Equilibria on the Day-Ahead
Electricity Market
Margarida Carvalho
INESC Porto, Portugal
Faculdade de Ciências, Universidade do Porto, Portugal
[email protected]
João Pedro Pedroso
INESC Porto, Portugal
Faculdade de Ciências, Universidade do Porto, Portugal
[email protected]
Technical Report Series: DCC-2012-06
Departamento de Ciência de Computadores
Faculdade de Ciências da Universidade do Porto
Rua do Campo Alegre, 1021/1055,
4169-007 PORTO,
PORTUGAL
Tel: 220 402 900 Fax: 220 402 950
http://www.dcc.fc.up.pt/Pubs/
Equilibria on the Day-Ahead Electricity Market
Margarida Carvalhoa,b , João Pedro Pedrosoa,b
a
Faculdade de Ciências, Universidade do Porto,
Rua do Campo Alegre, 4169-007 Porto, Portugal
b
INESC TEC, Rua Dr. Roberto Frias 378, 4200-465 Porto, Portugal
Abstract
In the energy sector, there has been a transition from monopolistic to oligopolistic situations (pool markets); each time more companies’ optimization revenues depend on the
strategies of their competitors. The market rules vary from country to country. In this
work, we model the Iberian Day-Ahead Duopoly Market and find exactly which are the
outcomes (Nash equilibria) of this auction using game theory.
Keywords: Duopoly, Nash Equilibria, Day-Ahead Market
1. Introduction
Over the last years the way electricity is produced and delivered has changed considerably. Market mechanisms were implemented in several countries and electricity markets
are no longer vertically integrated. Nowadays, in many countries the electricity markets are
based on a pool auction to purchase and sell power. Producing companies offer electricity
in a market and the buyers submit acquisition proposals.
An electricity pool market is characterized by a single price (market clearing price) for
electricity paid to all the proposals accepted in the market. The way in which the market
clearing price is determined varies from one country to another: the last accepted offer, the
first rejected offer, multiple unit Vickey (see Anderson and Xu (2004), Son et al. (2004)
and Zimmerman et al. (1999)). The Iberian market uses the last accepted offer mechanism,
which seems to provide a competitive price and an appropriate incentive for investment
and new entry. Indeed, most pool markets use the last accepted block rule (see Son et al.
(2004)).
To analyze the companies’ behavior in the electricity market, game theory has been used
as a generalization of decision theory (see Singh (1999)). The concept of Nash equilibrium
(NE) for a game is used as a solution. The NE leads to the strategies that maximize the
companies’ profit (see for example Hasan and Galiana (2010), Hobbs et al. (2000) and Son
Email addresses: [email protected] (Margarida Carvalho), [email protected] (João
Pedro Pedroso)
and Baldick (2004)). This means that in a NE nobody has advantage in moving unilaterally
from it.
As stressed in Baldick (2006), for a model to be tractable it must abstract away from
at least some of the detail. Solvable electricity market models do not use, for example,
transition constraints. First it is necessary to understand the effect changes have on market
rules or structures. In Anderson and Xu (2004), the Australian power market is considered
in detail, as we will do here for the Iberian case. There are some crucial differences between
the formulation in Anderson and Xu (2004) and ours, such as the pool auction structure
and the demand shape. In this context, the authors of Son et al. (2004) analyze the market
equilibria with the first rejected mechanism and the pay-as-bid pricing. Here, the amount
supplied by each generator is a discrete value; this also differs from our model. In Lee and
Baldick (2003) an electricity market with three companies is formulated, and the space of
strategies is discretized in order to find a NE.
Recently, in an attempt to predict market prices and market outcomes, more complex
models have been used. However, many times that does not allow the use of analytical
studies. Thus, techniques from evolutionary programming (see Barforoushi et al. (2010)
and Son and Baldick (2004)) and mathematical programming (see Hobbs et al. (2000) and
Pereira et al. (2005)) have been used in these new models.
In our Iberian market duopoly model, we do not consider network constraints. We will
provide a detailed application of non cooperative game theory in our formulation of the
electricity market. To the best of our knowledge, such theoretical treatment has not been
considered before. In our formulation, demand elasticity will be a parameter as this is
a realistic approach that only has been considered in the simulation of markets, but not
in a theoretical way. Apart of being the first detailed approach of the Iberian market, it
points out the existence of NE in pure strategies in all the instances of this game, showing
how the NE are conditioned by capacity, production costs and elasticity parameters. Some
potential properties about the oligopoly case are also highlighted.
In Section 2, the Iberian duopoly market model will be presented and the concept of
NE will be formalized. In Section 3, some remarks will be made which allow us to find the
NE in a constructive way. Section 4 concludes this detailed NE classification.
2. Iberian Duopoly Market Model
In this section we will begin by explaining our model and fixing notation. Then, game
theory will be introduced with the aim of giving us tools to analyze this market. We set up
a model for a game with two producing firms, which will represent the players, labeled as
Firm 1 and Firm 2. It is assumed that each Firm i owns a generating unit with marginal
cost ci and capacity Ei > 0. Both firms submit simultaneously a bid to the market, using
a pair (qi , pi ) ∈ Si = [0, Ei ] × [0, b] for Firm i, where qi is the proposal quantity, pi is the
bid price and Si is the space of strategies. Firm i’s payoff Πi (q1 , p1 , q2 , p2 ) is a function
that depends on the strategic choices of the rival firm and its own.
The demand is a segment P = mQ + b characterized by the real constants m < 0 and
b > 0. It is assumed that demand, the firms’ marginal costs and capacities are known
3
EQUILIBRIA ON THE DAY-AHEAD DUOPOLY MARKET
b =1.2
Firm 3
0.6
Pd =0.4
0.2
3
$/MWh
supply curve
Firm 1
Firm 2
MWh
100
Qd =160
190
240
Economic Dispatch
Figure Figure
2.1: 2.1.
Economic
Dispatch
2. Iberian Duopoly Market Model
In this section we will begin by explaining our model and fixing notation. Then, game
theory will be introduced with the aim of giving us tools to analyse this market. We set up
a model for a game with two producing firms, which will represent the players, labelled as
Firm 1 and Firm 2. It is assumed that each Firm i owns a generating unit with marginal
cost ci and capacity Ei > 0. Both firms submit simultaneously a bid in the market, through
a pair (qi , pi ) ∈ Si = [0, Ei ] × [0, b] for firm i, where qi is the proposal quantity, pi is the bid
price and Si is the space of strategies. Firm i’s payoff Πi (q1 , p1 , q2 , p2 ) is a function that
depends on the strategic choices of the rival firm.
The demand is a segment P = mQ + b characterized by the real constants m < 0 and
b > 0. It is assumed that the demand and firms’ marginal costs and capacities are known
by all agents. These assumptions can be justified by the knowledge of information about
the technology available for each firm, fuel costs, and precise forecasts of demand. Without
loss of generality let c1 < c2 < b.
dOnce, the market operator has thedfirms’ proposals and the demand, it finds the intersection between the demand representation and the supply curve, which gives the market
−1
clearing price Pd and quantity Qd . For example, suppose that m = 200
, b = 1.2 and Firm 1
bid (90, 0.4), Firm 2 bid (100, 0.2) and Firm 3 bid (60, 0.6). The Market Operator organizes
the proposals by ascending order of prices which gives the supply curve, see Figure 2.1.
Then, the Pd and Qd are found. The accepted bids are the ones in the left side of Qd .
Thereforedthe revenue of firm i is given by:
by all agents. These assumptions can be justified by the knowledge of information on the
technology available for each firm, fuel costs, and precise demand forecasts. Without loss
of generality c1 < c2 < b.
Once, the market operator has the firms’ proposals and the demand, it finds the
intersection between the demand representation and the supply curve, which gives the
−1
, b = 65 and
market clearing price P and quantity Q . For example, suppose that m = 200
Firm 1 bid (90, 0.4), Firm 2 bid (100, 0.2) and Firm 3 bid (60, 0.6). The Market Operator
organizes the proposals by ascending order of prices which gives the supply curve, see
Figure 2.1. Then, Pd and Q are found. The accepted bids are the ones in the left side of
Qd .
Therefore the revenue of Firm i is given by:
Πi = (Pd − ci ) gi
where gi ≤ qi is the accepted quantity in the economic dispatch.
Note that with this structure (linear demand and linear production cost) the firms’
profit is concave and thefore the optimum strategies are an extreme point or a stationary
point.
As a tie breaking rule, we proportionally divide the quantities proposed by the firms
declaring the same price, in case the total quantity is not fully required. If the demand
segment intersects the supply curve in a discontinuity, all the proposals with prices below
the intersection are fully accepted and the market clearing price is given by the last accepted
proposal.
Now we are able to define a NE in pure strategies.
E NE
NE
E
Definition 1. In a game with n players a point sN E = sN
, where sN
1 , s2 , . . . , sn
i
specifies a strategy over the set of strategies of player i, is a Nash Equilibrium if ∀i ∈
{1, 2, . . . , n}:
E
Πi sN E ≥ Πi si , sN
−i
∀si ∈ Si
E
NE
E
where Πi is the utility of player i, Si is his space of strategies and sN
except sN
−i is s
i .
4
Our task is to classify all the NE in this model when players choose their actions
deterministically. This is a hard task because the classical approach of reaction functions
is inadequate, since the payoff functions are non-continuous.
During this work, besides the NE, it was observed that -equilibria are also likely to be
market outcomes.
E N E
E
, where
, s2 , . . . , sN
Definition 2. In a game with n players a point sN E = sN
1
n
N E
si
specifies a strategy over the set of strategies of player i, is an -equilibrium if ∀i ∈
{1, 2, . . . , n} and for a real non-negative parameter :
E
−
∀si ∈ Si .
Πi sN E ≥ Πi si , sN
−i
In our duopoly case, is infinitesimal and just one of the firms has an advantage in
changing its strategy. Thus, we also found -equilibria.
3. Nash Equilibria Classification
The goal of this section is to describe Nash equilibria that may arise in the duopoly
case, characterizing Nash equilibria in terms of the parameters: c1 , c2 , E1 , E2 , m and
b. For that purpose, we first eliminate some cases where there cannot be NE. Recall the
assumption c1 < c2 < b.
Lemma 3. In the duopoly market model no Nash equilibrium in pure strategies has a tie
in the proposals’ prices, which means that p1 = p2 .
Proof. In the tied case both firms bid p1 = p2 = Pd which implies
Qd
Qd
tied
tied
Π1 = (Pd − c1 ) q1
and Π2 = (Pd − c2 ) q2
q1 + q 2
q1 + q2
with q1 + q2 > Qd . (Note that if q1 + q2 = Qd , the firms’ proposals are totally accept.)
If Pd < c1 , then Πtied
< 0 an thus, Firm 1 has incentive to change its strategy to p1 = c1 ,
1
since its payoff will increase to zero.
If Pd = c1 < c2 , then Πtied
< 0, thus Firm 2 has incentive to change its behavior as
2
Firm 1 did in the previous case.
If Pd > c1 , Firm 1 has stimulus to choose pnew
< Pd = p2 and
1
(
E1
if E1 < Pdm−b
q1new = Pd −b
− ε otherwise
m
with ε > 0 infinitesimal. The reason is that with this new choice Firm 1 has an accepted
Pd −b
m
quantity g1 = q1new > q1 q1Q+qd 2 = q1 q1 +q
and the market clearing price is maintained, up to
2
an infinitesimal quantity. Therefore, revenue of Firm 1 increased.
Since there is always at least one firm that benefits from changing its behavior unilaterally, this cannot be an equilibrium.
5
Note that this lemma can be easily generalized for the oligopoly case, as long as the
marginal costs are different for all firms.
We have just eliminated from the possible set of NE the cases with a tie in prices.
Another particular situation occurs when the demand intersects the supply curve in a
discontinuity. This possibility can also be discarded from the potential NE.
Lemma 4. In the duopoly market model, a Nash equilibrium in pure strategies always
intersects the supply curve.
Proof. Let us prove the lemma by contradiction. In the duopoly market model, suppose
that there is an equilibrium such that the demand curve intersects the supply curve in a
discontinuity. It suffices to note that the firm with the last proposal being accepted takes
advantage increasing the price of this proposal unilaterally up to the intersection point,
because the market clearing price increases and the market clearing quantity is maintained.
Therefore, this leads to a contradiction, since it was assumed that there was an equilibrium.
Potential equilibria will have: Firm 1 monopolizing the market with Pd = p1 < p2 = c2 ,
Firm 1 deciding Pd = p1 > p2 or Firm 2 deciding Pd = p2 > p1 . Firm 2 never monopolizes
the market since it is the less competitive company (c2 > c1 ).
The proposition below summarizes the interesting strategies in an equilibrium, that is,
the potential equilibria. In the following sections we will evaluate under which conditions
they are an equilibrium.
Proposition 5. In the duopoly market model the equilibria have
1. Firm 1 deciding the market clearing price and both firms producing; this means Pd =
p1 > p2 , in which case Firm 2 plays the largest possible quantity, as long as p1 remains
greater than c2 , and Firm 1 may play:
2m
2m
, p1 = c1 +b+q
– stationary point;
(a) the duopoly optimum q1 ≥ c1 −b−q
2m
2
(b) the duopoly optimum (q1 = E1 , p1 = (E1 + q2 ) m + b) – extreme point;
2. Firm 1 monopolizing; which means Pd = p1 < p2 = c2 and in this case Firm 1 may
play:
1 −b
, p1 = c12+b – stationary point;
(a) the monopoly optimum q1 ≥ c2m
(b) the monopoly optimum (q1 = E1 , p1 = E1 m + b) – extreme point; (c) a price close to Firm 2’s marginal
cost q1 ≥ c2 −ε−b,
, p1 = c2 − ε (or equivam
c2 −b
lently q1 = m − ε, p1 < c2 ) for ε > 0 arbitrary small; ;
3. Firm 2 deciding the market clearing price and both firms producing; this means Pd =
p2 > p1 , in which case Firm 1 plays the largest possible quantity, as long as Pd = p2 ,
and Firm 2 may play:
c2 +b+q1 m
1m
(a) the duopoly optimum q2 ≥ c2 −b−q
,
p
=
– stationary point;
2
2m
2
(b) the duopoly optimum (q2 = E2 , p2 = (E2 + q1 ) m + b) – extreme point.
