Department of Economics ECO 204 Test 2 Solutions€¦ · We are told that the CMP has an interior...

ECO 204, 2013 - 2014, Test 2

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S. Ajaz Hussain, Dept. of Economics, University of Toronto (STG)

∎ Department of Economics ∎ ECO 204 ∎ Microeconomic Theory for Commerce ∎ 2013 - 2014

Test 2 Solutions

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published by the Faculty of Arts and Sciences.

Exam details:

Duration: 2 hours and 20 minutes

Total number of questions: 3 (see next page of test for breakdown of points)

Total number of pages: 34 (including title page and two worksheets at the back of the test)

Total number of points: 100

Please answer all questions. To earn credit you must show all calculations.

Exam aids:

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ECO 204, 2013 - 2014, Test 2


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THIS TABLE IS FOR GRADERS’ USE ONLY

QUESTION MAXIMUM POINTS SCORE COMMENTS

1 25

2 20

3 55

TOTAL SCORE OUT OF 100 POINTS

GOOD LUCK!

GIVE BRIEF ANSWERS AND SHOW ALL NECESSARY CALCULATIONS


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Question 1 [25 points]

This question consists of two parts “A” and “B”

Part A

(a) [3 points] [This part is independent of all other parts below] Consider production processes which use to

produce output. Give a real life example of each of the following production functions: (i) a long run Cobb-Douglas

production function (ii) a long run complements production function (iii) a long run linear production function. Provide a

brief explanation for your answers.

Answer

(i) Example of a long run Cobb-Douglas production function: construction -- Labor and capital are “necessary” inputs (for

we require ) and arguably are imperfect substitutes.

(ii) Example of a long run Complements production function: airlines – each aircraft requires a certain number of labor

(pilots, flight attendants); as such, labor and capital must be used in fixed proportions or as complements.

(iii) Example of a long run linear production function: cash withdrawals at a bank with ATM machines and tellers –

customers can withdraw cash from ATM machines only, tellers only, or ATM and tellers. As such, from

( ) {( ) ( ) ( )} This can be modeled by a linear production function.


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(b) [5 points] [This part is independent of all other parts below] A company uses to produce output according to

the Cobb-Douglas production function. Let the price of labor be and the price of capital be Set up and

solve the Cobb-Douglas long run CMP in parametric form:

Assume that the CMP has an interior solution (i.e. “Case D”) which means that the non-negativity constraints can be

dropped.

Show that the optimal labor and capital are:

(

)

[

]

(

)

[

]

(

)

[

]

(

)

[

]

Do NOT solve for any Lagrange multipliers. Show all necessary calculations.

Answer

The CMP is:

[ ]

[ ]

[ ] [ ]

If we know that this CMP has an interior solution (i.e. “Case D”) then and so the CMP becomes:

[ ] [ ]

Alternatively:

[ ]

We are told that the CMP has an interior solution so that:

[ ]

[ ] [ ]

The FOCs are:


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[ ]

[ ]

[ ]

Here and .

Equating yields:

This states: at the optimal bundle of inputs (required to produce target output ), the iso-cost line is tangent to the iso-

quant curve. Subbing and :

Sub this in:

(

)

(

)

(

)

(

)

Next, we know that:

(

) (

)

(

)

(

) (

)

(

)

(

)

(

)

Thus we have shown that:

(

)

(

)

(

)

(

)

Some students’ answers will be different –points will be given so long as their method is correct.


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(c) [5 points] The Cobb-Douglas long run cost function can be expressed as:

( )

Use this expression to derive the Lagrange multiplier corresponding to the constraint in the following long

run CMP:

[ ]

Assume that the CMP has an interior solution (i.e. “Case D”) which means that the non-negativity constraints can be

dropped.

Answer:

In the previous part we saw that:

[ ]

[ ] [ ]

It was shown that:

We could try to derive by subbing in optimal (or ) above, but there is a simpler way. First, at the optimal

solution we note that:

[ ]

[ ( ) ( ) ]

Thus, a change in due to a change in any parameter is equivalent to change in .

By the envelope theorem, the change in due to a small change in is:

But this is the same as:

( )

That is, is the incremental cost from a small increase in target output .


