E-Compaio: Stakeholder Perspectives o E-waste Take-Back Legislatio Atalay Atasu atalayatasu@mgtgatechedu College of Maagemet, Georgia Istitute of Techology 800, West Peachtree Street, Atlata, GA Özur Özdemir, oozdemir@rsml RSM, Erasmus Uiversity Rotterdam, Postbus 1738, Rotterdam, Netherlads Luk N Va Wassehove, lukva-wassehove@iseadedu INSEAD, Boulevard de Costace, 77300 Fotaiebleau, Frace Jauary 1, 01 1 Tax or Rate? We cosider a market for a sigle product, with three sets of decisio makers: cosumers, maufacturers ad a social plaer The decisio makig process occurs sequetially First, the social plaer decides o the policy implemetatio model ad parameters The the maufacturers determie prices/sales quatities, give the costs they icur uder the selected policy parameters Fially, give the maufacturers decisios, cosumers make the purchasig decisio assume that cosumer utilities (υ) from purchasig a product are uiformly distributed betwee 0 ad 1 ad market size is ormalized to 1 Hece, a cosumer with utility υ j purchases the product if υ j p, the price of the product We assume idetical maufacturers competig i a Courot settig Maufacturers select their sales quatities (q i ), which determie the price i the market, ie, p =1 i=1 q i The social plaer makes a choice betwee the two differet models (see Figure EC1): We 1
Tax Model (T ): I this model, the social plaer udertakes the take-back task, ie, collectio ad recyclig, to divert a fractio c of products from ladfill, ad charges a recovery fee (tax) τ to the maufacturers Take-back Rate Model (R): I this model, the social plaer determies a take-back rate (c) ad requires the maufacturers to esure this rate SOCIAL PLANNER MANUFACTURER qc SOCIAL PLANNER c MANUFACTURER q CONSUMERS q CONSUMERS qc TAX MODEL RATE MODEL Figure EC1: Take-back Legislatio Models The maufacturers uit cost of productio is μ The take-back cost icurred by the maufacturers depeds o the policy tool The party resposible for product take-back (ie, the operatig party) icurs a uit take-back cost of χ>0 We assume that χ+μ <1 so that the maufacturers ca operate at a positive profit margi Give these costs ad k (T,R) deotig the policy implemetatio model of the social plaer, maufacturer i maximizes his profit max q i Π Mi (k) =q i (p μ) q i ρ k (τ,c) where, τ, if k = T ; ρ k (τ,c) = cχ, if k = R (1) is the effective per uit cost of take-back to the maufacturer From the objective fuctio of maufacturer i, it is easy to show that the equilibrium is at qi = 1 μ ρ k(τ,c),aditur, p = 1+(μ+ρ k(τ,c)) By aticipatig the maufacturers choices, the social plaer ca focus o its ow objective Sice a stadard approach to determiig the appropriateess of alterative
