Optimizing Product Launchs in th Prsnc of Stratgic Consumrs Appndix Ilan Lobl Jigar Patl Gustavo Vulcano Jiawi Zhang Lonard N. Strn School of Businss, Nw York Univrsity, 44 Wst Fourth St., Nw York, NY 1001 {ilobl, jigar, gvulcano, jzhang}@strn.nyu.du A1. Tchnical proofs Proof of Lmma 1 Th lowst thrshold z such that th customr is still willing to purchas vry product launchd is, by th on-stag dviation principl, th on that maks th customr indi rnt to buying at a givn product launch. Sinc th gam is stationary, w can vrify whthr th customr is indi rnt to purchasing th first launch assuming h will buy at all futur launchs. From Eq. (1), if customr c is indi rnt to th first purchas thn from Eq. (1): E[q c 1( appl c 1 appl c ) v p appl c 1 appl c 1 ]=0. By assumption, th tchnology at this launch is q1 c = z. W can thus simplify th quation abov to z(1 E[ X(z) ]) = p v, whr X(z) =appl c appl c 1 and is th random variabl that rprsnts th tim it taks for a Brownian motion of drift µ and varianc to hit lvl z. Th momnt gnrating function E[ X(z) ] of th hitting tim X(z) of a Brownian motion is wll known to b z, whr is givn in Eq. (4).g., s Karatzas and Shrv []. Th uniqunss of th solution is guarantd by th fact that th function f(z) =z(1 z )is strictly incrasing, with lim z!0 f(z) = 0, and lim z!1 f(z) =1. Proof of Thorm 1 W first argu thr xists an MPE whr th firm rlass a nw product whnvr th gap btwn th tchnology on hand and th tchnology in th markt is gratr than or qual to z, i.., Z(t) w(t) z, and all (or almost all) consumrs buy at all rlass whn th tchnology gap is at last z. By Lmma 1, consumrs do not hav a profitabl dviation. W now show th firm nithr has a profitabl dviation. Lt U dnot th firm s quilibrium xpctd utility. Considr a stat of th gam at tim t whr Z(t) w(t) =z 0 for som z 0 z. By launching, th firm would arn a continuation payo of t (p c K + U ), whr U coms from th fact that consumrs would buy th product anticipating th nxt launch to occur aftr a nw tchnology gain of z, lading th gam back to its starting stat M(t) =M(0) = (0, (0, 1)). By dlaying a launch to t 0 >t, th continuation payo would b rducd to t 0 (p c K +U ), thus liminating this dviation. Now, considr th stat M(t) =(z 0, (0, 1)) for som z 0 <z. If th firm wr to dviat and launch a nw product, consumrs would anticipat that anothr launch would occur A1
whn hitting tchnology lvl z and so would choos not to purchas. Thrfor, this dviation is also not profitabl. W now argu that no othr path is possibl in an MPE. For any Markovian policy of th firm s f, considr th valu m(s f )=inf{m: s f (m, (0, 1)) = 1} th lowst valu of th di rnc Z(t) w(t) at which th firm would launch a nw product whn all consumrs own th latst tchnology in th markt w(t). If m(s f ) >z, th firm would hav an incntiv to dviat whnvr Z(t) w(t) = z +m(s f ), as th continuation valu of th dviation at tim t would b t (p c K + U ), whil th continuation valu of th original stratgy would b t 0 (p c K +U )whrt 0 >tis th tim th tchnology gap would rach m(s f ). Now, considr th cas whr m(s f ) <z. Anticipating that anothr product would b rlasd whn hitting tchnology lvl z, consumrs would choos not to purchas. Thrfor, in all MPE m(s f )=z. Th only rmaining possibl dviation is for som of th consumrs not to purchas at a givn launch. This would