Computng a Cournot Equlbrum n Integers Mchael J. Todd December 6, 2013 Abstract We gve an effcent algorthm for computng a Cournot equlbrum when the producers are confned to ntegers, the nverse demand functon s lnear, and costs are quadratc. The method also establshes exstence, whch follows n much more generalty because the problem can be modelled as a potental game. 1 Introducton Consder n producers who produce a sngle good. If each producer produces q, we assume the prce at whch the good can be sold s p = (a bq) +, where Q := nx q, here and below, denotes the total producton, a and b are postve, and z + denotes the postve part max{0, z} of z. Producer also faces a producton cost of c q + d q 2, wth c and d nonnegatve wth postve sum. Then f the producton vector s q = (q ), producer makes proft π (q) := (a bq) +q c q d q 2. Wth a slght abuse of notaton, we also wrte π (q) as π (q, q ) or π (q, Q ), where q = (q j) j and Q := P j qj, to hghlght ts dependence on producer s decson as well as those of the other producers. Note that π (q, Q ) = (a bq bq ) +q c q d q 2. Our man nterest s n the case that each producer chooses her decson q from the nonnegatve ntegers, Z +, but we also consder the smpler case where q s chosen from the nonnegatve reals, R +. In ether case, a Cournot equlbrum s a producton profle where each producer chooses her best response to the decsons of the other producers: School of Operatons Research and Informaton Engneerng, Cornell Unversty, Ithaca, NY 14853, USA. E-mal mjt7@cornell.edu. 1
Defnton 1 q R n + (q Z n +) s a real (nteger) Cournot equlbrum for {π } f for each q maxmzes π (, Q ) over R +(Z +). Cournot equlbra, wth the related Bertrand equlbra (where the producers choose prces, not quanttes) and Stackelberg (leader-follower) varants of these, form a standard topc n olgopoly theory: see, for example Mas-Colell, Whnston, and Green [4], pp. 389 394, or Varan [11], pp. 285 291. Kreps and Schenkman [2] show how a Cournot equlbrum arses from a two-stage game: frst the producers choose capactes, and then, knowng these capactes, each chooses a prce as n the Bertrand model. Integer or dscrete versons are much less studed, although exstence s proved n for example Kukushkn [3] usng Tarsk s fxed-pont theorem, and Dubey, Hamanko, and Zapechelnyuk [1] usng a potental game, followng a less general treatment of Shapley [7]. Ths paper s concerned wth the effcent computaton of real and nteger Cournot equlbra, and as a byproduct, ther exstence, n the case of a lnear nverse demand functon and convex quadratc costs as above. Whle ths case s qute restrctve, we note that t s prototypcal, and that our technque can be used n an teratve scheme where at each step, the nverse demand functon and the cost functons are approxmated by a lnear functon, respectvely quadratc functons, at the current terates. In the next secton we recall the proof of the exstence of nteger Cournot equlbra usng a standard potental functon argument, whch s non-constructve. Then we show how they can be computed (and ncdentally shown to exst) usng an effcent drect or bnary search on the total producton Q. Computaton of Cournot equlbra n much more general settngs s dscussed for example n Sheral, Soyster, and Murphy [8] and Thorlund-Petersen [10]. 