1. Three different time estimates for various activities of a project are given in the below table. Construct a PERT network and find out (a) The earliest possible time (TE) to complete the different activities, (b) The latest allowable time (TL) for them, (c) The slack values, (d) The critical paths, and (e) The probability factor for completing the project in 30 weeks.

[pic]

[pic]

SOLUTION:

[pic][pic]

[pic]

(iv) Critical Path

1–2–4–6–7–9–10 = 28 weeks.

Vcp = 2 and σcp = 1.41

(v) The probability factor for completing the project in 30 weeks.

Z = (X – μ)/σ

where X = 30 weeks, μ = 28 weeks and σCP = 1.41

Hence,

Z = (30 28)/141 = 1.41

Z is the number of standard deviations by which X exceeds μ

μ = Te is the total project duration, and

X = due or scheduled date or time

From the table of Standard Normal Distribution Function, the corresponding percentage probabilities

are given as follows:

For Z = 1.41 the probability is 92.07 % and for Z = 1.42 the probability is 92.22 %.

Therefore, there is 92.13% probability that the project will be finished in 30 weeks.

0O0

2. A reactor and storage tank are interconnected by a 3” insulated process line that needs periodic replacement. There are valves along the lines and at the terminals and these need replacing as well. No pipe and valves are in stock. Accurate, as built, drawings exist and are available. You are the maintenance and construction superintendent responsible for this project. The works engineer has requested your plan and schedule for a review with the operating supervision. The plant methods and standards section has furnished the...

...and a comprehensive environment where the youth feel that they have a place to turn to where it is they are, is what will really solve the roots problem.
These were just some of the techniques and plans that have been drawn up. Many of them are similar. Where they differ is where to draw the lines as to what is the most productive and effective tool. By examining this aspect we would know where to concentrate our effort and resources. The National Crime Prevention Council is a leading authority on this issue, although they are only known to most people by the group that sponsors McGruff the crime dog. What does work, works on three different levels, identifying problems in an individual, engaging awareness of the issue, and promoting physical prevention features such as metal detectors. So what works is a balance of Teachers, students, principles, parents, Law enforcement, and the community engaging in the three levels mentioned previously. According to NCPC there is a direct correlation between in community involvement in crime prevention and reported teen crimes.
As far as this issue is concerned the solution to the problem will be identifying the root of the problem. Some people are quick to blame the media, while some blame parents, and others blame the schools themselves. The solution does depend on the problem and identifying is very critical. At...

...teenage girls of their personal image and appearance. This includes but is not limited to their weight, hair color, breast size, color of skin or simply how much makeup they apply. The body type portrayed in advertising as the ideal is possessed naturally by only 5% of American females, yet 47% of girls ages 11-18 grade reported wanting to lose weight because of magazine pictures and 69% of girls reported that magazine pictures influenced their idea of a perfect body shape (Mellin) . This distorted and incorrect view that young girls have of themselves have caused many to develop serious self-confidence issues and eating disorders such as anorexia and bulimia nervosa. This is a tough problem to find solutions to, but by getting the media to use more diverse actresses and models this problem may be helped. By magazines advertising the same tall, skinny, dark skinned woman on their covers, adolescent girls are pressured to feel the need to have to look this way as well. In our society today it can be extremely hard to feel accepted, but by forcing the media to change their close minded opinions, views and advertisements, we can not only change teenage girl’s lives but in some extremely severe cases save them as well.
In a world where 90% of every female celebrity is on the verge of being vastly underweight it is easy for young teenagers to feel pressured to be similar. Being overweight as a young girl can cause isolation, verbal...

...When looking around the world today at our global civilization there are many problems that leap to our attention and everyone has a vested interest, one such example is food. Science has been tinkering with nature for the past one-hundred years trying to come up with solutions to help fight famine, and as a result, man made products have created ever more problems. The easiest way in which an individual or individuals can create change is to effect change in their local communities, so others can see an image of what the world could be. In my local area of Peoria County the topic of food is major problem and all of the issues that correspond, like genetically modified organisms (GM), pesticides, herbicides, cost of food, health of people, and the health of our environment. A practical solution to all these enormous problems we face when we walk through a grocery store and around our community can be fixed through the practice of local organic farming and community participation. With the help of volunteers in Peoria County we could begin growing organic produce that will help create, sustain, connect, and educate our community to help the less fortunate and build a healthier, stronger community.
Experts say that nearly sixty to seventy percent of processed foods contain genetically modified ingredients (Biotech). GM foods are created using biotechnology to change their genetic material....

...zones it just seems like none of them work. The problem is that Findlay is getting more rain than the Blanchard River can hold water, which is making the city of Findlay flood. In my opinion we have a couple of choices of ways to get around ruining our homes and lives, which would be corralling the rain to different location, creating more reservoirs, and probably the best solution, building floodwalls. As a community we need to look at all our options and come up with a plan.
The first flood we hear was in 1917, however the first one they could document the levels and severness was back in 1959 and the community has the opinion that since no one has done anything thus far, nothing will be done. As Jackie Fuller explained in her article “Findlay, Your Flood Problem can be Solved” stated, “Over the last two years Findlay, Ohio has flooded 10 times. While flooding is not a new problem, the frequency of it is. There is a solution for the city’s flooding problem. With the help of the Northwest Ohio Flood Mitigation Partnership, the solution will be found and implemented.”(para. 1) Fuller went on to explains that as the weather changes its pattern so will the rainfall, as a result Findlay may still flood, however it won’t be as regular (Para. 2-6). As a community we can hope the weather changes its pattern very soon!
One possible solution to the flooding of...

