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zavuch27 [327]
3 years ago
14

Project Q has an initial cost of $257,412 and projected cash flows of $123,300 in Year 1 and $180,300 in Year 2. Project R has a

n initial cost of $345,000 and projected cash flows of $184,500 in Year 1 and $230,600 in Year 2. The discount rate is 12.2 percent and the projects are independent. Which project(s), if either, should be accepted based on its profitability index value?
a) Reject both Project Q and R

b) Accept Project R and reject Project Q

c) Accept either Project R or Project Q, but not both

d) Accept Project Q and reject Project R

e) Accept both Project Q and R
Business
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

b) Accept Project R and reject Project Q

Explanation:

We can use the following method to solve the given problem in the question

We are given

Project Q: Initial Cost = $ 257,412

Projected Cash Flows: Yr 1 : $ 123,300 Yr 2 : $ 180,300

Total Present Value of all the Future Cash Flows using 12.2% as Rate of Return

= 123,300/1.122 + 180,300/(1.122*1.122)

= 109,893 + 143,222

= $ 253,115

Profitability Index = Total Present Values of all Cash Inflows / Initial Investment

= 253,115 / 257142 = 0.98

Since the Initial Investment is greater than the Present Value of Cash Inflows, that is, l Profitability Index < 0 the Project should not be selected.

Project R: Initial Cost = $ 345,000

Projected Cash Flows: Yr 1 : $ 184,500 Yr 2 : $ 230,600

Total Present Value of all the Future Cash Flows using 12.2% as Rate of Return

= 184,500/1.122 + 230,600/(1.122*1.122)

= 164,438.5 + 183,178

= $ 347,616.5

Profitability Index = Total Present Values of all Cash Inflows / Initial Investment

= 347,616.5 / 345,000 = 1.01

Since the Initial Investment is lower that the Present Value of the Cash Inflows, that is, Profitability Index > 0 the Project should be selected.

Accept Project R and Reject Project Q, so option B is the correct answer

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Consider the following linear program: Min s.t. 8X + 12Y 1X + 3Y &gt;= 9 2X + 2Y &gt;= 10 6X + 2Y &gt;= 18 A, B &gt;= 0 a. Use t
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Answer: Graph of (A) (B) and {D) are attached accordingly.

Explanation:

A)

The critical region of the constraints can be seen in the following diagram -

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The intersection points are found by using these equations -

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 48

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So, we can see the minimum value of the objective function occurs at point (3,2) and the minimum value of the objective function is = 48.

------------------------------------------------------------------------------------------------------------------------------------------------------------------

B)

When we change the coefficients of the variables in the objective function, the optimal solution may or may not change as the weights (coefficient) are different for each constraints for both the variabls. So, it all depends on the coefficient of the variables in the constraints.

In this case, the optimal solution does not change on changing the coefficient of X from 8 to 6 in the objective function.

The critical region would remain same (as shown below) as it is defined by the constraints and not the objective function.

(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white

However, the optimal value of the objective function would change as shown below-

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 42

(9,0) x+3y = 9; y = 0 54

(2,3) 2x+2y = 10; 6x+2y = 18 48

(0,9) 6x+2y = 18; x = 0 108

So, we can see that the minimum value now has become 42 (which had to change obviously).

-------------------------------------------------------------------------------------------------------------------------------------------------------

C)

Now, when we change the coefficient of the variable Y from 12 to 6, again the critical region would remain same as earlier. But in this case, the optimal solution changes as shown below -

Vertex Lines Through Vertex Value of Objective

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(9,0) x+3y = 9; y = 0 72

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(0,9) 6x+2y = 18; x = 0 54

We can see that the minimum value now occurs at (2,3) which is 34, so both the optimal solution and optimal value have changed in this case.

----------------------------------------------------------------------------------------------------------------------------------------------------------

D)

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So, the new critical points are (4,12), (4,24), (8,24) and (8,12).

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Vertex Value of Objective

(4,12) 8*4+12*12 = 176

(4,24) 8*4+12*24 = 320

(8,24) 8*8+12*24 = 352

(8,12) 8*8+12*12 = 208

So, the new optimal solution is (4,12) and the optimal value is 176.

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