forms of regulatio is to focus o a first-best socially optimal solutio, we assume the social plaer maximizes the social welfare, which cosists of maufacturer profits, cosumer surplus (followig the traditioal defiitio i ecoomics, eg, Barett 1980, Fishma ad Rob 000, ad Plambeck ad Wag 009 to ame a few) ad other exteralities I particular, we cosider the followig: Total Maufacturer Profit: Π M (k) = i=1 Π M i (k) =(q i (p μ ρ k (τ,c)) = ( 1 μ ρ k(τ,c) ) Total Cosumer Surplus: Π C (k) =(1 p )q i / = ((1 μ ρ k(τ,c))) () Note that Π C (k) = Π M (k) Evirometal Impact: We focus o the level of ladfill aversio as a evirometal objective, which is oe of the mai goals of the curret practice of product take-back legislatio, alog with providig recyclable product desig icetives As our focus is o ladfill diversio, we assume there exists a uit evirometal cost associated with each ladfilled product, deoted by ɛ We assume that the total evirometal impact of ot takig back (1 c) proportio of total sales is Π E (k) = ɛ(1 c)qi = ɛ(1 c) 1 μ ρ k(τ,c) Depedig o the particular policy model chose, the social plaer will either collect taxes from the maufacturers ad take resposibility of take-back herself or impose a take-back rate o the maufacturers Hece, give the policy model (k), the social plaer s objective icludes the followig term; qi (τ cχ), if k=t; φ k (τ,c)= 0, if k=r () The total social welfare is the writte as; SW(k) =Π M (k)+π C (k)+π E (k)+φ k (τ,c) (3) The social plaer maximizes SW(k) by decidig o the tax (τ) ad the take-back rate (c) i the tax model ad oly the take-back rate i the rate model A summary of otatio is give i Table EC1 3
Parameters μ χ ɛ k Uit productio cost Uit take-back cost Uit evirometal cost Number of maufacturers Policy model, k {T,R} Decisio variables c τ q i q p Take-back rate Uit recovery fee (tax) Sales quatity of maufacturer i Total sales quatity Uit price Objectives Π M Π C Π E SW Total maufacturig profit Cosumer surplus Evirometal impact Social welfare Table EC1: Notatio Impact o Stakeholders 1 Social Plaer Perspective 11 Rate Model I this case, the objective fuctio of maufacturer i is Sice max q i Π Mi = (p μ)q i cχq i (4) where p = 1 q i (5) i=1 Π Mi q i =1 ( +1)q i μ cχ ad Π Mi q i = ( +1) the maufacturer i s objective fuctio is cocave i q i By the first order coditios, the optimal sales quatity for each maufacturer (qi ) ad the optimal price (p )aregiveas 1 μ cχ ad 4
1+(cχ+μ), respectively Give this, Π M = Π Mi = i=1 (cχ + μ 1) ( +1) Π C = (cχ + μ 1) ( +1) ɛ(c 1)(cχ + μ 1) Π E = +1 Propositio 1 (Rate Model) Table EC summarizes the optimal decisio of the social plaer for the take-back rate model The optimal take-back rate icreases i the degree of the competitio () ad the evirometal cost (ɛ) 0 <ɛ (1 μ)χ ) (1 μ)χ c 0 q (1 μ) ) <ɛ<χ( + ) ɛ χ( + ɛ(μ 1)() χ(ɛ()+(μ 1)(+)) χ(χ(+) ɛ()) 1 ɛ(χ+μ 1) χ(+) ɛ() ) ( χ μ+1) (μ 1) Π ɛ (χ+μ 1) (χ+μ 1) M () (χ(+) ɛ()) () Π E ɛ(1 μ) ɛ (χ+μ 1) (ɛ+ɛ χ(+)) χ(χ(+) ɛ()) 0 (μ 1) Π ɛ (χ+μ 1) (χ+μ 1) C () (χ(+) ɛ()) () SW (μ 1)(ɛ()+(μ 1)(+)) () ɛ (χ+μ 1) χ(χ(+) ɛ()) Table EC: The optimal solutio for the rate model (χ+μ 1) (+) () Proof Takig the sum of Π M,Π C ad Π E, the social plaer s problem is writte as; max c SW = (cχ + μ 1)(ɛ( +1)+(μ 1)( +)+c(χ( +) ɛ( + 1))) ( +1) (6) st 0 c 1 (7) The first ad the secod order derivatives of the objective with respect to c are; SW c SW c = = ( c( +)χ + ((1 c)ɛ( +1)+(μ 1)( +))χ ɛ(μ 1)( +1) ) ( +1) (8) χ(χ( +) ɛ( +1)) ( +1) (9) SW c < 0, if χ( +) ɛ> ( +1) ; (10) SW c =0, if χ( +) ɛ = ( +1) ; (11) SW c > 0, if χ( +) ɛ< ( +1) (1) 5
