Necessary and sufficient condition for equilibrium of the Hotelling model on a circle
aa r X i v : . [ ec on . T H ] O c t NECESSARY AND SUFFICIENT CONDITION FOREQUILIBRIUM OF THE HOTELLING MODEL ON A CIRCLE
SATOSHI HAYASHI † AND NAOKI TSUGE ‡ Abstract.
We are concerned with a model of vendors competing to sell ahomogeneous product to customers spread evenly along a circular city. Thismodel is based on Hotelling’s celebrated paper in 1929. Our aim in this pa-per is to show a necessary and sufficient condition for the equilibrium, whichdescribes geometric properties of the equilibrium. To achieve this, we firstformulate the model mathematically. Next, we prove that the condition holdsif and only if vendors are equilibrium. Introduction
We study a model in which a circular city lies on a circle with circumferencelength 1 and customers are uniformly distributed with density 1 along this circle.We consider n vendors moving on this circle. The model governs a competitionbetween vendors on a simple closed curve. Let the location of the vendor k ( k =1 , , , . . . , n ) be x k ∈ [0 , x ≤ x ≤ · · · ≤ x n and denote thelocation of n vendors ( x , x , x , . . . , x n ) by x . Since we study the competitionbetween vendors, we consider n ≥ l ( l = 1 , , , . . . , n ) vendors nearest to a customer, the customerpurchases 1 /l unit of product per unit of time from each of the l vendors respectively.Every vendor then seeks a location to maximize his profit.We then represent the profit of vendor k per unit of time by a mathematicalnotation. We first denote a distance on the circle between x, y ∈ [0 ,
1] by d ( x, y ) =min {| x − y + 1 | , | x − y | , | x − y − |} . Remark 1.1.
We regard x, y ∈ [0 , as points on a circle with circumference length . Then d ( x, y ) represents the length of bold-faced arcs in the following figure. xy x y yx y x Key words and phrases.
Competitive location problem, The Hotelling model on a circle, Equi-librium, Mathematical formulation, Necessary and sufficient condition.N. Tsuge’s research is partially supported by Grant-in-Aid for Scientific Research (C)17K05315, Japan. † AND NAOKI TSUGE ‡ That is, d ( x, y ) is the shortest distance between x and y on the circle. Given a vector ξ = ( ξ , ξ , . . . , ξ n ) ∈ [0 , n and 0 ≤ y ≤
1, we then define a set S ( ξ , y ) = { j ∈ { , , , . . . , n } : d ( ξ j , y ) = min i d ( ξ i , y ) } . By using a density function ρ k ( ξ , y ) = d ( ξ k , y ) > min i d ( ξ i , y )1 | S ( ξ , y ) | if d ( ξ k , y ) = min i d ( ξ i , y ) , we define f k ( ξ ) = Z ρ k ( ξ , y ) dy, where | S ( ξ , y ) | represents a number of elements in a set S ( ξ , y ). We call f k ( x ) theprofit of vendor k per unit of time for a location x . We then define equilibrium asfollows. Definition 1.1.
A location x ∗ = ( x ∗ , x ∗ , x ∗ , . . . , x ∗ n ) ∈ [0 , n is called equilibrium,if f k ( x ∗ ) ≥ f k ( x ∗ , x ∗ , x ∗ , . . . , x ∗ k − , x k , x ∗ k +1 , . . . , x ∗ n ) holds for any k ∈ { , , , . . . , n } and x k ∈ [0 , . We review the known results. The present model is based on the Hotelling’spioneer work [3]. Although we consider the circle as a product space, Hotellingdealed with a finite line. In [2], authors show a necessary and sufficient conditionfor the equilibrium of n vendors on the line.On the other hand, our model was introduced by Eaton and Lipsey [1], whodiscussed with the existence of equilibrium for the model without the price. Subse-quently, taking the price into account, Salop [4] studied the model for two vendors.In this paper, we are concerned with n vendors on the circle and investigate theirequilibrium. Our goal is to present a necessary and sufficient condition for theequilibrium.For convenience, we set x = 0 , x = x n , x − = x n − , x − = x n − , x n +1 = x , x n +2 = x and denote a interval [ x k , x k +1 ] by I k ( k = − , − , , , . . . , n + 1).Then our main theorem is as follows. Theorem 1.1. x is equilibrium, if and only if the following condition (1.1) hold.We define a set | ¯ I | = max l ∈{ , , ··· ,n } | I l | . | I k | + | I k +1 | ≥ | ¯ I | (1 ≤ k ≤ n ) , (1.1) where | I | represents the length of an interval I . Remark 1.2.
