IDENTIDAD-LENGUA-COMUNIDAD SORDA

Example of Forced Flow Law

Question: Suppose we are given the network shown in Figure6.12. What are the visit ratios? That is, what are_{E [V}a],E [Vb], andE [Vcpu]?

Disk a

Disk b CPU

N = 10

Figure 6.12. Calculating the visit ratios.

Answer: Although it may seem obvious for this example, let’s work it out formally

because later exercises may be more complicated. Looking at the figure, we see

Ca= Ccpu· 80/181

Cb = Ccpu· 100/181

C = Ccpu· 1/181

Ccpu= Ca+ Cb + C.

Dividing through byC(the number of system completions) yields the visit ratios. So we get

E [Va] = E [Vcpu]· 80/181 E [Vb] = E [Vcpu]· 100/181

1 = E [Vcpu]· 1/181

E [Vcpu] = E [Va] + E [Vb] + 1. Solving this system of simultaneous equations yields

E [Vcpu] = 181

E [Va] = 80 E [Vb] = 100.

6.9 Combining Operational Laws

Simple Example

Suppose we have an interactive system with the following characteristics: r _{25 terminals (N = 25)}

r _{18 seconds average think time (E [Z] = 18)}

r _{20 visits to a specific disk per interaction on average (E [V}

108 little’s law and other operational laws r _{30% utilization of that disk (ρ}

disk = .3)

r _{0.025 sec average service time per visit to that disk (E [Sdisk] = .025)}

That is all the information we have. We are not told anything else about the rest of the system.

Question: What is the mean response time,E [R]?

Answer:

1. _{E [R] =} N

X − E [Z], but we still needX.

2. X = Xdisk

E [Vdisk]

, but we still needXdisk. 3. Xdisk =

ρdisk

E [Sdisk], both of which we know.

Working backward, we calculate

Xdisk = ρdisk E [Sdisk] = 12requests/sec ⇒ X = Xdisk E [Vdisk] = .6interactions/sec ⇒ E [R] = N X − E [Z] = 23.7sec. Harder Example

This is a case study taken from Lazowska et al., p. 49 [117]. We are told the following information only:

r _{It is an interactive system.}

r _{The central subsystem consists of a CPU and three disks.}

r _{Swapping may occur between interactions, causing a user to lose her memory} partition. Thus a request sometimes has to queue up in the memory queue to get back its memory partition before entering the central subsystem, but sometimes can skip this queue.

Figure6.13is a sketch of what the system looks like based on this information. Observe that some jobs have to wait in the “get memory queue,” whereas others already have the prerequisite memory allocated and can skip over this part and go directly to the central subsystem. We are not given information as to the fraction of jobs that go each way. Here are the measurements that were collected about this system:

r _{number of time-sharing users (N = 23)}

r _{average think time per user (E [Z] = 21seconds)} r _{system throughput (X = 0.45interactions per second)}

r _{average number of requests trying to get memory (E [Ngetting memory] = 11.65)} r _{average number of visits to the CPU per interaction (E [V}

cpu] = 3)

6.9 combining operational laws 109

Disk a

Disk b

Disk c CPU

Get memory queue

Central subsystem

Figure 6.13. A system with a memory queue.

Question: What is the average amount of time that elapses between getting a memory

partition and completing the interaction?

Answer: This question asks us for the expected time that jobs spend in the central

subsystem. There are several ways to answer it. Here is one:

E [Time in central subsystem_{] = E [}Response Time]− E [Time to get memory]

It is true that not every job has to go get memory, but you can think of E[Time to get memory]as the expected time to go from the point right before the split in the figure to right after the join in the figure.

Now by the Response Time Law,

E [Response Time] = N

X − E [Z] =

23

0.45 − 21 = 30.11sec.

Furthermore,

E [Time to get memory] = E [Number getting memory]

X =

11.65

0.45 = 25.88sec.

Thus,

E [Time in central subsystem] = E [Response Time]− E [Time to get memory] = 30.11− 25.88

= 4.23sec.

Question: What is the CPU utilization? Answer:

110 little’s law and other operational laws

6.10 Device Demands

We end with one final law, called the Bottleneck Law. This law is very important in answering “what-if” type questions about systems, which come up in the next chapter. DefineDito be the total service demand on deviceifor all visits of a single job (i.e., a single interaction). That is,

Di= Vi  j = 1

S_i(j ),

whereS_i(j ) is the service time required by thejth visit of the job to serveri. We immediately see by (3.3) that

E [Di] = E [Vi]· E [Si] , provided that Vi and theS

(j )

i ’s are independent. That is, we are assuming that the number of visits a job makes to deviceiis not affected by its service demand at the device.

We will soon discuss the importance of theseDi’s. First, let’s observe how easy Di typically is to measure. Suppose we had to measure the Vi’s. This would be hard, because we would have to keep track of a particular job and count its visits to device

i. If deviceiis time-shared among jobs, it would be even harder. Luckily, we do not have to do this!

Question: How would you determineE [Di]in practice? Answer: Consider a long observation period. Observe that

E [Di] =

Bi

C,

whereBi is the busy time at deviceifor the duration of our observation period and

Cis the number of system completions during this observation period. These are very easy measurements to get.

The importance of_{E [D}i]lies in the following law, which we call the Bottleneck Law:

ρi = X· E [Di]

Question: Can you explain the Bottleneck Law intuitively?

Answer:Xis the jobs/sec arriving into the whole system. Each of those outside arrivals into the system contributesE [Di]seconds of work for devicei. So deviceiis busy for

X· E [Di]seconds out of every second (e.g., deviceimight be busy for half a second out of every second). ThusX· E [Di]represents the utilization of devicei.

Here is a proof of the Bottleneck Law:

6.12 exercises 111

Example

As a simple example, consider a system with an outside arrival rate of 3 jobs/second. Suppose that each job, on average, visits the disk 10 times. Suppose that each visit to the disk takes 0.01 second on average. Then, the per-job demand on the disk is

.1seconds, and the utilization of the disk is.3.