Sunk Cost and Entry: Appendix
Stephen Martin
Department of Economics
Krannert School of Management
Purdue University
403 West State Street
West Lafayette, Indiana 47907-2056
USA
[email protected]
November 2001
Contents
1 No depreciation, no resale
2
2 Depreciation, no resale
2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
9
11
3 nB = nA + 1
3.1 nB = nA + 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4 Entrant’s value, nB = nA + 1
17
5 Infinite horizon, partial sunk cost
21
6 First solution
25
7 Second solution
34
1
1
No depreciation, no resale
First (and highly stylized) example:
Linear inverse demand curve:
p=a−Q
Stone-Geary production function:
¶
µ
K −K L−L
,
q = min
aK
aL
Capital does not depreciate and cannot be resold.
Present-discounted value of a monopoly supplier (who, facing unchanging
demand conditions, will produce the same output in each period):
∙
¸
¢ £
¡
¢¤
¡
1
1
k
Vm = −p K + aK q + (a − q) q − w L + aL q
+ ...
+
1 + r (1 + r)2
¢
¡
¢
¡
a − waL − rpk aK − q q − rpk K + wL
=
r
(a − cH − q) q − FH
=
r
Monopoly output:
qm =
¢
1
1¡
(a − cH ) =
a − waL − rpk aK
2
2
If entry occurs, the entrant maximizes
V2 =
(a − cH − q1 − q2 ) q2 − FH
r
Taking into account the fact the equilibrium duopoly output of a single
firm is less than monopoly output, the incumbent, with surplus and nondepreciating capital, maximizes
V1 =
(a − cL − q1 − q2 ) q2 − FL
,
r
for
cL = waL
2
FL = wL
First-order conditions are
2q1 + q2 = a − cL
q1 + 2q2 = a − cH
and in the usual way, we find equilibrium outputs
¶µ
¶ µ
¶
µ
q1
a − cL
2 1
=
1 2
q2
a − cH
¶ µ
¶µ
¶
µ
2 −1
a − cL
q1
=
3
−1 2
q2
a − cH
¶
¶
µ
µ
1 a − 2cL + cH
q1
=
q2
3 a + cL − 2cH
¶
¶
µ
µ
1
a − 2waL + rpk aK + waL
q1
=
q2
3 a − 2rpk aK − 2waL + waL
¶
¶
µ
µ
1
a − waL + rpk aK
q1
=
.
q2
3 a − waL − 2rpk aK
The entrant’s equilibrium value is
V2 =
q22 − FH
r
¸
∙
¢2
1 1¡
k
a − waL − 2rp aK − FH
=
r 9
By analogy with the innovation literature, where a drastic innovation
is one that makes the post-innovation equilibrium option of the successful
innovator’s rivals to shut down, we can call sunk costs drastic if
¢2
1¡
a − waL − 2rpk aK < FH = rpk K + wL.
9
In this case, entry is blocked: not because of the impact the need to
make its own investment in sunk cost has on the decision of the entrant,
but because of the impact entry would have on the economic costs of the
incumbent. Further, if
¢2
¢2
1¡
1¡
a − rpk aK − waL ≥ FH >
a − 2rpk aK − waL ,
9
9
entry is blocked because of the sunk nature of the investments made by the
incumbent: if investments were not sunk, entry would be profitable.
3
2
Depreciation, no resale
Second (and less stylized) example: let capital depreciate at rate δ per period,
0 ≤ δ ≤ 1.
Monopoly:
¢
¡
Vm = −pk K + aK q +
¢¤
¡
1 £
(a − q) q − w L + aL q
1+r
¾
½
¢
¢¤
¡
¡
1
1 £
k
+
−δp K + aK q +
(a − q) q − w L + aL q
1+r
1+r
½
¾
¡
¢
¢¤
¡
1 £
1
k
−δp K + aK q +
(a − q) q − w L + aL q
+ ...
+
1+r
(1 + r)2
(collecting terms that are multiplied by 1/ (1 + r), by 1/ (1 + r)2 , etc.)
¢
¡
= −pk K + aK q +
+
¢
¢¤
¡
¡
1 £
(a − q) q − w L + aL q − δpk K + aK q
1+r
¢
¢¤
£
¡
¡
1
k
K + aK q + . . .
2 (a − q) − w L + aL q − δp
(1 + r)
(adding up the infinite series)
¢
¢¤
¡
£
¡
k
¢
¡
L
+
a
q
−
δp
K
+
a
q
a
−
q
−
w
L
K
= −pk K + aK q +
r
¢
¢
¡
¡
(a − q) q − w L + aL q − (r + δ) pk K + aK q
=
r
(separating out fixed costs)
¤
£
¤
£
a − waL − (r + δ) pk aK − q q − wL + (r + δ) pk K
=
r
=
(a − cH − q) q − FH
,
r
redefining
cH = waL + (r + δ) pk aK
and
FH = wL + (r + δ) pk K.
4
Monopoly output is
qm =
¤
1
1£
(a − cH ) =
a − waL − (r + δ) pk aK
2
2
and the incumbent purchases capital stock
1
Km = K + aK (a − cH )
2
at the start of the first period.
Entry occurs.
Let μ1t be the Lagrangian multiplier associated with the identify that
defines the incumbent’s period t capital stock,
Kit = (1 − δ) K1,t−1 + I1t
(where K10 = Km ).
Let λ1t be the Lagrangian multiplier associated with the period t capital
input constraint,
K1t ≥ K + aK q1t .
The Lagrangian that describes the incumbent’s constrained optimization
problem, from the moment of entry, is
V1 =
+
1
1+r
μ11 [(1 − δ) Km + I11 − K11 ] − pk I11
¡
¢
£
¡
¢¤
(a − q11 − q21 ) q11 − waL q11 + wL + λ11 K11 − K − aK q11
1 ©
μ [(1 − δ) K11 + I12 − K12 ] − pk I12
1 + r 12
¾
¡
¢
£
¡
¢¤
(a − q12 − q22 ) q12 − waL q12 + wL + λ12 K12 − K − aK q12
+
+
1
1+r
+
©
1
μ13 [(1 − δ) K12 + I13 − K13 ] − pk I13
(1 + r)2
¾
¡
¢
¡
¢¤
1 £
(a − q13 − q23 ) q13 − waL q13 + wL + λ13 K13 − K − aK q13
+
+...
1+r
5
or, more compactly and writing K10 = Km ,
V1 =
∞
X
t=1
©
1
k
t−1 μ1t [(1 − δ) K1,t−1 + I1t − K1t ] − p I1t
(1 + r)
¾
¡
¢¤
¡
¢
1 £
+
(a − q1t − q2t ) q1t − waL q1t + wL + λ1t K12 − K − aK q1t .
1+r
Kuhn-Tucker first-order conditions for this constrained optimization problem are
t = 1:
∂V1
= (1 − δ) Km + I11 − K11 = 0.
∂μ11
¡
¢
∂V1
= μ11 − pk ≤ 0
I11 μ11 − pk ≡ 0
I11 ≥ 0.
∂I11
∂V1
λ11 + (1 − δ) μ12
= −μ11 +
= 0.
∂K11
1+r
∂V1
= a − waL − λ11 aK − 2q11 − q21 = 0.
(1 + r)
∂q11
¡
¢
∂V1
(1+r)
= K11 −K −aK q11 ≥ 0
λ11 K11 − K − aK q11
λ11 ≥ 0.
∂λ11
t = 2:
∂V1
= (1 − δ) K11 + I12 − K12 = 0.
(1 + r)
∂μ12
¡
¢
∂V1
(1 + r)
= μ12 − pk ≤ 0
I12 μ12 − pk ≡ 0
I12 ≥ 0.
∂I12
∂V1
λ12 + (1 − δ) μ13
= 0.
= −μ12 +
∂K12
1+r
∂V1
= a − waL − λ12 aK − 2q12 − q22 = 0.