From (1.1), we notice that any location becomes equilibrium for n =2. On the other hand, we find that equilibrium for n = 3 is x ∗ = (cid:18) , , (cid:19) , (cid:18) , , (cid:19) . OTELLING MODEL 3 Preliminary
In this section, we prepare some lemmas and a proposition to prove our maintheorem in a next section. We first consider the profit of i vendors which locate atone point. We have the following lemma. Lemma 2.1.
We consider a location x = ( x , x , x , . . . , x n ) with x ≤ x ≤ x ≤ · · · ≤ x n . We assume that x l < x l +1 = · · · = x k = · · · = x l + i < x l + i +1 ( n ≥ , l ≥ , l + i + 1 ≤ n + 1 , ≤ i ≤ n ) . f k ( x ) = 12 i ( | I l | + | I l + i | ) . Proof. x l x k x l + i +1 ... i vendors We have f k ( x ) = 1 i (cid:18) | I l | + 12 | I l + i | (cid:19) = 12 i ( | I l | + | I l + i | ). (cid:3) Next, the following proposition play an important role.
Proposition 2.2.
If the location of n vendors x = ( x , x , x , . . . , x n ) with x ≤ x ≤ x ≤ · · · ≤ x n ( n ≥ is equilibrium, the following holds.No more than 2 vendors can occupy a location. (2.1) Proof.
We prove that x is not equilibrium, provided that i (3 ≤ i ≤ n ) vendors occupyat a point. We assume that x l < x l +1 = · · · = x k = · · · = x l + i < x l + i +1 ( l ≥ , l + i + 1 ≤ n + 1 , ≤ i ≤ n ). Therefore there exist x l and x l + i +1 at leastone respectively. We notice that x l < x k < x l + i +1 and there exists no vendor on( x l , x k ) and ( x k , x l + i +1 ). SATOSHI HAYASHI † AND NAOKI TSUGE ‡ x l x k x l + i +1 ... i vendors If | I l | ≥ | I l + i | , we notice that f k ( x ) = 12 i ( | I l | + | I l + i | ) ≤
16 ( | I l | + | I l + i | ) ≤
16 ( | I l | + | I l | ) = 13 | I l | . Setting x ′ k ∈ ( x l , x l +1 ), we have f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) =12 ( | [ x l , x ′ k ] | + | [ x ′ k , x k ] | ) = 12 | I l | > f k ( x ).For the other case | I l | < | I l + i | , from the symmetry, we can similarly show thatthere exists x ′ k such that f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) > f k ( x ).We can complete the proof of (2.1). (cid:3) Finally, we compare a location after the movement of a vendor with the originalone. To do this, we introduce the following notation.For a given location of n vendors x = ( x , x , x , . . . , x n ) with (0 =) x ≤ x ≤· · · ≤ x n , we move vendor k from x k to any fixed point in A ⊂ [0 , x k → A . We notice that x k → A represents the followingvector ( x , x , , . . . , x k − , x ′ k , x k +1 , . . . , x n ) , where x ′ k is a location of vendor k after movement and x ′ k ∈ A .3. Proof of Theorem 1.1
We are now position to prove our main theorem. We divide the proof into twocases, (I) n = 2 and (II) n ≥ Proof of Theorem 1.1 (I)We first treat with the case where n = 2 and prove that every x is equilibrium. OTELLING MODEL 5 x x Proof. (i) x = x We notice that f ( x ) = f ( x ) = 12 in this case.(ii) x = x x = x We notice that f ( x ) = f ( x ) = 12 in this case.It follows that f ( x ) = f ( x ) = 12 for every x . As a result, we showed that f k ( x ) ≥ f k ( x k → [0 , k = 1 , | I | + | I | ≥ | I | and | I | + | I | ≥ | I | . Therefore, we can prove (1.1). (cid:3) Proof of Theorem 1.1 (II)Finally, we are concerned with the case where n ≥ SATOSHI HAYASHI † AND NAOKI TSUGE ‡ Proof.