I (10), SW is cocave i c From the first order coditios, c = ɛ(μ 1)() χ(ɛ()+(μ 1)(+)) χ(χ(+) ɛ()) if ɛ χ( + ), c 1, thus the optimal c (c )issetto1 if (1 μ)χ if χ(+) () ) <ɛ<χ( + ), 0 <c<1, thus c = ɛ(μ 1)() χ(ɛ()+(μ 1)(+)) χ(χ(+) ɛ()) <ɛ (1 μ)χ ), c 0, thus c =0 I (11), SW is liear ad decreasig i c Hece, c =0 I (1), SW is covex i c We compare the two boudary solutios: SW c=0 ad SW c=1 The root of SW c=0 SW c=1 =0isɛ = χ(χ+μ )(+) (μ 1)() SW c=0 >SW c=1 whe ɛ is smaller tha this root This root is greater tha χ(+) () (the upper boud o ɛ for covexity): Sice by assumptio χ + μ<1, χ(χ +μ )( +) (μ 1)( +1) (χ +μ ) (μ 1) ( μ χ) (1 μ) χ (1 μ) χ( +) ( +1) (13) 1 (14) 1 (15) 1 (16) χ < 1 Hece, (17) (1 μ) χ (1 μ) > 1 (18) Thus, give the covexity coditio SW c=0 is always greater tha SW c=1 ad c =0 Give these results, the optimal decisios of the social plaer i the rate model ca be summarized as i Table EC To show the secod part of the propositio, first ote that as ɛ icreases, the optimal solutio occurs at higher take-back rate itervals, this is obvious from the table The we check the chage of the iterval bouds i Table EC with These bouds decrease with the degree of competitio (): Moreover; ( (1 μ)χ )) = χ(μ 1) (χ μ +1)( +1) ( (1 μ)χ )) > )) < 0 ad (χ( + (χ( + )) χ = ( +1) < 0 This implies that as icreases the zero ad the partial take-back itervals get smaller while the perfect take-back iterval elarges Also the iterior take-back rate icreases with ad ɛ: ( ) ɛ(μ 1)() χ(ɛ()+(μ 1)(+)) χ(χ(+) ɛ()) ɛ(1 χ μ) = (χ( +) ɛ( +1)) > 0 6
( ) ɛ(μ 1)() χ(ɛ()+(μ 1)(+)) χ(χ(+) ɛ()) (1 χ μ)( +1)( +) = ɛ (χ( +) ɛ( +1)) > 0 Hece, we ca coclude that the optimal take-back rate icreases with the degree of competitio () ad the evirometal impact (ɛ) 1 Tax Model The problem of each maufacturer is max q i Π Mi = (p μ τ)q i (19) where p = 1 q i (0) Sice Π Mi =1 ( +1)q i μ τ ad Π Mi q i qi = ( +1) the maufacturer i s objective fuctio is cocave i q i By the first order coditios, the optimal sales quatity for each maufacturer qi = 1 μ τ ad the optimal price p = 1+(μ+τ) Whewe substitute these expressios i Π M,Π C,adΠ E,weget Π M = Π Mi = i=1 i=1 (μ + τ 1) ( +1) Π C = (μ + τ 1) ( +1) (c 1)ɛ(μ + τ 1) Π E = +1 Propositio (Tax Model) The optimal decisios of the social plaer uder the tax model are give i Table EC3 The optimal tax is o-decreasig i the take-back cost (χ) adthe evirometal cost (ɛ), ad icreasig i the degree of competitio () Proof The social plaer s problem is writte as, max c,τ SW = SW is cocave i τ sice (μ + τ 1)(c(χ ɛ)( +1)+ɛ( +1)+(μ 1)( +) τ) ( +1) (1) SW τ = st 0 c 1 () 0 τ 1 μ (3) (ɛ + ɛ + μ + c(χ ɛ)( +1) τ 1) ( +1) (4) SW τ = ( +1) < 0 (5) 7
