Chapter 5.S, Problem 9P

a)

Summary Introduction

To draw: A decision tree for the given problem.

Introduction:

Decision tree is one of the methods used in decision-making process. It would graphically represent the available alternatives and states of nature. It would also mention the payoffs and probabilities of the alternatives. It helps to choose the best alternative that would give the best result among the alternatives.

Explanation of Solution

Given information:

Three decision alternatives:

• 1^st construct small stamping plant
• 2^nd construct medium stamping plant
• 3^rd construct stamping plant

Each decision has demand:

• Low demand which have 20% probability
• Low demand which have 80% probability
• If firm build small facility and demand turn out to be low than NPV is $42,000,000
• If firm build small facility and demand turn out to be high than either subcontract or expand greatly, if sub contract than NPV is $42,000,000 and if expand greatly than $48,000,000.
• If firm build medium facility and demand turn out to be low than NPV is $22,000,000.
• If firm build medium facility and demand turn out to be high than either it do nothing or expand, if sub contract than NPV is $46,000,000 and if expand greatly than $50,000,000.
• If firm build large facility and demand turn out to be low than NPV is -$20,000,000.
• If firm build large facility and demand turn out to be high than NPV is $72,000,000.

As per given information, construct the decision tree:

Now, calculate the value of payoff and monetary value in decision tree:

Analysis the decision from right to left:

• In the decision tree, there are two decisions making; first decision is between small, medium and large and second decision is between high demand of small and medium build facilities.
• If small facilities build with high demand then select expand greatly because the payoff of expand greatly is higher than subcontract. The payoff of expand greatly is $48,000,000 and place double slash (reject) on subcontract.
• If large facility build with high demand then select expand, because the payoff of expand greatly is higher than do nothing. The payoff of expand is $50,000,000 place double slash (reject) on do nothing.

Calculate the product of the chance probabilities and their respective payoffs for the remaining branches:

• If small facility build:

$\begin{array}{c}\text{Low}\text{demand=}\left(\text{Probability×}{\text{NPV}}_{\text{small,low}}\right)\\ =\left(20\%\times \$42,000,000\right)\\ =\$8,400,000\end{array}$

$\begin{array}{c}\text{High}\text{demand}=\left(\text{Probability×}{\text{NPV}}_{small,high}\right)\\ =\left(80\%\times \$48,000,000\right)\\ =\$38,400,000\end{array}$

• If medium facility IS build:

$\begin{array}{c}\text{Low}\text{demand=}\left(\text{Probability×}{\text{NPV}}_{\text{medium,low}}\right)\\ =\left(20\%\times \$22,000,000\right)\\ =\$4,400,000\end{array}$

$\begin{array}{c}\text{High}\text{demand=}\left(\text{Probability×}{\text{NPV}}_{medium,high}\right)\\ =\left(80\%\times \$50,000,000\right)\\ =\$40,000,000\end{array}$

• If large facility is build:

$\begin{array}{c}\text{Low}\text{demand=}\left(\text{Probability×}{\text{NPV}}_{\text{large,low}}\right)\\ =\left(20\%\times -\$20,000,000\right)\\ =-\$4,000,000\end{array}$

$\begin{array}{c}\text{High}\text{demand=}\left(\text{Probability×}{\text{NPV}}_{\text{medium,high}}\right)\\ =\left(80\%\times \$72,000,000\right)\\ =\$57,600,000\end{array}$

Calculation of the expected value in each alternative:

• If small facility build

$\begin{array}{c}\text{Expected}\text{value}=\left(\text{Low}\text{demand}\text{payoff}+\text{high}\text{demand}\text{payoff}\right)\\ =\left(\$8,400,000+\$38,400,000\right)\\ =\$46,800,000\end{array}$

• If medium facility build

$\begin{array}{c}\text{Expected}\text{value}=\left(\text{Low}\text{demand}\text{payoff}+\text{high}\text{demand}\text{payoff}\right)\\ =\left(\$4,400,000+\$40,400,000\right)\\ =\$44,400,000\end{array}$

• If large facility build

$\begin{array}{c}\text{Expected}\text{value}=\left(\text{Low}\text{demand}\text{payoff}+\text{high}\text{demand}\text{payoff}\right)\\ =\left(-\$4,00,000+\$57,600,000\right)\\ =\$53,600,000\end{array}$

Here, it is obtained that large facility build have higher expected value, hence it should be selected. And small and medium facilities have to be rejected.

b)

Summary Introduction

To determine: Maximin alternatives

Introduction:

Maximin is the decision making method which is used to make decision under uncertainty. This method will find an alternative that maximizes the minimum outcome of every alternative or we can say that calculating the minimum outcome within the each alternative.

