Production and Operations Analysis, Seventh Edition
7th Edition
ISBN: 9781478623069
Author: Steven Nahmias, Tava Lennon Olsen
Publisher: Waveland Press, Inc.
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Chapter 2.10, Problem 43P
a)
Summary Introduction
To determine: The resulted model using backshift notation with backshift operator B , considering ARIMA (0, 2, 2) process.
Introduction: ARIMA is based on exploiting the autocorrelation structure of a time series.Backshift operator B is a helpful notational tool when operating with lags in the time series used to create distinctions.
b)
Summary Introduction
To determine: The resulted model using backshift notation with backshift operator
Introduction: ARIMA is based on exploiting the autocorrelation structure of a time series. Backshift operator is a specifically helpful notational tool in case of combining differences.
(c)
Summary Introduction
To determine: The resulted model using no backshift notation, considering ARIMA process.
Introduction: ARIMA is based on exploiting the autocorrelation structure of a time series.The ARIMA process without backshift notation represents a stationary state.
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Consider an ARIMA(2,1,1) process. Write out the resulting model usinga. Backshift notation with the backshift operator B.b. Backshift notation with the operatorc. No backshift notation
Bad simulations Explain why each of the followingsimulations fails to model the real situation properly:a) Use a random integer from 0 through 9 to representthe number of heads when 9 coins are tossed.
b) A basketball player takes a foul shot. Look at a ran-dom digit, using an odd digit to represent a good shot
and an even digit to represent a miss.c) Use random numbers from 1 through 13 to represent thedenominations of the cards in a five-card poker hand.
An urn contains 10 red balls and 30 blue balls.a Suppose you draw 4 balls from the urn. Let Xi bethe number of red balls drawn on the ith ball (Xi 0 or1). After each ball is drawn, it is put back into the urn.
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Chapter 2 Solutions
Production and Operations Analysis, Seventh Edition
Ch. 2.4 - Prob. 1PCh. 2.4 - Prob. 2PCh. 2.4 - Prob. 3PCh. 2.4 - Prob. 4PCh. 2.4 - Prob. 5PCh. 2.4 - Prob. 6PCh. 2.4 - Prob. 7PCh. 2.4 - Prob. 8PCh. 2.4 - Prob. 9PCh. 2.6 - Prob. 10P
Ch. 2.6 - Prob. 11PCh. 2.6 - Prob. 12PCh. 2.6 - Prob. 13PCh. 2.6 - Prob. 14PCh. 2.6 - Prob. 15PCh. 2.7 - Prob. 16PCh. 2.7 - Prob. 17PCh. 2.7 - Prob. 18PCh. 2.7 - Prob. 19PCh. 2.7 - Prob. 20PCh. 2.7 - Prob. 21PCh. 2.7 - Prob. 22PCh. 2.7 - Prob. 23PCh. 2.7 - Prob. 24PCh. 2.7 - Prob. 25PCh. 2.7 - Prob. 26PCh. 2.7 - Prob. 27PCh. 2.8 - Prob. 28PCh. 2.8 - Prob. 29PCh. 2.8 - Prob. 30PCh. 2.8 - Prob. 31PCh. 2.8 - Prob. 32PCh. 2.9 - Prob. 33PCh. 2.9 - Prob. 34PCh. 2.9 - Prob. 35PCh. 2.9 - Prob. 36PCh. 2.9 - Prob. 37PCh. 2.10 - Prob. 38PCh. 2.10 - Prob. 42PCh. 2.10 - Prob. 43PCh. 2.10 - Prob. 44PCh. 2.10 - Prob. 45PCh. 2 - Prob. 47APCh. 2 - Prob. 48APCh. 2 - Prob. 49APCh. 2 - Prob. 50APCh. 2 - Prob. 51APCh. 2 - Prob. 52APCh. 2 - Prob. 53APCh. 2 - Prob. 54APCh. 2 - Prob. 55APCh. 2 - Prob. 56APCh. 2 - Prob. 57APCh. 2 - Prob. 58AP
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