0 <ɛ 1 μ ɛ χ 1 μ 1 μ <ɛ<1 μ 0 <χ ɛ>χ 1 μ <χ<1 μ c 0 0 1 1 τ 0 q (1 μ) 1 μ ɛ (μ 1) Π (1 ɛ μ) M () Π E 1+()ɛ+μ 0 (1 μ) 1+()χ+μ 1 μ χ (μ 1) (1 χ μ) () ɛ(1 μ) ɛ(1 ɛ μ) 0 0 Π (μ 1) (1 ɛ μ) C () SW (μ 1)(ɛ()+(μ 1)(+)) (1 ɛ μ) () (μ 1) (1 χ μ) () (μ 1)(χ()+(μ 1)(+)) (1 χ μ) () Table EC3: The optimal solutio for the tax model From the the first order coditios, SW τ =0 τ = SW is liear i c sice SW c = ɛ + ɛ + μ + c(χ ɛ)( +1) 1 (χ ɛ)(μ + τ 1) +1 SW c = 0 (8) Hece, the optimal c (c ) takes a boudary value (either 0 or 1) I the tax model, τ + μ 1 should hold so that the maufacturers have oegative profit τ =1 μ is a special case where there is o productio (qi =0)adSW = 0 But this solutio ca ever be optimal, hece we igore it Whe μ + τ < 1, the value of c depeds o the relatio betwee ɛ ad χ ad substitutig c i τ = ɛ+ɛ+μ+c(χ ɛ)() 1, we fid ( 0, 1+()ɛ+μ) (c, if χ ɛ;,τ)= ( 1, 1+()χ+μ) (9), if χ<ɛ Uit tax should always be oegative ad lower tha 1 μ so that there is always positive productio i the market as metioed above Hece, we eed to check whether the τ values foud above satisfy these coditios τ c=0 isgreatertha0ifɛ> 1 μ ad lower tha 1 μ if ɛ<1 μ Siceɛ χ i this case ad χ<1 μ by assumptio (see page ), ɛ<1 μ is always satisfied Hece, for c =0,optimalτ (τ )is0wheɛ 1 μ 1+()ɛ+μ,ad Similarly, τ c=1 isgreatertha0ifχ> 1 μ (6) (7) whe ɛ> 1 μ ad lower tha 1 μ whe χ<1 μ By assumptio χ<1 μ Thus,forc =1,optimalτ is 0 whe χ 1 μ 1+()χ+μ ad aresummarizeditableec3 To show the secod part of the propositio, ote that 8 whe χ> 1 μ These
i the iterval ɛ χ, uptothethresholdof 1 μ, the optimal tax is always 0, hece ot affected by ɛ After this threshold, sice τ ɛ =()/ > 0, the optimal tax is icreasig i ɛ These two observatios prove that the optimal tax is o-decreasig i the evirometal cost (ɛ) i the iterval ɛ>χ,uptothethresholdof 1 μ, the optimal tax is always 0, hece ot affected by χ After this threshold, sice τ χ =( +1)/ > 0 the optimal tax is icreasig i χ These observatios prove that the optimal tax is o-decreasig i the take-back cost (χ) the itervals where the optimal tax is 0 get smaller as icreases This is because the upper boud o these itervals ( 1 μ ) decreases as icreases Also i the itervals where the optimal tax is positive, τ icreases as icreases: ( ( 1+()ɛ)+μ ) ( ( 1+()χ)+μ ) = ɛ + μ 1 > 0siceɛ<1 μ i the associated iterval (30) = χ + μ 1 > 0siceχ<1 μ by assumptio (31) Give these, we ca coclude that τ is icreasig i the degree of competitio () Corollary 1 SW(T ) SW(R), ie, the social welfare uder the tax model is better tha or equal to that uder the rate model (see Table EC4) Furthermore, icreased competitio () reiforces the domiace of the tax model from the social plaer s perspective 0 <ɛ 1 μ SW(T )=SW(R) ɛ χ χ<ɛ< (+)(1 μ) 1 μ <ɛ<1 μ 0 <χ<1 μ SW(T ) >SW(R) Table EC4: Social plaer prefereces (+)(1 μ) ɛ Proof Cosiderig the solutio itervals of the two models, we evaluate the two models i terms of the social welfare uder three cases; whe ɛ χ, χ<ɛ< (+)(1 μ) ad (+)(1 μ) ɛ Whe ɛ χ: I the iterval of ɛ 1 μ, The optimal solutio for the tax model is c =0adτ = 0 ad the optimal social welfare is SW(T )= (μ 1)(ɛ()+(μ 1)(+)) () 9
