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Question 303 MCQ · 1.0 marks
The Cobb-Douglas production function F(K,L)=K\(^{p}\)L\(^{q}\)
6406534314247.
Never exhibits decreasing returns to scale
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Shows constant returns to scale if p+q=1
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Has constant marginal product of labour
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Has constant marginal product of capital
Published solution
**1. Scale both inputs:** \[F(tK,tL)=t^{p+q}F(K,L).\]
**2. Classify returns:** If \(p+q=1\), output scales exactly by \(t\), so returns to scale are constant.
**Answer: B — Shows constant returns to scale if p+q=1.**
**Review note:** This production-function question is from the preceding economics section and appears in the Game Theory export.
A: **Incorrect:** Decreasing returns occur when p+q<1.
B: **Correct:** The homogeneity degree is exactly 1 in this case.
C: **Incorrect:** The labour marginal product generally depends on K and L.
D: **Incorrect:** The capital marginal product generally depends on K and L.
When average product is decreasing in labor, marginal product is greater than average product.
6406534314251.
TRUE
6406534314252.
FALSE
Published solution
**1. Relate average and marginal product:** For output \(Q(L)\), \(AP=Q/L\) and \(MP=dQ/dL\), so
\[\frac{dAP}{dL}=\frac{MP-AP}{L}.\]
**2. Read the sign:** When average product is decreasing, marginal product is less than average product.
**Answer: B — False.**
**Review note:** This question is from the preceding economics section and appears in the Game Theory export.
A: **Incorrect:** This reverses the required marginal-average comparison.
B: **Correct:** The stated comparison is reversed, so the statement is false.
When average product is increasing in labor, marginal product is less than average product.
6406534314253.
TRUE
6406534314254.
FALSE
Published solution
**1. Relate average and marginal product:** For output \(Q(L)\), \(AP=Q/L\) and \(MP=dQ/dL\), so
\[\frac{dAP}{dL}=\frac{MP-AP}{L}.\]
**2. Read the sign:** When average product is increasing, marginal product is greater than average product.
**Answer: B — False.**
**Review note:** This question is from the preceding economics section and appears in the Game Theory export.
A: **Incorrect:** This reverses the required marginal-average comparison.
B: **Correct:** The stated comparison is reversed, so the statement is false.
When average product **AP**L is at a maximum, then marginal product is equal to average product
6406534314255.
TRUE
6406534314256.
FALSE
**Game Theory and Strategy**
**Section Id :** 64065391708
**Section Number :** 15
**Section type :** Online
**Mandatory or Optional :** Mandatory
**Number of Questions :** 5
**Number of Questions to be attempted :** 5
**Section Marks :** 20
**Display Number Panel :** Yes
**Section Negative Marks :** 0
**Group All Questions :** No
**Enable Mark as Answered Mark for Review and**
No
**Clear Response :**
**Section Maximum Duration :** 0
**Section Minimum Duration :** 0
**Section Time In :** Minutes
**Maximum Instruction Time :** 0
Published solution
**1. Relate average and marginal product:** For output \(Q(L)\), \(AP=Q/L\) and \(MP=dQ/dL\), so
\[\frac{dAP}{dL}=\frac{MP-AP}{L}.\]
**2. Read the sign:** When average product is at an interior maximum, marginal product is equal to average product.
**Answer: A — True.**
**Review note:** This question is from the preceding economics section and appears in the Game Theory export. The maximum-product equality assumes a differentiable interior maximum.
A: **Correct:** At a differentiable interior maximum, the derivative vanishes and MP=AP.
B: **Incorrect:** The equality follows from the zero derivative at an interior maximum.
**(Choosing a route)**Four people must drive from A to B at the same time. Two routes are available,
one via X and one via Y (Refer to the left panel of following Figure). The roads from A to X, and
from Y to B are both short and narrow; in each case, one car takes 6 minutes, and each additional
car increases the travel time per car by 3 minutes. (If two cars drive from A to X, for example, each
car takes 9 minutes.) The roads from A to Y, and from X to B are long and wide; on A to Y one car
takes 20 minutes, and each additional car increases the travel time per car by 1 minute; on X to B
one car takes 20 minutes, and each additional car increases the travel time per car by 0.9 minutes.