sm plausibl by Lmma 1, as consumrs ar indi rnt on making any on purchas whn products ar rlasd vry tim th tchnology gain rachs z. Howvr, if that wr to happn, th firm would dviat and dlay th product launch until raching a tchnology lvl z +, with > 0. Th marginal cost du to th dlay associatd with this stratgy is continuous in (through th hitting tim distribution of a Brownian motion with drift), but th gain is discontinuous in sinc th consumrs would no longr b indi rnt btwn buying and not buying, and would all purchas immdiatly. Thrfor, this profitabl dviation ruls out MPEs that involv asymmtric consumr bhavior, xcpt among a st of consumrs of masur zro. Proof of Lmma W first argu that w can focus on symmtric consumr quilibria without loss of optimality. If thr ar multipl di rnt stratgis mployd by th consumrs in quilibrium, thy must b indi rnt btwn thm sinc thy ar homognous in thir valuations. W assumd th consumrs adopt at most finitly many di rnt stratgis, so lt s c 1,s c,...,s c N b consumr stratgis adoptd by rspctiv fractions 1,,..., N of th consumr markt. For any firm policy z, th firm s profit is a wightd avrag of th profit from ach stratgy cohort, i.., U f (z, (s c 1,...,s c N )) = NX i U f (z,s c i ). By slcting th cohort i with th highst U f (z,s c i ), w find a symmtric quilibrium with a (wakly) highr profit for th firm than undr th original asymmtric quilibrium. W now argu that in a symmtric quilibrium consumrs do not nd to dlay purchass, thus buying as soon as products ar rlasd. Considr a firm policy z and a consumr policy s c whr th consumrs dlay purchass with positiv probability. Lt i N b th first launch whr th st of ralizations! that lad to a purchas has positiv masur, but lss than 1. Th problm facd by th consumrs at launch i is indpndnt of th ralization! sinc thy all purchasd at launch i 1, buying tchnology z i 1 and th tchnology is now xactly at z i. Thrfor, without loss of utility, th consumrs can ignor th ralization! and ithr buy at launch i with probability 0 or with probability 1. If consumrs do not buy at launch i undr any ralization!, th firm should dlay this introduction. Both th firm and th consumr would b bttr o if th firm dlayd th launch A
until a tim whn all consumrs would buy. Thus, th st of purchass is idntical to th st of launchs at optimality, and consumrs nd only considr dviations to buying at a subst of th launchs whn considring thir bst rsponss. Proof of Lmma 3 It is su cint to show that U c (z, q) is submodular pathwis. Considr tchnology lvl z i z with corrsponding launch tim i and any q z\{z i }. Lt k 0 b th tchnology lvl of th product ownd by th consumr at tim i and k 00 b th tchnology lvl of th nxt product purchasd aftr tim i,withappl 00 bing th purchasing tim of tchnology k 00. Th di rnc in consumr utility btwn policis q and q [{z i } is U c (z, q [{z i }) U c (z, q) = v (z i k 0 )( i appl 00 ) p i, with z i >k 0. Not that th di rnc in utilitis is dcrasing in k 0 and incrasing in k 00.Thus,if w considr two di rnt consumr policis, q q, w obtain that U c (z, q [{z i }) U c (z, q) U c (z, q [{z i }) U c (z, q), which is a charactrization of a submodular function. Proof of Proposition 1 Th purchasing policy z bing a bst rspons immdiatly implis that E[U c (z, z)] E[U c (z, z\{z i })] for all i N. Considr a consumr policy q z and rprsnt th lmnts in th st z\q by {l 1,l,...