(1 + r)2
∂q12
¡
¢
∂V1
(1+r)2
= K12 −K−aK q12 ≥ 0
λ12 K12 − K − aK q12
∂λ12
t = 3:
∂V1
= (1 − δ) K12 + I13 − K13 = 0.
(1 + r)2
∂μ13
(1 + r)
6
λ12 ≥ 0.
¡
¢
∂V1
= μ13 − pk ≤ 0
I13 μ13 − pk ≡ 0
I13 ≥ 0.
∂I13
∂V1
λ13 + (1 − δ) μ14
(1 + r)2
= 0.
= −μ13 +
∂K13
1+r
∂V1
= a − waL − λ13 aK − 2q13 − q23 = 0.
(1 + r)3
∂q13
¡
¢
∂V1
(1+r)3
= K13 −K−aK q13 ≥ 0
λ13 K13 − K − aK q13
λ13 ≥ 0.
∂λ13
(And so on.)
The entrant maximizes
(1 + r)2
V2 = μ21 (I21 − K21 ) − pk I21
¡
¢
¡
¢¤
1 £
(a − q11 − q12 ) q21 − waL q21 + wL + λ21 K21 − K − aK q21
1+r
1 ©
μ [(1 − δ) K21 + I22 − K22 ] − pk I22 +
+
1 + r 22
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q22 − waL q22 + wL + λ22 K22 − K − aK q22
1+r
©
1
μ23 [(1 − δ) K22 + I23 − K23 ] − pk I23
+
(1 + r)2
¾
¡
¢
¡
¢¤
1 £
(a − q13 − q23 ) q23 − waL q23 + wL + λ23 K23 − K − aK q23
+
+...
1+r
+
or, more compactly, with K20 = 0,
V2 =
∞
X
t=1
©
1
k
t−1 μ2t [(1 − δ) K2,t−1 + I2t − K2t ] − p I2t
(1 + r)
¾
¡
¢¤
¡
¢
1 £
+
(a − q1t − q2t ) q2t − waL q2t + wL + λ2t K2t − K − aK q2t .
1+r
μ2t is the Lagrangian multiplier for the identify that defines the capital
stock in period t. λ2t is the Lagrangian multiplier for the capital input
constraint in period t.
Kuhn-Tucker first-order conditions:
7
t = 1:
∂V2
= I21 − K21 = 0.
∂μ21
¡
¢
∂V2
= μ21 − pk ≤ 0
I21 μ21 − pk = 0
I21 ≥ 0.
∂I21
λ21 + (1 − δ) μ22
∂V2
= −μ21 +
= 0.
∂K21
1+r
∂V2
(1 + r)
= a − waL − λ21 aK − q11 − 2q21 = 0.
∂q21
¡
¢
∂V2
(1 + r)
= K21 −K−aK q21 ≥ 0
λ21 K21 − K − aK q21 = 0
λ21 ≥ 0.
∂λ21
t = 2:
∂V2
(1 + r)
= (1 − δ) K21 + I22 − K22 = 0.
∂μ22
¡
¢
∂V2
(1 + r)
= μ22 − pk ≤ 0
I22 μ22 − pk = 0
I22 ≥ 0.
∂I22
λ22 + (1 − δ) μ23
∂V2
= 0..
= −μ22 +
(1 + r)
∂K22
1+r
∂V2
(1 + r)2
= a − waL − λ22 aK − q12 − 2q22 = 0.
∂q22
¡
¢
∂V2
(1 + r)2
= K22 −K−aK q22 ≥ 0
λ22 K22 − K − aK q22 = 0
λ22 ≥ 0.
∂λ22
t = 3:
∂V2
(1 + r)2
= (1 − δ) K22 +I23 −K23 ≥ 0
μ23 (I23 − K23 ) = 0
μ23 ≥ 0.
∂μ23
¡
¢
∂V2
(1 + r)2
= μ23 − pk ≤ 0
I23 μ23 − pk = 0
I23 ≥ 0.
∂I23
∂V2
λ23 + (1 − δ) μ24
(1 + r)2
= −μ23 +
= 0.
∂K23
1+r
∂V2
(1 + r)3
= a − waL − λ23 aK − q13 − 2q23 = 0.
∂q23
¡
¢
∂V2
(1 + r)3
= K23 −K−aK q23 ≥ 0
λ23 K23 − K − aK q23 = 0
λ23 ≥ 0.
∂λ23
(And so on.)
We need to distinguish up to three types of time periods
(1 + r)
8
• the first nA £periods, when incumbent
¤ can produce its equilibrium out1
k
put qH = 3 a − waL + (r + δ) p aK without purchasing capital;
must purchase capital
to
• periods nB and after, when the incumbent
£
¤
produce its equilibrium output qD = 13 a − waL − (r + δ) pk aK ;
• if nB > nA +1, then for periods nA +1, nA +2,. . . , nB −1 the incumbent
produces just enough output to fully utilize the existing capital stock,
but does not purchase capital.
How many such capital-constrained periods exist, if any, depends on δ.
nA is the greatest integer in the value n1 that satisfies
¡
¢
(1 − δ)n1 K + aK qm = K + aK qH
¢
£
¤
¡
ln K + aK qH − ln K + aK qm
.
n1 =
ln (1 − δ)
nB is the least integer that is greater than the value n2 that satisfies
¡
¢
(1 − δ)n2 K + aK qm = K + aK qD
¤
¡
¢
£
ln K + aK qD − ln K + aK qm
.
n2 =
ln (1 − δ)
Remark:
n2 − n1 =
¤
¡
¢
¢
£
¤
£
¡
ln K + aK qD − ln K + aK qm
ln K + aK qH − ln K + aK qm
−
=
ln (1 − δ)
ln (1 − δ)
¢
£
¤
¡
ln K + aK qH − ln K + aK qD
> 0,
− ln (1 − δ)
noting that ln (1 − δ) < 0.
2.1
Example
Let w = aL = aK = pk = 1, r = δ = 1/10, K = 1, a = 100.
¶
µ
1
6
1
cH = 1 +
+
=
10 10
5
9
¢
¡
¢¢
¡
¡
ln K + 13 aK a − waL + (r + δ) pk aK − ln K + 12 aK (a − cH )
=
n
bA =
ln (1 − δ)
¡
¡
¡1
¡
¢
¢¢
¡
¢¢
1
ln 1 + 13 (1) 100 − (1) (1) + 10
+ 10
(1) (1) − ln 1 + 12 (1) 100 − 65
¡
¢
= 3.7174
1
ln 1 − 10
¤
£
¤
£
ln K + 13 aK (a − cH ) − ln K + 12 aK (a − cH )
n
bB =
=
ln (1 − δ)
¡
¡
¡
¢¢
¡
¢¢
ln 1 + 13 (1) 100 − 65 − ln 1 + 12 (1) 100 − 65
¡
¢
= 3.7547
1
ln 1 − 10
Monopoly output and capital stock:
µ
¶
1
6
247
qm =
100 −
=
2
5
5
252
247
=
= 50.4.
5
5
Duopoly outputs in first phase:
µ
¶
¶
µ
1
q1t
a − waL + (r + δ) pk aK
=
q2t
3 a − waL − 2 (r + δ) pk aK
¡1
¢
¶
¶
µ
µ
¶ µ 496 ¶ µ
1
1
33. 067
100 − 1 + ¡10
q1t
+ 10
(1) (1)
15
¢
=
=
=
493
1
1
32. 867
q2t
+ 10
(1) (1)
3 100 − 1 − 2 10
15
µ 496 ¶
Km = 1 +
15
493
15
Capital stocks:
¶ µ ¶ µ
µ
1
K1
=
+
1
K2
496
15
493
15
¶
=
µ
511
15
508
15
¶
=
µ
34. 067
33. 867
¶
.