Formulate this situation as a strategic game
Based on the above data, answer the given subquestions.
Choose the correct alternative .
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In Nash equilibrium two people take each route
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For the NE action profile, each person’s travel time is 28 minutes
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Both In Nash equilibrium two people take each route and For the NE action
profile, each person’s travel time is 28 minutes
6406534314262.
None
Published solution
**1. Express congestion:** If \(k\) drivers use A–X–B, their time is \(26+3.9(k-1)\); the \(4-k\) drivers on A–Y–B take \(26+4(3-k)\).
**2. Test deviations:** With two drivers on each route, times are 29.9 and 30 minutes. Switching would put three drivers on the other route and raise the switcher’s time to 34 or 33.8 minutes. Any other split gives a driver on the crowded route a profitable switch.
**Answer: A — Two drivers take each route.**
A: **Correct:** The equilibrium route loads are two and two.
B: **Incorrect:** The equilibrium times are 29.9 and 30, not 28 minutes.
C: **Incorrect:** The equal split is true but the 28-minute claim is false.
D: **Incorrect:** The equal-split statement is true.
**(Choosing a route)**Four people must drive from A to B at the same time. Two routes are available,
one via X and one via Y (Refer to the left panel of following Figure). The roads from A to X, and
from Y to B are both short and narrow; in each case, one car takes 6 minutes, and each additional
car increases the travel time per car by 3 minutes. (If two cars drive from A to X, for example, each
car takes 9 minutes.) The roads from A to Y, and from X to B are long and wide; on A to Y one car
takes 20 minutes, and each additional car increases the travel time per car by 1 minute; on X to B
one car takes 20 minutes, and each additional car increases the travel time per car by 0.9 minutes.
Formulate this situation as a strategic game
Based on the above data, answer the given subquestions.
Now suppose that a relatively short, wide road is built from X to Y, giving each person four options
for travel from A to B: A–X–B, A–Y–B, A–X–Y–B, and A–Y– X–B (As shown in the right panel of the
given figure). Assume that a person who takes A–X–Y–B travels the A–X portion at the same time as
someone who takes A–X–B, and the Y–B portion at the same time as someone who takes A–Y–B.
(Think of there being constant flows of traffic.) On the road between X and Y, one car takes 7
minutes and each additional car increases the travel time per car by 1 minute. Find the Nash
equilibrium in this new situation
Choose the correct alternative
6406534314263.
In any Nash equilibrium, one person takes A–X–B, two people take A–X–Y–B,
and one person takes A–Y–B
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In this equilibrium profile, every person’s travel time increases compared to
the case when new road was not there
6406534314265.
Both In any Nash equilibrium, one person takes A–X–B, two people take
A–X–Y–B, and one person takes A–Y–B and In this equilibrium profile, every person’s travel time
increases compared to the case when new road was not there
6406534314266.
None
Published solution
**1. Find the equilibrium split:** One driver uses A–X–B, one uses A–Y–B, and two use A–X–Y–B; nobody uses A–Y–X–B. Loads on A–X and Y–B are 3 each, and the new road has load 2.
**2. Calculate and check:** The route times are \(12+20=32\), \(20+12=32\), and \(12+8+12=32\) minutes. A direct-route driver switching to the shortcut takes 35 minutes; a shortcut driver switching to a direct route takes 32.9 or 33 minutes. The reverse shortcut is still longer. Checking the other integer route splits leaves only this equilibrium load pattern.
**3. Compare:** All drivers now take 32 minutes, exceeding the original 29.9 or 30 minutes: a Braess-type congestion effect.
**Answer: C — Both statements.**
A: **Incorrect:** This is the equilibrium load pattern, but the travel-time statement is also true.
B: **Incorrect:** All times rise to 32 minutes, but the equilibrium-pattern statement is also true.
C: **Correct:** Both the route-load pattern and the travel-time increase hold.
D: **Incorrect:** Both statements hold.
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
Choose the correct statement
6406534314267.
There is a unique Nash equilibrium in pure strategy
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There are finitely many Nash equilibria in this game in pure strategy
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There is no Nash equilibrium in this game in pure strategy
6406534314270.
