1. The problem at work


The shop made two kinds of products, the dressers and the cabinets that pass in the assembly and finishing department. The table 1 below shows the given data as well as the unknowns.


This data can be collected using the analyzation and measuring the time of every department and the number of units that produced.


Table 1.


 


Cabinets (x)


Dressers (y)


Time Available


Time in Assembly


4 hours


2 hours


60 hours


Time Finishing


2 hours


4 hours


48 hours


Profit


0/unit


/ unit


 


 


I. Decision Variables:


Let x = no. of the units that can be produced as well as sold in order to maximize the profit


      y = units that the dressers to be produced and sold so that it can maximize its profit


 


II. The objective function:


Max. P = 100x + 80y


III. Subjected to:


            Explicit Constraints:


            4x + 2y < 60


            2x + 4y < 48


            x > 0 and y > 0


 


IV. Assume that both the equations be equal.


4x + 2y < 60                                       2x + 4y < 48


4x + 2y = 60                                       2x + 4y = 48


The ordered pairs (0,30) and (15,0) for 4x + 2y = 60.


On the other hand, the ordered pairs for 2x + 4y = 48 are (0,12) and (24,0)


Get the point of intersection for the two equations.


2(4x + 2y) = (2)60


2x + 4y = 48


6x = 72


x = 12


Substitute the value of x to any equation


4x + 2y = 60


4(12) + 2y = 60


y = 12/2


y = 6 The point of intersection therefore is (12, 6)


 


 


 


 


 


 


 


 


V. Graph



VI. Test the corners


P (0,0)


Max P = 100(x) + 80(y)


Max P = (100) (0) + 80(0)


Max. P =


P (0,12)


Max P = 100(x) + 80(y)


Max P = (100) (0) + 80(12)


Max. P = 0


P (12, 6)


Max P = 100(x) + 80(y)


Max P = (100) (12) + 80(6)


Max P = 80


P (15, 0)


Max P = 100(x) + 80(y)


Max P = (100) (15) + 80(0)


Max P = 00


 


VII. Decision


For the company to maximize the maximum profit of ,680, it needs to manufacture and sell 12 units of the cabinets and 6 units of the dressers.


 


2. Break Even Analysis


The formula in break even point analysis is:


TR = TC


Where: TR = Total Revenue


              TC = Total Cost


TC = VC + FC


Where VC = Variable Cost


              FC = Fixed Cost


Then the Fixed Cost = 0,000


                Variable Cost is 0/passenger


To get the average sales per flight, get the 80% of the airplane capacity which is 240 and it is equal average of 192 passengers per flight.


Te variable cost therefore is (0) (192) = ,000


The equation for the break even therefore is:


TR = ,000 + 0,000


In getting the total revenue it needs to multiply the number of average passengers to the minimum amount to be paid.


Let x is the amount to be paid an it must be multiplied to the average passengers of 192.


192x = ,000 + 0,000


192x = 6,000


x = , 541.67


The company therefore has the minimum price which can be charge to its passengers of , 541.67 in order to break even.


 


3. Probability Normal Distribution


a.


Use the x equation which is:


z = (x-mean)/standard deviation


z = 6,000 – 5,200 / 600


z = 1.33



The probability for this area is 0.4082 or 40.82% which the probability of getting the egg-tart in between 5,200 and 6000. But, it also needs to add the other areas covered because it is part of the restrictions of less than 5,200. Thus,  0.4082 + 0.5 is equal to 0.9082 or 90.82% which is the probability of getting less than 5,200 egg-tart sold for the day.


 


b. Using the same formula:


z = 7,500 – 6,000 / 600


z = 2.5



The probability for this area is 0.4938 and this needs to subtract it from 0.5. Thus it has the probability of 0.0062 or 0.62% that the egg-tart sold will be greater that 7,500.


 


c.


Let x be the value greater that 95% chance


1.645 = 6,000 – x/600


x = 5,013 – 3,000


x = 2,013


 


 


This is the value that exceeds in order to have the 95% chance that the egg-tart can be sold in the day.


 


 


4. Probability in Market Decision


 


Market Reaction




Good


Medium


Poor




X


100 (0.30) = 30


110 (0.40) = 44


80 (0.30) = 24



 


Y


70 (0.30) = 21


90 (0.40) = 36


120 (0.30) = 36



 


Z


130 (0.30) = 39


100 (0.40) = 40


70 (0.30) = 21



 


 


For the product X, the company can have the ability to generate the total profit of 101.


In the product Y, it can generate have the ability to generate the total amount of 93.


For the product Z, the company can have the ability in generating the total income of 100.


Thus, the company needs to launch the version of product X in order to generate much profit which can be total to 101 as it compared to the products Y and Z.


 


5. Probability in Business Decision


a. Probability of a favorable market given a favorable study = 0.72 


b. Probability of an unfavorable market given a favorable study = 0. 28


c. Probability of a favorable market given an unfavorable study = 0. 22


d. Probability of an unfavorable market given an unfavorable study = 0. 78


e. Probability of a favorable research study = 0. 6


f. Probability of a favorable research study = 0.4


Answers:


a. (0,000 – ,000)0.72 = profit of 138,960


b. (-,000 – ,000)0.28 = loss of ,960


c. (0,000 – 7,000)0.22 = Profit of ,460


d. (-50,000 – ,000)0.28 = Loss of ,460


e. (-,000)0.6 = Loss of ,200


f. (-,000)0.4 = Loss of ,800


This means that the cook is willing to pay s much as ,460 in the market study.