6
Proof. Let us start with the simplest case:
(2) Consider an equilibrium with Pd = p1 < p2 = c2 . The optimal strategy in the
monopoly case is:
p1 − b
maximize Π1 (q1 , p1 ) =
(p1 − c1 )
m
subject to p1 ∈ [0, b]
A stationary point is at
∂Π1
c1 + b
= 0 ⇔ p∗1 =
∂p1
2
2
1 −b
2
1 −b
≥ E1 and c1 +c
< c2 . When E1 < c2m
the monopoly
with ∂∂pΠ21 < 0, q1∗ ≥ p1m−b = c2m
2
1
optimum is an extreme point: q1 = E1 and p1 = E1 m + b. Therefore, Firm 1’s best
strategy is one of the just derived if c2 > E1 m + b > c12+b . Otherwise, for Firm 1 to
monopolize, it has to bid a price below c2 . In this case, the strategy with the highest
payoff is q1 ≥ c2 −ε−b
, p1 = c2 − ε , for ε arbitrarily small. To make this bid Firm 1’s
m
c2 −b
capacity must satisfy E1 ≥ c2 −ε−b
.
Equivalently,
Firm
1
could
bid
q
≥
−
ε,
p
<
c
,
1
1
2
m
m
allowing Firm 2 to decide Pd = c2 , but Firm 2’s participation in the market would be as
small as ε.
(1) Considerer an equilibrium with Pd = p1 > p2 . In this case, Firm 2 is playing a
quantity bid as high as possible, as long as p1 > c2 . Now, we just have to see which is the
best strategy for Firm 1 when Pd = p1 > p2 (which implies q2 < p1m−b ):
p1 − b
− q2 (p1 − c1 )
maximize Π1 (q1 , p1 , q2 , p2 ) =
m
subject to p1 ∈ [0, b]
A stationary point is at
∂Π1
c1 + b + q 2 m
= 0 ⇔ p∗1 =
∂p1
2
p∗ −b
2m
2m
which implies q1 ≥ 1m − q2 = c1 −b−q
or if E1 < c1 −b−q
⇒ p∗1 = (E1 + q2 ) m + b
2m
2m
(extreme point). The bid quantity for Firm 2 makes sense since once Pd = p∗1 is fixed Firm
2’s profit increases with q2 .
(3) Completely analogous to the above case.
In Figure 3.1, the above proposition is summarized.
Using this proposition, we will compute the conditions in which the above equilibria
exist; this means that neither of the firms will have advantage in unilaterally moving from
1 −b
the chosen strategies. Note that g1 = c2m
and p1 = c12+b is the monopoly optimum.
Therefore, we will use these two values to start our equilibria classification. Figure 3.2
represents the initial division of the space of parameters which will allow us to start a
classifying the equilibria that may occur in this market.
Before the computation of equilibria, we prove the following theorem.
7
54
CHAPTER 4. IBERIAN MARKET MODEL
Potential equilibria
Firm 1 monopolizes
Both firms participate in the market
EQUILIBRIA ON THE DAY-AHEAD DUOPOLY MARKET
7
Parameters
c1 , c2 , E1 , E2 , m, b
Case (a):
The monopoly
optimum is a
stationary point.
Case (b):
The monopoly
optimum is an
extreme point.
Firm 1
decides
Pd = p1 > p2
E1 <
Firm 2
decides
Pd = p2 > p1
c1 +b
2
Case (c): The
duopoly optimum
is a stationary
point.
Case (d): The
duopoly optimum
is an extreme
point.
Case (e): The
duopoly optimum
is a stationary
point.
≥ c2
Both firms participate in the
market
c1 −b
2m
c1 +b
2
c1 −b
2m
E1 ≥
c1 +b
2
< c2
Firm 1 produces
at full capacity
c1 +b
2
≥ c2
Firm 2 becomes
competitive
Firm 1 monopolizes
Figure 4.3: Potential equilibria.
Figure 3.1. Decision tree
Figure 3.1: Potential equilibria.
Figure 3.2:
Decision
tree.
Strategies 3.4.
p∗1 −b
2m
2m
which implies q1 ≥ m − q2 = c1 −b−q
or if E1 < c1 −b−q
⇒ p∗1 = (E1 + q2 ) m + b
2m
2m
(extreme point). The bid quantity for Firm 2 makes sense since once Pd = p∗1 is fixed
Firm 2’s profit increases with q2 .
< c2
c1 − b
c1 + b
, E1 ,
2m
2
(3.1)
Firm 2: s2N E = (q2 ∈ [0, E2 ] , p2 ∈ [c2 , b])
(3.2)
E
Firm 1: sN
=
1
q1 ∈
Theorem
6. toThere
is always an equilibrium in the
Iberian duopoly market model. The
(3) Completely analogous
the above case.
or
Firm 2: s
= (0, p ∈ [0, b])
equilibrium
is
a
Nash
equilibrium
or
an
-equilibrium
with infinitesimal.
In Figure 4.3, the above proposition is summarized.
< c and quantity Q
With these bids, the market clears with price P =
NE
2
c1 +b
2
d
2
d
Firm 1 is at monopoly’s optimum, and Firm 2 has no influence in the market.
Using this proposition, we will compute the conditions in which the above equilibria
(3.3)
2
=
c1 −b
.
2m
3.2. Firm 1 produces at full capacity: E <
and
<c .
Proof.
that
there
is aninalgorithm
exist; this meansSuppose
that neither of the
firms will
have advantage
unilaterally movingA which given the proposals’ firms (q1 , p1 , q2 , p2 ),
from the chosen strategies. Recall from the last section that g =
and p =
iswasable
to output
the
strictly
for each ofLetthem.
Therefore,
, p2in) the=market with
1, p
1 , q2enter
us start by considering
E m + b <A
c ,1ie,(q
Firm
2 cannot
the monopoly
optimum. Thus
we will
use these twobest
values toreaction
start our equilibria
positive profit. Note that
<E <
. In this case, Firm 1 will monopolize the elec0
0
0 of parameters
FigureA
4.4 represents
the
initial
division
of(q
the0 space
(qclassification.
,
p
)
and
(q
,
p
,
q
,
p
)
=
,
p
)
are
the
bids
that
maximize
the
profit
for
Firm
and
2
1 1 2 2
tricity market, although its capacity is less than the optimum
.1
The
Nash equilibrium
1 will
1 allow us to start
2 may
2 occur in this
which
a classification of the equilibria that
is given by:
Firm
market. 2, respectively, given (q1 , p1 , q2 , p2 ).
Strategies 3.5.
1: s
=initial
(E , E m + b)
(3.4)
Our goal is to prove that if we iteratively apply algorithm A forFirmsome
proposals,
2: see A
Equation
it will find a fixed point of A. This is A1 (q1 , p1 , q2 , p2 ) = (q1 , p1 )Firm
and
, por13.3, q2 , p2 ) =
2 (q13.2
(q2 , p2 ), which is an equilibrium of the game.
Let the initial bids be (q1 , p1 , q2 , p2 ) = (E1 , p01 , E2 , c2 ), where p01 is the best bid price for
Firm 1 such that it is lower than c2 . Remember, as mentioned in Proposition 5, that in an
equilibrium the firms bid their entire production capacity.
Applying A1 (E1 , p01 , E2 , c2 ), we can obtain:
c1 −b
2m
1
1
c1 −b
2m
1
c1 +b
2
2
c1 +b
2
c2 −b
m
1
1
c1 −b
2m
NE
1
2
c1 −b
2m
1
1
1. A1 (E1 , p01 , E2 , c2 ) = (E1 , p01 ), meaning that Firm 1 is already making its best proposal
according to Firm 2’s strategy. Now, we apply A2 (E1 , p01 , E2 , c2 ) to see if Firm 2 has
advantage in increasing its price bid.
(a) If A2 (E1 , p01 , E2 , c2 ) = (E2 , c2 ), then (E1 , p01 , E2 , c2 ) is a fixed point of A and
thus it is an equilibrium. Case 2 of Proposition 5 describes this situation.
(b) If A2 (E1 , p01 , E2 , c2 ) = (E2 , p∗2 ), where p∗2 is equal to the one described in case 3
of Proposition 5, then (E1 , p01 , E2 , p∗2 ) is an equilibrium. Firm 1 does not have
advantage in changing from p01 to the price of case 1 in Proposition 5, since if it
has, Firm 1 had done that in the previous step.
2. A1 (E1 , p01 , E2 , c2 ) = (E1 , p∗1 ), where p∗1 is equal to the one of case 1 in Proposition 5.
Firm 1 had advantage in increasing its price bid to p∗1 . Computing A2 (E1 , p∗1 , E2 , c2 )
we obtain:
8
(a) A2 (E1 , p∗1 , E2 , c2 ) = (E2 , c2 ). Firm 2 keeps its bid which implies that (E1 , p∗1 , E2 , c2 )
is an equilibrium.
(b) A2 (E1 , p∗1 , E2 , c2 ) = (E2 , p∗2 ), where p∗2 is equal to the one of case 3 in Proposition 5. Since A, and in particular A2 , only changes to bids that strictly increase
profit, we can conclude for this case that p∗1 < p∗2 . By Proposition 5, p∗1 , is
the best strategy to Firm 1 when it decides on P d and Firm 2 is biding a
price lower than p∗1 . Therefore, A1 (E1 , p∗1 , E2 , p∗2 ) = (E1 , p∗1 ) which means that
(E1 , p∗1 , E2 , p∗2 ) is an equilibrium.
In AppendixI the existence of equilibria is proven, but using a merge of the cases
presented in the following sections and algebraic arguments.
3.1. Firm 1 monopolizes: E1 ≥
c1 −b
2m
and
c1 +b
2
< c2
Firm 1 monopolizes the market, which means that its capacity is high enough, and
its marginal cost low enough, to keep Firm 2 out of the market. This is the case (a) of
Figure 3.1. In this case, the Nash equilibrium is given by:
Strategies 7.
Firm 1:
E
sN
1
c1 + b
c1 − b
, E1 ,
= q1 ∈
2m
2
(3.1)
E
Firm 2: sN
= (q2 ∈ [0, E2 ] , p2 ∈ [c2 , b])
2
(3.2)
E
Firm 2: sN
= (0, p2 ∈ [0, b])
2
(3.3)
or
With these bids, the market clears with price Pd = c12+b < c2 and quantity Qd =
Firm 1 is at the monopoly’s optimum, and Firm 2 has no influence on the market.
3.2. Firm 1 produces at full capacity: E1 <
c1 −b
2m
and
c1 +b
2
c1 −b
.
2m
< c2
We are going to have cases (b), (e) and (d) of Figure 3.1 as equilibria.
Let us start by considering that E1 m + b < c2 , meaning that, Firm 2 cannot enter
1 −b
the market with positive profit. Note that c2m−b < E1 < c2m
. In this case, Firm 1 will
1 −b
.
monopolize the electricity market, although its capacity is lower than the optimum c2m
The Nash equilibrium is given by:
Strategies 8.
E
Firm 1: sN
= (E1 , E1 m + b)
1
Firm 2: see Equations 3.2 or 3.3
9
(3.4)
As before, Firm 1 does not have an incentive to change its strategy, as this will mean that
(q1 , p1 ) = (E1 , E1 m + b) is not an optimum.
Let us now consider E1 m + b ≥ c2 . Firm 1 cannot monopolize the market and, in
this case, the marginal cost of Firm 2 is lower than the monopoly price E1 m + b, i.e.,
E1 m + b ≥ c2 . Moreover, in an equilibrium for this case, Firm 1 never decides the price,
that is., Pd = p2 > p1 . Otherwise, by Proposition 5, if Pd = p1 > p2 , Firm 1 would be
2m
bidding the duopoly optimum price Pd = p∗1 = c1 +b+q
requiring:
2
p∗1 =
c1 + b + q 2 m
> c2
2
but this would imply a negative quantity for Firm 2, which is absurd:
q2 <
2c2 − c1 − b
< 0.
m
1 −b
The inequality 2c2 −c
< 0 is equivalent to c2 > c12+b , which holds by assumption.
m
Therefore, we discarded the possibility of Firm 1 deciding Pd = p1 > p2 (case (c) of
Figure 3.1). The cases (d) and (e) in Figure 3.1 remain as potential NE, which we will
discuss below.
Suppose that Firm 1’s bidding price is p1 < c2 ≤ p2 = Pd . The best reaction for Firm
1m
and the corresponding quantity
2 is the duopoly optimum in Proposition 5: p∗2 = c2 +b+q
2
c2 −b−q1 m
∗
∗
. Indeed p2 ≥ c2 :
is q2 ≥
2m
p∗2 =
c2 + b + q 1 m
c2 + b + E 1 m
≥
≥ c2 ⇔ E1 m + b ≥ c2
2
2
therefore, the bid price p∗2 makes sense, since p∗2 > c2 . Furthermore, Firm 1 bids q1∗ = E1
(and p∗1 < c2 ), otherwise it would have advantage in increasing its quantity and this would
not be an equilibrium.
c2 +b+E1 m
∗
1m
Firm 2 does not have incentive to change its move q2∗ ≥ c2 −b−E
=
,
p
, as
2
2m
2
this is the optimum (stationary point) when p1 < c2 . In order to have an equilibrium,
neither of the firms can benefit from unilaterally changing their behavior.
When Firm 2 decides Pd = p2 > p1 , this firm will not benefit from changing its behavior,
because that would mean decreasing the price p2 , p1 > p2 , and this does not increase Firm
2’s profit, since p1 is lower than c2 .
May Firm 1 be encouraged to reconsider its strategy? In other words, would Firm 1
be interested in increasing price p1 ? The answer is no, because as we
have seen at the
c1 +b+q2∗ m
∗
beginning: if Firm 1 picks the price Pd = p1 > p2 it will be p1 =
but p∗1 < c2 ≤ p∗2 .
2
Therefore, now we just have to distinguish the duopoly optimum as a stationary point and
the duopoly optimum as an extreme point:
1m
• In the case E2 ≥ c2 −b−E
, Firm 2’s optimum is a stationary point. As we assume
2m
E1 m + b > c2 , Firm 1 is not monopolizing , and the Nash Equilibrium is given by
10
Strategies 9.
Firm 2:
E
sN
2
E
Firm 1: sN
= (E1 , p1 ∈ [0, c2 [)
1
(3.5)
c2 − b − E 1 m
c2 + b + E1 m
= q2 ∈
, E2 ,
2m
2
(3.6)
For E1 m + b = c2 and ε, ε0 > 0 arbitrary small, the equilibrium is given by
Strategies 10.