None
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Check each pure profile:** At matching locations, the dealer prefers to move. At different locations, the police prefer to move to the dealer’s location.
**2. Conclude:** Every pure profile allows a profitable unilateral deviation, so none is a pure-strategy Nash equilibrium.
**Answer: C — There is no Nash equilibrium in this game in pure strategy.**
A: **Incorrect:** Every pure profile admits a profitable deviation.
B: **Incorrect:** There are zero pure equilibria; this wording ordinarily suggests a positive number.
C: **Correct:** No pure profile is a mutual best response.
D: **Incorrect:** The no-pure-equilibrium statement is correct.
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
Choose the correct statement .
6406534314271.
There is a unique Nash equilibrium in this game
6406534314272.
There are finitely many Nash equilibria in this game
6406534314273.
There is no Nash equilibrium in this game
6406534314274.
None
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Establish uniqueness:** No pure equilibrium exists, and both indifference equations have one interior solution. Thus the mixed equilibrium is unique.
**Answer: A — There is a unique Nash equilibrium in this game.**
**Review note:** The option “finitely many Nash equilibria” also literally includes one. The selected option is the more precise intended answer; the alternatives overlap logically.
A: **Correct:** The unique equilibrium has police Street probability 1/3 and dealer Street probability 1/2.
B: **Incorrect:** One equilibrium is mathematically a finite number, so this is also literally true but less precise.
C: **Incorrect:** An interior mixed equilibrium exists.
D: **Incorrect:** The unique-equilibrium statement is correct.
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
If randomization probability of a police officer for patrolling the streets is p/q, then p =____________
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Read the requested component:** \(x=1/3\), so the numerator p in lowest terms is 1.
**Answer: 1.**
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
If randomization probability of a police officer for patrolling the streets is p/q, then q
=______________
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Read the requested component:** \(x=1/3\), so the denominator q in lowest terms is 3.
**Answer: 3.**
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
If randomization probability of a police officer for patrolling the park is r/s, then r =_____________
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Read the requested component:** The Park probability is \(1-x=2/3\), so its numerator r in lowest terms is 2.
**Answer: 2.**
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
If randomization probability of a police officer for patrolling the park is r/s, then s =_______________
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Read the requested component:** The Park probability is \(2/3\), so its denominator s in lowest terms is 3.
**Answer: 3.**
Determined to crack down on the drug trade, a city mayor puts more officers out on patrol to
disrupt the business of drug dealers. A drug dealer in a neighborhood can work his trade either on
a street corner or in the park. Each day, he decides where to set up shop, knowing that word
about his location will travel among users. Because a good snitch is lacking, word does not travel
to the police. The police officer on the beat then needs to decide whether she will patrol the park
or the street corner, while not knowing where the drug dealer is hanging out that day.
The decision of the officer and the dealer determine the extent of drug trades that day. Let
suppose the total number of potential trades in the market is 500. A dealer's payoff is determined
by the number of trades he consummates, while the officer’s payoff is based on the number of
trades she disrupts. The table below illustrates the payoffs for both parties according to their
respective strategies in the game.
Based on the above data, answer the given subquestions.
If randomization probability of the drug dealer choosing street is β, then β= _______________
Published solution
Police owns rows; dealer owns columns. Each pair is (police payoff, dealer payoff).
\[\begin{array}{c|cc} & Street & Park \\ Street & (400,100) & (0,500) \\ Park & (100,400) & (300,200)\end{array}\]
**1. Indifferent dealer:** Let \(x\) be the probability the police patrol Street. The dealer’s payoffs are \(100x+400(1-x)\) on Street and \(500x+200(1-x)\) in Park. Equating them gives \(x=1/3\).
**2. Indifferent police:** Let \(y\) be the probability the dealer chooses Street. The police payoffs are \(400y\) on Street and \(100y+300(1-y)\) in Park. Equating them gives \(y=1/2\).
**3. Read the requested component:** The dealer’s Street probability is \(y=1/2=0.5\).
**Answer: 0.5.**
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
This is a game of
6406534314280.
Perfect information
6406534314281.
