Question 1

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

Accepted Answer

To determine the best project, calculate the expected value of each project by multiplying the EUAW by its probability and summing the results. Project A has the highest expected value, making it the best alternative.

Question 2

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?
$$\begin{array} { | l | l | l | l | l | l | } 
\hline \text { Life, Years } & 5 & 6 & 7 & 8 & 9 \
\hline \text { Probability } & 0.3 & 0.1 & 0.15 & 0.20 & 0.25 \
\hline
\end{array}$$
A) $66,460
B) $55,970
C) $62,960
D) $58,206.2

Accepted Answer

D

Question 3

Table 10
 $$\begin{array} { | l | l | l | l | l | } 
\hline \text { First Cost, } \$ & \$ 60,000 & \$ 80,000 & \$ 100,000 & \$ 120,000 \
\hline \text { Probability } & 0.25 & 0.35 & 0.30 & 0.10 \
\hline
\end{array}$$
-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%
A) $2977
B) $5157
C) $3957
D) $4628

Accepted Answer

B

Question 4

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

Accepted Answer

The standard deviation is calculated using the formula for a discrete probability distribution. First, calculate the expected value (mean) of the costs, then find the variance by summing the squared differences from the mean weighted by their probabilities, and finally take the square root to find the standard deviation. The correct calculation results in approximately $4437.8.

Question 5

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

Accepted Answer

The answer of The probabilities that the life of a...

Question 6

What is standard deviation of the equipment life ____________.
A) 1.73
B) 1.29
C) 1.88
D) 1.54

Accepted Answer

The answer of What is standard deviation of the equipment...

Question 7

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.
$\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \text { vendor }&\text { Prob.Defect. } & 0.10 & 0.25 & 0.10 & 0.12 & 0.13 & 0.16 & 0.14 \
\hline A&\text { Mfg. Cost } & \$ 20 & 25 & 19 & 21 & 23 & 22 & 20.80 \
\hline \text { vendor }&\text { Prob.Defect } & 0.14 & 0.12 & 0.19 & 0.06 & 0.24 & 0.08 & 0.17 \
\hline B&\text { Mfg. Cost } & \$ 20 & 18 & 25 & 14 & 30 & 16 & 22 \
\hline
\end{array}$
A) Vendor A
B) Vendor B

Accepted Answer

Vendor A has a lower expected manufacturing cost when considering the probabilities of defects. Calculating the expected cost for both vendors, Vendor A's expected cost is lower than Vendor B's.

Question 8

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%
 $$\begin{array} { | l | l | } 
\hline \text { Annual Benefit } & \text { Probability } \
\hline \$ 15.000 & 0.25 \
\hline \$ 20,000 & 0.45 \
\hline \$ 25,000 & 0.3 \
\hline
\end{array}$$

Accepted Answer

The answer of A continuous improvement has helped a company...

Question 9

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.

Accepted Answer

The rule of thumb states that if the expected worth is greater than 2 standard deviations of the present worth, then the project is safe to invest.

Question 10

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.

Accepted Answer

The expected annual savings should be calculated using the probabilities and savings amounts: (0.4 * $12,000) + (0.5 * $15,000) + (0.1 * $20,000) = $14,300, not $15,667.

Question 11

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.

Accepted Answer

The expected sales revenue is calculated by multiplying each sales figure by its probability and summing the results: (80M * 0.5) + (75M * 0.2) + (62M * 0.1) + (85M * 0.2) = 40M + 15M + 6.2M + 17M = 78.2M.

Question 12

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

Accepted Answer

The expected annual cash flow is calculated by multiplying each cash flow by its probability and summing the results: (0.1 * $1,000) + (0.7 * $1,500) + (0.2 * $2,000) = $1,550, not $1,800.

Question 13

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

Accepted Answer

The probability of a loss is a fundamental aspect of risk assessment, as it quantifies the likelihood of an adverse event occurring.

Question 14

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

Accepted Answer

The most likely estimate is the best estimate of the life of the asset.

Quiz 10: Uncertainty in Future Events

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.

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.

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 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

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

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