E
Firm 1: sN
= (E1 − ε, p1 ∈ [0, c2 [)
1
(3.7)
E
Firm 2: sN
= (q2 ∈ [ε0 , E2 ] , c2 )
2
(3.8)
1m
• In the case E2 < c2 −b−E
both firms bid at full capacity. Here, Firm 2 does not
2m
1m
have enough capacity to produce the quantity c2 −b−E
required, so q2∗ = E2 and
2m
p∗2 = (q1 + E2 ) m + b. The best reaction quantity for Firm 1 is q1∗ = E1 as before.
Firm 2 is playing (q2 = E2 , p2 = (E1 + E2 ) m + b) which, by construction, is the best
reaction when Firm 1 plays (q1 , p1 ) = (E1 , p1 < c2 ). These strategies are a Nash
equilibrium.
Strategies 11.
E
Firm 1: sN
= (E1 , p1 ∈ [0, (E1 + E2 ) m + b])
1
E
Firm 2: sN
= (E2 , (E1 + E2 ) m + b)
2
(3.9)
(3.10)
Clearly, we can invert the prices of each firm, reaching the Nash equilibrium:
Strategies 12.
E
Firm 1: sN
= (E1 , (E1 + E2 ) m + b)
1
(3.11)
E
Firm 2: sN
= (E2 , p2 ∈ [0, (E1 + E2 ) m + b])
2
(3.12)
The conclusions of this section are summarized in the decision tree of Figure 3.3.
3.3. Firm 2 becomes competitive: E1 ≥
c1 −b
2m
and
c1 +b
2
≥ c2
In this case, we will have cases (c), (b) and (e) of Figure 3.1 as equilibria.
11
E1 <
c1 −b
2m
and
c1 +b
2
< c2
E1 m + b ≥ c2
E1 m + b < c2
Both Firms
participate in
the Market
Firm 1 monopolizes the market
see the NE 3.2 to 3.4
E2 <
c2 −b−E1 m
2m
E2 ≥
c2 −b−E1 m
2m
see the NE
3.9, 3.10,
3.11 and 3.12
E1 m + b > c2
see the NE
3.5 to 3.6
E1 m + b = c2
see the NE
3.7 to 3.8
Figure 3.3: Equilibria for Section 3.2.
Figure 3.3: Equilibria for Section 3.2.
3.3.1. Firm 1 decides Pd = p1 > p2
We are interested in finding under which conditions case (c) in Figure 3.1 is a NE.
Clearly p1 > c2 and therefore, we assume p∗2 = c2 . Under the results of Proposition 5, Firm
2m
2m
and q1∗ ≥ c1 −b−q
. This requires
1 must be bidding the stationary point p∗1 = c1 +b+q
2
2m
c
+b+q
m
2c2 −c1 −b
∗
1
2
p1 =
≥ c2 and thus, we must have q2 ≤
. Similarly, the quantity produced
2
m
8
c1 −b−q2 m
by Firm 1 ( 2m ) must be positive. This leads to the inequality q2 ≤ c1m−b , which is
weaker than the former one, since b > c2 > c1 . The best strategy for Firm 2 is:
q2∗ =
(
2c2 −c1 −b
m
E2
1 −b
− ε if E2 ≥ 2c2 −c
m
otherwise.
(3.13)
Note that the quantity q2∗ mentioned above is positive as long as c2 ≤ c12+b , which holds.
Therefore, Firm 2 is playing the largest possible quantity, as stated in Proposition 5.
c −b−q ∗ m
For the sake of simplicity, let us first consider the case of Firm 1 playing q1∗ = 1 2m 2 ,
c −b−q ∗ m
1 −b
rather than q1∗ > 1 2m 2 . Hence, q1∗ ≤ c2m
which by assumption is lower than E1 and
therefore, it makes sense to bid this quantity at price p∗1 .
For the profile of strategies s1 = (q1∗ , p∗1 ) and s2 = (q2∗ , c2 ) to be an equilibrium, it is
required that neither firm changes its strategy.
If Firm 2 has an incentive to choose another strategy it would be p˜2 > p∗1 . The
question now is whether there is a strategy s˜2 = (q˜2 , p˜2 ) such that Π2 (q˜2 , p˜2 , q1∗ , p∗1 ) >
Π2 (q2∗ , p∗2 , q1∗ , p∗1 ). When Firm 2 increases the price, the best strategy is to choose p˜2 =
12
c2 +b+q1∗ m
2
=
b+c1 +2c2 −q2∗ m
.
4
In order to have p˜2 > p∗1 , the following inequality must hold
q2∗ >
2c2 − c1 − b
.
3m
(3.14)
1 −b
1 −b
Inequality 3.14 depends on our instance of the problem. Notice that 2c2 −c
> 2c2 −c
,
m
3m
2c2 −c1 −b
so Inequality 3.14 holds whenever E2 > 3m . Our goal is to see under which conditions
Firm 2 does not change the price to p˜2 .
1 −b
In this context, if E2 ≤ 2c2 −c
then q2∗ = E2 and Firm 2 does not have advantage in
3m
picking a strategy different from (E2 , c2 ).
c1 +b+q2∗ m
b+c1 +2c2 −q2∗ m
∗
1 −b
Suppose E2 > 2c2 −c
=
;
then
p
˜
=
>
p
. Can this be the case
2
1
3m
4
2
that
c1 + b + q2∗ m
− c2 q2∗
Π2 (q2∗ , c2 , q1∗ , p∗1 ) =
2
is larger than
p˜2 − b
∗
∗ ∗
Π2
− q1 , p˜2 , q1 , p1 =
m
b + c1 + 2c2 − q2∗ m
b + c1 + 2c2 − q2∗ m − 4b c1 − b − q2∗ m
− c2
−
=
?
4
4m
2m
The answer is no. This proof falls naturally, see in AppendixA. Consequently, we are only
1 −b
interested in the case where q2∗ = E2 ≤ 2c2 −c
.
3m
Now, we have to study under which conditions Firm 1 does not have advantage in
picking p˜1 < p2 = c2 .
If E1 < c2m−b then Firm 1 bids p˜1 < c2 and q˜1 = E1 . Otherwise, if E1 ≥ c2m−b , Firm 1
bids p˜1 < c2 and q˜1 = c2m−b − ε or p˜1 = c2 − ε and q˜1 ≥ c2 −ε−b
. Obviously in the second
m
c2 −b
case (E1 ≥ m ) the Firm 1’s profit is higher, so we use it to compare with its profit of
our possible NE:
c1 + b + E2 m
− (b + E2 m − c1 )2
c1 − b − E2 m
∗ ∗
Π1 (q1 , p1 , E2 , c2 ) =
− c1
=
.
2
2m
4m
So
Π1 (q1∗ , p∗1 , E2 , c2 ) ≥ lim Π1 (q˜1 , p˜1 , E2 , c2 )
ε→0
when
"
E2 ∈ 0, min
c1 − b + 2
!#
p
(c1 − c2 ) (c2 − b) 2c2 − c1 − b
,
m
3m
see AppendixB.
Thus, when Equation 3.15 holds and E1 ≥
c2 −b
m
(3.15)
we have the equilibrium:
Strategies 13.
Firm 1:
E
sN
1
c1 + b + E 2 m
c1 − b − E2 m
= q1 ∈
, E1 ,
2m
2
13
(3.16)
c1 −b
2m
E1 ≥
and
c1 +b
2
≥ c2
Firm 1 decides Pd = p1 > p2
E1 <
c2 −b
m
E1 ≥
Firm 1 never
monopolizes
the market
E1 ≥
(b+c2 −2c1 )2
9m(c1 −c2 )
see the
NE 3.16 to
3.17
Firm 1 can
monopolize
the market
E1 <
A≤B
E2 ≤ A
(b+c2 −2c1 )2
9m(c1 −c2 )
c2 ≥
B<A
E2 > B
E2 > A
c2 −b
m
No NE under these
conditions
E2 ≤ B
c2 <
A0 ≤ B
E2 ≤ A0
see the
NE 3.16 to
3.17
b+4c1
5
b+4c1
5
A0 > B
E2 > A0
E2 > B
No NE under these
conditions
E2 ≤ B
see the
NE 3.16 to
3.17
Figure 3.4: Firm 1 decides Pd = p1 > p2 .
Figure 3.4: Firm 1 decides Pd = p1 > p2 .
E
Firm 2: sN
= (E2 , p2 ∈ [0, c2 ])
2
(3.17)
Note that we have relaxed Firm 2’s biding price. This is possible because the profits do
not depend on it.
When E1 < c2m−b and Equation 3.18 holds and we also have the above equilibrium (see
10
Equations 3.16 and 3.17).
!#
"
p
c1 − b + 2 E1 m (c1 − c2 ) 2c2 − c1 − b
,
(3.18)
E2 ∈ 0, min
m
3m
√
√
c1 −b+2 E1 m(c1 −c2 )
c1 −b+2 (c1 −c2 )(c2 −b)
0
1 −b
Let A =
,
A
=
and B = 2c2 −c
. We built the
m
m
3m
decision tree in Figure 3.4. In order to verify that all the decisions in the tree make sense,
that is, that all the regions in its leafs are non-empty, let us observe the following:
(b+c2 −2c1 )2
9m(c1 −c2 )
•
c1 −b
2m
•
(b+c2 −2c1 )2
9m(c1 −c2 )
≤
<
c2 −b
m
when c2 ≤
c1 +b
;
2
when c2 ∈
4c1 +b
5
, c12+b .
This supports the fact that the decision tree makes sense or, in other words, that decisions
do not lead us to empty spaces.
c −b−q ∗ m
Remember, that we used q1∗ = 1 2m 2 . In this case, the capacity of Firm 2 had to be
1 −b
lower than 2c2 −c
, otherwise, Firm 2 would change its strategy with benefit.
3m
14
It could be possible to find more general conditions for the Nash equilibrium with Firm
1 deciding Pd = p1 > p2 .
c −b−q ∗ m
q1∗ > 1 2m 2 . Note that with bids (s1 , s2 ) = (q1∗ , p∗1 , q2∗ , p2 ) =
We did not try the strategy c −b−q ∗ m c +b+q ∗ m
c −b−q ∗ m
q1∗ > 1 2m 2 , 1 2 2 , q2∗ , c2 , Firm 1 only produces the quantity g1 = 1 2m 2 . However Firm 1 may play a quantity q1 larger than g1 , in order to reduce Firm 2’s incentive in
1 −b
.
changing the price p2 . So, our goal is to find a lower bound for q1∗ when E2 ≥ 2c2 −c
3m
c2 +b+q1 m
∗
Let us see how larger q1 has to be. If Firm 2 increases the price c2 to p˜2 =
2
this implies:
c2 + b + q1∗ m
c1 + b + q2∗ m
c1 − c2
c2 − b
> p∗1 =
⇔ q1∗ <
+ q2∗ <
2
2
m
m
and the new quantity dispatched should be positive:
c2 +b+q1∗ m
2
m
−b
− q1∗ > 0 ⇔ q1∗ <
c2 − b
m
−c2
therefore q1∗ < c1m
+ q2∗ is the strongest condition until now.
We still have to add the condition that leads Firm 2 to a higher profit
c2 + b + q1∗ m
c2 − b − q1∗ m
∗ ∗ p˜2 − b
∗
Π2 q1 , p1 ,
− q1 , p˜2 =
− c2
m
2
2m
is larger than
Π2 (q1∗ , p∗1 , q2∗ , c2 ) =
q2∗
(c1 + b + q2∗ m − 2c2 )
2
√
√
K2
K2
, see AppendixC. If q1∗ ≥ c2 −b+
Firm 2 does not have advantage in
when q1∗ < c2 −b+
m
m
changing its strategy.
Now, we merely need the conditions for Firm 1 to keep the strategy (q1∗ , p∗1 ). Proceeding
as before:
√
K2
∗
,
p
1. Let E1 ≥ c2m−b . Firm 1 does not change its strategy ( q1∗ > c2 −b+
1 ) if
m
"
q2∗ ∈ 0,
Note that
c1 − b + 2
# "
"
p
p
(c2 − b) (c1 − c2 )
c1 − b − 2 (c2 − b) (c1 − c2 )
∪
,∞ .
m
m
p
c1 − b − 2 (c2 − b) (c1 − c2 )
2c2 − c1 − b
>
m
m
p
⇔ −2 (c2 − b) (c1 − c2 ) < 2 (c2 − c1 )
|
{z
} | {z }
<0
and
c1 − b + 2
p
>0
(c2 − b) (c1 − c2 )
2c2 − c1 − b
<
m
m
15
⇔ −2c22 + c2 (3c1 + b) − c1 b − c21 > 0
c1 + b
⇔ c2 ∈ c1 ,
2
thus
q2∗
= E2 ≤
c1 − b + 2
p
(c2 − b) (c1 − c2 )
.
m
The Nash equilibrium is given by:
Strategies 14.
Firm 1:
E
sN
1
=
c1 + b + E 2 m
q1 ,
2
c2 − b +
with q1 ∈
m
√
K2
, E1
Firm 2: see Equation 3.17
2. Let E1 <
c2 −b
.
m
"
q2∗ ∈ 0,
Note that:
Firm 1 does not change its strategy if
c1 − b + 2
# "
"
p
p
E1 m (c1 − c2 )
c1 − b − 2 E1 m (c1 − c2 )
∪
,∞ .
m
m
p
c1 − b − 2 E1 m (c1 − c2 )
2c2 − c1 − b
>
m
m
and
c1 − b + 2
p
E1 m (c1 − c2 )
2c2 − c1 − b
<
m
m
⇔ c2 ∈ [c1 , c1 − E1 m]
and
c1 − b
c1 + b
c1 + b
c1 − E1 m > c1 −
m=
⇒ c1 ,
⊂ [c1 , c1 − E1 m]
2m
2
2
thus for
q2∗ = E2 <
c1 − b + 2
p
E1 m (c1 − c2 )
m
we have the Nash equilibrium of 3.19 and 3.17.
The decision tree is in Figure 3.5.