First, we show that (1.1) is a sufficient condition for equilibrium.We recall that we defined | ¯ I | = max l ∈{ , , ··· ,n } | I l | .We show that f k ( x ) ≥ | ¯ I | .(i) x k − = x k and x k = x k +1 We notice that f k ( x ) = 12 ( | I k − | + | I k | ) ≥ | ¯ I | . x k − x k x k +1 (ii) x k − = x k (resp. x k = x k +1 ) x k − x k − = x k x k +1 If | I k − | = 0, since | I k − | + | I k − | ≥ | ¯ I | and | I k − | + | I k | ≥ | ¯ I | , we have | I k − | = | I k | = | ¯ I | . (resp. If | I k | = 0, since | I k − | + | I k | ≥ | ¯ I | and | I k | + | I k +1 | ≥ | ¯ I | , we have | I k − | = | I k +1 | = | ¯ I | .) Therefore, we obtain f k ( x ) = 12 · | ¯ I | + | ¯ I | ) = 12 | ¯ I | in this case.From (i) and (ii), we conclude that f k ( x ) ≥ | ¯ I | . OTELLING MODEL 7
Next, we show that f k ( x ) ≥ f k ( x k → [0 , x k → ( x k − , x k +1 ) (1 ≤ k ≤ n )(a) x k − = x k and x k = x k +1 We have f k ( x k → ( x k − , x k +1 )) = f k ( x ).(b) x k − = x k or x k = x k +1 We have f k ( x k → ( x k − , x k +1 )) ≤ | ¯ I | ≤ f k ( x ).(ii) x k → ( x l , x l +1 ) ( x l < x l +1 and l = k − , k and 1 ≤ l ≤ n )We have f k ( x k → ( x l , x l +1 )) = 12 | I l | ≤ | ¯ I | ≤ f k ( x ).(iii) x k → { x l } (1 ≤ k ≤ n and 1 ≤ l ≤ n )(a) l = k It clearly holds that f k ( x k → { x k } ) = f k ( x ).(b) The case where only vendor k occupies at x k and moves next to x k x k − )( x k − ) x k − x k x k +1 ( x k − ) x ′ k In this case, we deduce that f k ( x k → { x l } ) ≤ · | ¯ I | + 2 f k ( x )) ≤
14 (2 f k ( x ) + 2 f k ( x )) = f k ( x ).(c) The case where more than two vendors occupy at x k and vendor k moves next to x k , or vendor k moves to a location that is not next to x k In this case, we deduce that f k ( x k → { x l } ) ≤ · | ¯ I | + | ¯ I | ) = 12 | ¯ I | ≤ f k ( x ).From (i)–(iii), we conclude that f k ( x ) ≥ f k ( x k → [0 , x is not equilibrium. That is, assuming thereexists k such that | I k | + | I k +1 | < | ¯ I | , we prove that x is not equilibrium. ObservingProposition 2.2, it suffices to consider the case where only one vendor occupies alocation. SATOSHI HAYASHI † AND NAOKI TSUGE ‡ (i) x k = x k +1 and x k +1 = x k +2 We have f k +1 ( x ) = 12 ( | I k | + | I k +1 | ) < | ¯ I | . Let ( x j , x j +1 ) be a maximumopen interval such that | ¯ I | = | ( x j , x j +1 ) | . If x k +1 → ( x j , x j +1 ), we have f k +1 ( x k +1 → ( x j , x j +1 )) = 12 | ¯ I | > f k +1 ( x ).Thus x with this case is not equilibrium.(ii) x k = x k +1 (resp. x k − = x k )Since | I k | + | I k +1 | < | ¯ I | and | I k | = 0, we drive | I k +1 | < | ¯ I | . Forthermore, wenotice | I k − | ≤ | ¯ I | . Thus we deduce that f k ( x ) = 12 · | I k − | + | I k +1 | ) < · | ¯ I | + | ¯ I | ) = 12 | ¯ I | . Let ( x j , x j +1 ) be a maximum open interval suchthat | ¯ I | = | ( x j , x j +1 ) | . If x k → ( x j , x j +1 ), we have f k ( x k → ( x j , x j +1 )) =12 | ¯ I | > f k ( x ). Thus x is not equilibrium in this case.From (i)–(ii), we have showed that (1.1) is a necessary condition for equilibrium.Therefore, we can complete the proof of Theorem 1.1. (cid:3) References [1] Eaton, B. C. and Lipsey, R. G., 1975, The principle of minimum differentiation reconsidered: some new developments in the theory of spatial cometition, Review of Economic Studies,42, 27–49.[2] Hayashi, S. and Tsuge, N., Necessary and sufficient condition for equilibrium of the Hotellingmodel (preprint).[3] Hotelling, H., 1929, Stability in competition, Economic Journal, 39, 41–57.[4] Salop, S. C., 1979, Monopolistic competition with outside goods, Bell J. Econ. 10, 141-156. †‡ Department of Mathematics Education, Faculty of Education, Gifu University, 1-1Yanagido, Gifu Gifu 501-1193 Japan.
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