Quantitative Analysis Essay Sample
- Word count: 1838
- Category: management
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Quantitative Analysis Essay Sample
Nocaf Drinks, Inc., a producer of decaffeinated coffee, bottles Nocaf. Each bottle should have a net weight of 4 ounces. The machine that fills the bottles with coffee is new, and the operation manager wants to make sure that it is properly adjusted. The operation manager takes a sample of n = 8 bottles and records the average range in ounces for each sample. The data for several samples is given in the table below. Is the machine properly adjusted and in control?
|SAMPLE||SAMPLE RANGE||SAMPLE AVERAGE|
This scenario represents the problem where there is one population, some population statistics are unknown, we have collected sample statistics and we need to find that whether the sample statistics represent the characteristics of population or not. Since the sample size is small (less than 30), we would use t-test with following assumption.
- Population is approximately normally distributed
- Population standard deviation is unknown
Null hypothesis is the hypothesis to be tested.
H0: µ = 4
Ha: µ ≠ 4 (two tail test)
(two tail test is taken because the weight less than four ounces and the weight more than 4 ounces, both are not acceptable in quality control)
Level of Significance:
α = 5% = 0.05
Critical Value where n = 8 ( df = n – 1 = 7) and α = 0.05
Reject Ho if t-test statistics greater than 1.895 or less than -1.895
For given data, sample mean = 4.03 and sample standard deviation = 0.087
Putting these values:
T = ( 4.03 – 4)
T = 0.03 / 0.031
T = 0.97
Since t-test statistic is within the non-rejectable region, (between +1.895 and -1.895), therefore we fail to reject the null hypothesis that is the weight of the new bottles is equal to 4 ounces.
Thus, at 5% level of significance we do not have sufficient evidence to warrant the rejection of the claim that the weight in new bottles is equal to 4 ounces.
Therefore, there is not significant difference in the actual and desired weight of the bottles.
The Flair Furniture company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labor hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hour in painting and varnishing. During the current production period, 240 hours of carpentry time are available and 100 hours in painting and varnishing tie are available. Each table sold yields a profit of $7 and each chair produced is sold for $5 profit. Formulate a Linear programming problem to find the best possible combination of tables and chairs to be manufactured so that Flair Furniture makes maximum profit.
The first step towards solving the liner programming problem is to understand the question. The second step is to determine the objective function.
Definition of the decision variables:
T = units of tables produced
C= units of chairs produced
In the given scenario, the objective is to produce the tables and chairs in such a combination, that the profit out of these two productions would be maximum. Mathematically,
7T + 5C
The next step is to determine the constraints in the problem. Besides the non-negativity constraints, there are two other major constraints. First one is that of carpentry while the second one is that of painting and varnishing. As defined in the question, the carpentry can take no more than 240 hours provided that each table required 4 hours of carpentry and each chair required 3 hours of carpentry. Moreover, the painting and varnishing can take no more than 100 hours provided that each table required 2 hours of painting and varnishing and each chair required 1 hour of carpentry. Mathematically,
4T + 3C ≤ 240
Painting and Varnishing Constraints
2T + 1C ≤ 100
T ≥ 0 (Non-Negativity Constraints)
C ≥ 0 (Non-Negativity Constraints)
The next step is to draw a line on the graph using each of the constraint. This can be done by putting 0 instead of each variable one by one, for each constraint.
In case of first constraint, when t = 0, c= 80
4(0) + 3 C = 240
3c = 240
When C = 0, t = 60
4T + 3 (0) = 240
4T = 240
T = 60
Therefore, its ordered pair are (0, 80) and (60, 0)
In case of second constraint, when T = 0, C = 100
2(0) + C = 100
C = 100
When C = 0, t = 50
2T + 0 = 100
T = 50
Therefore, its ordered pair are (0,100) and (50, 0)
Putting all the abovementioned constraints on graph:
In order to determine the optimal solution, the corner point approach would be used.
In this method, each of the corner point’s value of T and C will be plugged in the objective function:
(0,0) Profit = 7(0) + 5(0) = 0$
(0,80) Profit = 7(0) + 5(80) = 400$
(50,0) Profit = 7(50) + 5(0) = 350$
(30,40) Profit = 7(30) + 5(40) = 410 $*
Since the profit is maximum at 30 unit of Table and 40 units of chairs (410$), therefore it is the optimal solution.
*The combination of 30 and 40 is obtained at the intersection of the two constraints, since graph is there; this ordered pair can be obtaining by simply observing. Otherwise, it can be calculated by solving the two constraints simultaneously.
- After six months of study, Dr. Starr, president of Southwestern University, had reached a decision. To the delight of the students, SWU would not be relocating to a new football site but would expand the capacity at its on campus stadium.
Adding 21000 seats, including dozens of luxury skyboxes would not please everyone. The football coach had long argued the need for first-class stadium, 1 with built-in dormitory rooms for his players and a palatial office appropriate for the coach of a champion team.
The job was to get construction going immediately after the 2002 season ended. This would allow exactly 270 days until the next season opening game. The contractor Hill Construction signed the contract. Bob Hill guaranteed to finish the work in 270 days. The contract penalty of $10000 per day for running late was agreed.
Back in the office, Hill reviewed the data given in the table below and worked out the target completion date.
1) Develop a network drawing for Hill Construction and determine the critical path. How long is the project expected to take?
2) What is the probability of finishing in 270 days?
|Time estimates in days||
Crash cost $/day
|A||Bonding Insurance, etc.||–||20||30||40||1500|
|B||Foundation, Concrete footing for boxes||A||20||65||80||3500|
|C||Upgrading skyboxes, stadium sitting||A||50||60||100||4000|
|D||Upgrading walkways, stairwells, elevators||C||50||50||100||1900|
|I||Hardware, Air-conditioning, Metal Working||H||20||25||60||2000|
|J||Tile, Carpeting, windows||H||8||10||12||6000|
|L||Final Detail Work, clean-up||I,K||20||25||60||4500|
The Networking Diagram:
Critical Path is the path on the network diagram whose tasks can be not delayed without delaying total time of the project. This means that delaying any activity on the critical path would result in the delay in the total project completion time. This means that those activities would not have any slack time available. (Slack time is the difference between the earliest start time and earliest ending time for that activity)
Applying earliest start and earliest end time on the given activities:
|Task||Early Start||Early Finish||Late Start||Late Finish||Slack|
As it is obvious from the above table that the path constituting of activity A, C, D, G, H , I and activity L is the critical path, as all these activities have zero slack and any delay in these activities would ultimately delay the project.
There is also identification for critical path that it is the longest path on the network diagram. Looking even this way, the answer would be same, that is, the path constituting of activity A, C, D, G, H , I and activity L.
The total project completion time is 260 days.
For determining the probability of ending the project in 270 days, we need to know the project variance.
The project variance is the sum of individual variances = ∑((pessimist – optimist) /6)2
Thus , project standard deviation is the square root of project variance.
In our case, the project standard deviation is 17.87
Z = due date – expected date of completion
Project standard deviation
Z score for probability of project completion in 270 days
Referring to the normal table, there is 71.23% probability that the project will be completed in 270 days.
Weiss, N., (1997). Introductory Statistics. Boston: Addison-Wesley.
Freund, J., & Walpole, R. (1987). Mathematical Statistics. Englewood Cliffs: Prentice-Hall.
Render, B., & Stair, R. (1997). Quantitative Analysis for Management. Englewood Cliffs: Prentice Hall.
Quinn, S., Bowser, K., & Flaherty, E. (1993). Budnick’s Applied Mathematics for Business, Economics and Social Sciences. Boston: McGraw-Hill College.