For firm 1, this is the minimum capital stock required to produce its equilibrium output.
Duopoly outputs in second phase:
µ
µ
¶ ¶
1
1
1
494
100 − 1 −
+
(1) =
= 32.933.
3
10 10
15
Capital stocks in second phase:
10
1+
494
509
=
= 33.933
15
15
Incumbent’s capital stock
252
= 50.4
0 ¡
5¢
1 252
1
1 − 10 5 = 45.36
¢
¡
1 2 252
= 40.824
2 1 − 10
¢ 5
¡
1 3 252
= 36.742
3 1 − 10
¢ 5
¡
1 4 252
= 33.067
4 1 − 10
5
For these parameters, there is no “capital constrained” transition phase.
The incumbent has excess capital for periods 1, 2, and 3, and purchases
capital in period 4.
If there is a capital constrained case, then during those periods, the
shadow value of a unit of capital is positive, but less than the cost of a
unit of capital.
3
nB = nA + 1
Consider the case nB = nA + 1.
Write out Kuhn-Tucker for periods nA and nB .
Incumbent:
nA :
(1 + r)nA −1
(1 + r)nA −1
∂V1
= (1 − δ) K1,nA −1 + I1nA − K1nA = 0.
∂μ1nA
¡
¢
I1nA μ1nA − pk ≡ 0
∂V1
= μ1nA − pk ≤ 0
∂I1nA
(1 + r)nA −1
I1nA ≥ 0.
λ1nA + (1 − δ) μ1nB
∂V1
= −μ1nA +
= 0.
∂K1nA
1+r
∂V1
= a − waL − λ1nA aK − 2q1nA − q2nA = 0.
∂q1nA
¡
¢
= K1nA −K−aK q1nA ≥ 0
λ1nA K1nA − K − aK q1nA
(1 + r)nA
(1+r)nA
∂V1
∂λ1nA
11
λ1nA ≥ 0.
nB :
(1 + r)nA
(1 + r)nA
∂V1
= (1 − δ) K1nA + I1nB − K1nB = 0.
∂μ1nB
¡
¢
I1nB μ1nB − pk ≡ 0
∂V1
= μ1nB − pk ≤ 0
∂I1nB
(1 + r)nA −1
I1nB ≥ 0.
λ1nB + (1 − δ) μ1,nB +1
∂V1
= −μ1nB +
= 0.
∂K1nB
1+r
∂V1
= a − waL − λ1nB aK − 2q1nB − q2nA = 0.
∂q1nB
¡
¢
= K1nB −K−aK q1nB ≥ 0
λ1nB K1nB − K − aK q1nB
(1 + r)nA
(1+r)nA
∂V1
∂λ1nB
That firm 1’s investment in periods nB and nB + 1 is positive allows us
to determine μ1nB and μ1nB +1 :
I1nB > 0 ⇒ μ1nB = pk
I1,nB +1 > 0 ⇒ μ1,nB +1 = pk .
Knowing μ1nB and μ1,nB +1 gives λ1nB , and by the same sort of argument
this is the value of λ1t for all t ≥ nB :
−μ1nB +
λ1nB + (1 − δ) μ1,nB +1
=0
1+r
− (1 + r) pk + λ1nB + (1 − δ) pk = 0
λ1nB = (r + δ) pk .
In period nA , capital is still in excess supply,
K1nA − K − aK q1nA > 0 ⇒ λ1nA = 0;
by the same argument,
λ1t = 0 ∀t ≤ nA .
We then have the first-order condition for the incumbent’s output for
period nA , with corresponding expressions for all periods before nA :
2q1nA + q2nA = a − waL .
12
λ1nB ≥ 0.
We also have the first-order condition for the incumbent’s output for
period nB , with corresponding expressions for all periods after period nB :
2q1nB + q2nB = a − waL − (r + δ) pk aK .
Knowing that λ1nA = 0 and μ1nB = pk , we are able to find μ1nA :
μ1nA =
λ1nA + (1 − δ) μ1nB
0 + (1 − δ) pk
1−δ k
=
=
p .
1+r
1+r
1+r
A consistency condition for this solution to be valid is
1−δ k
p − pk ≤ 0
1+r
1≥
1−δ
1+r
and this condition evidently is met.
The Kuhn-Tucker condition for firm 1’s capital stock for period nA − 1 is
(1 + r)nA −2
λ1,nA −1 + (1 − δ) μ1nA
∂V1
= 0.
= −μ1,nA −1 +
∂K1,nA −1
1+r
We know that λ1,nA −1 = 0 and that μ1nA = (1 − δ) pk ; hence
0 + (1 − δ) 1−δ
pk
1+r
=0
1+r
µ
¶2
1−δ
pk .
μ1,nA −1 =
1+r
−μ1,nA −1 +
In the same way,
¡ ¢2 k µ
¶3
p
0 + (1 − δ) 1−δ
1−δ
1+r
μ1,nA −2 =
=
pk ...
1+r
1+r
¶nA
µ
1−δ
pk .
μ11 = μ1,nA −(nA −1) =
1+r
For periods t = 1, 2, ..., nA − 1, nA .
¶n +1−t
µ
1−δ A
μ1t =
pk .
1+r
13
Entrant:
nA :
(1 + r)nA −1
(1 + r)nA −1
∂V2
= (1 − δ) K2,nA −1 + I2nA − K2nA = 0.
∂μ2nA
¡
¢
I2nA μ2nA − pk = 0
∂V2
= μ2nA − pk ≤ 0
∂I2nA
(1 + r)nA −1
I2nA ≥ 0.
λ2nA + (1 − δ) μ2nB
∂V2
= −μ2nA +
= 0.
∂K2nA
1+r
∂V2
= a − waL − λ2nA aK − q1nA − 2q2nA = 0.
∂q2nA
¡
¢
= K2nA −K−aK q2nA ≥ 0
λ2nA K2nA − K − aK q2nA = 0
(1 + r)nA
(1 + r)nA
∂V2
∂λ2nA
nB :
(1 + r)nA
(1 + r)nA
∂V2
= (1 − δ) K2,nB −1 + I2nB − K2nB = 0.
∂μ2nB
¡
¢
I2nB μ2nB − pk = 0
∂V2
= μ2nB − pk ≤ 0
∂I2nB
(1 + r)nA
λ2nA ≥ 0.
I2nB ≥ 0.
λ2nB + (1 − δ) μ2,nB +1
∂V2
= −μ2nB +
= 0.
∂K2nB
1+r
∂V2
= a − waL − λ2nB aK − q1nA − 2q2nB = 0.
∂q2nB
¡
¢
= K2nB −K−aK q2nB ≥ 0
λ2nB K2nB − K − aK q2nB = 0
(1 + r)nB
(1 + r)nB
∂V2
∂λ2nB
The solution for the entrant is more straightforward, since the entrant
makes positive investment in every period.
μ2t = pk ∀t.
λ2t = (r + δ) pk ∀t.
q1t + 2q2nB = a − waL − λ2t aK = a − waL − (r + δ) pk aK .
14
λ2nB ≥ 0.
3.1
nB = nA + 2
Consider briefly the case nB = nA + 2, so there is one period in which the
incumbent is capital constrained.
nA :
(1 + r)nA −1
(1 + r)nA −1
∂V1
= (1 − δ) K1,nA −1 + I1nA − K1nA = 0.
∂μ1nA
¡
¢
I1nA μ1nA − pk ≡ 0
∂V1
= μ1nA − pk ≤ 0
∂I1nA
(1 + r)nA −1
I1nA ≥ 0.
λ1nA + (1 − δ) μ1nB
∂V1
= −μ1nA +
= 0.
∂K1nA
1+r
∂V1
= a − waL − λ1nA aK − 2q1nA − q2nA = 0.
