Solution manual engineering economic analysis 9th edition ch10

The telephone lines are removed from the pole and put underground.. in California are: Optimistic Life: 59 years Most Likely Life: 28 years Pessimistic Life: 2.5 years Recognizing there

Chapter 10: Uncertainty in Future Events 10-1

(a) Some reasons why a pole might be removed from useful service:

1 The pole has deteriorated and can no longer perform its function of safely

supporting the telephone lines

2 The telephone lines are removed from the pole and put underground The poles,

no longer being needed, are removed

3 Poles are destroyed by damage from fire, automobiles, etc

4 The street is widened and the pole no longer is in a suitable street location

5 The pole is where someone wants to construct a driveway

(b) Telephone poles face varying weather and soil conditions; hence there may be large variations in their useful lives Typical values for Pacific Telephone Co in California are:

Optimistic Life: 59 years Most Likely Life: 28 years Pessimistic Life: 2.5 years Recognizing there is a mortality dispersion it would be possible, but impractical, to define optimistic life as the point where the last one from a large group of telephone poles is

removed (for Pacific Telephone this would be 83.5 years) This is not the accepted

practice Instead, the optimum life is where only a small percentage (often 5%) of the group remains in service Similarly, pessimistic life is when, say, 5% of the original group

of poles have been removed from the group

10-2

If 16,000 km per year, then fuel cost = oil/tires/repair = $990/year, and salvage value = 9,000

- 5x16,000x.05 = 9,000 – 4,000 = 5,000

= 9,000x.2505 + 1,980 – 5,000x.1705

= 2,254.5 + 1,980 - 852.5 = $3,382 Increasing annual mileage to 24,000 is a 50% increase so it increases operating costs by 50% The salvage value drops by 5x8,000x.05 = 2,000

= 9,000x.2505 + 1.5x1,980 - 3000x.1705

= 2,254.5 + 2,970 - 511.5 = $4,713 Decreasing annual mileage to 8,000 is a 50% decrease so it decreases operating costs by 50% The salvage value increases by 5x8,000x.05 = 2,000

= 9,000x.2505 + 5x1,980 – 7,000x.1705

= 2,254.5 + 990 - 1193.5 = $2,051

Mean Life = (12 + 4 x 5 + 4)/6 = 6 years

PW of Cost = PW of Benefits

$80,000 = $20,000 (P/A, i%, 6)

Rate of Return is between 12% and 15%

Rate of Return ≈ 13%

10-4

Since the pessimistic and optimistic answers are symmetric about the most likely value of 16,000, the weighted average is 16,000 km If 16,000 km per year, then fuel cost = oil/tires/repair = $990/year, and salvage value = 8,000 - 5x16,000x.05 = 9,000 – 4,000 = 5,000

= 9,000x.2505 + 1980 – 5,000x.1705

= 2,254.5 + 1,980 - 852.5 = $3,382

10-5

There are six ways to roll a 7: 1 & 6, 2 & 5, 3 & 4, 4 & 3, 5 & 2, 6 & 1

There are two ways to roll an 11: 5 & 6 or 6 & 5

Probability of rolling a 7 or 11 = (6 + 2)/36 = 8/36

10-6

Since the Ps must sum to 1: P(30K) = 1 - 2 - 3 = 5

E(savings) = 3(20K) + 5(30K) + 2(40K) = $29K

10-7

Since the Ps must sum to 1: P(20%) = 1 - 2/10 - 3/10 = 5

E(i) = 2(10%) + 3(15%) + 5(20%) = 16.5%

10-8

In Between Weather 300 Days 0.5 = 1 – 0.2 – 0.3

E(days) = 20(250) + 5(300) + 3(350) = 305 days

If you have another accident or a violation this year, which has a 2 probability, it is assumed

to occur near the end of the year so that it affects insurance rates for years 1-3 A violation

in year 1 affects the rates in years 2 and 3 only if there was no additional violation in this year, which is P(none in 0)P(occur in 1) = 8.2 = 16 So the total probability of higher rates for year 2 is 2 + 16 or 36 This also equals 1 - P(no violation in 0 or 1) = 1 - 82