2 Exstence Here we prove exstence of both real and nteger Cournot equlbra usng a potental functon approach. Ths replaces the mult-player crteron for an equlbrum wth a sngle objectve whose maxmzers yeld equlbra; games whch permt such a substtuton were ntroduced by Monderer and Shapley [5] and termed potental games. Frst we smplfy the payoff functons. Whle negatve prces have no nterpretaton n our settng, we can defne payoff functons as f the prce were gven by a bq nstead of ts postve part. Consder (agan wth a slght abuse of notaton) π (q, q ) := π (q, Q ) := (a bq bq )q c q d q 2. Proposton 1 The real (nteger) Cournot equlbra for { π } concde wth those for {π }. Proof: Note that n any real or nteger Cournot equlbrum q for ether set of payoff functons, every producer gets a nonnegatve payoff, snce producng nothng s always an alternatve. But then we cannot have 2
a bq nonpostve, for then any producer wth postve q (and there must be one) gets negatve payoff. Snce the two payoff functons only dffer for a bq negatve, and no producer has an ncentve to move to such a pont, an equlbrum for one set of payoff functons s also one for the other. We can now use the payoff functons { π } n our equlbrum arguments. Consder the potental functon f(q) := (a 1 n 2 bq)q X (c q + (d + 1 2 b)q2 ). Vewed as a functon of q and q, ths can alternatvely be wrtten as f(q, q ) = aq + aq 1 2 b(q + Q )2 c q (d + 1 2 b)q2 X j (c jq j + (d j + 1 2 b)q2 j ) = aq b(q + Q )q c q d q 2 +aq 1 2 bq2 X j (c jq j + (d j + 1 2 b)q2 j ) = π (q, Q ) + f (q ), where (as the notaton ndcates) f s ndependent of q. Ths gves the exstence of Cournot equlbra. The proof below s due to Monderer and Shapley [5] n the real case, and Shapley [7] for ntegers. It seems worthwhle to gve t for the sake of completeness. Theorem 1 Any maxmzer of f over R n + (Z n +) s a real (nteger) Cournot equlbrum for {π }, and hence such equlbra exst. Proof: Indeed, any maxmzer q of f has the property that q s a maxmzer of π (, Q ) by the equaton above, and so provdes a Cournot equlbrum for { π } and hence for {π } by Proposton 1. Ths holds over both the nonnegatve reals and the nonnegatve ntegers, and proves the frst part. For the exstence, we note f(0) = 0, and f(q) < 0 f Q > 2a/b, whence we can confne our search for the maxmzer of the contnuous functon f to the compact set of nonnegatve reals or ntegers wth Q 2a/b, so that exstence of a maxmzer s assured. The exstence proof above s bascally non-constructve (although there are algorthms to maxmze the strctly concave quadratc functon f over the nonnegatve orthant, and one could enumerate the exponentally large set of ponts n {q Z n + : Q 2a/b}), and so we develop effcent algorthms to compute a Cournot equlbrum (and ncdentally prove ts exstence) n the next secton. 3 Computaton The strctly concave quadratc π (q, Q ) = (a c bq )q (b + d )q 2 3
for any nonnegatve Q s maxmzed n q over the reals unquely at and over the nonnegatve reals at a c bq, 2(b + d ) q := (a c bq )+, (1) 2(b + d ) as s easly seen from the frst-order condtons or drectly. Snce quadratcs are symmetrc about ther maxmzers, the maxmum over the ntegers s attaned at any nteger n [ˇq, ˆq ], where ˇq := ˆq := a c bq 2(b+d 1 ) 2 a c bq 2(b+d ) + 1 2 + + = (a c b d bq ) + 2(b+d ), = (a c +b+d bq ) + 2(b+d ). In both cases above, the optmal value or bounds for q are gven n terms of Q, but t would be much more useful f they were defned (n a somewhat crcular way) as functons of the total producton Q. Such functons were frst consdered by Szdarovszky and Yakowtz [9] (see also Novshek [6], who called them backward reacton mappngs ). For ths reformulaton the followng result s helpful: Lemma 1 For real λ and 0 < µ < 1, λ x R (λ µy) + ff x R 1 µ µ «(y + x) 1 µ where R stands for less than or equal to, greater than or equal to, or equal to. Proof: We prove the lemma when R s and ; the thrd case then follows. For the frst case, the left-hand sde s true ff x 0 or 0 < x λ µy, whle the rght hand-sde s true ff x 0 or 0 < x λ 1 µ µ (y + x). 