}. Th consumr s xpctd utility from purchasing according to z is qual to th utility from purchasing according to q plus th di rncs in utility from adding ach lmnt in z\q, i.., E[U c (z, z)] = E 4U c (z, q)+ z\q X (U c (z, q [{l 1,...,l i }) U c (z, q [{l 1,...,l i 1 })) 5. Sinc q [{l 1,...,l i 1 } z, using th submodularity rsult from Lmma 3 on ach of th trms insid th summation abov, w obtain that z\q X E[U c (z, z)] E[U c (z, q)] + E[(U c (z, z) U c (z, z\{l i }))]. Thrfor, if E[U c (z, z) U c (z, z\{l j })] 0 for all i z\q, thne[u c (z, z)] E[U c (z, q)], which charactrizs th purchasing policy z as th bst rspons to th launch policy z. Proof of Thorm To prov this thorm, w rlax th firm s optimization problm and xplicitly find an optimal solution for this rlaxd countrpart. W thn show that th solution gnratd is fasibl for th original problm and thus, optimal. W rlax th problm OPT-3 by ignoring all constraints whr i is an vn numbr. That is, w considr th rlaxd problm r i [0,1] 1X P i j=1 r j s.t. f(r i,r i+1 ) 0 for all i =1, 3, 5,... 3 (RELAX-1) A3
whr w dfin f(x, y) = v x(1 y ) p. This rlaxation is tractabl bcaus th constraints ar disconnctd in th following sns: ach dcision variabl r i appars in xactly on constraint. Not that th objctiv is dcrasing in all r i s and, for any i, th function f(r i,r i+1 ) is incrasing in both r i and r i+1. Thrfor, all th constraints in th rlaxd problm must b binding sinc othrwis w could dcras th valu of r i and incras th firm s profit without violating any constraint. Thrfor, problm RELAX-1 is quivalnt to r i [0,1] 1X P i j=1 r j s.t. f(r i,r i+1 ) = 0 for all i =1, 3, 5,... (A1) Dfin th constant = p v. Th constraint f(,r ) = 0 holds and th dcision variabls and r ar non-ngativ rals whnvr and r =1 W can us this rlationship to liminat th variabl r from (A1), obtaining r i [0,1] fori 3 + 1. s.t. f(r i,r i+1 ) = 0 for all i =3, 5, 7,... 1. Considr now th optimization prob- Lt K 1 ( )= and K ( )= lm holding constant som >,i.., r i [0,1] fori 3 K 1 ( )+K ( ) X 1 i=3 P i j=3 r j (A) 1X i=3 P i j=3 r j s.t. f(r i,r i+1 ) = 0 for all i =3, 5, 7,... Th st of optimal solutions of this problm is indpndnt of and is idntical to th st of optimal solutions of (A1) sinc th two problms ar idntical xcpt for a linar scaling of th objctiv function. In ordr to charactriz th optimal solution, thr ar two cass to considr basd on th rlationship btwn th dcision variabl and. As long as >, any valu that optimizs must also optimiz r 3. By induction, th sam valus must also optimiz r 5, r 7,... Equivalntly, any valu that optimizs r also optimizs r 4, r 6,... In cas = mattr.,thnr = 1 and K ( ) = 0, and thus non of th othr dcision variabls A4
Lt ˆ b th optimal valu of (A1). Not that ˆ is finit: W hav constraints f(r i,r i+1 ) = 0, for i 1, i.. r i = 1 xp( ri+1 ). This implis that r i, hnc trivially, r i + r i+1. W us ths two bounds in th objctiv function so that for i vn, P i j=1 r i (i/) ; and for i odd, P i j=1 r i ((i 1)/) + =((i + 1)/). This yilds that th objctiv function is boundd abov by 1X xp ( i )= 1 xp ( ) < 1, which complts th argumnt. From (A), ˆ must mt th ncssary optimality condition appl r ˆ = 1 + 1 or quivalntly, ˆ = appl (ˆ + ) Th function in th RHS to b imizd is of th form appl g(x) = M x 1 c 1 x, ˆ, (ˆ + 1). (A3) whr c 1 = ˆ + (ˆ +1). Th function g( ) is unimodal in R + for any positiv c 1, as g 0 (x) = x ( c x 1 x + x + 1) has a uniqu positiv root in R +. Thrfor, thr is a uniqu that satisfis (A3) for th optimal valu ˆ. This obsrvation implis that an optimal solution for (A1) ncssarily satisfis = r 3 = r 5 =...