The probabilities that the life of a machine will vary from 6 to 12 years are given in the table below. The expected life of the machine is ____________. \[\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline \text { Life, Years } & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Probability } & 0.10 & 0.15 & 0.2 & 0.15 & 0.20 & 0.10 & 0.05 & 0.05 \\ \hline \end{array}\]

A) 8.70 years B) 9 years C) 10.15 years D) 7.95 A) 8.70 years B) 9 years C) 10.15 years D) 7.95

MKBIC Systems is evaluating four projects A, B, C and D that have risks associated with the producing benefits. Based on the data given in the table below, which project is the best alternative? \[\begin{array} { | l | l | l | l | l | l | l | l | l | } \hline { \text { Praject A}}&&\text{project B } &&{ \text { Project C } } && { \text { Project D } } \\ \hline \text { EUAW } & \text { Prob. } & \text { EUAW } & \text { Prob. } & \text { EUAW } & \text { Prob. } & \text { EUAW } & \text { Prob. } \\ \hline \$ 2,500 & 0.3 & \$ 3,000 & 0.1 & - \$ 5,000 & 0.25 & \$ 4,000 & 0.35 \\ \hline \$ 1,800 & 0.45 & - \$ 2,500 & 0.3 & \$ 6,500 & 0.45 & \$ 2,500 & 0.4 \\ \hline \$ 3,200 & 0.25 & \$ 4,000 & 0.6 & \$ 2,000 & 0.3 & - \$ 1,500 & 0.25 \\ \hline \end{array}\]

A) Project A B) Project B C) Project C D) Project D A) Project A B) Project B C) Project C D) Project D

Table 10 \[\begin{array} { | l | l | l | l | l | } \hline \text { First Cost, } \$ & \$ 60,000 & \$ 80,000 & \$ 100,000 & \$ 120,000 \\ \hline \text { Probability } & 0.25 & 0.35 & 0.30 & 0.10 \\ \hline \end{array}\] -Determine the associated risk measure in this equipment investment in terms of standard deviation.

A) $4,923 B) $6,123 C) $4437.8 D) $8,523 A) $4,923 B) $6,123 C) $4437.8 D) $8,523

Exam 10: Uncertainty in Future Events

A continuous improvement has helped a company to raise savings with associated probabilities shown in the table below. The useful life is 5 years with a probability of0.6 and 3 years with probability of 0.4. i. Determine the joint probability distribution for savings per year and useful life. ii. Determine the expected NPW if an investment of $50,000 is required. No salvage is expected. Use a MARR of 10% Annual Benefit Probability \ 15.000 0.25 \ 20,000 0.45 \ 25,000 0.3

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Question 1

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$\begin{array} { | c | c | c | c | c | l | } \hline \text { Arrual Benefit } & \text { Prob. } & \text { Life } & \text { Prob. } & \text { Joirt Prob. } & { \mathrm { NPW } } \\\hline \$ 15.000 & 0.25 & 5 & 0.6 & 0.15 & \begin{array} { l } \{ 15,000 ( \mathrm { P } / \mathrm { A } , 10 \% , 5 ) - 50,000 \} [ 0.15 ] \\= \{ 15,000 ( 3.791 ) - 50,000 \} [ 0.15 ] \\= \$ 1,029.75\end{array} \\\hline \$ 20,000 & 0.45 & 5 & 0.6 & 0.27 & \begin{array} { l } = \{ 20,000 ( 3.791 ) - 50,000 \} [ 0.27 ] \\= \$ 6,971.4\end{array} \\\hline \$ 25,000 & 0.3 & 5 & 0.6 & 0.18 & \begin{array} { l } = \{ 25,000 ( 3.791 ) - 50,000 \} [ 0.18 ] \\= \$ 8,059.5\end{array} \\\hline \$ 15.000 & 0.25 & 3 & 0.4 & 0.10 & \begin{array} { l } \{ 15,000 ( \mathrm { P } / A , 10 \% , 3 ) - 50,000 \} [ 0.10 ] \\= \{ 15,000 ( 2.487 ) - 50,000 \} [ 0.10 ] \\= - \$ 1,269.5\end{array} \\\hline \$ 20,000 & 0.45 & 3 & 0.4 & 0.18 & \begin{array} { l } = \{ 20,000 ( 2.487 ) - 50,000 \} [ 0.18 ] \\= - \$ 46.8\end{array} \\\hline \$ 25,000 & 0.3 & 3 & 0.4 & 0.12 & \begin{array} { l } = \{ 25,000 ( 2.487 ) - 50,000 \} [ 0.12 ] \\= \$ 1,461\end{array} \\\hline\end{array}$ Expected NPW = $1,029.75 + 6,971.4 + 8,059.5 -1,269.5-46.80+1,461= $16,205.25

A new equipment is being considered at a local company at a cost of $200,000. The operation and maintenance costs of this equipment are estimated to be about $20,000 per year. Salvage value is expected to be $40,000 at the end of its useful life. The life of this equipment is estimated to vary anywhere from 5 to 9 years with the associated probabilities as shown in the table below. If an interest rate of 10% is used, what is the expected EUAC for this equipment? Life, Years 5 6 7 8 9 Probability 0.3 0.1 0.15 0.20 0.25

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D

An E-Commerce business has estimated the sales revenue for this Christmas season during a four-week period from Black Friday at $80M, $75M, $62M and $85M with probabilities of 0.5,0,2,0.1 and 0.2 respectively. The expected sales then may be estimated to be for this Christmas season is $76.5 M.