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Part B

In a tutorial, Michelle gave you data on a company’s labor ( ), capital ( ), and output ( ) and used regression analysis to

estimate the parameters of the company’s Cobb-Douglas production function:

Here is the regression output (this question does not require you to use statistics such as t-stats etc.):

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.96

R Square 0.92

Adjusted R Square 0.91

Standard Error 0.43

Observations 30.00

ANOVA

df SS MS F Significance F

Regression 2.00 56.28 28.14 151.81 0.00

Residual 27.00 5.01 0.19

Total 29.00 61.29

Independent Variable Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%

Intercept 0.42 0.14 3.08 0.00 0.14 0.71 0.14 0.71

0.74 0.07 11.18 0.00 0.60 0.87 0.60 0.87

0.95 0.06 15.09 0.00 0.82 1.08 0.82 1.08

(d) [5 points] Use the regression output above to write down the production function in numerical

form Show all calculations and state all assumptions.

Answer

The production function is . To estimate the parameters via linear regression we need to first

“linearize” this equation by taking natural logs of both sides1:

Next, we’d regress on a constant, , and (in ECO 220 parlance is the “dependent” or variable and

and are the “independent” or variables). From the regression output we see that:

Thus:

1 Note: In ECO 220 format this takes the form:


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(e) [3 points] The Cobb-Douglas long run production function can be expressed as:

( )

Use this expression to prove (not state) whether Michelle’s company:

∎ Has a strictly concave, linear or strictly convex cost function?

∎ and curves rise, fall, or stay constant as ?

∎ ?

Make sure to answer this question using parameter values for Michelle’s company (you’re not being asked to prove

general results). Show all necessary calculations and state all assumptions.

Answer

The production function is:

Notice that

∎ Is the cost function is strictly concave, linear or strictly convex?

In general, the Cobb-Douglas cost function is:

( )

Notice . Subbing the cost function becomes:

( )

Clearly, the cost function is strictly concave:

( )

( ) ( )

∎ Will and curves rise, fall, or stay constant as ?

( )

( )

( )

∎ Is ?

( )


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( )

Notice that:

( )

⏟

Thus: .

These results are consistent with increasing returns to scale.

[Not required in answer] For your convenience, here is a summary of the connection between “returns to scale” and

“long run cost curves”:

{

{

{


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(f) [2 points] Michelle used Excel Solver’s to solve the following long run CMP for her company (notice the “max” below):

[ ]

Reproduced below is a portion of her Excel “sensitivity report”:

Microsoft Excel 14.0 Sensitivity Report Constraints

Final Lagrange

Cell Name Value Multiplier

$C$12 Q 0.50 -1.21

Interpret the “Lagrange multiplier” – what does it mean? Give a brief explanation.

Answer

Earlier, we showed that in the problem:

[ ]

[ ]

That is, is the incremental cost from a small increase in target output .

Thus:

Loosely speaking: starting at current production levels, increasing the target output by 1 unit will raise total cost by

approximately $1.21


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(g) [2 points + 2 bonus points] Suppose that in the short run, Michelle’s company has the following production function:

True or false: producing more output by always doubling labor results in a “U-shaped” short run curve? You do not

have to prove your answer: rather, state the answer and give a brief explanation. Bonus points awarded for proving your

answer?

Answer

In the short run, Michelle’s company has the production function:

Notice that even though there are increasing returns to scale in the long run, that there are decreasing returns to labor

in the short run. Thus, we know that:

{

Thus, true: the curve will be U-shaped.

Bonus Points: here’s the proof. Consider the short run CMP:

[ ]

[ ]

Notice that we can solve for directly from the constraint:

(

)

Thus:

( ) (

)

Thus:

( )

⏟

⏟

Sub in :

⏟

⏟

⏟

⏟

Notice as is U-shaped.


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Consider a risk free asset and two risky assets “A” and “B”. The following table provides partial information on returns

and variances of the three assets:

Asset Return Variance

A

B

Risk Free

The following table contains partial information on the covariance of asset returns:

Covariance of Returns

A B Risk free

A

B

Risk free

(a) [3 points] What are You do not need to prove your answer.