For year 3, the result can be found as P(higher in year 2) + P(not higher in year 2) P(viol in year 2) = 36 + 64.2 = 488 This also equals either 1 - P(no violation in 0 to 2) =

1 - 83

Rates for Year 0 1 2 3

36 488

10-10

Distribution

Expected Grade Point

Grade Distribution

Expected Grade Point

To minimize the Expected Grade Point, choose instructor A

10-11

Expected outcome= $2,000 (0.3) + $1,500 (0.1) + $1,000 (0.2)

+ $500 (0.3) + $0 (0.1) = $1,100

10-12

The sum of probabilities for all possible outcomes is one

An inspection of the Regular Season situation reveals that the sum of the probabilities for the outcomes enumerated is 0.95 Thus one outcome (win less than three games), with probability 0.05, has not been tabulated This is not a faulty problem statement The

student is expected to observe this difficulty

Similarly, the complete probabilities concerning a post-season Bowl Game are:

Probability of playing = 0.10

Probability of not playing = 0.90

Expected Net Income for the team

= (0.05 + 0.10 + 0.15 + 0.20) ($250,000) + (0.15 + 0.15 + 0.10) ($400,000)

+ (0.07 + 0.03) ($600,000) + (0.10) ($100,000)

= 0.50 ($250,000) + 0.40 ($400,000) + 0.10 ($600,000) + 0.10 ($100,000)

+ 0.90 ($0)

= $355.00

10-13

Determine the different ways of throwing an 8 with a pair of dice

Die 1 Die 2

The five ways of throwing an 8 have equal probability of 0.20

The probability of winning is 0.20

The probability of losing is 0.80

The outcome of a $1 bet = 0.20 ($4) + 0.80 ($0) = $0.80

This means a $0.20 loss

10-14

E(PWextra costs) = 2600(P/F,8%,1) +.36600(P/F,8%,2) +.488600(P/F,8%,3)

= 2600.9259 + 36600.8573 + 488600.7938 = $528.7

10-15

Leave the Valve as it is

Expected PW of Cost = 0.60 ($10,000) + 0.50 ($20,000)

+ 0.40 ($30,000)

= $28,000

Repair the Valve

Expected PW of Cost = $10,000 repair + 0.40 ($10,000)

+ 0.30 ($20,000) + 0.20 ($30,000)

= $26,000

Replace the Valve

Expected PW of Cost = $20,000 replacement + 0.30 ($10,000)

+ 0.20 ($20,000) + 0.10 ($30,000)

= $30,000

To minimize Expected PW of Cost, repair the valve

Expected number of wins in 100 attempts = 100/38 = 2.6316

Results of a win = 35 x $5 + $5 bet return = $180.00

Expected winnings = $180.00 (2.6313) = $473.69

Expected loss = $500.00 - $473.69 = $26.31

10-17

Do Nothing

EUAC = Expected Annual Damage

= 0.20 ($10,000) + 0.10 ($25,000) = $4,500

$15,000 Building Alteration

Expected Annual Damage = 0.10 ($10,000) = $1,000

Annual Floodproofing Cost = $15,000 (A/P, 15%, 15) = $2,565

EUAC = $3,565

$20,000 Building Alteration

Annual Floodproofing Cost = $20,000 (A/P, 15%, 15) = $3,420

EUAC = $3,420

To minimize expected EUAC, recommend $20,000 building alteration

10-18

Height above

roadway

Annual Probability

of Flood Damage x Damage

= Expected Annual Damage

Height

above

roadway Initial Cost

x (A/P, 12%, 50)

= EUAC of Embankment

Expected Annual Damage

Total Expected Annual Cost

Select the 3-metre embankment to minimize total Expected Annual Cost

E(first cost) = 300,000(.2) + 400,000(.5) + 600,000(.3) = $440K

E(net revenue)= 70,000(.3) + 90,000(.5) + 100,000(.2) = $86K

E(PW) = -440K + 86K(P/A,12%,10) = $45.9K, do the project

10-20

E(savings) = 2(18K) + 7(20K) + 10(22K) = $19,800

E(life) = (1/6)(4) + (2/3)(5) + (1/6)(12) = 6 years

0 = -81,000 + 19,800(P/A,i,6)