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An equipment proposal from two vendors A and B have been received from a local company. The quality of these equipment have been varying and consequently the cost of manufacturing the product using these two different equipment has also varied. The varying costs based on the probabilities of defects are given in table below. Determine the better vendor. vendor Prob.Defect. 0.10 0.25 0.10 0.12 0.13 0.16 0.14 A Mfg. Cost \ 20 25 19 21 23 22 20.80 vendor Prob.Defect 0.14 0.12 0.19 0.06 0.24 0.08 0.17 B Mfg. Cost \ 20 18 25 14 30 16 22

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Question 4

If the optimistic, most likely and pessimistic estimates of the life of an asset are 6, 8 and 10 respectively, then the estimates of the life is 8 years

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Question 5

What is standard deviation of the equipment life ____________.

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Question 6

A simple rule of thumb for comparing a project's risk and return by Joe Hardman is : If the expected worth is ? 2 standard deviation of the present worth, then the project is safe to invest.

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Question 7

The probabilities that the life of a machine will vary from 6 to 12 years are given in the table below. The expected life of the machine is ____________. Life, Years 5 6 7 8 9 10 11 12 Probability 0.10 0.15 0.2 0.15 0.20 0.10 0.05 0.05

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Question 8

MKBIC Systems is evaluating four projects A, B, C and D that have risks associated with the producing benefits. Based on the data given in the table below, which project is the best alternative? Praject A project B Project C Project D EUAW Prob. EUAW Prob. EUAW Prob. EUAW Prob. \ 2,500 0.3 \ 3,000 0.1 -\ 5,000 0.25 \ 4,000 0.35 \ 1,800 0.45 -\ 2,500 0.3 \ 6,500 0.45 \ 2,500 0.4 \ 3,200 0.25 \ 4,000 0.6 \ 2,000 0.3 -\ 1,500 0.25

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Question 9

One common measure of risk is the probability of a loss.

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Question 10

If the annual cash flows of an investments are: $1,000, $1,500 and $2,000 with probabilities of 0.1, 0.7 and 0.2 respectively. Then expected annual cash flow is $1,800. The expected sales then may be estimated to be for this Christmas season is $50.2 M

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Question 11

Table 10 First Cost, \ \ 60,000 \ 80,000 \ 100,000 \ 120,000 Probability 0.25 0.35 0.30 0.10 -Find the expected EUAW from the financial data provided in the table for a new equipment. Because of the uncertainty of technology being used in this equipment, it has not been possible to get the initial cost accurately. The annual benefit, however, is estimated to be $25,000 with a possible equipment life of 5 years. The salvage value is expected to be 10% of the initial cost. MARR =8%

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Question 12

At Tech Engineering, a new equipment is expected to cost $100,000. The annual savings are estimated to be $12,000, $15,000, and $20,000 respectively for 10 years of its life with probabilities of 0.4, 0.5, and 0.1 respectively. The annual savings for this equipment is $15,667.

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Question 13

Table 10 First Cost, \ \ 60,000 \ 80,000 \ 100,000 \ 120,000 Probability 0.25 0.35 0.30 0.10 -Determine the associated risk measure in this equipment investment in terms of standard deviation.

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Question 14

An E-Commerce business has estimated the sales revenue for this Christmas season during a four-week period from Black Friday at $80M, $75M, $62M and $85M with probabilities of 0.5,0,2,0.1 and 0.2 respectively. The expected sales then may be estimated to be for this Christmas season is $76.5 M.

If the optimistic, most likely and pessimistic estimates of the life of an asset are 6, 8 and 10 respectively, then the estimates of the life is 8 years

What is standard deviation of the equipment life ____________.

A simple rule of thumb for comparing a project's risk and return by Joe Hardman is : If the expected worth is ? 2 standard deviation of the present worth, then the project is safe to invest.

The probabilities that the life of a machine will vary from 6 to 12 years are given in the table below. The expected life of the machine is ____________. Life, Years 5 6 7 8 9 10 11 12 Probability 0.10 0.15 0.2 0.15 0.20 0.10 0.05 0.05

One common measure of risk is the probability of a loss.

If the annual cash flows of an investments are: $1,000, $1,500 and $2,000 with probabilities of 0.1, 0.7 and 0.2 respectively. Then expected annual cash flow is $1,800. The expected sales then may be estimated to be for this Christmas season is $50.2 M

At Tech Engineering, a new equipment is expected to cost $100,000. The annual savings are estimated to be $12,000, $15,000, and $20,000 respectively for 10 years of its life with probabilities of 0.4, 0.5, and 0.1 respectively. The annual savings for this equipment is $15,667.

Table 10 First Cost, \ \ 60,000 \ 80,000 \ 100,000 \ 120,000 Probability 0.25 0.35 0.30 0.10 -Determine the associated risk measure in this equipment investment in terms of standard deviation.

Making Economic Decisions

Estimating Engineering Costs and Benefits

Interest and Equivalence

Equivalence for Repeated Cash Flows

Present Worth Analysis

Annual Cash Flow Analysis

Rate of Return Analysis

Choosing the Best Alternative

Other Analysis Techniques

Depreciation

Income Taxes for Corporations

Replacement Analysis

Inflation and Price Change

Selection of a Minimum Attractive Rate of Return

Economic Analysis in the Public Sector

Accounting and Engineering Economy

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