Question

Carl's Construction Inc. will buy a machine that has an initial cost of $1,200,000 and is expected to be useful for 8 years. Several estimations have been created but the company is going to focus on 3 main scenarios. One scenario estimates yearly benefits of $310,000 and O&M costs of $60,000 per year with a 30% chance of occurrence. The second scenario involves benefits of $280,000 per year and O&M costs of $70,000 annually with a 50% chance of occurring. The last scenario involves annual benefits of $240,000 and annual O&M costs of $80,000 with a probability of 20%. The company uses an interest rate of 7%. What is the standard deviation of this project?

Answer 1

First of All we should find NPV of project. on the basis of NPV we find out Standard deviations

initial outflow = $210000

r = 7%

n = 8 years

Net present worth = Initial outflow - Net cash flow of n years

=> Net present worth of scenario 1 :-

NPV = -$1200000 + $310000(P/A 7% 8 years) - $60000(P/A 7% 8 years)

= -$1200000 + $310000 (5.971) -$60000(5.971)

= -$1200000 + $1,851,010 - $358260

= $292750

=> Net present value of scenario 2:-

NPV = -1200000 + $280000( 5.971) - $70000(5.971)

= -$1200000 + $1671880 - $417970

= $53910

=> Net present value of scenario 3 :-

NPV = -$1200000 + $240000(5.971) - $80000(5.971)

= -$1200000 + $1433040 - $477680

= -$244640

=> Expected Net present value = sum of ( probability * NPV of scenario)

Expected NPV = ($292750*0.30 ) + ($53910*0.50 ) + (-$244640*0.20)

= $87825 + $26955 - $48928

= $65852

so Expected NPV of project is $65852

=> Standard deviations :-

SD = √ pi ( NPV - Expected NPV)^2

= √[( 0.30(292750-65852)^2 )+ (0.50 (53910-65852)^2) + (0.20 ( -244640 - 65852) ^2)]

= √ [15444810721.2 + 71305682 + 19281056412.8 ]

= √ [ 34797172816]

= $186540

so the standard deviations of project is $186540