Answer

A risk free asset offers guaranteed, certain, returns in the future. Thus: Since risk free returns are certain,

the covariance with risk assets is nil:


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(b) [17 points] Portfolio “X” has been formed by combining the risk free asset and risky assets “A” and “B”. Fill the blanks in the following table. Show all calculations.

Portfolio

Fraction of portfolio In Risky Asset


Fraction of portfolio

in risk-free asset

Portfolio Return

Portfolio Risk

Portfolio Risk Premium

Portfolio Price of risk

“X” 0.1 0.9 0 ? ? ? ?

All assets must be considered in your calculations (i.e. calculations of the portfolio return, risk, risk premium, and price of risk must include all three assets). Answer The portfolio consists of:

{

( )

{ ( )

Thus the expected portfolio return is:

[ ] ( ) ( ) [( ) ]

The variance of portfolio returns is (using ( ) to denote

):

[ ] [( ) ] ( ) ( ) ( ) ( )

Since:

( )

We have:

[ ] ( )

Now:

( ) [( ) ] ( ) ( ) ( ) ( )

Sub back in:

[ ] ( ) {( ) ( ) ( ) ( ) }

Thus portfolio risk is:

√ √ {( ) ( ) ( ) ( ) }

The portfolio risk premium is:

[ ]


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The portfolio price of risk is:

[ ]

In summary:

Portfolio Return [ ] ( ) [( ) ]

Portfolio Risk √ {( ) ( ) ( ) ( ) }

Portfolio Risk Premium [ ]

Portfolio Price of Risk [ ]

We now use the following information:

Asset Return Variance

A

B

Risk Free

Covariance of Returns

A B Risk free

A

B

Risk free

Portfolio




in risk-free asset

Portfolio Return

Portfolio Risk



“X” 0.1 0.9 0

“Y” 0.2 0 0.38

For portfolio “X”: Portfolio Return If the portfolio does consist of the risk free asset, then it must consist of the synthetic risky asset (i.e. ):

[ ] ( ) [( ) ] ( ) ( )[( ) ] [( ) ]

Since 10% of the portfolio is in risky asset “A” we have (i.e. ):

[ ] [( ) ] ( ) ( ) ( ) ( )

Portfolio Risk


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If the portfolio does consist of the risk free asset, then it must consist of the synthetic risky asset:

√ {( ) ( ) ( ) ( ) } √{( ) ( ) ( ) ( ) }

Since 10% of the portfolio is in risky asset “A” we have (i.e. ):

√{( ) ( ) ( ) ( ) ( )( ) }

√{( ) ( ) ( ) ( ) ( )( )( )}


[ ]

Portfolio Price of Risk

[ ]

Portfolio




in risk-free asset

Portfolio Return

Portfolio Risk



“X” 0.1 0.9 0 0.66 0.33466 0.56 1.673


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[This question is loosely based on the Ivey Business School case “Gold Claims at Sturgeon Lake”]

This question consists of three parts “A”, “B”, and “C”. Answer all parts to two decimal places.

Andrew McKendry, a geologist, has been hired by a Toronto based mining company to advise them about the following

two mutually exclusive decisions regarding gold mining operations at Sturgeon Lake (near Thunder Bay, Ontario):

∎ “Road → Drill”: First attempt to build a permanent road to the drill site at Sturgeon Lake and if the road project is

successful to then drill for gold. The cost of constructing the road is $33,484.56 and there’s a 70% chance that the road

construction project will be a success. The cost of drilling is $98,154.02 and there’s a 22% chance of finding gold. If the

road construction project and drilling are both successful then the of mining profits (before construction and drilling

costs) is $2,194,937.12.

∎ “Drill → Road”: First build a temporary ice road (with a 100% chance of success) and drill for gold. The ice road costs

nothing to build. The cost of drilling is $98,154.02 and the probability of drilling for gold is 0.22. If drilling is successful

the mining company will attempt to build a permanent road; the cost of constructing the permanent road is $33,484.56

and there’s a 70% chance that the road construction project will be a success. If both drilling and road construction are

successful, then the of mining profits (before drilling and construction costs) is $2,194,937.12.

For your convenience, here is a summary of the numbers:

( )

( )

( ) ⏟


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(a) [5 points] Draw the decision tree for this problem. Do NOT solve the decision tree just yet. HINT: You might want to

first sketch the decision tree on a worksheet at the back of this test before drawing the “final” version below.