16
(3.19)
E1 ≥
c1 −b
2m
and
c1 +b
2
≥ c2 and E2 ≥
2c2 −c1 −b
3m
Firm 1 decides Pd = p1 > p2
c2 −b
m
E1 <
√
c2 −b+ K2
m
E2 ≤ A ∧ E1 ≥
E1 ≥
E2 > A ∨ E1 <
c2 −b
m
√
c2 −b+ K2
m
E2 > A0
No NE under
these conditions
See the NE
3.19 and 3.17
No NE under
these conditions
E2 ≤ A0
see the NE
3.19 and 3.17
Figure 3.5: Firm 1 decides Pd = p1 > p2 .
Figure 3.5: Firm 1 decides Pd = p1 > p2 .
4. Conclusions
5. Acknowledgments
3.3.2. Firm 1 monopolizes
This work was supported by a INESC Porto fellowship in the setting of the Optimization
Now we areInterunit
goingLine.
to establish the conditions that make Firm 1 monopolize the market
as an equilibrium (case (b) in Figure 3.1). Here, the market outcome is an -equilibrium.
AppendixA. Firm 2 changes its strategy
Notice that in this case,c p−b1 c<+b c2 (otherwise,
Firm 2 would have
advantage
in partici
c −b−q m
c +b+q m
−b
Assume that E1 ≥ 2m , 2 ≥ c2 and E2 > 2c −c
, and that
Firm 1 is cplaying
q1∗ =
, p∗1 =
.
c1 −b
1 −b
3m c2 −b
2m
2
(
<
),
otherwise
the
demand
is
pating in the market)
and
it
is
required
E
≥
1
−b
−b
m
By Equation 3.13, since E2 > 2c −c
then q2∗ > 2c −c
. Let2m
us prove thatmFirm 2 will change
3m
3m
c
−ε−b
2
b+c
+2c
−q
m
p
˜
−b
∗
∗
not intersecteditsby
Firm
bid. Hence the
best bid
for
, c2 − ε
strategy
(q2∗ , p1’s
: Firm 1 is (q1 , p1 ) =
2 = c2 ) to q˜2 = m − q1 , p˜2 =
4
m
with ε > 0. Firm 2 bids (q2 , p2 ) Π=2 (q(E
strategy from Firm 2 leads to the
∗ ∗2 ,∗c2∗). Any other
∗ ∗
2 , p2 , q1 , p1 ) < Π2 (q˜2 , p˜2 , q1 , p1 )
same or less profit.
q2∗
−1
2
⇔
(c + b + q ∗ m − 2c ) <
(b + c − 2c − q ∗ m)
1
1
2
2
2
1
1
2
2
∗
2
1
∗
2
1
1
2
2
1
1
2
16m
∗
2
1
2
2
1. Consider E2 ≥ c2m−b . ⇔If9 mFirm
1 chooses p˜ 1 > c , the2 produced quantity is zero, and
2
(q2∗ ) + 3q2∗ (c1 + b − 2c2 ) + 1 (c1 +2b − 2c2 ) < 0
2
2m
therefore, Firm 1 will not choose to make
a bid different from p1 = c2 − ε. Hence the
2c2 − b − c1
∗
⇔ q2 6=
3m
equilibrium is:
for this reason, Firm 2 has benefit in modifying its strategy.
Strategies 15.
Firm 1:
E
sN
1
c2 − ε − b
, E1 , c2 − ε
= q12
1 ∈
m
E
= (E2 , c2 )
Firm 2: sN
2
(3.20)
(3.21)
or
Strategies 16.
Firm 1:
E
sN
1
=
c2 − b
− ε, p1 ∈ [c1 , c2 − ε]
m
Firm 2: see Equation 3.21
17
(3.22)
2. Consider E2 < c2m−b . Then Firm 1 can bid p˜1 > c2 , producing a non zero quantity. If
Firm 1 has incentive to change its strategy it will be to:
c1 − b − E2 m c1 + b + E2 m
(q˜1 , p˜1 ) =
,
.
2m
2
However, for this new bid to make sense, we need
p˜1 =
c1 + b + E 2 m
> c2 ,
2
which depends on the instance of our problem.
1 −b
< c2m−b ⇔ c2 > c1 . Using this the following cases are possible.
Remark: 2c2 −c
m
1 −b
2m
(a) Consider E2 < 2c2 −c
⇔ p1 = c1 +b+E
> c2 . Is
m
2
p˜1 − b
c1 + b + E2 m
c1 − b − E 2 m
Π1
− E2 , p˜1 , E2 , c2 =
− c1
m
2
2m
higher than
lim Π1
ε→0
c2 − ε − b
, c2 − ε, E2 , c2
m
= (c2 − c1 )
c2 − b
?
m
Not when Equation 3.23 holds (see AppendixD for a proof).
#
"
p
c1 − b + 2 (c1 − c2 ) (c2 − b) 2c2 − c1 − b
,
.
E2 ∈
m
m
(3.23)
Hence, the equilibrium is given by:
Strategies 17.
Firm 1: see Equation 3.20
Firm 2: see Equation 3.21
or
Strategies 18.
Firm 1: see Equation 3.22
Firm 2: see Equation 3.21
1 −b
2m
(b) Consider E2 ≥ 2c2 −c
⇔ p1 = c1 +b+E
≤ c2 . In this case, Firm 1 does not
m
2
have stimulus to change its behavior and hence, the equilibrium is given by
Equations 3.20 to 3.22.
1 −b
Let B 0 = 2c2 −c
. We have the decision tree of Figure 3.6 corresponding to this case.
m
1 −b
Since , with the assumptions of this section, c2m−b ≥ 2c2 −c
≥ A0 yields, the decision
m
tree makes sense.
18
E1 ≥
c1 −b
2m
c1 +b
2
and
≥ c2
Firm 1 monopolizes
E1 ≥
E2 ≥
E1 <
c2 −b
m
c2 −b
m
No NE under
these conditions
c2 −b
m
c2 −b
m
E2 <
See the NE 3.20
to 3.22
E2 ≥
2c2 −c1 −b
m
E2 <
2c2 −c1 −b
m
See the NE 3.20
to 3.22
E2 ≥ A0
E2 < A0
See the NE 3.20
to 3.22
No NE under
these conditions
Figure 3.6: Firm 1 monopolizes.
Figure 3.6: Firm 1 monopolizes.
AppendixB. Firm 1 does not change its strategy
c1 −b
Assume that E1 ≥
,
c1 +b
2c2 −c1 −b
≥ c2 , E2 ≤
and E1 ≥
c2 −b
, and that Firm 2 is
2m
2
3m
m
∗
3.3.3. Firm 2 decides
p1 us
playing (q2∗P=d E=
conditions Firm
1 does not change its strategy
2 , p2p=
2 ). Let
2 c>
prove in cwhich
c −b−E m
c +b+E m
−ε−b
∗
∗
q =
, p1 =
to q˜1 = m , p˜1 = c2 − ε :
2m
2
Finally, we will1 search
for equilibria
where Firm 2 decides the market clearing price, in
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) ≥ lim Π1 (q˜1 , p˜1 , q2∗ , p∗2 )
other words, p1 < p2 = Pd (case (d) and (e) ofε→0Figure 3.1).
2
c2 −b
c2c−b
− (b + E2 m −
c2 −fact
b
1)
Note that we have to impose
⇔ E1 < m ,≥ due
lim (c2 to
− c1 the
)
− ε that if E1 ≥ m , Firm 1 has
ε→0
4m
m
capacity
incentive to monopolize the−1market
(since it has sufficient
for that purpose). So,
c2 − b
⇔
>0
c2 −b (b − c1 )2 + 2E2 m (b − c1 ) + E22 m2 − (c2 − c1 )
c1 −b
4m
m
we consider 2m ≤ E1 < m .
c2 +b+q1∗ m
∗
−1
1
1
c2 − b
2
Let us assume p1 < c2 .⇒From
Proposition
the
for
E22 m
− E2 (b − c1 ) − 5,
(b −
c1 ) best
− (c2 − strategy
c1 )
=
0 Firm 2 is p2 =
2
4
2
4m
m
c −b−q ∗ m
p
and q2∗ ≥ 2 2m 1 , where q1∗ = E1 .
c1 − b ± 2 (c2 − b) (c1 − c2 )
⇔ E2 =
m
For this strategy there are the following
requirements
(the price of the duopoly optimum
In what follows it is study this solution:
makes sense and Firm 2√has the production
capacity
of
playing
the stationary point of the
√
c −b−2 (c −b)(c −c )
c −b+2 (c −b)(c −c )
•
>
⇔
−2
<
2;
m
m
duopoly optimum):
√
1
1
c1 −b+2
2
1
2
1
2
2
2
1
2
(c2 −b)(c1 −c2 )
1
2
p
(b−c1 )2
b−c1
•
0c2⇔
⇔
−b (c1 − c2 ) (c2 − b) > 2 ⇔ (c1 − c2 ) (c2 − b) >
1m
m
4
≥ c (c⇔
E1 ≤<(b−c
by+ b)−c
ourb−assumption;
1. p∗2 = c2 +b+E
) , which holds
(b−c )
c +b
2
2
m
>
0,
note
that
−c
+c
(c
=
0
⇔
c
=
−c22 +c2
2 1 + b)−c1 b−
2
1
1
2
2
4
4
2
√
√
1m
c −b+2 (c −b)(c −c )
c −b−2 (c −b)(c −c )
2. E2 ≥ c2 −b−E
which
depends
on≤ the
instance
thus,
0≤
; of our problem.
m
m
2m
1
1
2
1
2
2
1
1
We have to study each of these cases.
1. Suppose E2 ≥
2
1
2
1
2
14
c2 −b−E1 m
,
2m
then Firm 2 plays the stationary optimum
c2 − b − E1 m ∗ c2 + b + E1 m
∗
q2 ≥
, p2 =
2m
2
and Firm 1 (E1 , p1 < c2 ).
Obviously Firm 2 will not have incentive in reconsidering another ∗ proposal, but
c +b+q m
Firm 1 may have advantage in choosing a higher price p˜1 = 1 2 2 > p∗2 , such
c +b+q ∗ m
that Π1 (q1∗ , p1 < c2 , q2∗ , p∗2 ) < Π1 p˜1m−b − q2∗ , p˜1 > p∗2 , q2∗ , p∗2 . Is p˜1 = 1 2 2 > p∗2 =
c2 +b+E1 m
?
2
19
c1 + b + q2∗ m
c2 + b + E1 m
c2 − c1 + E1 m
>
⇔ q2∗ <
2
2
m
which depends on the instance of our problem. Thus, these cases have to be considered separately:
2 −b
1 −b
1m
≥ c2 −c1m+E1 m ⇔ E1 ≤ 2c1 −c
. However, E1 ≥ c2m
≥
(a) Suppose c2 −b−E
2m
3m
2c1 −c2 −b
c1 +b
⇔
c
≥
,
which
by
assumption
does
not
hold.
So
this
case
never
2
3m
2
happens.
2 −b
1m
(b) Suppose E1 > 2c1 −c
and q2∗ = c2 −b−E
, then Firm 1 has benefit in increasing
3m
2m
its proposal price since:
∗
p1 − b
∗ c1 + b + q 2 m
∗ ∗
− q2 ,
, q2 , p2 > Π1 (E1 , p1 < c2 , q2∗ , p∗1 )
Π1
m
2
−1
E2
(2c1 − c2 − b + E1 m)2 > (c2 + b + E1 m − 2c1 )
16m
2
2c1 − b − c2
⇔ E1 6=
.
3m
2 −b
1m
(c) Suppose E1 > 2c1 −c
and q√2∗ > c2 −b−E
, then as we already did before, there is
3m
2m
c1 −b+ K1
, where K1 = −2m2 E12 +(−2mc2 − 2bm + 4mc1 ) E1 :
an equilibrium if E2 ≥
m
Strategies 19.
Firm 1: see Equation 3.5
√
c1 − b + K 1
c2 + b + E 1 m
NE
, E2 ,
Firm 2: s2 = q2 ∈
(3.24)
m
2
1m
2. Suppose E2 < c2 −b−E
. In this case (q1∗ , p∗1 ) = (E1 , p1 < c2 ) and (q2∗ , p∗2 ) = (E2 , (E1 + E2 ) m + b)
2m
.
Firm 2 will not change this behavior, so let us see when Firm 1 has advantage in
increasing p∗1 to p˜1 . For the purpose we need
⇔
p˜1 =
and E2 >
c1 −b
m
c1 + b + E 2 m
c1 − b
> p∗2 = (E1 + E2 ) m + b ⇔ E2 >
− 2E1
2
m
− 2E1 is true, since E2 > 0 and
c1 − b
c1 − b
− 2E1 ≤ 0 ⇔ E1 ≥
.
m
2m
So, p˜1 =
Is
c1 +b+E2 m
2
Π1
higher than
> p∗2 = (E1 + E2 ) m + b.
c1 + b + E2 m
p˜1 − b
− E2 ,
, E2 , p∗2
m
2
=
−1
(−c1 + b + E2 m)2
4m
Π1 (E1 , p1 < c2 , E2 , p∗2 ) = E1 ((E1 + E2 ) m + b − c1 )?
Yes, see AppendixE. Therefore Firm 1 has stimulus in changing its strategy.
In short, we can summarize this Nash equilibria with the decision tree of Figure 3.7.
20
E1 ≥
c1 −b
2m
and
c1 +b
2
≥ c2
Firm 2 decides Pd = p2 > p1
E1 <
E1 ≥
c2 −b
m
c2 −b
m
No NE under these conditions
E2 <
c2 −b−E1 m
2m
E2 ≥
c2 −b−E1 m
2m
No NE under these
conditions
E2 ≥
√
c1 −b+ K1
m
E2 <
See the NE
3.5 and
3.24
√
c1 −b+ K1
m
No NE under
these conditions
Figure 3.7: Firm 2 decides Pd = p2 > p1 .
Figure 3.7: Firm 2 decides Pd = p2 > p1 .
1m
2. Suppose E2 < c2 −b−E
. In this case (q1∗ , p∗1 ) = (E1 , p1 < c2 ) and (q2∗ , p∗2 ) = (E2 , (E1 + E2 ) m + b)
2m
.
Firm 2 will not change this behavior, so let us see cwhen
in increasing
c1advantage
+b
1 −b Firm 1 has
1
2
p∗1 to p˜1 . For the purpose we need
2m
2
3.4. Both firms participate in the market: E <
and
≥c
cgoing
c1 − b(c), (d) and (e) of Figure 3.1.