∂q1nA
¡
¢
= K1nA −K−aK q1nA ≥ 0
λ1nA K1nA − K − aK q1nA
(1 + r)nA
(1+r)nA
∂V1
∂λ1nA
nA + 1:
(1 + r)nA
(1 + r)nA
∂V1
= (1 − δ) K1,nA + I1,nA +1 − K1,nA +1 = 0.
∂μ1,nA +1
¡
¢
I1,nA +1 μ1,nA +1 − pk ≡ 0
∂V1
= μ1,nA +1 −pk ≤ 0
∂I1,nA +1
(1 + r)nA
(1 + r)nA +1
λ1nA ≥ 0.
I1,nA +1 ≥ 0.
λ1,nA +1 + (1 − δ) μ1,nA +2
∂V1
= 0.
= −μ1,nA +1 +
∂K1,nA +1
1+r
∂V1
= a − waL − λ1,nA +1 aK − 2q1,nA +1 − q2nA = 0.
∂q1,nA +1
∂V1
= K1,nA +1 − K − aK q1,nA +1 ≥ 0.
∂λ1,nA +1
¡
¢
λ1,nA +1 K1,nA +1 − K − aK q1,nA +1
λ1,nA +1 ≥ 0.
(1 + r)nA +1
In period nA + 1, the incumbent’s output is determined by its capital
stock, i.e.
(1 − δ)nA +1 Km − K
q1,nA +1 =
.
aK
15
The first-order condition for the entrant’s output in period nA + 1 is
q1,nA +1 + 2q2,nA +1 = a − waL − (r + δ) aK .
Since we know the incumbent’s output, we can solve for the entrant’s
output:
¸
∙
1
(1 − δ)nA Km − K
.
q2,nA +1 =
a − waL − (r + δ) aK −
2
aK
Knowing the entrant’s output, we can solve the first-order condition for
the incumbent’s output for λ1,nA +1 . First evaluate
2q1,nA +1 + q2,nA +1 =
1
[a − waL − (r + δ) aK − q1,nA +1 ] =
2
3
1
q1,nA +1 + [a − waL − (r + δ) aK ] =
2
2
nA
3 (1 − δ) Km − K 1
+ [a − waL − (r + δ) aK ]
2
aK
2
Then substitute in
2q1,nA +1 +
a − waL − λ1,nA +1 aK − 2q1,nA +1 − q2,nA +1 = 0.
λ1,nA +1 = a − waL − (2q1,nA +1 + q2,nA +1 )
½
¾
3 (1 − δ)nA Km − K 1
= a − waL −
+ [a − waL − (r + δ) aK ]
2
aK
2
¾
½
3 (1 − δ)nA Km − K 1
3
− (r + δ) aK
= (a − waL ) −
2
2
aK
2
¸
∙
3
1
(1 − δ)nA Km − K
=
+ (r + δ) aK
a − waL −
2
aK
2
A consistency condition for this solution to be valid is that this value of
λ1,nA +1 must be positive.
Knowing λ1,nA +1 and knowing that
μ1nB = pk
16
we can solve the Kuhn-Tucker condition for K1,nA +1
(1 + r)nA
λ1,nA +1 + (1 − δ) μ1,nA +2
∂V1
=0
= −μ1,nA +1 +
∂K1,nA +1
1+r
for μ1,nA +1 :
λ1,nA +1 + (1 − δ) μ1,nA +2
1+r
λ1,nA +1 + (1 − δ) pk
.
=
1+r
Knowing equilibrium values for period nA + 1, we can solve backward and
forward in time for equilibrium values in other periods. For the first nA periods, the incumbent does not purchase capital and is not capital constrained;
its rental cost of capital services is zero. In and after period nB , the incumbent purchases capital and its rental cost of capital services is (r + δ) pk .
If there is more than one period between periods nA and nB , solve first for
equilibrium values in the capital constrained periods, beginning with period
nB − 1 and working backward to period nA + 1. This will yield consistency
conditions that must be satisfied for the solution to be valid, and it will
permit finding the equilibrium values for earlier and later periods.
μ1,nA +1 =
4
Entrant’s value, nB = nA + 1
Return to the case nB = nA + 1. The solution is that for the first nA periods,
µ
¶
¶
µ
1
q1t
a − waL + (r + δ) pk aK
=
q2t
3 a − waL − 2 (r + δ) pk aK
µ
¶ µ
¶
λ1t
0
=
(r + δ) pk
λ2t
¶ µ ¡ 1−δ ¢nA +1−t ¶
µ
μ1t
1+r
=
pk
μ2t
1
µ
¶ µ
¶
0
I11
=
I21
K + aK q20
¶
¶ µ
µ
0
I1t
¡
¢ , t = 2, 3, ..., nA ;
=
I2t
δ K + aK q2t
17
after which
1
(a − cH )
3
λ1t = λ2t = (r + δ) pk
q1t = q2t =
µ
μ1t = μ2t = pk
µ
¶
I1n
=
I2n
¤
£
¶
K + 13 aK [a − waL − (r + δ) aK ] −£ (1 − δ)n ¡K + 12 aK (a − cH )
¢¤
K + 13 aK [a − waL − (r + δ) aK ] − (1 − δ) K + 13 aK a − waL − 2 (r + δ) pk aK
¡
¢
I1t = I2t = δ K + aK q2t , t = nA + 2, n + 3, ...
Now find the entrant’s equilibrium value, noting that the expressions multiplied by Lagrangian multipliers drop out by the Kuhn-Tucker conditions:
V2 = −pk I21 +
¤
1 £
(a − waL − q11 − q21 ) q21 − wL
1+r
¾
½
¤
1 £
1
k
−p I22 +
(a − waL − q12 − q22 ) q22 − wL
+
1+r
1+r
½
¾
¤
1
1 £
k
+
−p I23 +
(a − waL − q13 − q23 ) q23 − wL + ...
1+r
(1 + r)2
½
¾
¤
1 £
1
k
−p I2,nA +
(a − waL − q1,nA − q2,nA ) q2,nA − wL +
+
(1 + r)nA
1+r
½
¾
¤
1 £
1
k
−p I2,nA +1 +
(a − waL − q1,nA +1 − q2,nA +1 ) q2,nA +1 − wL +
+
1+r
(1 + r)nA +1
½
¾
¤
1 £
1
k
−p I2,nA +2 +
+
(a − waL − q1,nA +2 − q2,nA +2 ) q2,nA +2 − wL +...
1+r
(1 + r)nA +2
(expressing investment in terms of the capital stock)
£
¤
= −pk K + aK qL +
1
+
1+r
¤
1 £
(a − waL − q11 − q21 ) q21 − wL
1+r
½
£
¤
−δpk K + aK qL +
¾
¤
1 £
(a − waL − q12 − q22 ) q22 − wL
1+r
18
1
+
(1 + r)2
½
£
¤
−δpk K + aK qL +
1
+
(1 + r)nA
¾
¤ª
1 ©£
+...
(a − waL − q13 − q23 ) q23 − wL
1+r
½
£
¤
−δpk K + aK qL +
¾
¤ª
1 ©£
(a − waL − q1,nA − q2,nA ) q2,nA − wL
1+r
¡
© k£
¢¤
−p K + aK qH − (1 − δ) K + aK qL
1
(1 + r)nA +1
¾
¤
1 £
+
(a − waL − q1,nA +1 − q2,nA +1 ) q2,nA +1 − wL
1+r
½
¾
¤ª
£
¤
1
1 ©£
k
−δp K + aK qH +
+
+...
(a − waL − q1,nA +2 − q2,nA +2 ) q2,nA +2 − wL
1+r
(1 + r)nA +2
+
(collecting terms multiplied by 1/ (1 + r), by 1/ (1 + r)2 , and so on)
£
¤
= −pk K + aK qL +
¤
¢
1 £¡
a − waL − δpk aK − qH − qL qL − wL − δpk K
1+r
¤
£¡
¢
1
a − waL − δpk aK − qH − qL qL − wL − δpk K + ...