Imperfect information
6406534314282.
None
6406534314283.
Can’t be determined
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Identify observations:** Every player observes the preceding actions; each decision node is its own information set.
**Answer: A — Perfect information.**
A: **Correct:** Every decision maker knows the current history.
B: **Incorrect:** No distinct decision nodes are joined by an information set.
C: **Incorrect:** Perfect information applies.
D: **Incorrect:** The observations are specified.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
How many terminal nodes does this game have?
6406534314284.
3
6406534314285.
2
6406534314286.
4
6406534314287.
5
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Count endings:** The four terminal histories are A, BC, BDE, and BDF.
**Answer: C — 4 terminal nodes.**
A: **Incorrect:** This omits one terminal history.
B: **Incorrect:** There are more than two endings.
C: **Correct:** There are four terminal histories.
D: **Incorrect:** No fifth ending exists.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
How many information sets does this game have?
6406534314288.
2
6406534314289.
3
6406534314290.
4
6406534314291.
1
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Count decision nodes:** Player 1 acts at the root and after BD; player 2 acts after B. Perfect information makes these three singleton information sets.
**Answer: B — 3 information sets.**
A: **Incorrect:** This counts players rather than decision information sets.
B: **Correct:** There are three singleton information sets.
C: **Incorrect:** Terminal nodes are not information sets.
D: **Incorrect:** The whole game is not one information set.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
How many pure strategies does player 1 have?
6406534314292.
1
6406534314293.
2
6406534314294.
3
6406534314295.
4
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Specify complete plans:** Player 1 chooses A or B initially and E or F if BD occurs. A pure strategy specifies both choices, even off the realized path.
**2. Count:** \(2\times2=4\): AE, AF, BE, BF.
**Answer: D — 4 pure strategies.**
A: **Incorrect:** A pure strategy is not just the realized path.
B: **Incorrect:** This counts only one of player 1’s decisions.
C: **Incorrect:** All four combinations are feasible.
D: **Correct:** Two binary decisions give four strategies.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
How many pure strategies does player 2 have?
6406534314296.
1
6406534314297.
2
6406534314298.
3
6406534314299.
4
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Count complete plans:** Player 2 has one information set with actions C and D, hence two pure strategies.
**Answer: B — 2 pure strategies.**
A: **Incorrect:** Both C and D are available.
B: **Correct:** One binary decision gives two strategies.
C: **Incorrect:** There is no third action.
D: **Incorrect:** Player 2 has only one binary decision.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
How many pure strategy Nash equilibria does this game have?
6406534314300.
1
6406534314301.
2
6406534314302.
3
6406534314303.
4
Published solution
Player 1 owns rows; player 2 owns columns. Each pair is (player 1 payoff, player 2 payoff).
\[\begin{array}{c|cc} & C & D \\ AE & (2,0) & (2,0) \\ AF & (2,0) & (2,0) \\ BE & (3,1) & (0,0) \\ BF & (3,1) & (1,2)\end{array}\]
**1. Check mutual best responses:** Against C, player 1 chooses BE or BF; against D, player 1 chooses AE or AF. Player 2 prefers C after BE and D after BF, and is indifferent when A is chosen.
**2. List intersections:** The pure Nash equilibria are (AE,D), (AF,D), and (BE,C).
**Answer: C — 3 pure Nash equilibria.**
A: **Incorrect:** There are three intersections of best responses.
B: **Incorrect:** There are three, not two, equilibria.
C: **Correct:** The three listed profiles are mutual best responses.
D: **Incorrect:** (BF,C) fails because player 2 prefers D.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
Which one of the Nash equilibria is also a subgame perfect equilibrium?
6406534314304.
(AE, D)
6406534314305.
(AF, D)
6406534314306.
(BE, C)
6406534314307.
(BF, C)
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Work backwards:** At the last node, player 1 chooses F because 1 exceeds 0. Player 2 then chooses D because 2 exceeds 1.
**2. Choose at the root:** Player 1 chooses A because its payoff 2 exceeds the payoff 1 from continuing through BF. The complete strategy is AF, including the off-path F action.