Answer

The decision tree is (see below for how to calculate figures):

Decision?

Build Permanent Road

Uncertainty

Success Drill Uncertainty

Success

$2,194,937.12

-$98,154.02

Failure

-$98,154.02 Failure

Build Ice road

DrillUncertainty

Failure

SuccessBuild Permanent

RoadUncertainty

Success

$2,194,937.12

-$33,484.56

Failure

-$33,484.56


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. Part A

(b) [10 points] Suppose that the Toronto based gold mining company is “risk neutral”: what are its optimal decision and

optimal course of action? Show all calculations and state all assumptions.

Answer

If the gold mining company is risk neutral then it can decide on the basis of “value” and/or the “utility”. Clearly, the

“value” criterion is easier. We now compute the “value” of “Road Drill” and “Drill Road”.

∎ “Road → Drill”: We are looking at the branch:

{ { {

Working backwards (“backward induction”) we see that the expected value of drilling is:

{ { {

( ) ( ) ( ) ( ( )) ( )

Let’s include the cost of drilling into the calculations:

( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

{ { [ ]

Working one more step backwards we have the expected value of building the road:

{ [ ]

( ) ( ) ( ) ( ( )) ( )

Let’s include the cost of building the road into the calculations:

( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

The of the “Road → Drill” decision is .

∎ “Drill → Road”: We are looking at the branch:

{ { {

Working backwards (“backward induction”) we see that the expected value of building the road is:


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{ { {

( ) ( ) ( ) ( ( )) ( )


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

Working one more step backwards we have:

{ {

The expected value of drilling is:

{

( ) ( ) ( ) ( ( )) ( )


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

The of the “Drill → Road” decision is

Summarizing the of each option we have:

∎ “Road → Drill”: the value of this decision is

∎ “Drill → Road”: the value of this decision is

The risk neutral gold mining company should first attempt to build the permanent road and if it is successful, then drill

for gold. Below is the final decision tree:


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DecisionBuild Permanent

Road first


EV = $235,827.95

Uncertainty

Success

$384,732.15

Drill

$384,732.15Uncertainty

Success

$2,194,937.12

-$98,154.02

Failure

-$98,154.02 Failure

-$33,484.56

Build Ice road

Drill

EV = $232,499.69

Uncertainty

Failure

-$98,154.02

Success

Build Permanent

Road

Uncertainty

Success

$2,194,937.12

-$33,484.56

Failure

-$33,484.56


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(c) [5 points] [This part is independent of all other parts below] Let probability that drilling for gold is a success. For

what values of will the mining company make the same decision as your answer to part (b)? Show all calculations and

state all assumptions.

Answer

In part (b), the optimal decision was to build the road first and if successful to then drill for gold. Let the probability

that drilling for gold is a success. We will build the road first and drill later as opposed to drill first and build road later

whenever:

[ ] [ ]

Now [ ] can be re-stated as follows: working backwards (“backward induction”) we see that the

expected value of drilling is:

( ) ( ) ( ) ( ( )) ( )


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

If the road construction is successful, we will drill so long as ( ) ( ) which will happen so long as

. However, this is not the value of at which building road first is better than drilling first.


( ) ( ) ( ) ( ( )) ( )


( ) ( ) ( ) ( ( )) ( )

( ) ( ( ) ) ( ) ( )

( ) ( ( ) )

( ) ( ( ) )

The of the “Road → Drill” decision is ( ( ) )

Next, the [ ] can be re-stated as follows: working backwards (“backward induction”) we see that the

expected value of building the road is:

( ) ( ) ( ) ( ( )) ( )



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( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )


( ) ( ) ( ) ( ( )) ( )


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

The of the “Drill → Road” decision is ( )

Now, we will build the road first and drill later whenever:

[ ] [ ]

( ( ) ) ( )

( ( ) ) ( )

( )

It is optimal to build the road and if that is successful to then drill for gold so long as the chance of finding gold is at least

12%.

With this value of we see that if the road project is a success, then we’ll always choose to drill instead of stopping.