2
(1 + r)
¤
£¡
¢
1
+
a − waL − δpk aK − qH − qL qL − wL − δpk K
nA +1
(1 + r)
1
+
pk aK (qD − qL )
(1 + r)nA +1
¤
£¡
¢
1
k
k
+
−
δp
a
−
q
−
q
−
wL
−
δp
K
+
a
−
wa
q
L
K
D
D
D
(1 + r)nA +2
¤
£¡
¢
1
k
k
−
δp
a
−
q
−
q
−
wL
−
δp
K
+ ...
+
a
−
wa
q
L
K
D
D
D
(1 + r)nA +3
+
(factoring where possible)
£
¤
= −pk K + aK qL
∙
¸
£¡
¤
¢
1
1
+
a − waL − δpk aK − qH − qL qL − wL − δpk K
+ ... +
nA +1
1+r
(1 + r)
1
+
pk aK (qD − qL )
(1 + r)nA +1
19
∙
¸
£¡
¤
¢
1
+
a − waL − δpk aK − qD − qD qD − wL − δpk K
nA +2 + ...
(1 + r)
(making summations)
¸
∙
¤
1 k£
1
1
= − rp K + aK qL 1 −
+
r
(1 + r)nA +1 (1 + r)nA +1
¸
∙
£¡
¤
¢
1
1
+ 1−
a − waL − δpk aK − qH − qL qL − wL − δpk K
nA +1
r
(1 + r)
1
k
nA +1 p aK (qD − qL )
(1 + r)
¤
¢
1
1 £¡
+
a − waL − δpk aK − qD − qD qD − wL − δpk K
nA +1
r
(1 + r)
+
(rearranging terms involving K)
¸
∙
£¡
¤
¢
1
1
=
a − waL − (r + δ) pk aK − qH − qL qL − wL − (r + δ) pk K
1−
nA +1
r
(1 + r)
+
¤
¢
1
1 £¡
k
k
a
−
wa
−
(r
+
δ)
p
a
−
q
−
q
−
wL
−
(r
+
δ)
p
K
q
L
K
D
D
D
(1 + r)nA +1 r
(using first-order conditions for output to express gross profit per period as
the square of output)
¸
¾
½∙
£ 2
¤
¤
£ 2
1
1
1
k
k
qL − wL − (r + δ) p K +
=
1−
qD − wL − (r + δ) p K
r
(1 + r)nA +1
(1 + r)nA +1
(rearranging terms)
¸
∙
¡ 2
¢
1 2
1
2
k
=
q +
q − qL − wL − (r + δ) p K .
r L (1 + r)nA +1 D
Then if
qL2 +
entry is blocked.
¡ 2
¢
1
2
k
nA +1 qD − qL − wL − (r + δ) p K < 0,
(1 + r)
20
5
Infinite horizon, partial sunk cost
Linear inverse demand curve:
p=a−Q
Stone-Geary production function:
µ
¶
K −K L−L
q = min
,
aK
aL
Capital depreciate at rate δ, 0 ≤ δ ≤ 1.
Monopoly:
(a − cH ) q − FH
,
Vm =
r
For
cH = waL + (r + δ) pk aK
and
FH = wL + (r + δ) pk K.
Monopoly output is
qm =
¤
1
1£
(a − cH ) =
a − waL − (r + δ) pk aK .
2
2
The incumbent’s capital stock is
1
Km = K + aK (a − cH ) ,
2
and at the start of the first period after entry the incumbent inherits capital
stock
(1 − δ) Km
from the past.
Let μ1t be the Lagrangian multiplier associated with the identify that
defines the incumbent’s period t capital stock,
Kit = (1 − δ) K1,t−1 + I1t − Jit
(where K10 = Km ).
21
Let λ1t be the Lagrangian multiplier associated with the period t capital
input constraint,
K1t ≥ K + aK q1t .
The Lagrangian that describes the incumbent’s constrained optimization
problem, from the moment of entry, is
V1 = μ11 [(1 − δ) Km + I11 − J11 − K11 ] − pk I11 + αpk J11
¡
¢
¡
¢¤
1 £
(a − q11 − q21 ) q11 − waL q11 + wL + λ11 K11 − K − aK q11
1+r
1 ©
μ12 [(1 − δ) K11 + I12 − J12 − K12 ] − pk I12 + αpk J12 +
+
1+r
¾
¡
¢
¡
¢¤
1 £
+
(a − q12 − q22 ) q12 − waL q12 + wL + λ12 K12 − K − aK q12
+
1+r
©
1
+
μ13 [(1 − δ) K13 + I13 − J13 − K13 ] − pk I13 + αpk J13 +
(1 + r)2
¾
¡
¢
¡
¢¤
1 £
(a − q13 − q23 ) q13 − waL q13 + wL + λ13 K13 − K − aK q13
+
+...
1+r
+
or, with K10 = Km ,
V1 =
∞
X
t=1
©
1
k
t−1 μ1t [(1 − δ) K1,t−1 + I1t − J1t − K1t ] − p I1t
(1 + r)
¾
¡
¢¤
¡
¢
1 £
(a − q1t − q2t ) q1t − waL q1t + wL + λ1t K12 − K − aK q1t .
+
1+r
Kuhn-Tucker first-order conditions:
Period 1:
∂V1
= (1 − δ) Km + I11 − J11 − K11 = 0.
∂μ11
¡
¢
∂V1
= μ11 − pk ≤ 0
I11 μ11 − pk = 0
I11 ≥ 0
∂I11
¡
¢
∂V1
= −μ11 + αpk ≤ 0
J11 −μ11 + αpk = 0
J11 ≥ 0
∂J11
∂V1
= a − 2q11 − q21 − waL − λ11 aK = 0
(1 + r)
∂q11
22
∂V1
λ11 + (1 − δ) μ12
=0
= −μ11 +
∂K11
1+r
¡
¢
∂V1
= K11 −K−aK q11 ≥ 0
λ11 K11 − K − aK q11 = 0
(1 + r)
∂λ11
Period t:
(1 + r)t−1
(1 + r)t−1
(1 + r)t−1
λ11 ≥ 0
∂V1
= (1 − δ) K1,t−1 + I1t − J1t − K1t = 0.
∂μ1t
¡
¢
I1t μ1t − pk = 0
∂V1
= μ1t − pk ≤ 0
∂I1t
I1t ≥ 0
¡
¢
∂V1
= −μ1t + αpk ≤ 0
J11 −μ1t + αpk = 0
∂J1t
∂V1
(1 + r)t
= a − 2q11 − q21 − waL − λ1t aK = 0
∂q1t
λ1t + (1 − δ) μ1,t+1
∂V1
= −μ1t +
=0
∂K1t
1+r
¡
¢
∂V1
= K1t −K−aK q1t ≥ 0
λ1t K1t − K − aK q1t = 0
(1 + r)t
∂λ1t
The entrant maximizes
J1t ≥ 0
(1 + r)t
λ1t ≥ 0
V2 = μ21 (I21 − J21 + K21 ) − pk I21 + αpk J21
¡
¢
¡
¢¤
1 £
(a − q11 − q12 ) q21 − waL q21 + wL + λ21 K21 − K − aK q21
1+r
1 ©
μ [(1 − δ) K21 + I22 − J22 − K22 ] − pk I22 + αpk J22
+
1 + r 22
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q22 − waL q22 + wL + λ22 K22 − K − aK q22
1+r
©
1
k
k
+
2 μ23 [(1 − δ) K22 + I23 + J23 − K23 ] − p I23 + αp J23
(1 + r)
¾
¡
¢
¡
¢¤
1 £
(a − q13 − q23 ) q23 − waL q23 + wL + λ23 K23 − K − aK q23
+
+...
