91Ó°ÊÓ

Consider a computer system with a CPU and one disk drive. After a burst at the \(\mathrm{CPU}\) the job completes execution with probability \(0.1\) and requests a disk I/O with probability \(0.9\). The time of a single CPU burst is exponentially distributed with mean \(0.01 \mathrm{~s}\). The disk service time is broken up into three phases: exponentially distributed seek time with mean \(0.03 \mathrm{~s}\), uniformly distributed latency time with mean \(0.01 \mathrm{~s}\), and a constant transfer time equal to \(0.01 \mathrm{~s}\). After a service completion at the disk, the job always requires a CPU burst. The average arrival rate of jobs is \(0.8 \mathrm{job} / \mathrm{s}\) and the system does not have enough main memory to support multiprogramming. Solve for the average response time using the \(M / G / 1\) model. In order to compute the mean and the variance of the service time distribution, you may need the results of the section on random sums in Chapter 5.

Step by step solution

01

Determine CPU Service Time

The CPU burst time is exponentially distributed with a mean of 0.01s. Therefore, the first service time mean is \[ E[S_{cpu}] = 0.01s \].

02

Determine Disk Service Time Components

The disk service time consists of seek time, latency time, and transfer time. The seek time is exponentially distributed with a mean of 0.03s, uniformly distributed latency with mean 0.01s, and a constant transfer time of 0.01s.

03

Calculate Mean Disk Service Time

Total mean disk time is the sum of the individual times:\[ E[S_{seek}] + E[S_{latency}] + E[S_{transfer}] = 0.03s + 0.01s + 0.01s = 0.05s \].

04

Calculate Variances for Each Time Component

The variance for seek time (exponential):\[ \text{Var}[S_{seek}] = (0.03)^2 \].For latency time (uniformly distributed from 0 to 2x0.01):\[ \text{Var}[S_{latency}] = \frac{(0.02 - 0)^2}{12} = 0.00001667 \].The variance for transfer time is 0 since it is a constant.

05

Calculate Total Variance for Disk Service Time

The total variance is the sum of the individual variances:\[ \text{Var}[S_{disk}] = \text{Var}[S_{seek}] + \text{Var}[S_{latency}] + \text{Var}[S_{transfer}] \].\[ \text{Var}[S_{disk}] = 0.0009 + 0.00001667 + 0 = 0.00091667 \].

06

Calculate Mean and Variance of Service Time

The expected service time for each cycle (CPU + Disk I/O):\[ E[S] = E[S_{cpu}] + E[S_{disk}] = 0.01s + 0.05s = 0.06s \].Variance of the cycle:\[ \text{Var}[S] = \text{Var}[S_{cpu}] + \text{Var}[S_{disk}] \].\[ \text{Var}[S] = (0.01)^2 + 0.00091667 = 0.0001 + 0.00091667 = 0.00101667 \].

07

Calculate Traffic Intensity \( \rho \)

Traffic intensity is given by \( \rho = \lambda E[S] \). With \( \lambda = 0.8 \) jobs/second and \( E[S] = 0.06 \)s, we have:\[ \rho = 0.8 \times 0.06 = 0.048 \].

08

Calculate the Mean Response Time with \( M/G/1 \) Queue

The mean response time \( R \) is given by: \[ R = \frac{E[S]}{1 - \rho} + \frac{\text{Var}[S]}{2 E[S] (1 - \rho)} \].Substitute the values:\[ R = \frac{0.06}{1 - 0.048} + \frac{0.00101667}{2\times 0.06 \times (1 - 0.048)} \].Calculate each part separately and sum them:\[ R \approx 0.06284 + 0.00878 \approx 0.07162 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

queueing theory

Queueing theory is the mathematical study of waiting lines, or queues. In computer science and operations research, it models systems like CPU processes or network servers. This theory predicts queue lengths and waiting times, helping optimize system performance.
In our exercise, we're dealing with a queue composed of a CPU and a single disk drive. By understanding how jobs (tasks) move between the CPU and the disk or exit the system, we can improve efficiency and predict how long tasks will take.
Queueing theory gives us formulas and models, like the M/G/1 queue, to find these performance indicators. The M/G/1 model deals with a system having a single server where arrivals are memoryless (Markovian, M), service times follow a general distribution (G), and there is one server (1).
Using this model helps us understand the average response time, the average number of jobs in the system, and the probability distributions of these quantities.

service time distribution

Service time distribution refers to how long it takes to serve a job in the system, such as processing data through the CPU or completing transfer operations with the disk.
Different tasks require different times, which can follow various probability distributions. Common distributions include exponential, uniform, and constant times as seen in our exercise.
The CPU burst time follows an exponential distribution with a mean of 0.01s. This means the probability of longer service times decreases exponentially. In contrast, the seek time of the disk also follows an exponential distribution with a mean of 0.03s. The latency time is uniformly distributed between 0 and 0.02s, with a mean of 0.01s, meaning each possible time within the range has equal probability.
Constant times, like the disk's transfer time of 0.01s, don't vary, making them easy to handle. Combining these, we get the overall service time distribution, crucial for calculating performance metrics.

traffic intensity

Traffic intensity, represented by \(\rho\), measures how busy a system is. It's the ratio of the arrival rate of jobs multiplied by the mean service time. This value lies between 0 (an idle system) and 1 (a fully occupied system).
In our exercise, with an arrival rate \(\lambda\) of 0.8 jobs/second and a mean service time \(E[S] = 0.06s\), we calculate traffic intensity as \(\rho = \lambda E[S] = 0.8 \times 0.06 = 0.048\).
A lower \(\rho\) indicates a system with lower waiting times and less congestion. However, a higher \(\rho\) suggests that the system is busy most of the time, potentially leading to longer waits and more jobs queueing up.
Knowing the traffic intensity helps predict system behavior under different loads and ensures that it operates efficiently without overloading.

mean response time

The mean response time, represented by \( R \), is the average time it takes for a job to move through the system from arrival to completion.
For the M/G/1 queue, \( R \) depends on both the mean service time and traffic intensity, given by the formula: \( R = \frac{E[S]}{1 - \rho} + \frac{\text{Var}[S]}{2 E[S] (1 - \rho)} \).
In our case, substituting the values \( E[S] = 0.06s \), \( \rho = 0.048 \), and \( \text{Var}[S] = 0.00101667 \), we get: \( R \approx 0.07162s \).
Breaking down, \( \frac{0.06}{1 - 0.048} \approx 0.06284 \) and \( \frac{0.00101667}{2 \times 0.06 \times (1 - 0.048)} \approx 0.00878 \), adding them gives the mean response time.
Understanding \( R \) aids in evaluating system performance and ensuring that the response time is acceptable for intended operations.