1. a. The quantitative methods are research procedures which have numbers anything that are measurable. This is also the scientific method for solving the problems and has the goal of achieving the best possible solution which is under the certain restraining conditions. This is also applied in different field and its applications can be found in economic sciences and the business.


  b. The other two names which can be applied to quantitative methods are the operations research and management science.


 


2. (C4*A2)/A1


3. C4 * (C5 – c6)


4.


 


 


 


Marginal


 


A


B


Probability


X


.10


0.40


.50


  Y


0.30


0.20


0.50


Marginal


 


 


 


Probability


0.40


0.60


1.00


 


5. a. Pr. (A1 and A2) = (0.6 x 0.40 = 0.24


    b. Pr. (A2 and A3) = (0.4 x 0.4) = 0.16


    c. Pr. (A1 and A2 and A3) =(0.6 x 0.3 x 0.3) =  0.054


6. a. Since 60 percent are women then the probability that they will take all the prices is 0.6.


   b. Since the remaining percentage of men is 40% then the probability that all the men can get all prices is 0.40.


   c. (1/3) for the prices and (0.40) for the percentage of men, then the probability that at least one price can be won by man is 0.13333


  d. (2/3) for the prices and (0.60) for the percentage of girls is 0.4


7. a. In solving the payoff measure and the profit the faucet A, B and C needs to determine the total cost and its total revenue.


For the faucet A, if its sells (2,000 x 20 as the variable cost) = 40,000 plus the 20, 000 fixed cost which is equal to 60,000. If they sell the product by per faucet, then the profit will be ,000. Same is true to the other number of units to be sold.


The payoff table is given below with the profit as the measurement.


 


 


        Acts


 


    Events


2,000


7,000


12,000


Faucet A


20,000


50,000


100,000


Faucet B


10,000


90,000


190,000


Faucet C


-50,000


75,000


200,000


     


      b. The decision tree diagram



 


 


8. Minimize C = 6X1 + 3X2


      Subject to            4X1 + 3X2> 18          (restriction A)


                        2X1 + 3X2> 12          (restriction B)


      where             X1, X2  >  0


Graph the explicit equation and assume it was equal.


Let the X1 = x


            X2 = y


Determine the 2 equations and graph.


4x + 3y = 18


(0, 6) and (4 ½ , 0)


2x + 3y = 12


(0, 4) and (6, 0)


Determine the point of intersection also,


4x + 3y = 18


[2x + 3y = 12] -2


2x = 6


   x = 3


Substitute the value of x


4(x) + 3y = 18


4(3) + 3y = 18


12 + 3y = 18


y = 2


 


 



To determine the minimum point, substitute the coordinated to the objective function.


a. P (0, 6)


Min. Cost = 6x + 3y


                  = 6(0) + 3(2)


                  = 18


b. P (6,0)


Min. Cost = 6x + 3y


                  = 6(3) + 3(2)


                  = 24


c. P (6,0)


Min. Cost = 6x + 3y


                  = 6(6) + 3(0)


                  = 36


The minimum Cost is the 18 units of currency and at the point of P (0,6)


 


 


9. . Solve the following linear program graphically:      


 


      Maximize P = 4X1 + 1X2                                              


      Subject to              2X1 + 4X2< 24   (resource A)                  


                          6X1 + 3X2< 36        (resource B)                        


                            Xl          <   5   (resource C)                                                         


                                    2X2< 10   (resource D)                                           


      where                           X1, X2> 0          


 


Graph the equations and assumed it is equal.


Let the X1 = x


             X2 = y


Graph the given equations:


2x + 4y = 24


6x + 3y = 36


            x = 5


            2y = 10 or y = 5


Get the point of intersection of 2x + 4y = 24   and   6x + 3y = 36.


[2x + 4y = 24] -3       


6x + 3y = 36


            -9y = -36


            y = 4


2x + 4y = 24


2x + 4(4) = 24


2x = 8


 x = 4


 


 



Determine the maximum P by substituting the value of the point of intersection.


Maximize


P = 4x + y


P (0, 5)


P = 4(0) + 5


P = 5


 


P (4, 4)


P = 4(4) + 4


P = 20


 


P (5,0)


 = 4(5) + 0


P = 20


 


P (5,2)


P = 4(5) + 2


P = 22


 


P = 4(2) + 5


P = 13


 


The maximum P therefore is 22 at the point of P (5,2).


 


 


 


 


 


 


 


 


 


 


 


 


 


10. a.  The column and row numbers


          To


From


 


J


 


K


 


L


 


M


 


N


 


Capacity


 


 


5


6


7


8


7


 


 


A


 


 


 


100


 


 


100


rA = 1


 


6


5


8


9


7


 


 


B


 


100


 


100


 


0


200


rB = 2


 


3


4


8


9


6


 


 


C


 


 


 


 


250


 


250


rC = 3


 


6


7


4


5


6


 


 


D


50


 


150


0


 


 


200


rD = 4


Demand


50


100


150


200


250


750


 


 


kJ =  1


kK = 2


kL = 3


kM = 4


kN = 5


 


 


 


The best entering cell is row c and column n or row column 4 which has the total of 250.


The total cost will be:


50 (6) + 100(5) + 150(4) + 100(8) + 100(9) + 0(5) + 0 + 250 (6) = 4,600 unit currency


 


 


b. The new feasible solution is:


          To


From


 


J


 


K


 


L


 


M


 


N


 


Capacity


 


 


5


6


7


8


7


 


 


A


50


50


 


 


 


 


100


 


 


6


5


8


9


7


 


 


B


 


50


150


 


 


 


200


 


 


3


4


8


9


6


 


 


C


 


 


 


200


50


 


250


 


 


6


7


4


5


6


 


 


D


 


 


 


 


 


200


200


 


Demand


50


100


150


200


250


750


 


 


 


 


 


 


 


 


 


 


c. The total cost of the shipment will be:


50 (5) + 50 (6) + 50 (5) + 150 (8) + 200 (9) + 50 (6) + 200 (6) = 5,300 currency units


This is not yet the feasible solution because it has higher cost as compared to the first one this can also be evaluated further.


 


Extra Credit


1. 95% of 20 is 19 then the probability that all of the chosen will be satisfactory is 3/19.


2.


3. The faucet C of 12,000 and probability of 0.4 will maximize its profit for up to 80,000 unit currency


4. Let X1 = x and X2 = y



 


Test the given points using the objective function:


Maximize P = 4x + 6y


P (0,5)


                     = 4(0) + 6(5)  = 30


P (1, 5)


                      = 4 (1) + 6(5) = 34


P (3,3)


                       = 4(3) + 6(3) = 30


P (4, 3/2)


                        = 4 (4) + 6 (3/2) = 25


P (4, 0)


                        = 4(4) + 0 = 16


Therefore the highest P is 34 in the point (1,5).


 


 


 



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