1+r
+
or, more compactly, with K20 = 0, (and taking note that J21 = 0)
23
∞
X
©
1
k
k
t−1 μ2t [(1 − δ) K2,t−1 + I2t − J2t − Ktt ] − p I2t + αp J2t
(1 + r)
t=1
¾
¡
¢¤
¡
¢
1 £
(a − q1t − q2t ) q1t − waL q1t + wL + λ1t K12 − K − aK q1t .
+
1+r
Kuhn-Tucker first-order conditions.
Period 1:
∂V2
= I21 − J21 − K21 = 0.
∂μ21
¡
¢
∂V2
= −pk + μ21 ≤ 0
I21 −pk + μ21 = 0
I21 ≥ 0.
∂I21
¡
¢
∂V2
= αpk − μ21 ≤ 0
J21 αpk − μ21 = 0
J21 ≥ 0.
∂J21
∂V2
(1 + r)
= a − q11 − 2q21 − waL − λ21 aK = 0.
∂q21
λ21 + (1 − δ) μ22
∂V2
= −μ21 +
(1 + r)
= 0.
∂K21
1+r
£
¤
∂V2
(1 + r)
= K21 −K−aK q21 ≥ 0
λ21 K21 − K − aK q21 = 0
λ21 ≥ 0.
∂λ21
Period t:
V2 =
(1 + r)t−1
∂V2
= (1 − δ) K2,t−1 + I2t − J2t − K2t = 0.
∂μ2t
¡
¢
∂V2
= μ2t − pk ≤ 0
I2t μ2t − pk = 0
I2t ≥ 0.
∂I2t
¡
¢
∂V2
(1 + r)t−1
= αpk − μ2t ≤ 0
J2t αpk − μ2t = 0
J2t ≥ 0.
∂J2t
∂V2
= a − q1t − 2q2t − waL − λ2t aK = 0.
(1 + r)t
∂q2t
λ2t + (1 − δ) μ2,t+1
∂V2
(1 + r)t
= 0.
= −μ2t +
∂K2t
1+r
£
¤
∂V2
= K2t −K−aK q2t ≥ 0
λ2t K2t − K − aK q2t = 0
(1 + r)t
λ2t ≥ 0.
∂λ2t
(1 + r)t−1
24
6
First solution
Incumbent sells capital in the first period, thereafter buys.
J11 > 0 ⇒ μ11 = αpk .
I12 > 0 ⇒ μ12 = pk .
Then
λ11 + (1 − δ) μ12
=0
1+r
λ11 + (1 − δ) pk
=0
−αpk +
1+r
−μ11 +
− (1 + r) αpk + λ11 + (1 − δ) pk = 0
λ11 = [(1 + r) α − (1 − δ)] pk
Note that
(1 + r) α − (1 − δ) = (r + δ) − (1 + r) (1 − α) ;
thus λ11 can be rewritten as
λ11 = [(r + δ) − (1 + r) (1 − α)] pk < (r + δ) pk .
A consistency condition for this solution to apply is λ11 ≥ 0; this requires
(1 + r) α − (1 − δ) ≥ 0
(1 + r) α ≥ (1 − δ)
1−δ
.
1+r
αpk is what the firm can sell a unit of excess capital for at the start of the
period entry occurs, and after one period invested at interest rate r, this rises
to (1 + r) αpk . If the firm does not sell a unit of excess capital at the start
of the period, the value remaining at the end of the period is (1 − δ) pk . If
α≥
(1 + r) αpk > (1 − δ) pk ,
the firm is better off selling excess capital at the start of the period than
holding it and allowing it to depreciate.
25
λ11 > 0 ⇒
K11 = K + aK q11 .
Provided α ≥ (1 − δ) / (1 + r), the incumbent sells unneeded capital at the
moment of entry, and never carries excess capacity.
Firm 1’s period 1 output first-order condition is
2q11 + q21 = a − [waL + λ11 aK ] .
Firm 2 buys an initial capital stock in the first period, buys replacement
capital each period thereafter.
I21 > 0 ⇒ μ21 = pk
I22 > 0 ⇒ μ22 = pk
Then
−μ21 +
λ21 + (1 − δ) μ22
=0
1+r
−pk +
λ21 + (1 − δ) pk
=0
1+r
λ21 = (r + δ) pk .
Firm 2’s period 1 output first-order condition is
£
¤
q11 + 2q21 = a − waL + (r + δ) pk aK .
We have seen that
λ11 = [(r + δ) − (1 + r) (1 − α)] pk < (r + δ) pk ;
in the first period, the incumbent’s rental cost of capital services is less than
the entrant’s rental cost of capital services, because in the first period the
incumbent’s opportunity cost of capital is tied to the resale value of capital,
while the entrant’s rental cost of capital services is tied to the purchase price
of capital.
Solve for equilibrium outputs in the first period:
µ
¶µ
¶ µ ¶
µ ¶
£
¤ª
2 1
q11
1 ©
1
k
=
a − waL + (r + δ) p aK +
(1 + r) (1 − α) pk
1 2
1
0
q21
26
3
µ
q11
q21
¶
=
¶µ ¶
¶µ ¶
µ
£
¤ª
1
1 ©
2 −1
2 −1
k
(1 + r) (1 − α) pk aK
a − waL + (r + δ) p aK +
0
1
−1 2
−1 2
¶ µ ¶
µ
µ
¶
£
¤ª
1 ©
q11
2
k
=
a − waL + (r + δ) p aK +
3
(1 + r) (1 − α) pk aK
1
−1
q21
£
¤
ª
1©
a − waL + (r + δ) pk aK + 2 (1 + r) (1 − α) pk aK
q11 =
3
©
ªª
1©
=
a − waL + [(r + δ) − 2 (1 + r) (1 − α)] pk aK
3
£
¤
ª
1©
q21 =
a − waL + (r + δ) pk aK − (1 + r) (1 − α) pk aK
3
©
ªª
1©
a − waL + [r + δ + (1 + r) (1 − α)] pk aK
=
3
The amount of capital sold by the incumbent at the start of period 1 is
¡
¢ ¡
¢
J11 = (1 − δ) Km − K11 = (1 − δ) K + aK qm − K + aK q11 .
µ
In the second and later periods, both firms purchase replacement capital
in each period: I1t > 0, I2t > 0. Then
μ1t = μ2t = pk , t = 2, 3, ...
This implies that
J1t = J2t = 0, t = 2, 3, ...
The first-order conditions for K1t and K2t imply
λ1t = λ2t = (r + δ) pk , t = 2, 3, ...
In and after period 2, the market is a Cournot duopoly in which firms
have identical marginal costs; equilibrium output per period is
qD =
1
(a − cH ) .
3
Note that
q11 =
¤
1£
a − cH + 2 (1 + r) (1 − α) pk aK
3
27
= qD +
2
(1 + r) (1 − α) pk aK
3
and
©
ªª
1©
a − waL + [r + δ + (1 + r) (1 − α)] pk aK
3
©
£
¤
ª
1
=
a − waL + (r + δ) pk aK − (1 + r) (1 − α) pk aK
3
1
= qD − (1 + r) (1 − α) pk aK .
3
The incumbent produces more, and the entrant less, in the first period
than in later periods.
Now evaluate firm 1’s equilibrium value. In equilibrium, the capital stock
constraint terms in the expression for firm 1’s value (those multiplied by the
Lagrangian multipliers μ1t ) drop out:
q21 =
V1 =
¡
¢
¡
¢ª
1 ©
(a − q11 − q21 ) q11 − waL q11 + wL + λ11 K11 − K − aK q11
1+r
1 © k
−p I12 + αpk J12 +
+
1+r
¾
£
¡
¢
¡
¢¤
(a − q12 − q22 ) q12 − waL q12 + wL + λ12 K12 − K − aK q12
−pk I11 +αpk J11 +
1
1+r
© k
1
k
2 −p I13 + αp J13 +
(1 + r)
¾
£
¡
¢
¡
¢¤
(a − q13 − q23 ) q13 − waL q13 + wL + λ13 K13 − K − aK q13
+ ...
