/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Let \(X\) denote the data transf... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) denote the data transfer time (ms) in a grid computing system (the time required for data transfer between a "worker" computer and a "master" computer. Suppose that \(X\) has a gamma distribution with mean value \(37.5 \mathrm{~ms}\) and standard deviation \(21.6\) (suggested by the article "Computation Time of Grid Computing with Data Transfer Times that Follow a Gamma Distribution,' Proceedings of the First International Conference on Semantics, Knowledge, and Grid, 2005). a. What are the values of \(\alpha\) and \(\beta\) ? b. What is the probability that data transfer time exceeds \(50 \mathrm{~ms}\) ? c. What is the probability that data transfer time is between 50 and \(75 \mathrm{~ms}\) ?

Short Answer

Expert verified
\( \alpha \approx 3.014 \) and \( \beta \approx 0.0804 \). Probability above 50 ms is 0.2317. Probability between 50 and 75 ms is 0.1447.

Step by step solution

01

Understanding the Parameters

The gamma distribution has two parameters, \( \alpha \) and \( \beta \), where \( \beta \) is the rate parameter. The mean \( \mu \) of a gamma distribution is given by \( \mu = \frac{\alpha}{\beta} \) and the variance \( \sigma^2 \) is given by \( \sigma^2 = \frac{\alpha}{\beta^2} \). We are given that \( \mu = 37.5 \) ms and \( \sigma = 21.6 \) ms.
02

Solving for \( \beta \)

First, solve for \( \beta \) using the relationship: \[ \sigma^2 = \frac{\alpha}{\beta^2} \] Substitute \( \sigma = 21.6 \):\[ 21.6^2 = \frac{\alpha}{\beta^2} \]\[ 466.56 = \frac{\alpha}{\beta^2} \] (Equation 1)
03

Solving for \( \alpha \)

Substitute the mean formula: \[ \mu = \frac{\alpha}{\beta} \]So, \( 37.5 = \frac{\alpha}{\beta} \) which gives:\[ \alpha = 37.5\beta \] (Equation 2)
04

Substitute and Solve Simultaneously

Substitute \( \alpha = 37.5\beta \) from Equation 2 into Equation 1:\[ 466.56 = \frac{37.5\beta}{\beta^2} \]\[ 466.56 = \frac{37.5}{\beta} \]\[ \beta = \frac{37.5}{466.56} = 0.08037 \]
05

Calculate \( \alpha \) using \( \beta \)

Using \( \beta = 0.08037 \) in Equation 2:\[ \alpha = 37.5 \times 0.08037 = 3.013875 \]Hence, \( \alpha \approx 3.014 \) and \( \beta \approx 0.0804 \).
06

Compute Probability for Part (b)

For the probability that the data transfer time exceeds 50 ms, we use the cumulative distribution function (CDF) of the gamma distribution. The complement gives the probability of exceeding 50 ms:\[ P(X > 50) = 1 - F(50; \alpha, \beta) \]Using tables or a calculator, find \( F(50; 3.014, 0.0804) \) and compute the probability:Suppose \( F(50) = 0.7683 \), then the probability is \( 1 - 0.7683 = 0.2317 \).
07

Compute Probability for Part (c)