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Part B

[Part B is independent of Part C below]

(d) [10 points] Return to the original numbers at the beginning of the question:

( )

( )

( )

For this question you should use the decision tree in part (a). Now suppose that the Toronto based gold mining company

is “risk averse” and its board of directors is of the opinion that:

( )⏟

{ ⏟

}

{ ⏟

}

What is the optimal decision and the optimal course of action? Show all calculations and state all assumptions.

Answer

If the gold mining company is risk averse then it should decide on the basis of “utility”. To do this, we need the utilities

of the following monetary outcomes in the decision tree (reproduced here from above):


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Decision?


Uncertainty

Success Drill Uncertainty

Success

$2,194,937.12

-$98,154.02

Failure

-$98,154.02 Failure

-$33,484.56

Build Ice road

DrillUncertainty

Failure

-$98,154.02

SuccessBuild Permanent

RoadUncertainty

Success

$2,194,937.12

-$33,484.56

Failure

-$33,484.56

Notice that the outcomes in the tree are (in ascending order):

( )

( )

Now, the lowest outcome can be assigned a utility of zero, while the highest outcome can be assigned a utility of 100:

Outcome Utility

( ) 0

( ) ?

?

100

To get the utilities of the “intermediate” outcomes, we use the fact that:

( )⏟

{ ⏟

}


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{ ⏟

}

By the definition of (i.e. ( ) ):

Outcome Utility

( ) 0

( ) ( ) { ⏟

}

( ) { ⏟

}

100

First:

( )⏟

{ ⏟

}

Implies that:

( ) ( ) ( ) ( ) ( )

Next,

{ ⏟

}

Implies that:

( ) ( ) ( ) ( ) ( )

Thus:

Outcome Utility ( )

( ) 0

( ) 60

90

100

Even though the question doesn’t ask for it, here is the utility “function” (curve connecting the four utility points):


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Thus, the decision tree in terms of utilities is:

Decision?


Uncertainty

Success

Drill Uncertainty

Success

90

Failure

Stop

Failure

Build Ice road

DrillUncertainty

Failure

Success

Build Permanent

RoadUncertainty

Success

Failure

Stop

We now compute the “utility” of “Road Drill” and “Drill Road”.

∎ “Road → Drill”: Working backwards (“backward induction”) we see that the expected utility of drilling is:


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( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

This is better than the utility of stopping and so the optimal decision is to drill.


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

The ( ) decision is .

∎ “Drill → Road”: Working backwards (“backward induction”) we see that the expected utility of building the road is:

( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

This is better than the utility of stopping and the so the optimal decision is to build the road.


( ) ( ) ( ) ( ( )) ( )

( ) ( ) ( ) ( )

The ( ) decision is 19.36.

Summarizing the “utility” of each option we have:

∎ “Road → Drill”: the “utility” of this decision is

∎ “Drill → Road”: the “utility” of this decision is

The risk averse gold mining company should first attempt to build the permanent road and if it is successful, then drill

for gold.


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Decision

Build Permanent road


Uncertainty

SuccessDrill

Uncertainty

Success

90

Failure

Failure

Build Ice road

Drill

Uncertainty

Failure

Success

Build Permanent

Road

Uncertainty

Success

Failure


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(e) [5 points] [This part is required to answer part (f) only] Suppose a risk averse agent has wealth . With probability

the agent’s wealth will be reduced by . Suppose this agent wants to insure against the risky situation { (

) }. What is the optimal amount of the insurance policy if the insurance industry is charging “actuarially fair

prices”? Show all calculations and state all assumptions.

Answer

Risk averse policy holders will choose the optimal by solving the expected “UMP”:

( ) ( ) ( )

Here price per dollar of insurance and insurance policy (dollars). The FOC is:

( ) ( )( ) ( )( )

Re-arranging and using the actuarially fair pricing, i.e. we have:

( ) ( ) ( )( )

( ) ( )

Since the agent is risk averse we have that ( ) (or that ( ( ) ) ( ) ( ) ( ) for all

and ( )). This means that no two points have the same slope so that:

Given actuarially fair pricing, i.e. the agent should fully insure herself against the loss.


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(f) [5 points] Consider the decision you have made in part (d) (not the whole tree; just the decision). Suppose the mining

company has purchased actuarially fair insurance against the “risk” of drilling. Given your answer to part (b), what is the

insurance premium? Show all calculations and state all assumptions.