+
1
1+r
Substitute I11 = J12 = J13 = ... = 0.
V1 = αpk J11 +
¡
¢
¡
¢ª
1 ©
(a − q11 − q21 ) q11 − waL q11 + wL + λ11 K11 − K − aK q11
1+r
½
−pk I12 +
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q12 − waL q12 + wL + λ12 K12 − K − aK q12
1+r
½
¾
¡
¢
¡
¢¤
1 £
1
k
−p I13 +
+
(a − q13 − q23 ) q13 − waL q13 + wL + λ13 K13 − K − aK q13
1+r
(1 + r)2
+...
1
+
1+r
28
Collect terms in q1t :
V1 = αpk J11 +
1
+
1+r
¢¤
¡
1 £
(a − waL − λ11 aK − q11 − q21 ) q11 − wL + λ11 K11 − K
1+r
½
−pk I12 +
1
+
(1 + r)2
¾
¢¤
¡
1 £
(a − waL − λ12 aK − q12 − q22 ) q12 − wL + λ12 K12 − K
1+r
½
−pk I13 +
¾
¢¤
¡
1 £
(a − waL − λ13 aK − q13 − q23 ) q13 − wL + λ13 K13 − K
1+r
+...
Now substitute
J11 = (1 − δ) Km − K11
I1t = K1t − (1 − δ)K1,t−1 , t = 2, 3, ...
V1 =
£
¢¤
¡
(a − waL − λ11 aK − q11 − q21 ) q11 − wL + λ11 K11 − K
1
1+r
1 © k
−p [K12 − (1 − δ)K11 ] +
+
1+r
αpk [(1 − δ) Km − K11 ]+
¾
¢¤
¡
1 £
(a − waL − λ12 aK − q12 − q22 ) q12 − wL + λ12 K12 − K
1+r
© k
1
+
2 −p [K13 − (1 − δ)K12 ] +
(1 + r)
¾
¢¤
¡
1 £
+ ...
(a − waL − λ13 aK − q13 − q23 ) q13 − wL + λ13 K13 − K
1+r
Now collect terms in K1t :
V1 = (1 − δ) αpk Km +
£
¤
ª
1 ©
(a − waL − λ11 aK − q11 − q21 ) q11 − wL − λ11 K + λ11 − (1 + r)αpk + (1 − δ)pk K11
1+r
©
1
+
(a − waL − λ12 aK − q12 − q22 ) q12 − wL − λ12 K
(1 + r)2
£
¤
ª
+ λ12 − (1 + r)pk + (1 − δ)pk K12
29
+
©
1
(a − waL − λ13 aK − q13 − q23 ) q13 − wL − λ13 K
(1 + r)3
£
¤
ª
+ λ13 − (1 + r)pk + (1 − δ)pk K13 + ...
Collect terms in pk in coefficients of K1t , t = 2, 3, 4...:
V1 = (1 − δ) αpk Km +
£
¤
ª
1 ©
(a − waL − λ11 aK − q11 − q21 ) q11 − wL − λ11 K + λ11 − (1 + r)αpk + (1 − δ)pk K11
1+r
£
©
¤
ª
1
k
K12
+
2 (a − waL q12 − λ12 aK − q12 − q22 ) q12 − wL − λ12 K + λ12 − (r + δ)p
(1 + r)
£
©
¤
ª
1
k
+
K13 +...
3 (a − waL − λ13 aK − q13 − q23 ) q13 − wL − λ13 K + λ13 − (r + δ)p
(1 + r)
Using
λ11 = (1 + r) αpk − (1 − δ) pk , λ1t = (r + δ) pk , t = 2, 3, ...,
the terms multiplied by K1t drop out:
V1 = (1 − δ) αpk Km +
¤
1 £
(a − waL − λ11 aK − q11 − q21 ) q11 − wL − λ11 K
1+r
¤
£
1
q
−
λ
a
−
q
−
q
)
q
−
wL
−
λ
K
(a
−
wa
L
12
12
K
12
22
12
12
(1 + r)2
¤
£
1
−
λ
a
−
q
−
q
)
q
−
wL
−
λ
K
+ ...
+
(a
−
wa
L
13
K
13
23
13
13
(1 + r)3
+
Now use the Kuhn-Tucker first-order conditions for output to express
gross profit in each period as the square of output:
V1 = (1 − δ) αpk Km +
+
¢
1 ¡ 2
q11 − wL − λ11 K
1+r
¢
¢
¡ 2
¡ 2
1
1
−
wL
−
λ
K
+
−
wL
−
λ
K
+ ...
q
q
12
13
12
13
(1 + r)2
(1 + r)3
Finally substituting equilibrium outputs:
V1 =
30
¢
¡ 2
¢
¡ 2
¢
1 ¡ 2
1
1
q11 − wL − λ11 K +
−
F
−
F
q
+
q
+...
H
H
D
D
1+r
(1 + r)2
(1 + r)3
¢
¤
1 ¡ 2
1 £ 2
q11 − wL − λ11 K −
qD − wL − (r + δ) pk K
= (1 − δ) αpk Km +
1+r
1+r
¢
¡ 2
¢
¡ 2
¢
1
1
1 ¡ 2
qD − FH +
+
2 qD − FH +
3 qD − FH + ...
1+r
(1 + r)
(1 + r)
£
¤ ª 1¡ 2
¢
1 © 2
2
+ (r + δ) pk − λ11 K +
= (1 − δ) αpk Km +
q11 − qD
qD − FH
1+r
r
¢
ª 1¡ 2
1 © 2
2
q11 − qD
qD − FH
+ (1 + r) (1 − α) pk K +
= (1 − δ) αpk Km +
1+r
r
∙
¸
¢
¢
1 ¡ 2
1¡ 2
2
= (1 − δ) αpk Km + (1 − α) pk K +
q11 − qD
qD − FH .
+
1+r
r
(1 − δ) αpk Km +
The equilibrium value of the incumbent is the sum of three terms:
• the resale value of capital inherited from the past that becomes excess
in the post-entry market;
• the extra economic profit the incumbent earns in the first period because its rental cost of capital services is λ11 < (r + δ)pk ;
• the value of a firm in a Cournot duopoly where both firms have marginal
cost cH and fixed cost FH in all periods.
Proceeding in the same general way, find the entrant’s equilibrium value.
The entrant maximizes
V2 = μ21 (I21 − J21 + K21 ) − pk I21 + αpk J21
¡
¢
¡
¢¤
1 £
(a − q11 − q12 ) q21 − waL q21 + wL + λ21 K21 − K − aK q21
1+r
1 ©
μ [(1 − δ) K21 + I22 − J22 − K22 ] − pk I22 + +αpk J22
+
1 + r 22
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q22 − waL q22 + wL + λ22 K22 − K − aK q22
1+r
©
1
k
k
+
2 μ23 [(1 − δ) K22 + I23 + J23 − K23 ] − p I23 + αp J23
(1 + r)
+
31
¾
¡
¢
¡
¢¤
1 £
+
(a − q13 − q23 ) q23 − waL q23 + wL + λ23 K23 − K − aK q23
+...
1+r
Terms multiplied by the μ2t drop out in equilibrium:
V2 = −pk I21 + αpk J21
¡
¢
¡
¢¤
1 £
(a − q11 − q12 ) q21 − waL q21 + wL + λ21 K21 − K − aK q21
1+r
1 © k
+
−p I22 + αpk J22 +
1+r
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q22 − waL q22 + wL + λ22 K22 − K − aK q22
1+r
© k
1
−p I23 + αpk J23 +
+
(1 + r)2
¾
¡
¢
¡
¢¤
1 £
(a − q13 − q23 ) q23 − waL q23 + wL + λ23 K23 − K − aK q23
+ ...