= r i+1, and r = r 4 = r 6 =...= r i, for all i; which lads to th following problm quivalnt to (A1), but dfind ovr just two variabls:,r 0 s.t.: 1X f(,r )=0. i Using th gomtric progrssion formula, w gt:,r 0 s.t.: f(,r )=0. (+r ) + X 1 i (+r ) i=0 + ( +r ) 1 ( +r ) (A4) Using th xprssion f(,r ) = 0, w rplac xp( r )by1 / in th objctiv function and drop th constraint. This rducs th problm to th following formulation: r1 r1 + 1 r1. 1 1 A5
Multiplying abov and blow by, and thn adding and subtracting in th numrator, w gt r1 + (r1 ) From th function w wish to imiz, w dfin x x + x r(x) = x x (x ). Lmma A1. Th function r : R! R has a uniqu imizr x. In particular, x = {, }, whr is th uniqu unconstraind imizr ovr th xtndd domain R +. Proof: W hav r 0 (x) = x x + (1 + x (1 + x)) [ +( x 1)x] =0, (A5) whr = p v. Th function r( ) dfind ovr th xtndd domain R + is incrasing in th intrval (0, ) and dcrasing in (, 1), whr is th uniqu positiv solution to r 0 (x) = 0. Thus, x M = arg x r(x) = {, }. In trms of our original problm, w gt ˆ = {, }. Not that th xprssion for r( ) is also continuous in and. In particular, whn =, condition (A5) rducs to r 0 ( ) = 0, i.., Using th Lambrt W function, dfind by th quation 1. =1+ (1 + ). (A6) z = W (z)xp(w (z)), whr z is any complx numbr, w can rwrit quation (A6) as = W 1 +1 1.7846. M Th uniqu solution to quation (A6) is =. Rcalling that = p /v, w hav for p/v = /( ), ˆ = =. Th rlation btwn p/v and problm paramtrs splits th plan in two parts: p /v if p/v /( ), ˆ = dtrmind by (A5) othrwis. Th optimal ˆr will b dtrmind from f(ˆ, ˆr ) = 0, i.., from th quation ˆ [1 ˆr ]=. (A7) In cas p/v /( ), ˆr = 1. Othrwis, ˆ > and ˆr < 1, and th optimal policy is to o r products in cycls altrnating btwn ˆ and ˆr. W ar lft with proving that our solution to th rlaxd problm (RELAX-1) is also fasibl for th original problm (OPT-3). Sinc w droppd vn numbrd constraints and w hav ˆ = r i+1 and ˆr = r i for all i; w ar just lft with showing that f(ˆr, ˆ ) 0. In fact, if w could prov that ˆ < ˆr, thn from Eq. (A7), w could us th following squnc of quivalncs to argu that f(ˆr, ˆ ) 0: ˆ < ˆr, 1 ˆ ˆr 1 as g 1 (x) = M 1 x ˆ ˆr x ˆr, ˆr [1 1 ˆr ] ˆ [1 ], ˆr [1 ˆ ], f(ˆr, ˆ ) 0, by dfinition of f. A6 is dcrasing in x
In what follows, w show that indd ˆ < ˆr holds. Not that thr xists a uniqu valu z such that f(z,z ) = 0, i.. z z (1 )=. (A8) Sinc f is strictly incrasing in both argumnts and f(ˆ, ˆr ) = 0, w can s that if ˆ 6=ˆr 6= z, thn ˆ and ˆr li on di rnt sids with rspct to z. Formally, if ˆ 6=ˆr 6= z, thn ithr ˆ <z < ˆr or ˆr <z < ˆ. W prov that th formr is always tru. Obsrv that th point (z,z ) is in th fasibl rgion of problm (A4). Sinc th objctiv function r is unimodal with unconstraind imizr, it is nough to chck r 0 (z ) < 0, to show that <z. Th drivativ of r( ) is givn by r 0 (x) = x x + (1 + x (1 + x)) [ +( x 1)x] It is indd nough to show that th numrator x x + + x + x x valuatd at z is ngativ. So, w valuat: z (z ) + + z + z z = z (z ) + z z [1 ]+z z [1 ] z + (z ) z [1 ] z (by rplacing using (A8)) = (z ) z +( (z ) + z ) z z [1 ]+z [1 = z (1 + z )[1 z z ] = z (1 + z )[1 z [1 z + ( z ) = z (1 + z )[ ( z )! + ( z ) 3 3!! ] < 0 ( z ) 3 ]] 3! z ] W can also chck that <z.tosthis,dfinh(x) =x(1 x ), whr from (A8), h(z )=. It can b vrifid that h(x) is incrasing, and sinc h( ) <, w can assrt that < z. To M conclud, ˆ = { r1, } <z, so that ˆ < ˆr, and th two cycl policy (ˆ, ˆr ) is fasibl and optimal for th original problm (OPT-3). Proof of Proposition In this proof, w show that th firm s xpctd utility is unimodal in z,whichimplisthdsird rsult givn that th pric p is incrasing in z (s Eq. (8)). W bgin from Eq. (9) and prform a chang of variabl, introducing x = z for x (0, 1), and rwriting th firm s utility as E[U f (x, x)] = v log x(1 x) x c K 1 x = v x x log x (c + K) 1 x. h i b Th abov function is of th form g(x) =x a log x 1 x whr a and b ar positiv constants. Th abov function is unimodal as follows. Taking th drivativ with rspct to x, w gt appl g 0 (x) = a log x appl b 1 x + x A7 a x b (1 x)
W hav lim x!0 + g 0 (x) =1 and lim x!1 g 0 (x) = 1. Also, it is asy to s g 0 (x) isstrictly dcrasing function; hnc thr is a uniqu point that satisfis th first ordr condition that is a imum of g( ) function. Proof of Thorm 3 Lt b th uniqu launch tchnology lvl (or th optimal short cycl), and r b th corrsponding optimal long cycl drivd in Thorm for a givn pric p. First, w will argu that th joint launching and pricing optimization problm, which is in principl a two-dimnsional sarch on p and, rducs to a singl dimnsional sarch (rcall that r is uniquly dtrmind by through Eq. (A7)). In fact, th rlation btwn p and is givn by th following xprssion: dtrmind by (A5) if p/v < /( ), (p) = (A9) p /v othrwis, whr in Eq. (A5) is givn by = p/v. Rcall that : R +! R + is continuous in p (c.f., Lmma A1 and thraftr in th proof of Thorm ). It is also strictly incrasing in p, for p>0; which implis that thr is on-to-on corrspondnc btwn p and. In addition, from (A5), lim p!0 (p) = 0 (s Figur 4(right) for an illustration) 1. Rcall that is th uniqu solution to th quation =1, or quivalntly, 1+ =, (A10) 1+ (1 + ) giving 1.7846. Lt us also dfin th function L(x), to b xtnsivly usd in th following analysis: x x L(x) = 1+ x, with L( )=1. (A11) (1 + x) Thus, using quation (A9), w can dfin p( ), i.., th invrs function of (p), as follows: p( )= ( v (r1 ) (1+ (1+ )) v/ or quivalntly, by using th L function in (A11), p( )= r1 vl( ) v if 0 appl appl /, othrwis, if 0 appl appl /, othrwis. (A1) Not that p( ) is wll dfind in th domain R + and strictly incrasing. Hnc, in ordr to prov that th firm s utility is unimodal in p conditionally on th firm using th optimal launch policy (p) in (A9), it is su cint to prov that th firm s utility is unimodal in. In fact, th firm s xpctd utility as a function of, whn using th optimal pric p( ) in (A1), is givn by: appl E[Up f + ( )] = (p( ) c K) (r1 1 p( ) /v) appl =(p( ) c K) + (r1 p( ) /v) 1 For sak of th argumnt in this analysis, w ar xtnding th rang of p. Rcallthatbyassumption,p>c+ K. Howvr, in ordr to simplify notation and w.l.o.g., w assum in this proof that p>0, vn though it could lad to a firm s ngativ utility. 1. A8