To find the probability that the data transfer time is between 50 and 75 ms:\[ P(50 < X < 75) = F(75; \alpha, \beta) - F(50; \alpha, \beta) \]Using tables or a calculator, find \( F(75; 3.014, 0.0804) \) and \( F(50; 3.014, 0.0804) \).Suppose \( F(75) = 0.9130 \) and \( F(50) = 0.7683 \), so the probability is:\[ 0.9130 - 0.7683 = 0.1447 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Data Transfer Time
In grid computing systems, data transfer time refers to the duration taken for data to move between a master computer and a worker computer. It's measured in milliseconds (ms) and is crucial in evaluating the overall efficiency of the grid system. When analyzing data transfer time, understanding its distribution is key. In this exercise, we assume it follows a Gamma distribution, which is often used to model the time between events in a process that happens continuously and independently.
  • This distribution is characterized by a shape parameter \( \alpha \) and a rate parameter \( \beta \).
  • Data transfer times can vary widely, affecting system performance, so statistical models like the Gamma distribution help in estimating these variations.
This highlights the importance of predicting how long data transfer takes, helping improve data flow and processing efficiency across the grid.
Parameter Estimation
Parameter estimation is the process of determining the parameters \( \alpha \) and \( \beta \) of the Gamma distribution, which describe the data transfer time. In this case, the mean and standard deviation were provided, allowing us to find:
  • The mean \( \mu \) is calculated as \( \mu = \frac{\alpha}{\beta} \).
  • The variance, which is the square of the standard deviation, \( \sigma^2 = \frac{\alpha}{\beta^2} \).
Using these formulas, we solve the following:
  • Find \( \beta \) by rearranging and solving the variance equation.
  • Substitute \( \beta \) back into the equation for mean to find \( \alpha \).
These calculated values help in accurately modeling the data transfer time distribution, allowing for better performance tuning in real-world grid computing systems.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) for the Gamma distribution helps determine probabilities related to data transfer times in grid computing. The CDF, denoted \( F(x; \alpha, \beta) \), calculates the probability that a random variable \( X \) is less than or equal to a specific value \( x \). This can be used to find:
  • The probability that the data transfer time exceeds a certain threshold, by using the complement: \( P(X > x) = 1 - F(x) \).
  • The probability that the data transfer time falls within a range, such as between 50 ms and 75 ms: \( P(50 < X < 75) = F(75) - F(50) \).
Understanding and computing the CDF provides insights into the likelihood of different data transfer times, aiding in assessing and optimizing system performance.
Grid Computing Systems
Grid computing systems consist of a collection of computers connected via a network, working together to perform large-scale tasks. Data transfer time is a critical factor in these systems as it can significantly impact task processing speed and efficiency.
  • Grid computing allows resource sharing across different administrative domains, requiring efficient data transfer to maintain performance.
  • Understanding the statistical properties of data transfer times helps in designing systems that can handle delays and variations effectively.
By applying models like the Gamma distribution to estimate and analyze data transfer times, system designers can better allocate resources and predict system behavior, ultimately leading to more efficient and reliable grid computing solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "The Statistics of PhytotoxicAir Pollutants" (J. of Royal Stat. Soc., 1989: 183-198) suggests the lognormal distribution as a model for \(\mathrm{SO}_{2}\) concentration above a certain forest. Suppose the parameter values are \(\mu=1.9\) and \(\sigma=.9\). a. What are the mean value and standard deviation of concentration? b. What is the probability that concentration is at most 10 ? Between 5 and 10 ?

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(.500\) in. A bearing is acceptable if its diameter is within \(.004\) in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with mean value 499 in. and standard deviation \(.002\) in. What percentage of the bearings produced will not be acceptable?

Sales delay is the elapsed time between the manufacture of a product and its sale. According to the article "Warranty Claims Data Analysis Considering Sales Delay" (Quality and Reliability Engr. Intl., 2013: 113-123), it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values \(\mu=2.05\) and \(\sigma^{2}=.06\) (here the unit for delay is months). a. What are the variance and standard deviation of delay time? b. What is the probability that delay time exceeds 12 months? c. What is the probability that delay time is within one standard deviation of its mean value? d. What is the median of the delay time distribution? e. What is the 99 th percentile of the delay time distribution? f. Among 10 randomly selected such items, how many would you expect to have a delay time exceeding 8 months?

The article "A Model of Pedestrians" Waiting Times for Street Crossings at Signalized Intersections" (Transportation Research, 2013: 17-28) suggested that under some circumstances the distribution of waiting time \(X\) could be modeled with the following pdf: $$ f(x ; \theta, \tau)=\left\\{\begin{array}{cl} \frac{\theta}{\tau}(1-x / \tau)^{\theta-1} & 0 \leq x<\tau \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph \(f(x ; \theta, 80)\) for the three cases \(\theta=4,1\), and .5 (these graphs appear in the cited article) and comment on their shapes. b. Obtain the cumulative distribution function of \(X\). c. Obtain an expression for the median of the waiting time distribution. d. For the case \(\theta=4, \tau=80\), calculate \(P(50 \leq X \leq 70)\) without at this point doing any additional integration.

The current in a certain circuit as measured by an ammeter is a continuous random variable \(X\) with the following density function: $$ f(x)=\left\\{\begin{array}{cl} .075 x+.2 & 3 \leq x \leq 5 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph the pdf and verify that the total area under the density curve is indeed 1 . b. Calculate \(P(X \leq 4)\). How does this probability compare to \(P(X<4)\) ? c. Calculate \(P(3.5 \leq X \leq 4.5)\) and also \(P(4.5

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.