system is non-determinate it is still possible to talk about itstime-average performance measures. Suppose, the system has been under observation for a period of T secs and we observe the following:the total number of customer arrival during T: )(TAthe total number of completions observed during T: )(TCthe busy time for the system during T: )(TBAvg. service time received by a client (or a task), )()(TCTBS Avg. system throughput, TTCX)( System utilization during time T, TTBU)(From these, we see that XSU  (1)  The first fundamental law serverserverClient Queue(s) CompletionsClient arrivalsSystem SAlso, assume that when the system is in equilibrium the number of customers or clients trapped within the system (awaiting for services, getting services, etc.) is .N Secondly, we assume that on an average a client spends Rsecs in the system, the residency time. Little’s Law (Residency Law)Consider a system at equilibrium with a count of N clients inside it on the average. Suppose a client, say C, now arrives the system at time at. If he looks back he will see no other client behind him since he is the last one to join the queue. Suppose, after Runits time at Rttac he gets out of the system. As he now takes a look behind him he will see on an average N customers behind him which took all these Runits of time to congregate. Since the system is in equilibrium, the arrival rate of the customers into it is precisely its throughput Xcustomers per unit time. Therefore, XRN  (2)  The second fundamental lawThis is Little’s law that connects customer’s residency time with customer-density within the system. This law is very robust and works in all types of queuing organizations (independent of queue disciplines employed within the system).Little’s Law extended to Interactive systemsConsider an interactive system as depicted below where response from a terminal must be considered part of the overall process of interaction. At the terminal, we assume the average user spends Zunits of sleep-time. InteractiveUser at thisTerminalSystem user Interacts withSleep time = ZAn average customer spends all total R units of residency time. This could be seen as )( timeResidence systemRZNXR Therefore, the response time at the system ZNXsystemR )(. If there is no confusion, we could write it as ZNXR  (3) Interactive Response Time remembering though that Rhere is the response time of a system. Forced Flow laws We will expand the system to include a number of devices within it. Each device may be a queuing device, may be just a counter through which a customer just transits through, etc.Suppose, for the device k, )(TCk= total number of completions within the observed T units of time. We consider visit count of the kth device to be )()(TCTCvkk(4)And the device throughput to be TTCXkk)(. This showskkXvX  (5) Forced Flow LawEvery time a client visits a device it does receive a service from it lasting, we assume, ksunits of time per visit. Thus, the total service received by the client at that device during its tenure in the system amounts to somekkksvD  (6a)Multiplying by Xfrom both sides, we getkkkkUsXXD  (6b)as the utilization rate at the device k.Examples. 1. Each job generates 50 disk requests. The disk utilization rate is 35%. The mean service time at the disk is to be determined given that the overall interactive response time is 1 sec with 25 terminals in the system and an average sleep time of 19 secs. Let s = disk-service time per visit. We have sUXXvdiskdiskdiskTherefore, Xs 50/35.0. Now from response time law: 19251 X and .20/25XSubstituting this above, we get ss 0056.0 Bounds on performance measureConsider a device k in an interactive system servicing clients. Since its utilization is limited from above, we have1kkUXDAnd this implies the system throughput X to be bounded askDX1min or max1DX  (7a)Let kkDDbe the total service received by an average client from all the devices in the system. Now, minimally a customer needs to spend a total of )( ZD units of time within the system. If during this time the system could deliver service to all its Nclients (via maximum service parallelism the system can provide), we could receive maximum system throughput in that case. Therefore, system throughput X will always be boundedfrom above byZDNX (7b)Combining (7a) and (7b), we get an upper bound for system throughput asZDNDX ,1minmax (7c)Notice that the lower bound for Xcould be ascertained as follows. The worst possiblesystem configuration for throughput would be the situation where all jobs or tasks are processed in sequence without any service overlap or any built-in system parallelism. In that case, the total set of N clients would spend )( ZND  units of time to flush out N clients resulting in a throughput ZNDN. Therefore,  ZNDNZDNDX ,1minmax (8a)Taking reciprocals, we get NZDDXNZND,max 1 maxMultiplying this by N and then subtracting Z from it, we get ),max( maxDZNDRND  (8b)as the interactive response time bound.Note that the optimum number of active clients (beyond which system performance deteriorates exponentially) is *Nwhere max*DZDNExample. NXmax/1 D)/( ZDN 1. Consider a system with a single CPU and a single disk serving a set of terminals where the average sleep time is Z=35 secs. Some possible configurations are envisaged as shown below. The three pairs of (diskcpuDD ,) are considered: (system