Solutions to Chapter 12 Exercises
SOLVED EXERCISES
S1.
(a)
The number choosing X should decrease to move the population division between X and
Y away from the unstable center equilibrium.
(b)
Because the line for action X is above the line for action Y when 100 people choose X,
and because we expect the number choosing X to decrease from this point [see part
(a)
], it must be the
case that the lines in the diagram represent
costs
associated with each action. Thus, when 100 people are
choosing X, the cost of X exceeds the cost of Y, so people will want to switch from X to Y.
S2.
(a)
This game is a prisoners’ dilemma since
s
(
n
) is greater than
p
(
n
+ 1) for all
n
. Any single
player can always raise his payoff by switching from participating to shirking.
(b)
T
(
n
) =
n
*
p
(
n
) + (100 –
n
)*
s
(
n
) =
n
2
+ (100 –
n
)(4 + 3
n
).
(c)
Plugging the appropriate values into the text’s Equation (12.1) yields
T
(
n
+ 1) –
T
(
n
) =
n
+ 1 – 4 – 3
n
+
n
(
n
+ 1 –
n
) + (100 –
n
– 1)(4 + 3
n
+ 3 – 4 – 3
n
)
= –3 – 2
n
+
n
+ 3(100 –
n
– 1)
= 294 – 4
n.
This formula is positive for all
n
up to
n
= 73, but turns negative for
n
= 74 and higher. Thus, the
last time the change in total social payoff expression is positive is for
T
(74) –
T
(73). Thus
T
is maximized
at
n
= 74.
Using calculus, you can find the same answer by differentiating
T
(
n
) from part b. T´(
n
) = 2
n
+
3(100 –
n
) – (4 + 3
n
) = 296 – 4
n
. Set this derivative equal to zero to find that
T
(
n
) is maximized at
n
= 74.
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 Spring '08
 Charness,G
 Equilibrium, Supply And Demand, Economic equilibrium, Alphaville

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