Econ

4. Solution is in the Excel file named Problem 4

4.

Let

p= 1+ the rate of return

Step 1:

Convert all terms to Future Worth. Every two years the  values must equal the original amount invested since that is the minimum rate of return required for the process to continue on forever. Every value is converted to future worth so that the all values are comparable with the 3820 in the future.

The 3820 in year 0 is a present value which gets interest for 2 years

The 500 in year 1 if considered a present value gets interest for 1 year

The 250 is a future value and so it is left alone

Step 2:

Add the values together and set them equal to the -3820 which is what is required for the process to continue.

-3820p2+500p+250=-3820

Step 3:

The equation is rearranged

-3820p2+500p +4070=0

Step 4:

The equation is solved for p using quadratic formula

p=1.099721622

Step 5:

One is subtracted from p as defined above in order to get the rate of return

Therefore, the rate of return is equal to 0.099721622

6. Solution is in the Excel file named Problem 5

7.

Option A

An present cost: $500,000

An annual cost: $20,000

n= 12

An annual inflow: $100,000

i=11

Annual payment= P1-1(1+i)ni =P / [ ( 1 - (1/(1+i)^n) ) / i ]

Annual equivalent of the present cost=500000 1-1(1+0.11)120.11 =77,013.64

Total Annual equivalent amount=$100,000-$20,000-$77,013.64

                                              =$2,986.36

Option B

An present cost: $750,000

An annual cost: $40,000

n= 20

An annual inflow: $120,000

i=11

Annual payment=  P1-1(1+i)ni

Annual equivalent of the present cost=7500001-1(1+.11)200.11 =94,181.73

Total Annual equivalent amount=$120,000-$40,000-$94,181.73

                                            = -$14,181.73

b)

IRR should not be used as the time periods being compared are different which may result in the IRR being misleading.

See the excel sheet for calculation using Goal Seek

Option A:

IRR: 0.1180758Option B: IRR: 0.086291 |

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