...student the student to do well, and without putting too much pressure on him/her.
Procedures
These are the procedures I have taken in order to find the solution to this problem:
1. Research
I have created a poll for students in high school to take (30 people), and these were the results:
Are parents and teachers putting too much pressure on teenagers?
Yes, they should encourage the student without overwhelming them will all the work
15 People
50%
Yes, but it’s good preparation for college
10 People
30%
No, there isn’t any pressure
5 People
20%
Articles about students that have commit suicide because of so much pressure:
a. About 40 students commit suicide every year because of pressure during examinations
Stress levels with high amounts of work
b. One in every five high school students are constantly stressed due to too much work
c. Movie- “Race to Nowhere” is a film about the high stress levels students are faced with at school everyday
d. Most school districts have now came up with the fair rule that each class should give out a reasonable amount of half an hour of homework, no more than that, in order for students to be less stressed about high amounts of work, and to give students some free time for themselves.
2. Ideas for how I can change this/inform adults and teachers about this problem:
Poster
a. At first I had the idea to make a few posters and hang them up around the school while parents...

...Problem Set on Chapter 9.
CAPM, beta, and WACC
1. Bradshaw Steel has a capital structure with 30 percent debt (all long-term bonds) and 70 percent common equity. The yield to maturity on the company’s long-term bonds is 8 percent, and the firm estimates that its overall composite WACC is 10 percent. The risk-free rate of interest is 5.5 percent, the market risk premium is 5 percent, and the company’s tax rate is 40 percent. Bradshaw uses the CAPM to determine its cost of equity. What is the beta on Bradshaw’s stock?
Beta risk
2. Sun State Mining Inc., an all-equity firm, is considering the formation of a new division that will increase the assets of the firm by 50 percent. Sun State currently has a required rate of return of 18 percent, U. S. Treasury bonds yield 7 percent, and the market risk premium is 5 percent. If Sun State wants to reduce its required rate of return to 16 percent, what is the maximum beta coefficient the new division could have?
Risk-adjusted discount rate
3. Assume you are the director of capital budgeting for an all-equity firm. The firm’s current cost of equity is 16 percent; the risk-free rate is 10 percent; and the market risk premium is 5 percent. You are considering a new project that has 50 percent more beta risk than your firm’s assets currently have, that is, its beta is 50 percent larger than the firm’s existing beta. The expected return on the new project is 18 percent. Should the project be accepted if beta risk is...

...Aib Homework Wk3
Problem 1
The following data were obtained from a project to create a new portable electronic.
Activity Duration Predecessors
A 5 Days ---
B 6 Days ---
C 8 Days ---
D 4 Days A, B
E 3 Days C
F 5 Days D
G 5 Days E, F
H 9 Days D
I 12 Days G
Step 1: Construct a network diagram for the project.
Step 2: Answer the following questions: (15 points total)
a) What is the Scheduled Completion of the Project? (2 points)
32 -> CORRECT
b) What is the Critical Path of the Project? (3 points)
B,D,F,G,I -> CORRECT
2,4,6,7,9
c) What is the ES for Activity D? (2 points)
6 -> Correct
d) What is the LS for Activity G? (2 points)
15
e) What is the EF for Activity B? (2 points)
6 -> Correct
f) What is the LF for Activity H? (2 points)
32 -> Correct
g) What is the float for Activity I? (2 points)
0
Problem 2
The following data were obtained from a project to build a pressure vessel:
Activity Duration Predecessors
A 6 weeks ---
B 6 weeks ---
C 5 weeks B
D 4 weeks A, C
E 5 weeks B
F 7 weeks D, E, G
G 4 weeks B
H 8 weeks F
I 5 weeks G
J 3 week I
Step 1: Construct a network diagram for the project.
Step 2: Answer the following questions: (15 points total)
a) Calculate the scheduled completion time. (3 points)
30
b) Identify the critical path (4 points)
B,C,D,F,H
c) What is the slack time (float) for activity A? (2 points)
5
d) What is the slack time (float) for activity D? (2 points)
0
e) What...

...Solution to Case Problem Specialty Toys
10/24/2012
I. Introduction:
The Specialty Toys Company faces a challenge of deciding how many units of a new toy should be purchased to meet anticipated sales demand. If too few are purchased, sales will be lost; if too many are purchased, profits will be reduced because of low prices realized in clearance sales. Here, I will help to analyze an appropriate order quantity for the company.
II. Data Analysis:
1.
20,0
00
.025
10,0
00
30,0
00
.025
.95
20,0
00
.025
10,0
00
30,0
00
.025
.95
Since the expected demand is 2000, thus, the mean µ is 2000. Through Excel, we get the z value given a 95% probability is 1.96. Thus, we have: z= (x-µ)/ σ=(30000-20000)/ σ=1.96, so we get the standard deviation σ=(30000-20000)/1.96=5102.
The sketch of distribution is above. 95.4% of the values of a normal random variable are within plus or minus two standard deviations of its mean.
2. At order quantity of 15,000, z= (15000-20000)/5102=-0.98,
P(stockout) = 0.3365 + 0.5 = 0.8365
At order quantity of 18,000, z= (18000-20000)/5102=-0.39,
P(stockout) = 0.1517 + 0.5= 0.6517
At order quantity of 24,000, z= (24000-20000)/5102=0.78,
P (stockout) = 0.5 - 0.2823 = 0.2177
At order quantity of 28,000, z= (28000-20000)/5102=1.57,
P (stockout) = 0.5 - 0.4418 = 0.0582
3.
Order Quantity = 15,000 |
Unit Sales | Total Cost | Sales at $24 | Sales at $5 | Profit |
10,000 |...