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Crash the project to July What activities would you crash What C05[5 crashing What are the gains from crashing? worth for youCrash the project to June 27. What acti...

Question

Crash the project to July What activities would you crash What C05[5 crashing What are the gains from crashing? worth for youCrash the project to June 27. What activities would You crash ? What the cost crashing What are the gains from crashing? Is it worth it for youBONUS QUESTIONS points Looking at the estimatec (mes completion, you are litcle concerned DocJusc Iccks Iike the variance for some ofthe cntical taskis significant and you might meet your most ambitious deadline You want t0 be 95# c

Crash the project to July What activities would you crash What C05[5 crashing What are the gains from crashing? worth for you Crash the project to June 27. What activities would You crash ? What the cost crashing What are the gains from crashing? Is it worth it for you BONUS QUESTIONS points Looking at the estimatec (mes completion, you are litcle concerned DocJusc Iccks Iike the variance for some ofthe cntical taskis significant and you might meet your most ambitious deadline You want t0 be 95# certain t0 meet the June 27 s0 you want Kno Dkho Many extna days should plan crash the project reach that level of cenainty, The_ score torione tailed) 95" probability = 1.645 What thc variance of thc projoct? 12. What F the standaro deviation the project? 13. By how many Hannd days should you crash the projoct for _ 95* probability - comploting thc projcct by Juno 27?



Answers

In cost estimation, the total cost of a project is the sum of component task costs. Each of these
costs is a random variable with a probability distribution. It is customary to obtain informationabout the total cost distribution by adding together characteristics of the individual component
cost distributions - this is called the "roll-up" procedure. For example, $E\left(X_{1}+\cdots+X_{n}\right)=$$E\left(X_{1}\right)+\cdots+E\left(X_{n}\right),$ so the roll-up procedure is valid for mean cost. Suppose that there are two
component tasks and that $X_{1}$ and $X_{2}$ are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 $\mathrm{th}$ percentile? That is, is the 75 th percentile of the
distribution of $X_{1}+X_{2}$ the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum
of percentiles? For what percentiles is the roll-up procedure valid in this case?

Part one. The Poland L s estimate of beta one oh is 0.36 zero. If the change in concentrate concentration is 0.1, then the change in the log of fair would be Beijing one head times the change in concentration and that would be 0.36 times 0.1, which is 0.36 That implies airfare is estimated to be about 3.6% higher. Part two, The 95% confidence interval obtain using the usual L s standard error is 0.301 2.419 And if we use the fully robust standard Iran's we will get point 245 and 2450.475 which is wider than the one above. The wider confidence interval is appropriate as the neglected serial correlation introduced uncertainty into our parameter estimation. Yeah, Part three. The quadratic has a use shape form, and the turning point is calculated by mhm taking partial derivative of lock of airfare with respect to lock of distance. And you will set that derivative equal zero. You wouldn't be able to find the value of lack of distance where the slope becomes positive, sir. the value of a lot of distance at the turning point is you will take 0.902 divided by two times 20.103 and you can get 4.38 This is the lock of distance, sir. When you convert it back, the value of distance is exponential of 4.38 Okay, about 80. And the shortest distance in the sample is 95 miles. So the turning point is outside the range of the data, which is a good thing in this case, what is being captured in an increasing elasticity affair with respect your distance As fare increases hard for the random effect, estimate of data one is 10.209 which is a bit smaller than the parent LS estimate. This estimate still implies a positive relationship between fair and concentration. The estimate is also very significant, with a T statistic of 7.88 Part five. The fixed effects estimate of beta one is 10.169 which is lower but not so different from the random effect estimate. And this is so because the value of, um, a perimeter in Equation 11 equation 14.11. Yeah, let's say it's, um, Fate. A hat. The Fed ahead is about 0.9, so random effects and fixed effects as meats are fairly similar. Remember, random effect uses a quasi demeaning. That depends on the estimate of this fada, I suggest in equation 14.11. Hard six. Heterogeneous effect. A supply could capture two types of factors that might correlate with concentration. Variable mhm. First, it could be factors about cities mhm near the two airports, for example, population, education level and type of employers. These factors could affect the demand for air travel, and the second set of factors could be factors relate you geographical features and infrastructure condition, such as highway qualities and whether the city locates near a river. So these factors are able to change over time. But in a short time period, let's say, um, the length of the time study in their sample. They are roughly time constant course, Yeah, and so they are able to be captured by a sub I. There are various factors like that, and it's better if we are able to control for them. So in part seven, it is more appropriate to choose to fix effect, estimate

23. Given standard Division one is seven. Standard division two is 12. Some of division three is 15 on the formula off. See, I is that one over. Signal. I squared over one over. Single on squared plus one. Oversee men square, so see one is equal to one over. Signal one square over one over. Signal one squared, plus one over signal to square. Waas one overseeing a three square, which is equal to 4.6418 Similarly, C two is equal toe open toe 184 and C three is equal to 1.23 98 So, um, the ignition rate this is the ignition rate should be as consistent as possible because if the ignition rate is unusually high, then it is possible to run out off the feeling there. So the more consistent that ignition rate are the list. The unusually hi ignition rate will Okay, so a signal w square. She's see one square times. C'mon square Waas. Seen two squared times. Sigma to Square Secret L Square. I'm sitting CNN Square Times Square Square. So the signal it's never the square is April to 31.45 and the Cendant deviation is the square root off his wedding, which is equal to 5.6 or 80 So the blinded jet fair will have a more consistent technician rate than each separate fellow because the summer deviation off value is smaller than each separate standard deviation.

Full of a total cost of a project is the sum of the component task costs. We're told that each of these component has costs as they were in a variable with a probability distribution and were asked to find information about the total cost distribution using a roll up procedure. So we're told that there are two component tasks on the X one and x two are independent, normally distributed random variables where asked, the royal procedure is valid for the 75th percentile. So is the 75th percentile of the distribution of excellent classics to the same as some of these 75 75% houses to individual distributions. He saw that it was it was valid for the mean cost. Well, we have at least some of the percentiles. This is going to be new one plus z times sigma one times or plus in you too. Yeah, plus Z times signature, which is the same as new one. Plus Mewtwo plus z times signal one plus sigma too. And we have that the percentile of sums. On the other hand, this is going to be new one, plus mewtwo, the mean of the some plus z times the standard deviation of this some which is the square root of sigma one squared plus sigma to square. So we see that in fact, some of the percentiles is going to be the same as the percentile of sums. We would have to have these two expressions equal. So these air equal if and only if we have that sigma one plus sigma too is equal to the square root of sigma one squared plus sigma two squared. So in other words, we have that or we have that Z is equal to zero mhm. In this first case, we get signal in squared plus two sigma one sigma two plus Sigma two squared equals signal and squared plus sigma two squared. So we have that two times Sigma one Sigma two equals zero. So, in other words, Sigma one asked equals zero or a sigma to has to be equal to zero. Either of them could be zero. So it's the role procedure is not valid or 75th percentile because in this case, Z is not equal to zero. Unless, of course, one of the staring deviations zero


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