1 µ It s easy to see that these are equvalent. For the second case, the left-hand sde s true ff x 0 and x λ µy, whle the rght hand-sde s true ff x 0 and x λ 1 µ µ (y + x), 1 µ and once agan these are easly seen to be equvalent. We now apply the lemma to the condtons for optmal q s above. It s convenent to defne e := a c for each. We may also assume that these are all postve, snce any producer wth a nonpostve e wll choose to produce nothng no matter what the others do., + (2) 4
Theorem 2 a) A producton profle q R n s a real Cournot equlbrum ff q = q (e bq)+ (Q) := for each. b) A producton profle q Z n s an nteger Cournot equlbrum ff q [ˇq (Q), ˆq (Q)], where ˇq (Q) := ˆq (Q) := (e b d bq)+, (e + b + d bq)+ for each. Proof: For part (a), we apply the lemma to (1) wth R = and λ = e /[2(b + d )], µ = b/[2(b + d )]. For part (b), we apply the lemma to (2) wth R and λ = [e b d ]/[2(b + d )], µ = b/[2(b + d )], and wth R and λ = [e + b + d ]/[2(b + d )], µ = b/[2(b + d )]. We can now nvestgate effcent algorthms to compute Cournot equlbra. In the real case, t suffces to fnd Q R + satsfyng Q = φ(q) := X (e bq) +. Ths provdes an alternatve proof of exstence, snce φ s contnuous and nonncreasng, postve at zero, and zero for large enough Q, and hence has a fxed pont by the ntermedate-value theorem. It seems that to fnd the fxed pont we need to do a bnary search on Q to determne for whch ndces the numerator s postve, but f the e s are sorted (whch takes O(n log n) tme), ths can be avoded by computng the desred sums ncrementally. Let us assume the e s are n nonncreasng order. Then for some 1 j n, Q = = jx e bq P j e/(b + 2d) 1 + P j b/(b + 2d) and e j/b Q e j+1/b, where we take e n+1 := 0. Ths gves our algorthm, where we successvely compute the numerator (ν) and the denomnator (δ), termnatng when the rato exceeds e j+1/b. Algorthm 1 (Computng a real Cournot equlbrum) Intalze ν = 0 and δ = 1. For j = 1, 2,..., n, Next j. e j b b+2d j. ν ν + b+2d j, δ δ + If ν e j+1, set Q = ν and δ b δ q = q (Q), = 1,..., n, and STOP. 5
It s easy to see nductvely that ν/δ e j/b. Indeed, the prevous ν/δ was less than e j/b ether because t s ntally zero or because the test at the prevous j faled, and the current value s a convex combnaton of the old value and e j/b. Hence Algorthm 1 fnds a real Cournot equlbrum n O(n) tme (or O(n log n) tme f the e s need to be sorted). The stuaton s more complcated n the nteger case. We can proceed as n the real case to obtan bounds on Q for any nteger Cournot equlbrum. Indeed, f (ebd []) denotes the components of (ej b dj) arranged n nonncreasng order, and (d [] ) denotes the components of (d j) n the same order, then for some j = 1,..., n, Q jx ebd [] bq, so b + 2d [] P j ebd [] /(b + 2d []) Q 1 + P j b/(b + 2d []) (3) where ebd [j] /b Q ebd [j+1] /b and agan we take ebd [n+1] := 0. Smlarly, f (ebd + {}) denotes the components of (ej + b + dj) arranged n nonncreasng order, and (d {} ) denotes the components of (d j) n the same order, then for some j = 1,..., n, Q jx ebd + {} bq, so b + 2d {} P j ebd+ {} /(b + 2d {}) Q 1 + P j b/(b + 2d {}) (4) where ebd + {j} /b Q ebd+ {j+1} /b wth ebd+ [n+1] := 0. The dffcultes are several. Frst, these condtons are necessary but not suffcent, snce for ease of analyss we have neglected the requrement that each