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Competition Before Sunset: The Case of the Finnish ATM Market

  • Maria Kopsakangas-Savolainen and Tuomas Takalo EMAIL logo


We study service fee and network size competition in an ATM market between an incumbent and an independent deployer, and its optimal regulation. We also analyze an actual regulation of such a market by competition authorities in Finland. Compared with the first-best regulation, we find unregulated foreign and interchange fees too high and an unregulated size of the incumbent’s ATM network too small. However, if network sizes cannot be directly regulated, then competitive fees may also be too low from the welfare point of view. The Finnish regulation caps the incumbent’s foreign fee which, according to our model, results in an increased interchange fee and a larger ATM network. In contrast with the actual regulation, it would also be essential to regulate the interchange fee.

Corresponding author: Tuomas Takalo, Monetary Policy and Research Department, Bank of Finland, Helsinki, Finland; and Department of Managerial Economics, Strategy, and Innovation, Katholieke Universiteit Leuven, Leuven, Belgium, e-mail:


We thank an anonymous referee for perceptive comments. We also thank Tom Björkroth, Monika Hartmann, Ari Hyytinen, Kari Kemppainen, Klaus Kultti, Mauri Lehtinen, Kari Takala, Juha Tarkka, Otto Toivanen, Sinikka Salo, Matti Virén and the seminar participants at the Bank of Finland and the ECB-Bank of Italy Workshop on Interchange Fees for helpful comments and discussions.


Proof of Proposition 2. For the moment, let us assume that the non-negativity constraints on network sizes do not bind. Then the first-order condition for the problem of choosing the socially optimal f reads as


Using equations (3) and (33) and simplifying gives

(35)dWdf=Δcft1k[12[ΔcΔM+t(1sIsB)]ksB]=0. (35)

The first-order condition for the socially optimal a is analogously given by


After using equations (3) and (34), this can be simplified to

(36)dWda=t(1sIsB)ksBksI=0. (36)

Plugging equation (36) into equation (35) yields


Next, substituting equations (34) and (33) for sI and sB gives


Solving this equation for f yields after some manipulation

(37)f=fΔc+2t(ΔMΔc)4k+t. (37)

Inserting equations (33) and (34) into equation (36) yields


Solving this equation for a gives

(38)a(f)=ktt+k+f+cI+cB2. (38)

Substituting equation (37) for f in equation (38) then gives

(39)a(f)a=ktt+k+cI+t4k+t(ΔMΔc). (39)

Finally, substitution of equations (37) and (39) for f and a in equations (33) and (34) results in the equilibrium network sizes of

(40)sB=t2k(kt+kΔMΔc4k+t) (40)


(41)sI=t2k(kt+k+ΔMΔc4k+t). (41)

Comparing equations (40) and (41) with equations (31) and (32) produces the following observations: i) If ΔM≥Δc, then sIsI>0 and sBsB. ii) If Δc≥ΔM, then sBsB>0 and sIsI. Thus, because we work in the range of parameter values where sB>0 and sI>0 we also have that sB>0 and sI>0, satisfying our initial assumption of positive network sizes.      ▀

Proof of Proposition 3. Note first from equation (5) that since 0≤XB(f, sB, sI)≤1 (we later verify that this holds in equilibrium) and since cB≥0, a necessary condition for πB(f, sB, a, sI)≥0 is that f>a. However, if f>a and cB≥0, then afcB<0. Equation (33) then implies that the constraint sB≥0 is violated, leading to sBR=0.

Let us for the moment assume that acI implying that πI(f, a, sI, sB)≥0 and Δs=sI=(acI)/2k≥0. Then, after substituting equation (3) with Δs=sI=(acI)/2k for equation (5) and some straightforward calculations, we can re-express the constraint πB(f, a)≥0 as


The left-hand side is a quadratic polynomial in f. Its two roots are given by

(42)fR(a)=f(a)±f(a)2[ΔM(acB)+t(a+cB)]t(acB)(acI)2k (42)



can be found after substitution of Δs=sI=(acI)/2k for equation (7).

To pin down a, we maximize social welfare from equation (28) with respect to a when sB=0. The first-order condition reads as


After using equations (3) and (34), and simplifying we get

(43)sI=sIRt+ΔMΔc2k+t. (43)

To determine the optimal interchange fee, we substitute equation (34) for sIR in equation (43), and solve for a. This results in

(44)a=aR2k(t+ΔMΔc)2k+t+cI. (44)

Substituting equation (44) for a in equation (42) gives


Note that both roots are larger than Δc, which would constitute the optimal foreign fee in the absence of zero profit constraint in the case where sB=0. By the concavity of the welfare function, it is therefore optimal to implement as small f as possible compatible with the constraint πB(f)≥0. This is given by the lower root of the above equation. As a result, we have πB(fR,aR,sIR,sBR)=0


(45)fR=t+ΔM(t2k+t)2(t+ΔMΔc)22cBt. (45)

From equations (43) and (44) we see that under restriction (19), sIR0 and aRcI, as we initially assumed. We should also assume that SR=sIR<1, which is equivalent to 2kM–Δc. Then, after substitution of sIR and fR from equations (43) and (45) for Δs and f in equation (3), it is also straightforward to show that XB(fR,sIR,sBR)[0,1] as we initially assumed.

Given (19) and Proposition 3, it is immediate that fRf*c, aRcIas, and sBsBR=sBc=0. Claim sIRsIc if kt/3 follows from the comparison of equations (43) and (17).      ▀

Proof of Proposition 4. We show in the main text that the cap on the foreign fee binds. Then, solving the deployer I’s problem (8) under the assumption that the foreign fee is given by fFCA(a)=acB yields that

(46)a=aFCA(sB,sI)[t(1+Δs)+ΔM+cB+cI]2=a(sB,sI), (46)

which where the last equality follows from the comparison of aFCA(sB, sI) with equation (10). This implies that the regulated foreign fee becomes


It is immediate from the deployer B’s objective function (5) that ∂πB/∂sI<0 if f(a)=acB. Thus sBFCA=sBc=0.

After substituting fFCA(sB, sI) for equation (4), we may write the deployer I’s market share as

(47)XIFCA(sB,sI)=1+Δs4+ΔMΔc4t=2XI(sB,sI) (47)

where the last equality comes from the comparison of XIFCA(sB,sI) with equation (13). Noting equations (46) and (47) and inspecting the deployer I’s objective functions from (8) and (16) shows that we may write the deployer I’s problem as


Using sBFCA=sBc=0 we quickly get that the deployer I’s optimal network size is given by


It is easy to verify that the second-order condition for the deployer I’s maximization problem is satisfied when


[This is also the second-order condition for a monopoly’s problem in Alexandrov (2008).] Under this assumption, we have sIFCAsIc=(t+ΔMΔc)/(8kt) [see equation (17)]. As a result, the total size of the ATM network increases since SFCA=sIFCASc=sIc.

Next, we substitute ΔsFCAsIFCAsBFCA=sIFCA=(t+ΔMΔc)/(4kt) for equation (46). This yields


As a result, we have


Comparing fFCA and aFCA with fc and ac of equations (22) and (21), respectively, shows that under the restriction (19), aFCAac and fcfFCA.      ▀


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Published Online: 2014-11-20
Published in Print: 2014-3-1

©2014 by De Gruyter

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