The NE of this section are
the ones from cases
1 + b + Eto
2 m be
p˜1 =
> p∗2 = (E1 + E2 ) m + b ⇔ E2 >
− 2E1
2
m
1 −b
Since E1 < c2m
< c2m−b , Firm
1
does
not
have
capacity
to
monopolize
the market.
c −b
and E2 >
1
m
− 2E1 is true, since E2 > 0 and
3.4.1. Firm 2 decides Pd = p2 > p1 c1m− b − 2E1 ≤ 0 ⇔ E1 ≥ c12m− b .
We start with the
case in mwhich
P = p > p1 (case (e) of Figure
So, p˜1 = c +b+E
> p∗2 = (E1d+ E2 ) m2+ b.
3.1).∗ As Proposition
5
2
c2 −b−q1 m c2 +b+q1∗ m
Is
∗ ∗
,
and
states, Firm 2 plays the stationary
duopoly
optimum
c1 + b + E 2 m
−1(q2 , p2 ) = 2
p˜1 − b
2m
2
Π1
− E2 ,
, E2 , p∗2 =
(−c1 + b + E2 m)
m
2
4m
q1∗ will be as large as
possible such that:
higher than
1
2
Π1 (E1 , p1 < c2 , E2 , p∗2 ) = E1 ((E1 + E2 ) m + b − c1 )?
c + b + q∗m
c −b
≥ c2 ⇔ q 1 ≤
In short, we can summarize this 2
Nash equilibria with the decision tree m
of Figure 3.7.
2
2
Yes, see Appendix
AppendixE.
Therefore
Firm 1 has stimulus
its strategy.
1
∗
∗ in changing
p2 =
and
4. Conclusions
q∗ =
5. Acknowledgments
2
c2 − b − q1∗ m
c2 − b
≥ 0 ⇔ q1∗ ≤
2m
m
This work was supported by a INESC Porto fellowship in the setting of the Optimization
Interunitthat
Line. q ∗ > 0, since q ∗ = c2 −b−E1 m > c2 −b − c1 −b = 2c2 −b−c1 > 0 ⇔ c2 <
thus q1∗ = E1 . Note
2
2
2m
2m
4m
4m
c2 +b+E1∗ m
b+c1
∗
.
An
important
fact
is
that
p
=
>
c
,
since
this
is
equivalent
to E1 < c2m−b
2
2
2
2
c
−b−E
m
which is true. On the other hand q2∗ = 2 152m 1 ≤ E2 depends on the instance of our
problem, so Firm 2 may have to play the extreme point of the duopoly optimum.
1m
1m
1. Suppose E2 ≥ c2 −b−E
. Firm 1 plays (q1∗ = E1 , p∗1 < c2 ) and Firm 2 plays q2∗ = c2 −b−E
, p∗2 =
2m
2m
It is easy to see that Firm 2 does not have advantage in choosing other strategy. Let
1m
us see if Firm 1 will change its behavior to p˜1 = 2c1 +c2 +b−E
. In that case, p˜1 must
4
∗
be higher than p2 :
p˜1 =
and
2c1 −c2 −b
3m
2c1 + c2 + b − E1 m
c2 + b + E1 m
2c1 − c2 − b
>
= p∗2 ⇔ E1 >
4
2
3m
≤
c1 −b
,
2m
which depends on the instance of our problem.
21
c2 +b+E
2
2 −b
(a) Let E1 ≤ 2c1 −c
. Then, Firm 1 does not have stimulus in changing its behavior
3m
unilaterally. Here, the Nash equilibrium is given by:
Strategies 20.
Firm 1: see Equation 3.5
Firm 2: see Equation 3.6
(b) Let E1 >
2c1 −c2 −b
3m
Π1
⇔ p1 =
2c1 +c2 +b−E1 m
4
p˜1 − b
− q2∗ , p˜1 , q2∗ , p∗2
m
which is larger than the profit
>
=
c2 +b+E1 m
2
= p∗2 . We have:
−1
(c2 + b − 2c1 − E1 m)2
16m
E1
(c2 + b + E1 m − 2c1 )
2
2
if E1 6= 2c1 −b−c
. Therefore, Firm 1 has advantage in changing its strategy.
3m
1m
However, like we already did in the Section 3.3.3, Firm 2 can pick q2∗ > c2 −b−E
2m
sufficiently large such that Firm 1 does not have advantage in changing its
strategy. This case is completely analogous to the one treated √in Section 3.3.3,
K1
let K1 = −2E12 m2 + (−2c2 m + 4c1 m − 2bm)E1 . If E2 ≥ c1 −b+
we have the
m
Nash equilibrium:
Strategies 21.
Firm 1: see Equation 3.5
Π1 (E1 , p1 < c2 , q2∗ , p∗2 ) =
Firm 2: see Equation 3.24
1m
. Here Firm 2’s duopoly optimum is an extreme point,
2. Suppose E2 < c2 −b−E
2m
∗
∗
q2 = E2 , p2 = (E2 + E1 ) m + b and q1∗ = E1 . Notice that p∗2 = (E2 + E1 ) m + b > c2 ,
since both firms are playing smaller quantities than the last case (the market clearing
price increases when the market clearing quantity decreases).
Clearly, Firm 2 will not change its strategy, but Firm 1 may have incentive in choosing
a higher price p˜1 > p∗2 and that requires:
c1 − b − E2 m
c1 + b + E 2 m
> p∗2 = (E1 + E2 ) m + b ⇔ E1 >
2
2m
which depends on the instance of our problem.
p˜1 =
2m
(a) Let E1 ≤ c1 −b−E
. Then, Firm 1 will not change its behavior.
2m
The Nash equilibrium is given by:
Strategies 22.
Firm 1: see Equation 3.9
Firm 2: see Equation 3.10
22
E1 <
c1 −b
2m
and
c1 +b
2
≥ c2
Firm 2 decides Pd = p2 > p1
E2 ≥
E1 ≤
c2 −b−E1 m
2m
2c1 −c2 −b
3m
E2 <
E1 >
2c1 −c2 −b
3m
E1 >
c1 −b−E2 m
2m
No NE under
these conditions
See the NE
3.5 to 3.6
E2 <
c2 −b−E1 m
2m
√
c1 −b+ K1
m
E2 ≥
No NE under
these conditions
E1 ≤
c1 −b−E2 m
2m
See the NE
3.9 to 3.10
√
c1 −b+ K1
m
See the NE
3.5 and 3.24
Figure 3.8: Firm 2 decides Pd = p2 > p1 .
Figure 3.8: Firm 2 decides Pd = p2 > p1 .
Strategies 21.
(b) Let E1 >
c1 −b−E2 m
.
2m
(b) Let E1 >
Firm 1: see Equation 3.9
Is
Firm 2: see Equation 3.10
Is
c1 + b + E2 m
p1 − b
∗
,E ,p
Π1
2 2
p1 − b− E2c,1 + b + E2 m
− E2 ,
Π1m
2, E2, p∗2
m
2
c1 −b−E2 m
.
2m
higher thanhigher than
Π1 (E1 , p1 < c2 , E2 , (E1 + E2 ) m + b)?
Π (E , p1 < c2 , E2 , (E1 + E2 ) m + b)?
1 solve:
1
This is equivalent to
−1
2
This is equivalent to solving:
(−c1 + b + E2 m) < ((E1 + E2 ) m + b − c1 ) E1
4m
−1 ⇔ −mE12 + (c1 − b − E2m)2 E1 − 1 (−c1 + b + E2m)2 < 0
4m
(−c1 + b + E2 m) < ((E
1 + E2 ) m + b − c1 ) E1
c1 − b − E2 m
4m
⇔ E1 6=
,
2m
so Firm 1 has
2 advantage in changing its strategy.
1
(−c1 + b + E2 m)2 < 0
4m
c1 − b − E 2 m
4. Conclusions
⇔ E1 6=
,
2m
5. Acknowledgments
so FirmThis
1 work
has was
advantage
changing
its strategy.
supported by in
a INESC
Porto fellowship
in the setting of the Optimization
⇔ −mE + (c1 − b − E2 m) E1 −
From the above we1reach the decision tree of Figure 3.8.
Interunit Line.
From the above we reach the decision tree of Figure 3.8.
17
3.4.2. Firm 1 decides Pd = p1 > p2
Firm 1 wants to play the stationary point (q1∗ , p∗1 ) =
requires:
(
, which
c1 + b + q2∗ m
2c2 − c1 − b
> c2 ⇔ q2∗ <
2
m
1 −b
− ε E2 ≥ 2c2 −c
m
. Let us note that:
E2
otherwise
the instance of our problem.
thus q2∗ =
c1 −b−q2∗ m c1 +b+q2∗ m
,
2m
2
2c2 −c1 −b
m
23
c1 −b−q2∗ m
2m
≤ E1 depends on
2c2 −c1 −b
, then q2∗
m
∗
c −b−q m
−c2
Let E1 ≥ 1 2m 2 =c1m
1 −b
play 2c2 −c
− ε, c2 .
m
1. Suppose E2 ≥
(a)
=
−
2c2 −c1 −b
− ε.
m
ε
. Hence, Firm
2
1 can play (q1∗ , p∗1 ) and Firm 2 can
Will Firm 1 decrease the price p1 < c2 ? This means:
Π1 (E1 , c1 , q2∗ , c2 ) = (c2 − c1 ) E1 ≥ lim Π1 (q1∗ , p∗1 , q2∗ , c2 )
ε→0
−1 (c1 − c2 )2
m
c1 − c2
⇔ E1 ≥
.
m
⇔ (c2 − c1 ) E1 ≥
c −b−q ∗ m
−c2
Since E1 ≥ 1 2m 2 = c1m
− 2ε , Firm 1 will change its behavior.
−c2
then q1∗ = E1 , p∗1 = (E1 + q2 ) m + b and
(b) Let E1 < c1m
(
c2 −b
− E1 − ε E2 ≥ c2m−b − E1
m
q2 =
E2
otherwise.
If E2 ≥ c2m−b − E1 , Firm 2 will change its strategy as with the present one, its
1m
(> c2 ⇔ E1 <
profit is almost zero and increasing p2 from c2 to p˜2 = c2 +b+E
2
c2 −b
−1
which holds by assumption), Firm 2’s profit is higher: 4m (−c2 + b + E1 m)2 >
m
0. Hence, let E2 < c2m−b − E1 that implies q2∗ = E2 . In this case, it is easy to see
that Firm 1 does not have advantage in changing its strategy:
Π1 (E1 , (E1 + E2 ) m + b, E2 , c2 ) ≥ Π1 (E1 , p1 < c2 , E2 , c2 )
⇔ ((E1 + E2 ) m + b − c1 ) E1 ≥ (c2 − c1 ) E1
c2 − b
⇔ E2 ≤
− E1
m
which holds.
Now, we are going to see under which conditions Firm 2 does not have incentive
to change its behavior. First of all, if Firm 2 changes its strategy, it will be with
1m
, requiring
p˜2 = c2 +b+E
2
p˜2 > p∗1 = (E1 + E2 ) m + b ⇔ E2 >
c2 − b − E1 m
2m
which depends on our instance.
1m
If E2 > c2 −b−E
, Firm 2 chooses this new strategy, see AppendixF.
2m
c2 −b−E1 m
If E2 ≤
, we have the NE:
2m
Strategies 23.
Firm 1: see Equation 3.11
Firm 2: see Equation 3.12
24
2c2 −c1 −b
, which
m
2m
. So,
E1 ≥ c1 −b−E
2m
2. Suppose E2 <
implies q2∗ = E2 .
2 m c1 +b+E2 m
Firm 1 can play (q1∗ , p∗1 ) = q1∗ ≥ c1 −b−E
,
.
(a) Let
2m
2
Firm 1√does not have advantage in decreasing the price to p˜1 < c2 when E2 <
c1 −b+2 E1 m(c1 −c2 )
, see AppendixG.
m
Finally, we have to check if Firm 2 has advantage in increasing the price p˜2 > p∗1 .
c +b+q ∗ m
In that case, p˜2 = 2 2 1 which requires:
p˜2 =
c1 + b + E2 m
c1 − c2
c2 + b + q1∗ m
> p∗1 =
⇔ q1∗ <
+ E2 .
2
2
m
−c2
2m
1 −b
Note that c1m
+ E2 ≤ c1 −b−E
⇔ E2 ≤ 2c2 −c
. In this way, if E2 ≤
2m
3m
2c2 −c1 −b
c1 −c2
∗
⇒ q1 ≥ m + E2 and we have the NE:
3m
Strategies 24.
Firm 1: see Equation 3.16
Firm 2: see Equation 3.17
−c2
−c2
1 −b
2m
Let E2 > 2c2 −c
⇔ c1 m
+E2 > c1 −b−E
, thus q1∗ need to be q1∗ < c1m
+E2 ⇔
3m
2m
∗
c2 +b+q1 m
c1 +b+E2 m
∗
> p1 =
. Will this new proposal price for Firm 2 increases
p2 =
2
2
its profit?
√
K2
If E1 > c2 −b+
, the answer is no (see AppendixI) and thus we have the
m
following NE:
Strategies 25.
√
K2
c1 + b + E 2 m
c
−
b
+
2
NE
, E1 ,
(3.25)
Firm 1: s1 = q1 ∈
m
2
Firm 2: see Equation 3.17
(b) Let E1 <
c1 −b−mE2
,
2m
then q1∗ = E1 , p∗1 = (E1 + q2 ) m + b and
(
c2 −b
− E1 − ε E2 ≥ c2m−b − E1
m
q2 =
E2
otherwise.
2
− E1 then E1 ≥ c2m−b − E2 . Since E1 < c1 −b−mE
, then c2m−b − E2 <
2m
c1 −b−mE2
1 −b
1 −b
⇔ E2 > 2c2 −c
and we previously assumed E2 < 2c2 −c
. Thus
2m
m
m
c2 −b
∗
E2 < m − E1 , implies q2 = E2 . Firm 1 does not change its strategy, since:
If E2 ≥
c2 −b
m
Π1 (E1 , (E1 + E2 ) m + b, E2 , c2 ) ≥ Π1 (E1 , p1 < c2 , E2 , c2 )
⇔ ((E2 + E1 ) m + b − c1 ) E1 ≥ Π1 = (c2 − c1 ) E1
⇔ E2 ≤
25
c2 − b
− E1
m
E1 <
c1 −b
2m
and
c1 +b
2
≥ c2
Firm 1 decides Pd = p1 > p2
E2 ≥
E1 <
2c2 −c1 −b
m
E1 ≥
c1 −c2
m
c1 −c2
m
No NE under
these conditions
E2 <
E2 ≥
c2 −b
− E1
m
c2 −b
− E1
m
No NE
under these
c −b−E1 m
E2 < 2
2m
c −b−E1 m
E2 ≥ 2
2m
No NE
See the NE
under these
3.11 and
conditions
3.12
conditions
Figure 3.9: Firm 1 decides P = p > p .