Answer to Problem 9P

To suitable option is to build a small facility.

Explanation of Solution

Determine the worst possible alternatives from the given demand:

As definition stated above, Maximim is the selection from the best of the worst possible payoff for each alternatives.

Alternative	Next Year’s Demand		Worst Payoff	Decision
	Low	High
Small	$42,000,000.00	$48,000,000.00	$42,000,000.00	Best of the Worst
Medium	$22,000,000.00	$50,000,000.00	$22,000,000.00
Large	-$20,000,000.00	$72,000,000.00	-$20,000,000.00

Hence, the best decision according to Maximin decision is to select a small medium facility.

c)

Summary Introduction

To determine: Expected value of perfect information (EVPI) and interpret it.

Expected value of perfect information: It is the rate that a person is willing to pay to gain access to perfect information. A common area which uses expected value of perfect information is the healthcare economy. This value tries to evaluate the expected cost of the uncertainty, which can be interpreted as the expected value of perfect information

The expected value of perfect information can be calculated by using below given formula:

$\text{EVPI}\text{=}\text{Expected}\text{value}\text{with}\text{perfect}\text{information}-\text{Maximum EMV}$

Answer to Problem 9P

Theexpected value of perfect information is $12,400,000.

Explanation of Solution

Calculate the expected value with perfection information or Expected payoff under certainty:

$\begin{array}{c}\text{EVwPI}={\displaystyle \sum \text{Max}\left(\text{Alternative}\text{payoff}\right)}\times \text{Probability}\text{state of nature }\\ =\left(42,000,000\times 20\%\right)+\left(\$72,000,000\times 80\%\right)\\ =\$66,000,000\end{array}$

Calculate the expected value with perfect information:

$\begin{array}{c}\text{EVPI}\text{=}\text{Expected}\text{value}\text{with}\text{perfect}\text{information}-\text{Maximum EMV}\\ =\$66,000,000-\$53,600,000\\ =\$12,400,000\end{array}$

Therefore, EVPI is $12,400,000.

d)

Summary Introduction

To determine: The sensitive analysis on P (high).

Introduction

Decision table is formats or visual representations were data is expressed arranged, determined and calculated to make a effective decision making. A decision table is a tabular representation that is used to analyze decision alternatives and states of nature.

Explanation of Solution

Explanation

Here we draw each alternative to P (high), low demand should be on the left hand side and high demand value on the right hand side:

	Next Year’s Demand
Alternative	Low	High
Small	$42,000,000.00	$48,000,000.00
Medium	$22,000,000.00	$50,000,000.00
Large	-$20,000,000.00	$72,000,000.00

Graph plot:

From the above graph, we can determine the value of P (high) where each alternative are optimal.

From the graph, we can obtain that small build facility is best option, because of higher expected value. For low value of P(high) while for higher and intermediate value of P(high) large is the best option.

Calculate the range needed to determine upper part of line intersects:

For each line, b is the slope of the line and x = P(High). The slope of each line $\text{Right hand value}-\text{Left}\text{hand value}$.

Determination of equation:

Small build facility = 42+6P (because high demand $48 million which is subtracted with low demand value $42 million)

Large build facility =-20+92P (because high demand $72million which is subtracted with low demand value -$20million)

Determination of intersection between small and large

$\begin{array}{c}\text{42 }+\text{ 6P}=-\text{20 + 92P}\\ \text{6P – 92P = }-\text{20 – 42}\\ -\text{86P = }-\text{62}\\ \text{P }=\frac{-62}{-86}\\ =0.7209\end{array}$

Hence, the intersection between small and large is 0.7209.

Therefore, optimal range can be derived as:

Small:  P (High) = 0 to < .7209

Large: P (High) > .7209 to 1.00