q be nteger. Second, whle both P ˇq(Q) and P ˆq(Q) are nonncreasng n Q, and t s not hard to see that ths range always ncludes an nteger (see the proof of Theorem 4 below), t s a pror possble that the range jumps over Q as t moves from one nteger value to the next. Hence exstence does not follow drectly. We deal wth these problems ndvdually and sequentally. We use the necessary bounds above to restrct the range of possble Q s. Then we perform a bnary search over ths range, usng the correct bounds wth nteger parts for the ndvdual q s to fnd the equlbrum, and relyng on our earler exstence proof. Fnally, we provde a drect proof of exstence based on nteger changes n Q. The algorthm below provdes lower and upper bounds on Q n any nteger Cournot equlbrum: Algorthm 2 (Computng bounds for an nteger Cournot equlbrum) Intalze ν = 0 and δ = 1. For j = 1, 2,..., n, ν ν + ebd [j] b+2d [j], δ δ + b b+2d [j]. 6
If ν ebd [j+1], set ˇQ = ν and BREAK. δ b δ Next j. Intalze ν = 0 and δ = 1. For j = 1, 2,..., n, Next j. ν ν + ebd+ {j} b+2d {j}, δ δ + b b+2d {j}. If ν ebd+ {j+1}, set ˆQ = ν and STOP. δ b δ Ths algorthm requres some justfcaton. Consder frst the lower bound ˇQ. Whle the test fals, the bound n (3) s weaker than the bounds on Q (between ebd [j+1] /b and ebd [j]/b). However, when t frst apples, t gves a bound above the lower of these numbers. But why can we gnore hgher values of j? Note that, as above, the rato ν/δ s updated as a convex combnaton of the old rato and ebd [j]/b, so the bound for the next j s ether the same or outsde the approprate range, and smlarly for hgher j s. Next consder the upper bound ˆQ. Whle the test fals, the bound n (4) contradcts the bounds on Q, so the current range s mpossble. When the test holds, the bound les wthn the approprate range (usng agan the convex combnaton property), so provdes a vald upper bound f Q s n ths range. For subsequent j s, the range for Q s lower, thus showng that ˆQ s a vald upper bound for all ranges. Hence Algorthm 2 provdes correct bounds. We now provde a smple bnary search to obtan an nteger Cournot equlbrum based on Theorems 1 and 2. Algorthm 3 (Computng an nteger Cournot equlbrum) Use Algorthm 2 to fnd ˇQ and ˆQ. Set l := ˇQ, u := ˆQ. Whle l u > 0, Set Q = (l + u)/2. If P ˇq(Q) > Q, l := Q + 1; Elsef P ˆq(Q) < Q, u := Q 1; Else l := Q and BREAK. Endwhle Q := l, set q between ˇq (Q) and ˆq (Q) so that P q = Q. Theorem 3 Algorthm 3 correctly obtans an nteger Cournot equlbrum n O(n log(na/b)) tme. Proof: By Theorem 1, there exsts an nteger Cournot equlbrum q, and wth Q := P q, we then have P ˇq(Q) Q P ˆq(Q). Snce these two sums are nonncreasng functons of Q, t follows that the bnary search wll successfully fnd such a Q gven that the ntal bounds are vald. But ths was establshed below Algorthm 2. For the tme complexty, we requre O(n log n) tme to order the e b d s and e + b + d s, and then O(n) tme for Algorthm 2. Snce 0 ˇQ ˆQ a/b, we need O(log(a/b)) steps n the bnary search, each requrng O(n) tme. 7
Before presentng examples, we show drectly usng a method smlar to Algorthm 3 that an nteger Cournot equlbrum exsts. For ths, we show that teratvely tryng Q = 0, 1,... wll gve a value for whch the necessary and suffcent condtons of Theorem 2 are satsfed. Theorem 4 For some nonnegatve nteger Q, X ˇq (Q) Q X ˆq (Q), mplyng that an nteger Cournot equlbrum exsts. Proof: Note frst that both ˇq (Q) and ˆq (Q) are nonncreasng n Q, and also that there s always an nteger between them. Ths s trval f ˇq (Q) = 0, and f not, ˆq (Q) = ˇq (Q) + 