Question

In: Civil Engineering

c) Use your spreadsheet to estimate the position of the Risk line, using any necessary “reality...

c) Use your spreadsheet to estimate the position of the Risk line, using any necessary “reality
checks”. In other words, justify or even make a clear statement of the inadequacies of what
you have done and the need for further investigation. Estimate the Risk value.

Solutions

Expert Solution

Solution:

Concept of Risk

Broadly we say any situation / outcome is “risky” when we do not know in advance what could be the actual state(s) of the world for that situation / outcome

• For e.g. – Two friends, Humpty & Dumpty takes a bait among each other by tossing a coin. The bait is - if it turns up Head, Humpty will pay Dumpty Rs. 1000 and if it turns up Tail, Dumpty will give Humpty Rs. 1000

• Now we can define the situation / outcome in terms of the appearance of either Head or a Tail in tossing an unbiased coin

And we can define the states of the world based on the situation / out come as - either Humpty becomes richer than Dumpty (if it is Tail) or Dumpty is richer than Humpty (if it is Head)

Example: The heating, ventilating, and air-conditioning (HVAC) system in a commercial building has become unreliable and inefficient. Rental income is being hurt, and the annual expenses of the system continue to increase. Your engineering firm has been hired by the owners to (1) perform a technical analysis of the system, (2) develop a preliminary design for rebuilding the system, and (3) accomplish an engineering economic analysis to assist the owners in making a decision. The estimated capital-investment cost and annual savings in operating and maintenance expenses, based on the preliminary design, are shown in the following table. The estimated annual increase in rental revenue with a modern HVAC system has been developed by the owner’s marketing staff and is also provided in the following table. These estimates are considered reliable because of the extensive information available. The useful life of the rebuilt system, however, is quite uncertain. The estimated probabilities of various useful lives are provided. Assume that MARR = 12% per year and the estimated market value of the rebuilt system at the end of its useful life is zero. Based on this information, what are E(PW), V(PW), and SD(PW) of the project’s cash flows? Also, what is the probability of PW ≥ 0? What decision would you make regarding the project, and how would you justify your decision using the available information? Solve by spreadsheet.

a

b

c

d

e

f

f ^ 2

g

Series Present

X = (4 x b)

f(x)

x. f(x) = (c x d)

x – E(x)

f (x) . {(x – E (x)} ^ 2 =

Worth Factor

{(x – E (x)} ^ 2

d x f ^2

Square of

Useful

PW @ 12 % as

Deviation of PW @

Deviation of PW

Life

a fn of useful

Expected PW as a

12 % as a fn of

@ 12 % from its

(A/P,12%,N)

life

Probability

expected value

fn of useful life

useful life from

(Random

Expected Value

as a fn of useful

Variable x)

life from

Expected Value

12

6.1944

-27925.76

0.10

-2792.58

-37909.10

1437100014.45

143710001.44

13

6.4235

-9689.40

0.20

-1937.88

-19672.74

387016777.80

77403355.56

14

6.6282

6604.72

0.30

1981.42

-3378.62

11415086.62

3424525.99

15

6.8109

21147.64

0.20

4229.53

11164.30

124641549.83

24928309.97

16

6.9740

34130.40

0.10

3413.04

24147.06

583080410.06

58308041.01

17

7.1196

45720.16

0.05

2286.01

35736.82

1277120160.77

63856008.04

18

7.2497

56076.12

0.05

2803.81

46092.78

2124544183.76

106227209.19

Total Expected PW = E(x)

9983.34

Variance of PW = V (x)

477857451.19

Standard Deviation of PW = V(x) ^ 0.5

21859.95

Interpretation of the result

Based on the PW of the project as a function of N , the probability of the PW being ≥ 0 is Pr { PW ≥ 0} = 1 − (0.1 + 0.2) = 0.7= 0.3+0.2+0.1+0.05+0.05 .

The results of the engineering economic analysis indicate that the project is a questionable business action.

The E(PW) of the project is positive ($9,984) but small relative to the large capital investment.

Also, even though the probability of the PW being greater than zero is somewhat favorable (0.7), the SD(PW) value is large [over two times the E(PW) value]

Therefore, it is project with positive expected return but with high degree of volatility as well


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