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).
Credit:ivythesis.typepad.com
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