(P/A,i,6) = 81,000/19,800 = 4.0909

(P/A,12%,6) = 4.111 & (P/A,13%,6) = 3.998

i = 12 + 01(4.111 - 4.0909)/(4.111 - 3.998) = 12.18%

Because (P/A,i,N) is a non-linear function of N, the use of 6 years for the expected value of

N is an approximation The discounting by i has much more impact in the 12 year life, so

that we expect the true IRR to be less than 12.18%

0 = -81K + 19.8K(1/6)(P/A,i,4) + 19.8K(2/3)(P/A,i,5) + 19.8K(1/6)(P/A,i,12)

PW12% = -81K + 19.8K(1/6) 3.037 + 19.8K(2/3) 3.605 + 19.8K(1/6) 6.194

= -2952

PW11% = -81K + 19.8K(1/6) 3.102 + 19.8K(2/3) 3.696 + 19.8K(1/6) 6.492

= -553

i = 11 - 01(553)/(553 + 2952) = 10.84%

10-21

Since $250,000 of dam repairs must be done in all alternatives, this $250,000 can be included or ignored in the analysis Here it is ignored (Remember, only the differences between alternatives are relevant.)

Flood Probability of

damage in any year = 1/yr flood

Downstream Damage

Spillway Damage

Alternative I: Repair existing dam but make no other alterations

Spillway damage: Probability that spillway capacity equaled or exceeded in any year is

0.02 Damage if spillway capacity exceed: $250,000

Expected Annual Cost of Spillway Damage = $250,000 (0.02)

= $5,000

Downstream Damage during next 10 years:

Flood Probability

that flow*

will be equaled or exceeded

Damage ∆ Damage over

more frequent flood

Annual Cost

of Flood Risk

Next 10 year expected annual cost of downstream damage = $13,000

Downstream Damage after 10 years: Following the same logic as above,

Expected annual cost of downstream damage

= $2,000 + $3,000 + 0.1 ($2,000,000 - $200,000)

= $23,000

Present Worth of Expected Spillway and Downstream Damage

PW= $5,000 (P/A, 7%, 50) + $13,000 (P/A, 7%, 10)

+ $23,000 (P/A, 7%, 40) (P/F, 7%, 10)

= $5,000 (13.801) + $13,000 (7,024) + $23,000 (13.332) (0.5083)

= $316,180

Equivalent Uniform Annual Cost

Annual Cost = $316,180 (A/P, 7%, 50)

= $316,180 (0.0725)

= $22,920

* An N-year flood will be equaled or exceed at an average interval of N years

Alternative II: Repair the dam and redesign the spillway

Additional cost to redesign/reconstruct the spillway = $250,000

PW to Reconstruct Spillway and Expected Downstream Damage

Downstream Damage- same as alternative 1

PW = $250,000 + $13,000 (P/A, 7%, 10)

+ $23,000 (P/A, 7%, 40) (P/F, 7%, 10)

= $250,000 + $13,000 (7.024) + $23,000 (13.332) (0.5083)

= $497,180

EUAC = $497,180 (A/P, 7%, 50)

= $497,180 (0.0725)

= $36,050

Alternative III: Repair the dam and build flood control dam upstream

Cost of flood control dam = $1,000,000

EUAC = $1,000,000 (A/P, 7%, 50)

= $1,000,000 (0.7225)

= $72,500

Note: One must be careful not to confuse the frequency of a flood and when it might be expected to occur The occurrence of a 100-year flood this year is no guarantee that it won’t happen again next year In any 50-year period, for example, there are 4 chances in 10 that

a 100-year flood (or greater) will occur

Conclusion: Since we are dealing with conditions of risk, it is not possible to make an

absolute statement concerning which alternative will result in the least cost to the

community Using a probabilistic approach, however, Alternative I is most likely to result in the least equivalent uniform annual cost

10-22

The $35K is a sunk cost and should be ignored

a E(PW) = $5951

b P(PW<0) = 3 and σ = $65,686

Net

Revenue $-15,000 $15,000 $20,000

PW^2  Prob 2,263,491,770 360,778,191 1,725,760,288 $65,686 σPW

10-23

a The $35K is still a sunk cost and should be ignored Note: P(PW<0) = 3 and N = 1 used

for PWbad since termination allowed here This improves the EPW by 18,918 - 5951 =

