Question

Hi, can you shaw haw yau 50t the answer. 2-1f need to pay extra let me ase Know. Steph Problem 1 R. C. COLEMAN R. C. Coleman distributes a variety of food products that are sold through grocery store and supermarket outlets. The company receives orders directly from the individual out- lets, with a typical order requesting the delivery of several cases of anywhere from 20 to 50 different products. Under the company's current warehouse operation, warehouse clerks dispatch order-picking personnel to fill each order and have the goods moved to the warehouse shipping area. Because of the high labor costs and relatively low productivity of hand order-picking, management has decided to automate the warehouse operation by installing a computer-controlled order-picking system, along with a conveyor system for moving goods from storage to the warehouse shipping area. Immediate Predecessor Activity Description Determine equipment needs Obtain vendor proposals Select vendor A, B Order system Design new warehouse layout Design warehouse Design computer interface Interface computer Install system Train system operators Test system D, F, G D, F I. J Time (weeks) Time (weeks) Activity Optimistic Most Probable Pessimistic 16 10 24 13 10 20 14 gage Learning. All Righhs Reserved May not be copied, scanned u duplicated n whole o in pon, Due s ed that any suppressed coniens does nos materially affect the overal eali.erent at any tune-r Mbsequen nghes eetri.อเm rnga" i. Appendix 9.1 Finding Cumulative Probabilties for Normally Distributed Random Variables 455 R. C. Coleman's director of material management has been named the project man- ager in charge of the automated warehouse system. After consulting with members of the engineering staff and warehouse management personnel, the director compiled a list of activities associated with the project. The optimistic, most probable, and pessimistic times (in weeks) have also been provided for each activity Managerial Report Develop a report that presents the activity schedule and expected project completion time for the warehouse expansion project. Include a project network in the report. In addition, take into consideration the following issues: 1. R. C. Coleman's top management established a required 40-week completion time for the project. Can this completion time be achieved? Include probability infor- mation in your discussion. What recommendations do you have if the 40-week completion time is required? 2. Suppose that management requests that activity times be shortened to provide an 80% chance of meeting the 40-week completion time. If the variance in the project completion time is the same as you found in part (1). how much should the expect project completion time be shortened to achieve the goal of an completion within 40 weeks? 80% chance of Crashed Activity Time (weeks) Cost ($) Activity Normal 1,000 1,000 1,500 2,000 5,000 3,000 8,000 5,000 10,000 4,000 5,000 1,900 1,800 2,700 3,200 8,000 4,100 10,250 4 6,400 12,400 4,400 5,500 endix 9.1 FINDING CUMULATIVE PROBABILITIES FOR NORMALLY DISTRIBUTED RANDOM VARIABLES Excel can be used to find the probability a project with uncertain activity times will be com- pleted in some given completion time (assuming the project completion time is normally distributed). We demonstrate this on the Porta-Vac Project we considered in Section 9.2. Recall that management allotted 20 days to complete the project. We have found the z value that corresponds to T 20: 20 17 = 1.82 1.65 Chapter 9 Project Scheduling: PERT/CPM Now we will make use of the Excel function -NORM.S.DIST(z, TRUE) by substituting the value of z we have found into the function (entering "TRUE" for the second argument signifies that we desire the cumulative probability associated with z). Enter the following function into any empty cell in an Excel worksheet: NORM.S.DIST(1.82, TRUE) The resulting value is 0.96562, which is the probability that the completion time for the Porta-Vac project will be no more than 20 days.

Answer 1

The probability that the project will be completed in the allotted 20 days is

P(X< 20)

We convert to z scores and calculate the probability. The z-score is $z=\frac{20-17}{1.65}=1.82$ . That is

$P(X<20)=P\left ( Z<\frac{20-17}{1.65} \right )\\ P(X<20)=P\left ( Z<1.82 \right )\\ P(X<20)=\Phi \left ( 1.82 \right )$

Now you can use the cumulative standard normal distribution table corresponding to $z=1.82$ or use the shown EXCEL function.

$P(X<20)=\Phi \left ( 1.82 \right )\\ {\color{Blue} P(X<20)=0.96562}$