2
Figure 3.9: Firm 1 decides dPd =1 p1 >
p2 .
4. Conclusions
and the last inequality holds.
1m
Firm 2 does not change its strategy (note that c2 +b+E
> (E1 + E2 ) m + b ⇔
2
c
−b
E
c
−b−E
m
2
1
2
1
This work
supported
by a E
INESC
fellowship
E2 >was
− 2 ) when
, since in the setting of the Optimization
2 = Porto
2m
2m
5. Acknowledgments
Interunit Line.
Π2 (E1 , (E1 + E2 ) m + b, E2 , c2 ) ≥
AppendixA. Firm 2 changes its strategy
c +b+E m
p −b
−E ,
≥ Π E , (E + E ) m + b,
m
2
c −b−q ∗ m
c1 −b c1 +b
1 −b 2
≥ c2 and E2 > 2c2 −c
, and that Firm 12 is playing1 q1∗ = 1 2m 2 , p∗1
2m , 2 2
3m
1
1
1
2
1 −b
1 −b
By Equation 3.13, since E2 > 2c2 −c
then q2∗ > 2c2 −c
. Letus prove that Firm 2 will change
3m
3m
b+c1 +2c2 −q2∗ m
:
its strategy (q2∗ , p∗2 = c2 ) to q˜2 = p˜2m−b − q1∗ , p˜2 =
4
Assume that E1 ≥
⇔ ((E2 + E1 ) m + b − c2 ) E2 ≥
Π2 (q2∗ , p∗2 , q1∗ , p∗1 )
<
−1
(E1 m + b − c2 )2
4m
Π2 (q˜2 , p˜2 , q1∗ , p∗1 )
q2∗
−1c − b − E m ∗ 2
2 (b + c1 − 2c
1 2 − q2 m)
(c1 + b + q2∗ m − 2c2 ) <
⇔
E
=
2
2 16m
2m
9
1
2
2
⇔ m (qc2∗−b
) + 3qE2∗ (c1 + b − 2c2 ) +
(c1 + b − 2c2 ) < 0
2
Thus, for E22 ≤ 2m
− 21 , we have the 2m
NE:
2c2 − b − c1
Strategies 26.
⇔ q2∗ 6=
3m
Firm
1:
see
Equation 3.11
for this reason, Firm 2 has benefit in modifying its strategy.
⇔
AppendixB. Firm 1 does not change its strategy
Firm 2: see Equation 3.12
1 −b
1 −b
Assume that E1 ≥ c2m
, c12+b ≥ c2 , E2 ≤ 2c2 −c
and E1 ≥ c2m−b , and that Firm 2 is
3m
∗
∗
playing (q2as= before,
E2 , p2 =we
c2 ).have
Let us
prove
in which
conditions
Firm
1 does
change its strategy
Proceeding
decision
tree
of Figures
andnot
3.10.
the
3.9
c1 +b+E2 m
c2 −ε−b
∗
2m
q1∗ = c1 −b−E
,
p
=
to
q
˜
=
,
p
˜
=
c
−
ε
:
1
1
2
1
2m
2
m
In AppendixI there are the trees with all the possible equilibria in pure strategies.
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) ≥ lim Π1 (q˜1 , p˜1 , q2∗ , p∗2 )
ε→0
26
20
=
c1 +b+q2∗ m
2
.
Firm 1 decides Pd = p1 > p2
E2 ≥
2c2 −c1 −b
m
E1 ≥
c1 −c2
m
E1 <
c1 −c2
m
No NE under
these conditions
E2 <
E2 ≥
c2 −b
− E1
m
c2 −b
− E1
m
No NE
under these
c −b−E1 m
E2 < 2
2m
c −b−E1 m
E2 ≥ 2
2m
No NE
See the NE
under these
3.11 and
conditions
3.12
conditions
Figure 3.9: Firm 1 decides Pd = p1 > p2 .
Figure 3.9: Firm 1 decides Pd = p1 > p2 .
4. Conclusions
5. Acknowledgments
1 −b
and c12+b ≥ c2
E1 < c2m
This work was supported by a INESC
Porto fellowship in the setting of the Optimization
Interunit Line.
Firm 1 decides Pd = p1 > p2
E
<
2
AppendixA. Firm 2 changes its strategy
2c2 −c1 −b
m
c1 −b c1 +b
1 −b
≥ c2 and E2 > 2c2 −c
, and that Firm 1 is playing
2m , 2
3m
c −b
c1 −b
E2
E2
E1 < 1
−
2c2 −c1 −b
2c2 −c1E−b
∗
1 ≥ 2m − 2
2m
2
By Equation 3.13, since E2 >
then q2 >
. Letus prove that Firm
3m
3m
b+c1 +2c2 −q2∗ m
:
its strategy (q2∗ , p∗2 = c2 ) to q˜2 = p˜2m−b − q1∗ , p˜2 =
4
Assume that E1 ≥
E2 ≥
E1
c2 −b
−
2m
2
E2 <
c2 −b
E
− ∗1
2m 2
22
Π (q , p∗2 , q1∗ , p∗1 ) p
< Π2 (q˜2 , p˜2 , q1∗ , p∗1 )
E
>
c1 −b+2
See 2the NE
AppendixB. Firm 1 does not change its strategy
q1∗ =
c1 −b−q2∗ m
, p∗1
2m
2 will change
p
c −b+2 E1 m(c1 −c2 )
m
E1 m(c1 −c2 )
2
E2 ≤ 1
m
q∗
−1
2
⇔
(c1 + b + q2∗ m − 2c2 ) < No NE
(b under
+ c1 − 2c2 − q2∗ m)
under these
3.11
2 and
16m
these conditions
conditions
3.12
9
1
2
∗ 2
∗
−c1 −b
⇔ m (q2 ) + 3q2 (c1 + b − 2c2 ) + E2 ≤(c2c12+
b − 2c2 ) < 0E >
3m
2
2
2m
2c
−
b
−
c
2
1 See the NE
⇔ q2∗ 6=
3m
3.16 to 3.17
p
K2
for this reason, Firm 2 has benefit in modifying its strategy. E1 < c2 −b+
m
No NE
2c2 −c1 −b
3m
E1 ≥
p
c2 −b+ K2
m
No NE
See the NE
under these
3.25 and
conditions
c2 −b
3.17
1 −b
1 −b
Assume that E1 ≥ c2m
, c12+b ≥ c2 , E2 ≤ 2c2 −c
and E1 ≥ m , and that Firm 2 is
3m
∗
∗
playing (q2 = E2 , p2 = c2 ). Let
us
prove
in
which
conditions
not change its strategy
Figure
1 decides Pd =Firm
p1 > 1p2does
3.10: Firm
c1 +b+E
c2 −ε−b
∗
2m
2m
Figure
3.10:
Firm
1 decides
Pdε=
p1 >. p2 .
q1∗ = c1 −b−E
,
p
=
to
q
˜
=
,
p
˜
=
c
−
:
1
1
2
1
2m
2
m
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) ≥ lim Π1 (q˜1 , p˜1 , q2∗ , p∗2 )
ε→0
20
20
27
=
c1 +b+q2∗ m
2
.
4. Discussion and conclusions
In the Iberian duopoly market model, the demand and the production costs are linear.
As Proposition 5 suggests, there are five types of Nash equilibria. Instances where Firm 2
participates in the market with infinitesimal quantity ε will be considered as a monopoly
for Firm 1.
When Firm 1 monopolizes the market in an equilibrium, the selected prices may be
c1 +b
, E1 m + b or c2 − ε, depending on the efficiency/competitiveness of Firm 2. For a
2
high marginal cost c2 , Firm 1 bids the monopoly optimum: p1 = c12+b or p1 = E1 m + b.
Furthermore, for c2 closer to c1 , if Firm 2’s capacity E2 is large enough, Firm 1 monopolizes
bidding p1 = c2 − ε; otherwise, for a limited capacity E1 , Firm 1 may prefer to bid a higher
price, sharing the market with Firm 2. For c2 even closer to c1 , the market clearing price
in an equilibrium may be decided by either one of the firms. Firm 1 decides Pd , which
means Pd = p1 > p2 , if the capacity of Firm 2 is limited. The case of Firm 2 deciding Pd ,
meaning Pd = p2 > p1 , is analogous.
In some competitive situations, there are two NE: one case with Pd = p1 > p2 and
another with Pd = p2 > p1 . By simulating random instances it was observed that Firm 1
has a high profit in the equilibrium Pd = p2 > p1 , while Firm 2 benefits when Pd = p1 > p2
and there is no rational way of deciding among them. We have discretized the space of
strategies S1 and S2 to compute NE in mixed strategies in these cases. It was observed
that a combination of these two equilibria leads to a new NE in mixed strategies. However,
the new equilibrium does not benefit either of the firms comparatively to the pure NE.
In conclusion, this work completely classified the NE that may occur in a duopoly
day-ahead market. This helps understand the sensitivity of the outcomes to the instances’
parameters and the diversity of equilibria that may arise. Furthermore, it illustrates the
rational strategies of each firm.
5. Acknowledgments
This work was supported by a INESC TEC fellowship in the setting of the Optimization
Interunit Line.
References
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pp.
29
AppendixA. Firm 2 changes its strategy
1 −b
1 −b
E1 ≥ c2m
, c12+b ≥ c2 and E2 > 2c2 −c
, and that Firm 1 is playing
3m
Assume that
c1 −b−q2∗ m
c1 +b+q2∗ m
∗
∗
1 −b
1 −b
q1 =
, p1 =
. By Equation 3.13, since E2 > 2c2 −c
then q2∗ > 2c2 −c
.
2m
2
3m
3m
b+c1 +2c2 −q2∗ m
Let us prove that Firm 2 will change its strategy (q2∗ , p∗2 = c2 ) to q˜2 = p˜2m−b − q1∗ , p˜2 =
:
4
Π2 (q2∗ , p∗2 , q1∗ , p∗1 ) < Π2 (q˜2 , p˜2 , q1∗ , p∗1 )
−1
q2∗
(c1 + b + q2∗ m − 2c2 ) <
(b + c1 − 2c2 − q2∗ m)2
2
16m
9
1
⇔ m (q2∗ )2 + 3q2∗ (c1 + b − 2c2 ) +
(c1 + b − 2c2 )2 < 0
2
2m
2c2 − b − c1
⇔ q2∗ 6=
3m
for this reason, Firm 2 benefits from changing its strategy.
⇔
AppendixB. Firm 1 does not change its strategy
1 −b
1 −b
Assume that E1 ≥ c2m
, c12+b ≥ c2 , E2 ≤ 2c2 −c
and E1 ≥ c2m−b , and that Firm 2 is
3m
Firm 1 does not change its
playing (q2∗ = E2 , p∗2 = c2 ). Let us prove
in whichc conditions
c1 −b−E2 m
c1 +b+E2 m
∗
∗
2 −ε−b
strategy q1 =
, p1 =
to q˜1 = m , p˜1 = c2 − ε :
2m
2
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) ≥ lim Π1 (q˜1 , p˜1 , q2∗ , p∗2 )
ε→0
c2 − b
−ε
m
−1
c2 − b
2
2 2
⇔
(b − c1 ) + 2E2 m (b − c1 ) + E2 m − (c2 − c1 )
>0
4m
m
1
1
c2 − b
2
2 −1
− E2 (b − c1 ) −
(b − c1 ) − (c2 − c1 )
=0
⇒ E2 m
4
2
4m
m
p
c1 − b ± 2 (c2 − b) (c1 − c2 )
⇔ E2 =
m
Nextm this solution is studied:
√
√
c1 −b−2 (c2 −b)(c1 −c2 )
c1 −b+2 (c2 −b)(c1 −c2 )
•
>
⇔ −2 < 2;
m
m
√
p
c1 −b+2 (c2 −b)(c1 −c2 )
1
(c1 − c2 ) (c2 − b) > b−c
•
<
0
⇔
⇔ (c1 − c2 ) (c2 − b) >
m
2
− (b + E2 m − c1 )2
≥ lim (c2 − c1 )
⇔
ε→0
4m
2
+ c2 (c1 + b) − c1 b − (b−c4 1 ) > 0, note that −c22 + c2 (c1 + b) − c1 b −
√
√
c1 −b+2 (c2 −b)(c1 −c2 )
c1 −b−2 (c2 −b)(c1 −c2 )
c2 = c12+b thus, 0 ≤
≤
;
m
m
−c22
30
(b−c1 )2
4
(b−c1 )2
=
0
4
⇔
⇔
•
√
p
1 −b
> 2c2 −c
⇔
4c
−
2b
−
2c
−
6
(c1 − c2 ) (c2 − b) < 0 ⇔
1
2
3m
p
−3 (c1 − c2 ) (c2 − b) < b + c2 − 2c1 ;
{z
}
|
{z
} |
c1 −b−2
(c2 −b)(c1 −c2 )
m
>0
<0
•
√
p
1 −b
< 2c2 −c
⇔ 3 (c1 −c2 ) (c2 − b) > c2 + b − 2c1 ⇔ −10c22 +
3m
1 b+c1
c2 (7b + 13c1 ) − 5c1 b − b2 − 4c21 > 0 ⇒ c2 ∈ b+4c
, 2 .