2(b + d )/() > ˇq (Q) + 1, from whch the clam follows. Next, for Q = 0, P ˇq(Q) s ether zero, n whch case ths Q satsfes our nequaltes, or postve. In the latter case, we know that P ˇq(Q) s zero for Q a/b. Hence there s a frst Q for whch X ˇq (Q) Q, (5) so that X ˇq (Q 1) > Q 1. Now for any, f ˇq (Q 1) s postve, ˇq (Q 1) + 1 = e b d b(q 1) + 1 = e + b + d bq = ˆq (Q), and so ˇq (Q 1) q (Q 1)+1 = ˆq (Q), whence ˇq (Q 1) ˆq (Q). On the other hand, f ˇq (Q 1) s zero, then ths nequalty holds trvally. Hence X ˆq (Q) X ˇq (Q 1) Q. (6) But now (5) and (6) mply that the nequaltes n the theorem are satsfed, and hence an nteger Cournot equlbrum q exsts. Indeed, we can start wth q = ˆq (Q) for all and then ncrease q s wthn ther range untl a sum of Q s acheved. Example 1 Let us llustrate our algorthms wth a small example. Suppose a = 5, b = 1, and there are ten producers wth costs gven by Then c = (0, 0, 0, 1, 1, 1, 1, 2, 2, 2), d = (1, 1, 1, 1, 1, 1, 1, 0, 0, 0). e = (5, 5, 5, 4, 4, 4, 4, 3, 3, 3). Note that these are n nonncreasng order. Also, () = (3, 3, 3, 3, 3, 3, 3, 1, 1, 1). 8
For the real case, Algorthm 1 ncreases j to 7, and then ν = 31/3 and δ = 10/3, so ν/δ = 31/10 > e 8, so the loop ends. Thus Q = 31/10, and we have found the real Cournot equlbrum q = q (Q) = ( 19 30, 19 30, 19 30, 3 10, 3 10, 3 10, 3, 0, 0, 0). 10 It s easy to check that ths s ndeed an equlbrum, ether from Theorem 2 or drectly from (1). In the nteger case, we frst compute and (e b d ) = (3, 3, 3, 2, 2, 2, 2, 2, 2, 2) (e + b + d ) = (7, 7, 7, 6, 6, 6, 6, 4, 4, 4). Note that these are both n nonncreasng order. We frst apply Algorthm 2. For the lower bound, j ncreases to 10, and then ν = 35/3 and δ = 19/3, so ˇQ = 35/19 = 2. For the upper bound, j ncreases to 7, wth ν = 45/3 and δ = 10/3, so ν/δ = 9/2 > ebd + 8. So the loop ends, and ˆQ = 9/2 = 4. Now we do bnary search. The frst tral Q s 3. We fnd ˇq(Q) = (0, 0, 0, 0, 0, 0, 0, 0, 0, 0), ˆq(Q) = ( 4 3, 4 3, 4, 1, 1, 1, 1, 1, 1, 1). 3 So the lower bounds are all zero, and the upper bounds all 1. It follows that any (0, 1) vector wth three ones s an nteger Cournot equlbrum. In partcular, (0, 0, 0, 0, 0, 0, 0, 1, 1, 1) s an nteger equlbrum, whereas the last three producers always produce zero n the (unque) real Cournot equlbrum. Agan, t s easy to confrm that ths s ndeed an nteger Cournot equlbrum, ether from Theorem 2 or drectly from (2). We could also try Q equal to 2 or 4, but n both cases the nequaltes n Theorem 4 fal, so these do not lead to equlbra. To show the value of obtanng bounds n Algorthm 2, note that f we keep the data above except that we change a to 50, then the algorthm gves bounds of 40 and 42, so that only at most two values of Q need to be checked (and 41 and 42 both turn out to yeld equlbra, whereas 40 does not); by contrast, a bnary search on the range [0,50] would requre up to sx tests. A last pont llustrated by ths example s the followng: snce the payoff functons are strctly concave quadratcs, one mght expect that the output chosen by a producer n a Cournot equlbrum would ether be a unque nteger, or perhaps one of two consecutve ntegers. However, n ths example, for Q = 41, we obtan ˇq(Q) = (3, 3, 3, 2, 2, 2, 2, 6, 6, 6), ˆq(Q) = (3, 3, 3, 3, 3, 3, 3, 8, 8, 8). Hence there are equlbra where the last three producers choose ether 6, or 7, or 8, for example for producer 8, q = (3, 3, 3, 3, 3, 3, 3, 6, 8, 6), or q = (3, 3, 3, 3, 3, 3, 3, 7, 7, 6), or q = (3, 3, 3, 3, 3, 3, 3, 8, 6, 6). 9
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