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Alexxx [7]
1 year ago
6

A firm is considering projects s and l, whose cash flows are shown below. These projects are mutually exclusive, equally risky,

and not repeatable. The ceo wants to use the irr criterion, while the cfo favors the npv method. You were hired to advise the firm on the best procedure. If the wrong decision criterion is used, how much potential value would the firm lose?.
Mathematics
1 answer:
antiseptic1488 [7]1 year ago
6 0

If the wrong discion criterion is used, the potential lose in value for the firm is $209.07.

<h3>What is the correct decision criterion?</h3>

The  correct decision criterion is the net present value. Net present value is the present value of after-tax cash flows from an investment less the amount invested. The internal rate of return is the discount rate that equates the after-tax cash flows from an investment to the amount invested

<h3>What is the loss in value if the wrong decision criteria is used?</h3>

The net present vaue and the internal rate of return can be calculated using the financial calcuator.

Project S:

  • Cash flow in year 0 = $-1025
  • Cash flow each year from year 1 to 4 = $380

I = 6%

NPV = 291.74

IRR = 17.86%

Project I:

Cash flow in year 0 = $-2150

Cash flow each year from year 1 to 4 = 765

I = 6%

NPV = 500.81

IRR = 15.78%

If the wrong criteria is used, Project S would be chosen.

Loss in value = 500.81 - 291.74 = $209.07.

Here are the cash flows used:

WACC: 6.00%

Year 0 1 2 3 4

CF S -$1,025 $380 $380 $380 $380

CF I -$2,150 $765 $765 $765 $765

To learn more about net present value, please check: brainly.com/question/25748668

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a)    tan (157.5) = \frac{1-cos 315}{sin315}

b)

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c)

      sin^{2} (157.5) = \frac{1-cos (315) }{2}

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  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

<u><em>Step(i):-</em></u>

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

<u><em>Step(i):-</em></u>

Given

              tan\alpha = \frac{sin\alpha }{cos\alpha }

          we know that trigonometry formulas

        sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

         1- cos∝ =  2 sin² ∝/2

      Given

         tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }

put ∝ = 315

      tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }

     multiply with ' 2 sin (∝/2) both numerator and denominator

        tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2}  }{2sin(\frac{315}{2} cos(\frac{315}{2}) }

Apply formulas

 sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )

  1- cos∝ =  2 sin² ∝/2

now we get

 tan (157.5) = \frac{1-cos 315}{sin315}

       

b)

          sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 330° above formula

             sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}

            sin^{2} (165) = \frac{1-cos (330) }{2}

            sin (165) =\sqrt{ \frac{1-cos (330) }{2}}

c )

         sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}

               put ∝ = 315° above formula

             sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}

            sin^{2} (157.5) = \frac{1-cos (315) }{2}

           

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     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       cos 330 = 1 - 2sin^{2} (\frac{330}{2} )

      cos 330° = 1- 2 sin² (165°)

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