$12,967 This also equals the E(PW) of the avoided negative net revenue in years 2 –

5, which equals 3 (1/1.1)x15,000(P/A,.1,4).

b The P(loss) is unchanged at 3 However, the standard deviation improves by 65,686 - 47,957 = $17,709

Net Revenue $-15,000 $15,000 $20,000

PW^2  Prob 571,239,669 360,778,191 1,725,760,288 $47,95

Since the expected life is not an integer, it is easier to use a spreadsheet table to calculate

each PW For example, the first row's PW = -80K + 15K(P/A,9%,3)

Savings/

yr

Expected Value $7,627

10-25

If the savings were only $15K per year, spending $50K for 3 more years would not make sense For the two or three shift situations, the table from 10-24 can be modified for 3 extra years, and to include the $50K at the end of 3 or 5 years For example, the first and second

rows' PWs are unchanged The third row's PW = -80K + 15K(P/A,9%,6) - 50K(P/F,9%,3)

Savings/

yr

Expected Values

26,252

The option of extending the life is not used for single shift operations, but it increases the expected PW by 26,252 - 7,627 = $18,625

10-26

Al’s Score was x + (5/20) s = x + 0.25 s

Bill’s Score was x + (2/4) s = x + 0.50 x

Therefore, Bill ranked higher in his class

PW2 43,164,900 73,788,100 94,672,900 68,778,100

σPW = (68,778,100 - 82122)1/2 = $1158

10-28

PW1 = -25,000 + 7000(P/A,12%,4) = -$3739

PW2 = -25,000 + 8500(P/A,12%,4) = $817

PW3 = -25,000 + 9500(P/A,12%,4) = $3855

From the table the E(PW) = $361.9

σPW = (8,918,228 - 361.92)1/2 = $2964

Annual

Savings

0

668,256 14,859,62

8

8,918,228

10-29

To calculate the risk, it is necessary to state the outcomes based on the year in which the next accident or violation occurred

year of 2 nd

offence

σPW = (664,260 - 5292)1/2 = $620.0

10-30

For example, the first row's PW = -300K + 70K(P/A,12%,10)

First

Cost

P Net Revenue

-600 3 70 3 0.09 -204.5 -18.41 3,764

Expected Values

45.90 18,468

Risk can be measured using the P(loss), range, or the standard deviation of the PWs

P(loss) = 15 + 09 + 15 + 06 = 45

The range is -204.5K to $265K

The standard deviation is σPW = (18,468 - 45.902) = $127.9K

10-31

a The probability of a negative PW is 18 + 12 + 3 = 6

Savings/

yr

15,000 3 3 6 0.18 -42,031 -7,566 317,982,538

15,000 3 5 4 0.12 -21,655 -2,599 56,273,884

Expected Values

7,627 1,508,922,7

38 Risk can also be measured using the standard deviation of the PWs The standard deviation is σPW = (1,508,922,738 - 76272) = $38,089

b Extending the life for 2 & 3 shift operations reduces the probability of a negative PW by

3 to 3

Savings/

yr

15,000 3 3 6 0.18 -42,031 -7,566 317,982,538

15,000 3 5 4 0.12 -21,655 -2,599 56,273,884

45,000 2 8 4 0.08 136,570 10,926 1,492,115,5

47 Expected

Values

26,252 3,348,157,1

18

Risk can also be measured using the standard deviation of the PWs The standard deviation is increased by $13,477 This illustrates why standard deviation alone is not the best measure of risk Extending the life makes the project more attractive, and

increases the spread of the possible values The standard deviation is higher, but the P(loss) has dropped by half

σPW = (3,348,157,118 - 26,2522) = $51,565

10-32

(a) Expected fire loss in any year= 0.010 ($10,000) + 0.003 ($40,000)

+ 0.001 ($200,000)

= $420.00 (b) The engineer buys the fire insurance because

1 a catastrophic loss is an unacceptable risk

or 2 he has a loan on the home and fire insurance is required by the lender

10-33

Project IRR Std.Dev

10-34

Project IRR Std.Dev