 


 


 


6. Simple Linear Regression


For the given sales of the company for the last ten years, the graph of the data and the regression line can be computed as:


 


Year(x)


Sales(y)


(xy)


xsquared


 


1


9


9


1


 


2


14


28


4


 


3


20


60


9


 


4


28


112


16


 


5


40


200


25


 


6


60


360


36


 


7


90


630


49


 


8


130


1040


64


 


9


180


1620


81


 


10


250


2500


100


Total


55


821


6559


385


 


 


 


Using the formula  Y = a + bX  this can be solved.


xbar = 55/10 = 5.5


y bar = 821/10 = 82.1


b = ∑ xy – n (xbar) (ybar) / ∑x2 – n(ybar)2


b = 6559 – 10(5.5)(82.1) / 385 – 302.5


b = 2043.5 / 82.5


b = 24.77


To get the value of a it needs to follow the formula:


a = xbar – b(ybar)


a = 82.1 – 24.77(5.54)


a = -54.135


The simple linear regression line therefore which also graph below is equal to:


y = -54. 13 + 24. 77x 


 



 


a. The linear regression line is y = -54. 13 + 24.77x which can forecast to the upcoming number of years.


b. In order to get how strong the relationship of the two data, it is important to use the process of correlation which as the formula:



Px, y = r = (10)6559 – 55 (821)


              √ 10(385) – 3025 √10 (126561) – 674, 041


         r = 65,590 – 45,155


                28.72 • 769.135


         r = 20,435 / 22,089.56


         r = 0.93


 


This means that the having the given sales and the period or year as the variables, the they have the high or the strong relationship for having the 0.93 as the value of the pearson r. This implies that they have strong relationship.


 


c. Using the regression equation as the answer in the question a, which is y = -54. 13 + 24.77x the value of 11th year can be substitute to the x value so that the projected value of the sales can be computed.


y = -54. 13 + 24.77x


y = -54. 13 + 24.77(11)


y = 218.34 is the number of sales for the next year or the 11th year


 


7. Economic Order Quantity


 a.


Given:


A = Annual Total Sales = 1,200 pairs ( per pair) = , 000


P = ordering cost = per order


Qn  = optimum number of order at a time (EOQ) = 200 pairs


The carrying cost can be computed as


Qn2 = 2AP / C


To get rid of C


C = 2AP / Qn2


 


Solving simultaneously, we have to compute first the EOQ per year or the number of orders per year. Therfore 1200 pairs / 200 = 6 orders per year


C = 2(,200) (30)  / (200)2


C = 72,000 / 40,000


C = 1.8% or .8 per order per year


 


b. If the carrying cost is 5% then the optimal order quantity can be computed as:


C = 2AP / Qn2


Qn = 2(,200) (30)  / 5


Qn = 120  the EOQ therefore is 120 if the holding cost is 5%


 


8. Linear Programming


I. The decision variable


Let x = units of product in the standard testers


      y = the units of products in the normal testers


II. Objective Function


Max. P = 6x + 7y


III. Subjected to:


2x + 4y < 160


6x + 2y < 240


4x + 4y < 200


 


IV. Assume that the equations are equal


2x + 4y = 160 The values of ordered pairs are (0,40) and (80,0)


6x + 2y = 240 The values of the ordered pair are (0,120) and (40,0)


4x + 4y = 200 The values of the ordered pair are (0,50) and (50,0)


Get the point of intersection of 4x + 4y = 200 and 2x + 4y = 160 which is (20,30)


And the point of intersection of 4x + 4y = 200 and 6x + 2y = 240 which is (35,15)


V. Graph



 


VI. Test the corners points


P (0,40)


Max. P = 6x + 7y


         P = 6(0) + 7(40)


         P = 0


P (20,30)


Max. P = 6x + 7y


         P = 6(20) + 7 (30)


         P = 120 + 240


         P = 0


P(35,15)


Max. P = 6x + 7y


         P = 6(35) + 7(15)


         P = 210 + 105


         P = 5


 


P(40,0)


Max. P = 6x + 7y


         P = 6(40) + 7(0)


         P = 0


 


VI. Decision


a. The company therefore needs to produce 20 units of the standards testers and 30 units of the normal testers in order to allocate the maximum profit of 0.


b. There are no spare capacities to the production facilities since it has overcome the number of the units in the different processes.


 


 


 


 


 


 


 


9. Network Diagram


a. The network diagram can is given below:



b.  


 


Days


Required


Immediate


Predecessors


 


 


 


 


 


 


 


Activity


ES


EF (ES+t)


LS(LF-t)


LF(LS +t)


TS (LF - EF)


Critical Path?


A


5


 


0


5


1


6


1


No


B


3


 


0


3


0


3


0


Yes


C


3


B


3


6


3


6


0


Yes


D


7


A


5


12


6


13


1


No


E


10


B


3


13


6


16


3


No


F


14


A,C


6


20


6


20


0


Yes


G


7


D


12


19


15


22


3


No


H


4


E


13


17


18


22


5


No


I


5


D


12


17


15


20


3


No


J


2


F, G, H, I


20


22


20


22


0


Yes


 


The ES or the earliest activities start time. The ES for the activity J which has 20 days to finish and it will pass through the points B, C, F which has number of days of 3,3, and 14 respectively. The other variables can be computed and can be seen in the table above.


 


 


c.


The critical can be seen to the drawing below:



The project can be completed in 20 days.


   


 


 


 


 


 


 


 


 


 


 



Credit:ivythesis.typepad.com


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