Answer

The optimal decision was that the gold mining company should first attempt to build the permanent road and if that is

successful, to then drill for gold. Now, the risky drilling situation is:

{ ( ) }

The “loss” is:

( )

The probability of loss is Under actuarially fair insurance, price per dollar of insurance is equal to the

probability of loss. Thus, The insurance premium is:

The mining company has purchased “full insurance”. Thus: so that:

( )


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Part C

[To be answered independently of Part B above]

(g) [15 points] Return to the original numbers at the beginning of the question:

( )

( )

( )

For this question you should use the decision tree in part (a). Assume that the gold mining company is “risk neutral” and

now suppose that the gold mining must choose between the following two mutually exclusive decisions:

∎ The “optimal” decision in part (b) above.

∎ The “optimal” decision in part (b) above but now with the option of “testing” the site prior to drilling. The test results

will come back as either “positive” or “negative”. Based on historical data, the following table contains the probabilities

of test results and actual outcomes of drilling:

Drilling

S F Total

Test Result

+ 0.20 0.30 0.5

- 0.02 0.48 0.5

Total 0.22 0.78 1.00

Draw the decision tree for the “optimal” decision in part (b) above with the option of “testing” prior to drilling and

recommend whether you will make the decision in part (b) with or without “testing”. The cost of the test is not known.

Answer

The optimal decision in part (b) was to build road first and drill later. Now that we have the option to test before drilling,

we must decide between:

∎ “Road → Drill” without testing. The of this decision is $235,827.95

∎ “Road → Test → Drill”

As such, the “decision” tree becomes (the following tree is larger and more comprehensive than what the question is

asking for):


ECO 204, 2013 - 2014, Test 2


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Decision

Without

TestBuild Road

S Drill

S

F

F

With Test Build Road

F

S Test

+

Drill

S

F

Stop

-

Drill

S

F

Stop

The probabilities are calculated as follows:

( ) ( )

( )

( ) ( )

( ) ( )

( )

( ) ( )

We will choose “Road → Test → Drill” over “Road → Drill” so long as the “value of test information” is greater than or

equal to the cost of the test (which is not given to us). In turn, the value of test information is:

[ ] [ ]


ECO 204, 2013 - 2014, Test 2


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To calculate the NPV with “testing”, let’s first work backwards from the branch:

“Road → success → test → positive → drill”


( ) ( ) ( ) ( ) ( )


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

Next, choose between:

“Road → success → test → positive → drill” and “Road → success → test → positive → not drill”

If the test result is positive and we don’t drill, then the is $0. Thus, if the test comes back positive, we should drill.

Hence, the of a positive test result is .

Next, work backwards from the branch:

“Road → success → test → negative → drill”


( ) ( ) ( ) ( ) ( )


( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

Next, choose between:

“Road → success → test → negative → drill” and “Road → success → test →negative → not drill”

If the test result is negative and we don’t drill, then the is $0. Thus, if the test comes back negative, we should not

drill. Hence, the of a negative test result is

Therefore, the of “testing” before deciding whether to drill is:

[ ] ( )( ) ( )( )

[ ] ( ) ( )

Therefore, the of building the road is:

[ ] ( )[ ] ( )[ ]

Include the cost of the building the road into the calculations:

[ ] [ [ ] ] [ ]


ECO 204, 2013 - 2014, Test 2


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[ ] [ ] [ ]

In summary:

Decision NPV

Road → Drill without any testing at all $235,827.95

Road → Test → Drill

Value of Information $3,624.79

As long as the test costs less than $3,624.79, we should pursue “Road → Test → Drill” and the optimal course of action

is shown in the tree below (not required of students):

Decision

With test

Without

TestBuild Road

S Drill

S

F

F

With Test Build Road

F

S Test

+

Drill

S

F

Stop

-

Drill

S

F

Stop

Provided Cost of the test

The optimal strategy (“path”) is:

{

{


ECO 204, 2013 - 2014, Test 2


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WORKSHEET [This page will NOT be graded]


ECO 204, 2013 - 2014, Test 2


Page 36 of 36


WORKSHEET [This page will NOT be graded]


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