1+r
+
Substitute J21 = J22 = J23 = ... = 0.
V2 = −pk I21 +
¡
¢
¡
¢¤
1 £
(a − q11 − q12 ) q21 − waL q21 + wL + λ21 K21 − K − aK q21
1+r
½
−pk I22 +
¾
¡
¢
¡
¢¤
1 £
(a − q12 − q22 ) q22 − waL q22 + wL + λ22 K22 − K − aK q22
1+r
½
¾
¡
¢
¡
¢¤
1 £
1
k
−p I23 +
(a − q13 − q23 ) q23 − waL q23 + wL + λ23 K23 − K − aK q23
+
1+r
(1 + r)2
+...
1
+
1+r
Collect terms in q2t :
V2 = −pk I21 +
¡
¢¤
1 £
(a − waL − λ21 aK − q11 − q12 ) q21 − wL + λ21 K21 − K
1+r
½
−pk I22 +
¾
¡
¢¤
1 £
(a − waL − λ22 aK − q12 − q22 ) q22 − wL + λ22 K22 − K
1+r
½
¾
¢¤
¡
1 £
1
k
−p I23 +
(a − waL − λ23 aK − q13 − q23 ) q23 − wL + λ23 K23 − K
+
1+r
(1 + r)2
+...
1
+
1+r
32
Now substitute
I21 = K21
I2t = K2t − (1 − δ)K2,t−1 , t = 2, 3, ...
¢¤
¡
1 £
V2 = −pk K21 +
(a − waL − λ21 aK − q11 − q12 ) q21 − wL + λ21 K21 − K
1+r
1 © k
+
−p [K22 − (1 − δ)K21 ] +
1+r
¾
¢¤
¡
1 £
(a − waL − λ22 aK − q12 − q22 ) q22 − wL + λ22 K22 − K
1+r
© k
1
+
−p [K23 − (1 − δ)K22 ] +
(1 + r)2
¾
¢¤
¡
1 £
(a − waL − λ23 aK − q13 − q23 ) q23 − wL + λ23 K23 − K
+ ...
1+r
Now collect terms in K2t :
V2 =
£
¤
¤
1 £
(a − waL − λ21 aK − q11 − q12 ) q21 − wL − λ21 K + λ21 − (r + δ)pk K21
1+r
1
(1 + r)2
1
+
(1 + r)3
+
£
©
¤
ª
(a − waL − λ22 aK − q12 − q22 ) q22 − wL − λ22 K + λ22 − (r + δ)pk K22
£
£
¤
¤
(a − waL − λ23 aK − q13 − q23 ) q23 − wL − λ23 K + λ23 − (r + δ)pk K23
+...
Use the definition of λ2t to eliminate the terms in K2t and use the KuhnTucker first-order conditions for output to express the entrant’s gross profit
in each period as the square of its output:
V2 =
¢
¡ 2
¢
¡ 2
¢
1
1
1 ¡ 2
q21 − FH +
2 q22 − FH +
3 q23 − FH + ...
1+r
(1 + r)
(1 + r)
2
¢ qD
− FH
1 ¡ 2
2
qD − q21
.
+
1+r
r
The entrant’s equilibrium value is the value it would have in a Cournot
duopoly where both firms have marginal cost cH and fixed cost FH in all
periods, minus the present value of the profit it does not earn at the end of
V2 = −
33
the first period because the incumbent’s rental cost of capital services is less
than (r + δ)pk .
If
2
¢ qD
− FH
1 ¡ 2
2
−
qD − q21
< 0,
+
1+r
r
entry is blocked. If
2
¢
qD
− FH
1 ¡ 2
q 2 − FH
2
−
qD − q21
,
<0≤ D
r
1+r
r
entry is blocked that would be profitable if the incumbent’s costs were not
sunk.
7
Second solution
The incumbent does not sell capital in the first period. Since capital is in
excess supply, the incumbent lets capital depreciate until it is optimal to
purchase capital.
For convenience, repeat the incumbent’s first-order conditions:
Period 1:
∂V1
= (1 − δ) Km + I11 − J11 − K11 = 0.
∂μ11
¡
¢
∂V1
= μ11 − pk ≤ 0
I11 μ11 − pk = 0
I11 ≥ 0
∂I11
¡
¢
∂V1
= −μ11 + αpk ≤ 0
J11 −μ11 + αpk = 0
J11 ≥ 0
∂J11
∂V1
(1 + r)
= a − 2q11 − q21 − waL − λ11 aK = 0
∂q11
∂V1
λ11 + (1 − δ) μ12
=0
= −μ11 +
∂K11
1+r
£
¤
∂V1
= K11 −K−aK q11 ≥ 0
λ11 K11 − K − aK q11 = 0
(1 + r)
∂λ11
Period t:
(1 + r)t−1
∂V1
= (1 − δ) K1,t−1 + I1t − J1t − K1t = 0.
∂μ1t
34
λ11 ≥ 0
(1 + r)t−1
¡
¢
I1t μ1t − pk = 0
∂V1
= μ1t − pk ≤ 0
∂I1t
¡
¢
∂V1
= −μ1t + αpk ≤ 0
J11 −μ1t + αpk = 0
∂J1t
∂V1
(1 + r)t
= a − 2q11 − q21 − waL − λ1t aK = 0
∂q1t
λ1t + (1 − δ) μ1,t+1
∂V1
=0
= −μ1t +
(1 + r)t
∂K1t
1+r
= J11 = 0; hence
K11 = (1 − δ) Km .
(1 + r)t−1
I11
I1t ≥ 0
J1t ≥ 0
Capital is in excess supply,
K11 = (1 − δ) Km ≥ K + aK q11 , ⇒ λ11 = 0.
The first-order condition for firm 1’s output is
2q11 + q21 = a − waL
Suppose that the incumbent buys capital in period 2. Then
μ12 = pk .
Then from
∂V1
λ11 + (1 − δ) μ12
= −μ11 +
=0
∂K11
1+r
we find
λ11 + (1 − δ) μ12
0 + (1 − δ) pk
1−δ k
=
=
p ;
1+r
1+r
1+r
consistency conditions are
μ11 =
∂V1
= μ11 − pk ≤ 0 or μ11 ≤ pk
∂I11
∂V1
= αpk − μ11 ≤ 0 or μ11 ≥ αpk
∂J11
Thus for the incumbent to stand pat and let its capital depreciate in the
first period, we must have
αpk ≤ μ11 =
1−δ k
p ≤ pk
1+r
35
1−δ
≤ 1.
1+r
The right-hand inequality always holds, and also α ≤ 1. The left-hand
inequality may or may not hold. If it does not, it is optimal for the incumbent
to sell excess capital immediately if entry occurs; this is the first solution.
If the incumbent buys capital in period 2, it buys capital in every period
thereafter.
λ12 = (r + δ) pk .
α≤
The first-order condition for the incumbent’s period 2 output is
2q11 + q21 = a − cH .
If the incumbent neither buys nor sells capital in the second period but
does buy capital in the third period,
K12 = (1 − δ)2 Km ≥ .K + aK q12 , ⇒ λ12 = 0.
From
(1 + r)
λ12 + (1 − δ) μ13
∂V1
= −μ12 +
=0
∂K12
1+r
0 + (1 − δ) pk
1+r
At this point, knowing the incumbent’s rental cost of capital services, and
knowing that the entrant buys capital, we can solve for equilibrium output
and all other values that characterize period 2 equilibrium. Go forward in
the same way.
The entrant’s first-order conditions are as in the first solution: the entrant
buys capital in every period, always has rental cost of capital (r + δ) pk , has
in each period first-order condition for output
μ12 =
q1t + 2q2t = a − cH .
This solution ends up replicating that of the case in which resale is impossible.
36