5
c1 −b+2
(c1 −c2 )(c2 −b)
m
Therefore, Firm 1 does not change its strategy when
"
!#
p
c1 − b + 2 (c1 − c2 ) (c2 − b) 2c2 − c1 − b
E2 ∈ 0, min
,
.
m
3m
AppendixC. Firm 2 changes its strategy
1 −b
1 −b
E1 ≥ c2m
, c12+b ≥ c2 and E2 ≥ 2c2 −c
, and that Firm 1 is playing
3m
Assume that
∗m
∗m
c
−b−q
c
+b+q
1
1
2
2
q1∗ >
, where q2∗ is given by Equation 3.13. Let us prove in which
, p∗1 =
2m
2
c −b−q ∗ m
conditions Firm 2 changes its strategy (q2∗ , p∗2 = c2 ) to q˜2 = p˜2m−b − q1∗ , p˜2 = 2 2m 1 :
c2 + b + q1∗ m
c2 − b − q1∗ m
∗ ∗
Π2 (q1 , p1 , q˜2 , p˜2 ) =
− c2
2
2m
is higher than
q∗
Π2 (q1∗ , p∗1 , q2∗ , p∗2 ) = 2 (c1 + b + q2∗ m − 2c2 )
2
−1
q∗
(−c2 + b + q1∗ m)2 > 2 (c1 + b + q2∗ m − 2c2 )
4m
2
2
1
−m ∗ 2
c2 − b 1 b − c2
(q1 ) + q1∗
−
− q2∗ (c1 + b + q2∗ m − 2c2 ) > 0
⇔
4
2
4
m
2
⇔
let K2 = −2 (q2∗ m)2 + q2∗ (−2mc1 − 2bm + 4c2 m)
√ √
c
−
b
+
K
c
−
b
−
K2
2
2
2
∗
⇔ q1 ∈ 0,
∪
,∞
m
m
Note that for q2∗ ≤
2c2 −b−c1
,
m
K2 is non negative and :
√
c2 − b + K 2
c1 − c2
>
+ q2∗
m
m
⇔ −3 (mq2∗ )2 + q2∗ (−2bm + 4c2 m − 2mc1 − 2m (c1 − 2c2 + b)) − (c1 − 2c2 + b)2 < 0
2c2 − b − c1
2c2 − b − c1
∗
⇔ q2 ∈ 0,
∪
,∞
3m
m
but clearly
2c2 − b − c1
2c2 − b − c1
≤ q2∗ ≤
3m
m
√
c2 −b+ K2
∗
so q1 <
is the condition for Firm 2 to change its strategy.
m
31
AppendixD. Firm 1 does not change its strategy
1 −b
Assume that E1 ≥ c2m−b , c12+b ≥ c2 , E1 ≥ c2m−b , E2 < c2m−b and E2 < 2c2 −c
, and that
m
∗
∗
Firm 2 is playing (q2 = E2 , p2 = c2 ). Let us prove in which conditions Firm 1 does not
2m
change its strategy q1∗ = c2 −ε−b
:
, p∗1 = c2 − ε to q˜1 = p˜1m−b − E2 , p˜1 = c1 +b+E
m
2
c1 + b + E 2 m
c1 − b − E2 m
∗ ∗
Π1 (q˜1 , p˜1 , q2 , p2 ) =
− c1
2
2m
higher than
lim Π1
ε→0
is equivalent to:
c2 − ε − b
, c2 − ε, q2∗ , p∗2
m
= (c2 − c1 )
−1
(b + E2 m − c1 )2 > (c2 − c1 )
4m
2
2
c2 − b
m
⇔ (E2 m) + 2E2 m (b − c1 ) + (b − c1 ) − 4m (c1 − c2 )
"
⇒ E2 ∈ 0,
c1 − b + 2
p
c2 − b
m
c2 − b
m
>0
# "
#
p
(c1 − c2 ) (c2 − b)
c1 − b − 2 (c1 − c2 ) (c2 − b)
∪
,∞
(D.1)
m
m
Let us study this solution:
√
p
c1 −b−2 (c1 −c2 )(c2 −b)
1 −b
> 2c2 −c
⇔ c2 − c1 > − (c1 − c2 ) (c2 − b)
1.
m
m
| {z } |
{z
}
>0
<0
√
p
c1 −b+2 (c1 −c2 )(c2 −b)
2c2 −c1 −b
2.
<
⇔
(c1 − c2 ) (c2 − b) > c2 −c1 ⇔ −2c22 +c2 (3c1 + b)−
m
m bc1 − c21 > 0 ⇒ c2 ∈ c1 , c12+b
If
E2 ∈
"
c1 − b + 2
#
p
(c1 − c2 ) (c2 − b) 2c2 − c1 − b
,
m
m
holds then Firm 1 does not change its bid.
AppendixE. Firm 1 changes its strategy
1m
1 −b
Assume that c2m
≤ E1 < c2m−b , c12+b ≥ c2 and E2 < c2 −b−E
, and that Firm 2 is
2m
∗
∗
playing (q2 = E2 , p2 = (E2 + E1 ) m + b). Let us prove
that Firm 1 will change its strategy
p˜1 −b
c1 +b+E2 m
∗
∗
(q1 = E1 , p1 < c2 ) to q˜1 = m − E2 , p˜1 =
:
2
Π1 (q˜1 , p˜1 , q2∗ , p∗2 ) =
−1
(−c1 + b + E2 m)2
4m
higher than
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) = E1 ((E1 + E2 ) m + b − c1 )
32
is equivalent to
−1
(−c1 + b + E2 m)2 ≤ E1 ((E1 + E2 ) m + b)
4m
1
⇔ −mE12 + E1 (c1 − b − E2 m) −
(−c1 + b + E2 m)2 ≤ 0
4m
c1 − b − E2 m
c1 − b E2
⇒ E1 =
=
−
2m
2m
2
1 −b
which never occurs since E1 ≥ c2m
.
AppendixF. Firm 2 changes its strategy
−c2 c1 +b
1 −b
1m
Assume that E1 < c1m
, 2 ≥ c2 , E2 ≥ 2c2 −c
and c2 −b−E
< E2 < c2m−b − E1 , and
m
2m
∗
∗
that Firm 2 changes
that Firm 1 is playing (q1 = E1 , p1 = (E1 + E2 ) m + b). Let us prove
p˜2 −b
c2 +b+E1 m
∗
∗
its strategy (q2 = E2 , p2 = c2 ) to q˜2 = m − E1 , p˜2 =
:
2
Π2 (q1∗ , p∗1 , q2∗ , p∗2 ) ≥ Π2 (q1∗ , p∗1 , q˜2 , p˜2 )
−1
(−c2 + b + E1 m)2
4m
2
1 E1 m + b − c2
2
⇔ E2 m + (E1 m + b − c2 ) E2 +
≥0
4
m
⇔ (mE1 + E2 m + b − c2 ) E2 ≥
⇔ E2 =
c2 − b − E1 m
2m
thus Firm 2 will choose this new strategy.
AppendixG. Firm 1 does not change its strategy
2m
1 −b
1 −b
Assume that c1 −b−E
≤ E1 < c2m
, c12+b ≥ c2 and E2 < 2c2 −c
, and that Firm 2
2m
m
∗
∗
is playing (q2 = E2 , p2 = c2 ). Let us see
in
which
conditions
Firm
1
does
not change its
c1 −b−E2 m
c1 +b+E2 m
∗
∗
, p1 =
to (q˜1 = E1 , p˜1 < c2 ):
strategy q1 ≥
2m
2
Π1 (q1∗ , p∗1 , q2∗ , p∗2 ) ≤ Π1 (q˜1 , p˜1 , q2∗ , p∗2 )
−1
(−c1 + b + q2∗ m)2 ≤ (c2 − c1 ) E1
4m
"
#
p
p
c1 − b + 2 E1 m (c1 − c2 ) c1 − b − 2 E1 m (c1 − c2 )
∗
,
q2 ∈
m
m
⇔
Note:
•
√
c1 −b−2
E1 m(c1 −c2 )
m
>
2c2 −c1 −b
m
p
⇔ −2 E1 m (c1 − c2 ) < 2 (c2 − c2 );
|
{z
} | {z }
<0
33
>0
•
√
c1 −b+2
tion
E1 m(c1 −c2 )
−c2
1 −b
> 2c2 −c
⇔ E1 < c1m
which
m
m
c1 −b
E2
c1 −b
2c2 −c1 −b
c1 −c2
E1 ≥ 2m − 2 ≥ 2m − 2m = m .
So if E2 <
√
c1 −b+2
E1 m(c1 −c2 )
,
m
never happens because by assump-
Firm 1 does not change its strategy.
AppendixH. Firm 2 does not change its strategy
2m
1 −b
1 −b
1 −b
Assume that c1 −b−E
≤ E1 < c2m
, c12+b ≥ c2 and 2c2 −c
< E2 < 2c2 −c
, and that
2m
3m
m
c
−b−E
m
c
+b+E
m
c
−c
∗
∗
∗
1
2
1
2
1
2
Firm 1 is playing q1 ≥
, p1 =
with q1 < m + E2 . Let us see in which
2m
2
conditions Firm 2 does not change its (q2∗ = E2 , p∗2 = c2 ) to
c2 + b + q1∗ m
p˜2 − b
∗
− q1 , p˜2 =
,
q˜2 =
m
2
so:
Π2 (q1∗ , p∗1 , q˜2 , p˜2 ) ≥ Π2 (q1∗ , p∗1 , q2∗ , p∗2 )
c1 + b + E2 m
−1
2
∗
(−c2 + b + q1 m) ≥
− c2 E 2
⇔
4m
2
let K2 = −2E22 m2 + (−2mc1 − 2bm + 4mc2 ) E2
√
√ c2 − b − K 2
c2 − b + K 2
∗
,∞
∪
⇔ q1 ∈ 0,
m
m
Some facts about this solution:
•
c1 −c2
m
+ E2 <
√
c2 −b− K2
m
>0
2c2 + E2 m > 0 ⇔ E2 <
•
√
⇔ b + c1 − 2c2 + E2 m > − K2 which holds since b + c1 −
|
{z
}
√
2c2 −c1 −b
;
m
K2
+ E2 < c2 −b+
⇔ 3E22 m2 + (2 (−2c2 + c1 + b) m + 2mc1 + 2bm − 4mc2 ) E2 +
m
2c −c −b
2
1 −b
2
1
(−2c2 + c1 + b) > 0 ⇔ E2 ∈ 0, 2c2 −c
∪
,
∞
.
3m
m
c1 −c2
m
√
Then, q1∗ < c2 −b+
m
its strategy.
K2
is the strongest condition. If E1 >
√
c2 −b+ K2
m
Firm 2 does not change
AppendixI. There is always an equilibrium
First of all, we made a complete study of the Nash equilibria under the following
1 −b
1 −b
conditions: E1 ≥ c2m
∧ c12+b < c2 (Section 3.1) and E1 < c2m
∧ c12+b < c2 (Section 3.2 ).
1 −b
1 −b
In the two remaining cases, E1 ≥ c2m
∧ c12+b ≥ c2 (Section 3.3) and E1 < c2m
∧ c12+b ≥ c2
(Section 3.4), we still have to intersect the conditions of the equilibria found.
In order to achieve the purpose of summarizing all the classification made so far, the
leafs of the decision trees in Figures 3.4, 3.9 and 3.10 will be analyzed using the information
of the remaining trees:
34
2
2 −2c1 )
1 −b
1. Let E1 ≥ c2m
, c12+b ≥ c2 , E1 < c2m−b , E1 ≥ (b+c
, E2 ≤ A. The decision tree in
9m(c1 −c2 )
Figure 3.4 already provides an equilibrium in this case. Neither of the decision trees
in Figures 3.5 and 3.6 have an equilibrium under these conditions. However, tree 3.7
adds the equilibrium of Equations 3.5 and 3.24 to this case. Note that:
√
c1 − b + K 1
A≤
m
c2 − b
⇔ E1 ∈ {0} ∪
,∞
m
thus A >
√
c1 −b+ K1
.
m
c1 −b c1 +b
, 2
2m
√
c1 −b+ K1
depends on the instance;
m
2
2 −2c1 )
E1 < c2m−b , E1 ≥ (b+c
, E2 > A. Neither
9m(c1 −c2 )
So, E2 ≥
≥ c2 ,
of the trees
2. Let E1 ≥
in Figures 3.4, 3.5 and 3.6 have an equilibrium in this case. So, it is expected that
the decision tree of Figure
3.7 has an equilibrium under these conditions. Remember
√
c1 −b+ K1
that E2 > A >
, so the Equations 3.5 and 3.24 give us a NE;
m
2
2 −2c1 )
1 −b
3. Let E1 ≥ c2m
, c12+b ≥ c2 , E1 < c2m−b , E1 < (b+c
, E2 > B, E2 > A. As before,
9m(c1 −c2 )
there is only the NE of Equations 3.5 and 3.24;
2
2 −2c1 )
1 −b
, c12+b ≥ c2 , E1 < c2m−b , E1 < (b+c
, E2 > B, E2 ≤ A, E2 ≥
4. Let E1 ≥ c2m
9m(c1 −c2 )
√
c1 −b+ K1
,
m
√
K2
E1 ≥ c2 −b+
. It is obviously that in this case we have the NE of
m
Equations 3.19 and 3.17 and of Equations 3.5 and 3.24. It should be stressed that
there are instances with this conditions;
2
2 −2c1 )
1 −b
, c12+b ≥ c2 , E1 < c2m−b , E1 < (b+c
, E2 > B, E2 ≤ A, E2 ≥
5. Let E1 ≥ c2m
9m(c1 −c2 )
√
c1 −b+ K1
,
m
√
K2
E1 < c2 −b+
. Clearly the only NE in pure strategies is the one given by
m
Equations 3.5 and 3.24. Again, there is instances in this conditions.
2
2 −2c1 )
1 −b
, c12+b ≥ c2 , E1 < c2m−b , E1 < (b+c
, E2 > B, E2 ≤ A, E2 <
6. Let E1 ≥ c2m
9m(c1 −c2 )
√
c1 −b+ K1
.
m
We will prove that this conditions imply E1 ≥
Equations 3.19 and
√ 3.17, give us an equilibrium.
c1 −b+ K1
First, E2 <
≤ c2 −c1m+E1 m since:
m
which holds. Note that:
√
c1 − b + K 1
c2 − c1 + E 1 m
≤
m
m
2c1 − c2 − b 2c1 − c2 − b
⇔ E1 ∈
,
3m
m
√
c2 − c1 + E1 m
c2 − b + K 2
≥
m
m
2c2 − c1 − b
⇔ E2 ∈ B,
m
35
√
c2 −b+ K2
,
m
and thus,
2c2 −c1 −b
:
m
which holds, since E2 > B and E2 ≤ A <
A<
−c2
1 −b
and E1 ≥ c2m
> c1 m
.
c1 −c2 +E2 m
Second, E1 ≥
≥
m
c1 − c2
2c2 − c1 − b
⇔ E1 >
m
m
√
c2 −b+ K2
m
since:
E1 ≥
c1 − c2 + E2 m
m
⇔ E2 ≤
which holds;
1 −b
7. Let E1 ≥ c2m
,
c1 +b
√2
c1 −b+ K2
m
8.