**Answer: B — (AF, D).**
A: **Incorrect:** E is not optimal in the final subgame.
B: **Correct:** AF and D are optimal in every subgame.
C: **Incorrect:** C is not optimal when player 1 will choose F.
D: **Incorrect:** C is not optimal in player 2’s subgame.
Consider a two-player game in which player 1 can choose A or B. The game ends if he chooses A
while it continues to player 2 if he chooses B. Player 2 can either choose C or D, with the game
ending after C and continuing again with player 1 after D. Player 1 then can choose E or F, and
then the game ends after each of these choices.
Imagine that the payoffs following choice A by player 1 are (2, 0), following C by player 2 are (3, 1),
following E by player 1 are (0, 0) and following F by player 1 are (1, 2) Model this as an extensive
form game and answer the given subquestions:
No. of sub games this game has
6406534314308.
1
6406534314309.
2
6406534314310.
3
6406534314311.
4
Published solution
\[\begin{array}{ccccccc}&&\text{Player 1}&&&&\\ &\swarrow A&&\searrow B&&&\\ (2,0)&&&&\text{Player 2}&&\\ &&&\swarrow C&&\searrow D&\\ &&(3,1)&&&&\text{Player 1}\\ &&&&&\swarrow E&\searrow F\\ &&&&&(0,0)&(1,2)\end{array}\]
Each terminal pair is (player 1 payoff, player 2 payoff).
**1. Count singleton starting nodes:** Each of the three decision nodes starts a subgame, including the root, which starts the whole game.
**Answer: C — 3 subgames.**
A: **Incorrect:** This counts only the whole game.
B: **Incorrect:** This counts only proper subgames.
C: **Correct:** There are three subgames when the whole game is included.
D: **Incorrect:** Terminal nodes are not counted as decision subgames.
Based on the above data, answer the given subquestions.
6406534314312.
1
6406534314313.
1/2
6406534314314.
2
6406534314315.
1/3
Published solution
**1. Maximize payoff:** For \(v_i=(a+e_j)e_i-e_i^2\), differentiation gives \(a+e_j-2e_i=0\). With the nonnegativity constraint, \(BR_i(e_j)=\max\{0,(a+e_j)/2\}\).
**2. Evaluate:** For \(a>0\), \(BR_i(a)=a\), hence \(BR_i(a)/a=1\).
**Answer: A — 1.**
**Review note:** This ratio requires a>0. The supplied passage does not state that domain; at a=0 the ratio is undefined, and for a<0 the proposed rival effort a is infeasible.
A: **Correct:** Evaluating the best response at rival effort a gives a itself.
B: **Incorrect:** 1/2 is the best-response slope, not the requested ratio.
C: **Incorrect:** The best response is a, not 2a.
D: **Incorrect:** The ratio is 1, not 1/3.
Based on the above data, answer the given subquestions.
Choose the correct alternative.
6406534314316.
Under Nash equilibrium both players choose e1 = e2 = a .
6406534314317.
Equilibrium payoff of a player will be 2a\(^{2}\)
6406534314318.
Both Under Nash equilibrium both players choose Under Nash equilibrium
both players choose e1 = e2 = a and Equilibrium payoff of a player will be 2a\(^{2}\)
6406534314319.
None
Published solution
**1. Solve best responses:** For \(a\ge0\), the mutual conditions \(e_1=(a+e_2)/2\) and \(e_2=(a+e_1)/2\) imply \(e_1=e_2=a\).
**2. Substitute into payoff:** \[v_i(a,a)=(a+a)a-a^2=a^2.\]
**3. Identify the correct statement:** The effort assertion is correct under \(a\ge0\); the payoff assertion \(2a^2\) is incorrect.
**Answer: A — \(e_1=e_2=a\).**
**Review note:** The answer assumes a≥0. For a<0, nonnegative efforts instead give the equilibrium e1=e2=0, so the source should specify the parameter domain.
A: **Correct:** The equilibrium efforts equal a when a is nonnegative.
B: **Incorrect:** Subtracting the effort cost leaves a², not 2a².
C: **Incorrect:** The payoff claim is false.
D: **Incorrect:** The effort claim is true under the intended nonnegative-a assumption.