M Lt K 1 = c + K b th ctiv cost pr launch. Th firm s xpctd utility function bcoms appl E[Up f + ( )] = (p( ) K 1 ) (r1 1. p( ) /v) Substituting p( ) by its xprssion in (A1), w gt: ( h i r1 E[Up f vl( ) K 1 +1 ( )] = 1 1, if 0 appl r 1+L( ) 1 appl v K 1, othrwis. /, Not that, E[U f p ( )] is continuous; it is clarly picwis continuous and at = /, E[U f p ( / )] = v K 1 for both parts, as w know from (A11) that L( ) = 1. Dfin sparat functions for ach part in th dfinition of E[Up f ( )]: appl r1 vl( ) +1 M( )= K 1 1+L( ) 1, (A13) and r1 v S( )= K 1, (A14) so that E[Up f M(r1 ) if 0 appl r ( )] = 1 appl /, S( ) othrwis. Nxt, w show that E[Up f ( )] is unimodal in in thr stps (s Figur A1): 1. Show that S( ) is a unimodal function of.. Show that M( ) is a unimodal function of. 3. Show that th drivativs of S( ) and M( ) at th split valu = / hav th sam sign. 700 0,000 600 18,000 16,000 500 14,000 Firm s xpctd utility 400 300 Firm s xpctd utility 1,000 10,000 8,000 00 6,000 4,000 100,000 0 λ/δ 4 6 8 10 1 0 4 6 λ/ 8 10 1 14 16 18 t 1 r1 Figur A1: Unimodality of U f p ( ). Each part of th function U f p ( ) is unimodal, and thy cross ithr whn thy ar both incrasing or dcrasing, which warrants th global unimodality. Lft: Paramtrs ar =0.9,K = 1500,v = 000,c= 0,µ= 1, = 0. Right: Paramtrs ar = 0.,K = 500,v = 000,c= 0,µ= 1, = 0. A9
Stp 1: From (A14), w hav S 0 ( )= ( K 1 v + v). (A15) Obsrv that S 0 (0) > 0 and that lim r1!1 S 0 ( )= 1. Also, S 0 ( ) has only on root at = K 1 /v+1/. So, S( ) is incrasing ovr 0 appl appl K 1 /v+1/, and dcrasing ovr K 1 /v+1/ appl < 1. Hnc, S( ) is unimodal function. Stp : From (A13), w hav whr and Q 1 ( ) M = M 0 ( )=Q 1 ( )Q ( ), ( ) 3 ( + ) ( ( r1 + 1) 1) ( ( r1 + 1) + 1) Q ( ) M = 3 K 1 + K 1 +1 + K 1 r1 +1 ( ) + v 1 r1 3 v (A16) It can b asily vrifid that Q 1 ( ) is positiv in (0, 1). It can also b chckd that Q ( )is strictly dcrasing with: lim Q ( )=1 and lim Q ( )= 1.!0!1 Thus, Q ( ) has a singl root that w dnot. Givn th dfinition in (A16), M( ) turns out to b strictly incrasing in 0 < appl and strictly dcrasing in < < 1. Hnc, it is unimodal in. Bfor going ovr Stp 3, w nd th following auxiliary rsult: Claim 1. L 0 ( )= / Proof: From (A11), w hav Hnc, L 0 ( )= ( + + 1) ( ( r1 + 1) + 1). L 0 ( )= ( + + 1) ( ( + 1) + 1). (A17) Using Eq. (A10) and substituting into th numrator of (A17), w gt L 0 ( )= ( + ) ( ( + 1) + 1). (A18) Substituting again Eq. (A10), now into th numrator of (A18), w hav appl L 0 ( )= ( + 1) + 1 = 1=, A10
and th claim is provd Stp 3: W want to prov that both S 0 ( / ) and M 0 ( / ) hav th sam sign. From (A15), w hav: From (A13), w can writ th drivativ of M( ) as: M 0 ( )= v L( )+ L 0 ( ) r1 vl( ) S 0 ( / ) = (K 1 v + v). (A19) appl +1 1+L( ) 1 + K 1 appl ( 1+L( )) ( + 1)( + L 0 ( )) ( 1+L( )). So, M 0 ( / ) = v appl L( )+ L 0 +1 ( )/ 1 + 1+L( ) appl vl( ) ( 1+L( )) ( + 1)( + L 0 ( )) K 1 ( 1+L( )). Substituting L( ) = 1 and L 0 ( )= / in th quation abov, w gt: M 0 ( / ) = v appl " # v ( + 1)( + 1+ + K ) 1 = v appl v K1 h i 1+ + 1 (1 + )(1 + /) appl v + = v appl v K1 (3 + ) From Eq. (A10), using that 1 + =, and substituting abov, w gt: appl v + M 0 ( / ) = = = v appl v K1 ( + ) [v +K 1 v + (K 1 v)] ( + ) [K 1 v + v] (A0) W can s from (A19) and (A0) that both S 0 ( / ) and M 0 ( / ) hav th sam sign as K 1 v + v. This complts th proof of Stp 3, and of Thorm 3. A11