9.
10.
11.
12.
13.
≥ c2 , E1 <
c2 − c1 + E1 m
m
c2 −b
,
m
E1 <
(b+c2 −2c1 )2
,
9m(c1 −c2 )
E2 ≤ B < A. In this
case, E2 ≥
depends on the instance, and thus we can have the NE of
Equations 3.5 and 3.24, beyond the equilibrium of Equations 3.16 and 3.17;
1
1 −b
, c12+b ≥ c2 , E1 ≥ c2m−b , c2 ≥ b+4c
, E2 ≤ A0 ≤ B. In this case,
Let E1 ≥ c2m
5
Equations 3.16 and 3.17 give an equilibrium. If E2 = A0 , there is also the NE of
Equations 3.20 to 3.22;
1
1 −b
, c12+b ≥ c2 , E1 ≥ c2m−b , c2 ≥ b+4c
, E2 > A0 . Here, there is a single NE
Let E1 ≥ c2m
5
given by the Equations 3.20 to 3.22;
1 −b
1
Let E1 ≥ c2m
, c12+b ≥ c2 , E1 ≥ c2m−b , c2 < b+4c
, E2 > B. Since A0 > B depends on
5
the value of E2 , we can have at least one of the NE given by Equations 3.19 and 3.17
or 3.20 and 3.22;
1
1 −b
, c12+b ≥ c2 , E1 ≥ c2m−b , c2 < b+4c
, E2 ≤ B. In this case just one NE in
Let E1 ≥ c2m
5
pure strategies exists and it is given by Equations 3.16 and 3.17;
1 −b
1 −b
2 −b
1 −b
, c12+b ≥ c2 , E2 < 2c2 −c
, E1 < c2m
− E22 and E2 ≥ c2m
− E21 . In the
Let E1 < c2m
m
decision tree of Figure 3.10 there is no NE. So, the decision tree of Figure 3.8 must
have an equilibrium for this case. In this way we need to prove that this conditions
2 −b
.
imply E1 ≤ 2c1 −c
3m
Note that:
c2 − b E1
c2 − b
E2 ≥
−
⇔ E1 ≥
− 2E2
2m
2
m
in this way
c2 − b
c1 − b E2
2c2 − c1 − b
− 2E2 <
−
⇔ E2 >
m
2m
2
3m
thus
c1 − b E2
c1 − b 2c2 − c1 − b
2c1 − b − c2
E1 <
−
<
−
=
2m
2
2m
6m
3m
hence, we are in the conditions of the NE 3.5 to 3.6;
1 −b
1 −b
1 −b
2 −b
Let E1 < c2m
, c12+b ≥ c2 , E2 < 2c2 −c
, E1 < c2m
− E22 and E2 < c2m
− E21 . Apart
m
from the NE of Equations 3.11 and 3.12, there is also the equilibrium of Equations 3.9
and 3.10;
36
14. Let E1 <
c1 −b c1 +b
, 2
2m
≥ c2 , E2 <
2c2 −c1 −b
,
m
E1 ≥
c1 −b
E2
and E2 > A.
2
E1 m(c1 −c2 )
1m
> c2 −b−E
,
m
2m
√2m
c1 −b+2
−
First of all, it will be proved that A =
1m
:
E2 > c2 −b−E
2m
p
c1 − b + 2 E1 m (c1 − c2 )
c2 − b − E1 m
>
m
2m
p
⇔ 4 E1 m (c1 − c2 ) < c2 − 2c1 + b − E1 m
|
{z
}
which implies
>0
let γ = 2c21 − 5c2 c1 + b + c1 + 3c22 − c2 b
√ √
6c1 − 7c2 + b + 4 γ
6c1 − 7c2 + b − 4 γ
⇔ E1 ∈ 0,
,∞
∪
m
m
note that γ < 0 ⇔ c2 ∈ c1 , 2c13+b , so if c2 < 2c13+b our proof is over, on the other
hand, if c2 ≥ 2c13+b :
√
6c1 − 7c2 + b + 4 γ
c1 − b
≤
2m
m
√
⇔ 14c2 − 11c1 − 3b ≥ 8 γ
{z
}
|
>0
c1 + b
9b − 7c1
,∞
⇔ c2 ∈ 0,
∪
2
2
√
6c −7c +b+4 γ
1 −b
therefore, E1 < c2m
≤ 1 2m
which ends our proof.
2 −b
If E1 ≤ 2c1 −c
we
are
in
the
conditions
of the NE 3.5 to 3.6.
3m
2c1 −c2 −b
If E1 > 3m , note that:
E2 >
c1 − b + 2
since,
c1 − b + 2
p
E1 m (c1 − c2 )
c1 − b +
≥
m
m
p
E1 m (c1 − c2 )
c1 − b +
≥
m
m
√
K1
√
K1
⇔ 2E12 m2 + (4m (c1 − c2 ) + 2mcc + 2bm − 4mc1 ) E1 ≤ 0
c2 − b
⇔ E1 ∈ 0,
m
which holds. Thus, we are in the conditions of the NE 3.5 and 3.24;
1 −b
1 −b
1 −b
1 −b
15. Let E1 < c2m
, c12+b ≥ c2 , E2 < 2c2 −c
, E1 ≥ c2m
− E22 and E2 ≤ A, E2 ≤ 2c2 −c
.
m
3m
The decision tree of Figure 3.10 already provides the equilibrium of Equations 3.16
and 3.17. Depending on our instance, the Decision Tree 3.8 can give a new equilibrium;
37
1 −b
1 −b
1 −b
16. Let E1 < c2m
, c12+b ≥ c2 , E2 < 2c2 −c
, E1 ≥ c2m
− E22 and E2 ≤ A, E2 >
m
√
K2
E1 < c2 −b+
.
m
1m
.
As before, it will be proved that E2 ≥ c2 −b−E
2m
c1 −b
c2 −b
2c1 −c2 −b
If E1 ≤ 3m , then E2 ≥ m − 2E1 ≥ m − E21 since:
2c2 −c1 −b
,
3m
c1 − b
c2 − b E1
− 2E1 ≥
−
m
m
2
2c1 − c2 − b
≥ E1
3m
thus, we are in the conditions of the NE 3.5 to 3.6.
2 −b
1 −b
2 −b
If E1 > 2c1 −c
, then E2 > 2c2 −c
> c2m
− E21 since
3m
3m
⇔
2c2 − c1 − b
c2 − b E1
>
−
3m
2m
2
⇔ E1 >
2c1 − c2 − b
3m
√
K1
, since
as we intended to prove. We also have E2 ≥ c1 −b+
m
√
2c2 − c1 − b
c1 − b + K 1
<
m
3m
which holds since:
⇔ 9K1 > (2c2 − 4c1 + 2b)2
2c1 − c2 − b
2c1 − c2 − b
⇔ E1 ∈
,2
,
3m
3m
2
2c1 − c2 − b
3m
>
c1 − b
2m
⇔ 5c1 < 4c2 + b
So, we are in the conditions of the NE 3.5 and 3.24.
1 −b
1 −b
1 −b
1 −b
17. Let E1 < c2m
, c12+b ≥ c2 , E2 < 2c2 −c
, E1 ≥ c2m
− E22 and E2 ≤ A, E2 > 2c2 −c
,
m
3m
√
c2 −b+ K2
E1 ≥
. The decision tree of Figure 3.10 has an equilibrium under this
m
1m
conditions. Similarly to the last case, E2 ≥ c2 −b−E
, and thus, depending on the
2m
value of E1 , there is an equilibrium in Equations 3.5 and 3.6 or 3.5 and 3.24;
−c2
1 −b
1 −b
18. Let E1 < c2m
, c12+b ≥ c2 , E2 ≥ 2c2 −c
, E1 ≥ c1m
.
m
Note that:
2c2 − c1 − b
c2 − b − E1 m
2c1 − 3c2 + b
>
⇔ E1 >
,
m
2m
m
−c2
−c2
2 +b
which holds, since 2c1 −3c
≤ c1 m
⇔ c12+b ≥ c2 and E1 > c1m
. Thus E2 ≥
m
2c2 −c1 −b
c2 −b−E1 m
>
.
m
2m
2 −b
If E1 ≤ 2c1 −c
,
we
are in the conditions of the NE 3.5 to 3.6.
3m
38
If E1 >
2c1 −c2 −b
,
3m
since
E2 ≥
2c2 − c1 − b
c2 − c1
c2 − b
≥
+ E1 ⇔
≥ E1
m
m
m
which holds, and
2c2 − c1 − b
c2 − c1
c1 − b +
E2 ≥
≥
+ E1 ≥
m
m
m
we are in the conditions of the NE 3.5 and 3.24;
−c2
1 −b
1 −b
19. Let E1 < c2m
, c12+b ≥ c2 , E2 ≥ 2c2 −c
, E1 < c1m
, E2 ≥
m
E1 <
c2 −b
m
√
K1
,
− E1 . Note that:
2c1 − c2 − b
c1 − c2
≤
m
3m
and
c2 − b − E1 m
c2 − b − E1 m
≥
m
2m
so, we are in the conditions of the NE 3.5 to 3.6.
−c2
1 −b
1 −b
1m
20. Let E1 < c2m
, c12+b ≥ c2 , E2 ≥ 2c2 −c
, E1 < c1m
, E2 < c2m−b − E1 , E2 < c2 −b−E
.
m
2m
Apart from the NE of Equations 3.11 and 3.12, the decision tree of Figure 3.8 also
adds the NE of Equations 3.9 and 3.10. In order to demonstrate this we need to
2m
.
prove that this conditions imply E1 ≤ c1 −b−E
2m
c2 −b−E1 m
c2 −b−2mE2
Note that E2 <
⇔ E1 <
and:
2m
m
E2 ≥
c2 − b − 2mE2
c1 − b − E2 m
<
m
2m
⇔ E2 >
2c2 − c1 − b
3m
1 −b
which holds, since E2 ≥ 2c2 −c
;
m
c1 −b c1 +b
2c2 −c1 −b
−c2
1m
21. Let E1 < 2m , 2 ≥ c2 , E2 ≥
, E 1 < c1 m
, E2 < c2m−b − E1 , E2 ≥ c2 −b−E
.
m
2m
Note that:
c1 − c2
2c1 − c2 − b
c1 + b
≤
⇔
≥ c2
m
3m
2
−c2
2 −b
1m
thus, E1 < c1m
≤ 2c1 −c
. Since, by assumption E2 ≥ c2 −b−E
, we are in the
3m
2m
conditions of the NE 3.5 to 3.6.
In conclusion, the Nash equilibria were completely classified for this small sized example.
The resulting Global decision trees are in Figures I.1, I.2, I.3 and I.4.
39
Parameters
c1 , c2 , E1 , E2 , m, b
E1 <
c1 +b
2
≥ c2
c1 −b
2m
c1 +b
2
E1 ≥
c1 +b
2
< c2
See decision
tree I.3
c1 −b
2m
See decision
tree I.2
E1 m + b < c2
c1 +b
2
≥ c2
See the NE
3.1 to 3.3
E1 m + b ≥ c2
See the NE
3.2 to 3.4
E2 <
c2 −b−E1 m
2m
E2 ≥
c2 −b−E1 m
2m
See the NE 3.9 to
3.12
E1 m + b 6= c2
See the NE
3.5 and 3.6
E1 m + b = c2
See the NE
3.7 and 3.8
Figure I.1: Global Decision Tree.
Figure I.1: Global Decision Tree.
40
31
< c2
Figure I.2: Global Decision Tree - Part A.
E1 <
E2 <
E1 <
E2 ≥
c2 −b−E1 m
2m
c1 −b−E2 m
2m
E2 <
and
c1 +b
2
≥ c2
2c2 −c1 −b
m
E1 ≥
c2 −b−E1 m
2m
c1 −b
2m
E2 ≥
c1 −b−E2 m
2m
E1 <
E2 ≤ A
E2 > A
E2 <
2c2 −c1 −b
m
c1 −c2
m
E2 ≥
c2 −b−E1 m
m
E1 ≥
c2 −b−E1 m
2m
E1 >
c1 −c2
m
2c1 −c2 −b
3m
E1 ≤
2c1 −c2 −b
3m
See the NE 3.11
See the NE
See decision
See the NE 3.5
See the NE 3.5
See the NE 3.5
tree I.4
and 3.6
and 3.24
and 3.6
and 3.12;See the
3.5 and 3.6
NE 3.9 and 3.10
E1 >
E1 ≤
2c1 −b−c2
3m
2c1 −b−c2
3m
E2 ≥
E2 <
c2 −b−E1 m
2m
c2 −b−E1 m
2m
See the NE 3.11 and
See the NE
See the NE
See the NE
3.5 and 3.6
3.5 and 3.24
3.5 and 3.6
3.12;See the NE 3.9 and
3.10
Figure I.3: Global Decision Tree - Part B.
Figure I.3: Global Decision Tree - Part B.
1 −b
E1 < c2m
,
E2 ≤ A
E2 ≤
E2 <
c2 −b−E1 m
2m
c1 +b
2
≥ c2 , E2 <
2c2 −c1 −b
,
m
E1 ≥
c1 −b−E2 m
,
2m
2c2 −c1 −b
3m
E2 ≥
E2 >
c2 −b−E1 m
2m
E1 <
2c2 −c1 −b
3m
√
c2 −b+ K2
m
E1 ≥
√
c2 −b+ K2
m
See the NE 3.16
and 3.17
E1 =
2c1 −b−c2
3m
E1 6=
2c1 −b−c2
3m
E1 ≤
2c1 −c2 −b
3m
E1 >
2c1 −c2 −b
3m
See the NE
See the 3.5
See the NE 3.16 and
3.17; See the NE 3.5 and
3.5 and 3.6
3.6
E2 ≥
√
c1 −b+ K1
m
E2 <
√
c1 −b+ K1
m
See the NE 3.16 and
See the NE 3.16
3.17; See the NE 3.5 and
and 3.17
3.24
Figure I.4:
Global
Decision
TreeTree
- Part
C. C.
Figure
I.4: Global
Decision
- Part
E1 ≤
2c1 −c2 −b
3m
E1 >
2c1 −c2 −b
3m
See the NE 3.5
See the NE 3.5
and 3.6; See the
and 3.24; See the
NE 3.19 and 3.17
NE 3.19 and 3.17
and 3.24