The optimal solutio for the rate model is c = 0 ad the optimal social welfare is SW(R) = (μ 1)(ɛ()+(μ 1)(+)) () Hece, SW(T ) =SW(R) I the iterval of ɛ> 1 μ, The optimal solutio for the tax model is c =0adτ = 1+()ɛ+μ ad SW(T )= (1 ɛ μ) For the rate model there are two feasible optima: (1) whe ɛ (1 μ)χ ad SW(R) = (μ 1)(ɛ()+(μ 1)(+)),()whe (1 μ)χ () c (0, 1) ad SW(R) = ɛ (χ+μ 1) χ(χ(+) ɛ()) ), c =0 ) <ɛ<χ( + ), I case (1) we compare SW(T )= (1 ɛ μ) vs SW(R) = (μ 1)(ɛ()+(μ 1)(+)) () (1 ɛ μ) (μ 1)(ɛ( +1)+(μ 1)( +)) ( +1) (3) (ɛ + μ 1) ( +1) (μ 1)(ɛ( +1)+(μ 1)( + )) (33) ɛ ( +1) +(μ 1) ( +1) +ɛ(μ 1)( +1) (μ 1)ɛ( +1)+(μ 1) ( + ) (34) ( +1) ɛ +(μ 1) +ɛ(μ 1)( +1) 0 (35) (( +1)ɛ +(μ 1)) > 0 (36) Hece, we ca coclude that SW(T ) >SW(R) I case () we compare SW(T ) = (1 ɛ μ) vs SW(R) = ɛ (χ+μ 1) χ(χ(+) ɛ()) compariso iterval ca be writte as max( 1 μ, (1 μ)χ χ<χ( + (1 μ)χ )ad ) > 1 μ as (1 μ)χ The )) < ɛ mi(χ, χ( + )) Hece, the compariso iterval ca be summarized ) <ɛ χ IfwesolveSW(T ) SW(R) = 0, the roots are ɛ 1 = (μ 1) χ(+)(μ 1)+ (μ 1) ( 4(+)χ 4(μ 1)(+)χ+(μ 1) ) 4χ() (37) ɛ = (μ 1) χ(+)(μ 1) (μ 1) ( 4(+)χ 4(μ 1)(+)χ+(μ 1) ) 4χ() (38) ɛ 3 = χ (39) Note that ɛ 3 is already equal to the upper boud of the iterval Also ɛ < (1 μ)χ ) < χ<ɛ 1 give all the restrictios o parameters 1 Hece, oe of the roots exist i the compariso iterval This implies that SW(T ) SW(R) ever hits 0 i the give iterval ad the sig of SW(T ) SW(R) does ot chage withi the iterval Hece, to fid the sig of the differece throughout the iterval, it is sufficiet to determie the sig at a arbitrary poit Let s cosider the parameter istace, {χ =9/64,μ =1/8, =31} 1 The complexity of the expressios for ɛ 1 ad ɛ ad the iterval restrictios make the compariso betwee ɛ 1 ad ɛ tedious Hece, the required computatios for the proof have bee performed i Mathematica, which allows for symbolic computatios The Mathematica code is available from the authors 10
for which the value iterval for ɛ is computed as (079/16640, 9/64] We choose 1/8 (079/16640, 9/64] For this particular poit SW(T ) SW(R) = 1181/47680 > 0 This implies that SW(T ) > SW(R) i the etire iterval Whe χ<ɛ< (+)(1 μ) : I the iterval of 0 <χ 1 μ, The optimal solutio for the tax model is c =1adτ =0ad SW(T )= (μ 1)(χ()+(μ 1)(+)) () For the rate model there are three feasible optima: (1) whe χ ɛ() +, c =1 ad SW(R) = (χ+μ 1) (+),()whe ɛ() () + <χ< ɛ(1 μ)( 1) ɛ() (1 μ)(+), c (0, 1) ad SW(R) = ɛ (χ+μ 1) ɛ(1 μ)( 1), ad (3) whe χ, c =0ad χ(χ(+) ɛ()) ɛ() (1 μ)(+) SW(R) = (μ 1)(ɛ()+(μ 1)(+)) () I Case (1), we compare SW(T ) = (μ 1)(χ()+(μ 1)(+)) () vs SW(R) = (χ+μ 1) (+) () (μ 1)(χ( +1)+(μ 1)( +)) ( +1) (χ + μ 1) ( +) ( +1) (40) (μ 1)χ( +1)+(μ 1) ( +) (μ 1) ( +)+χ ( +)+(μ 1)χ( + ) (41) (1 μ) + > 1 μ (μ 1)χ( +1) χ ( +)+(μ 1)χ( + ) (4) (1 μ)χ χ ( + ) (43) (1 μ) + χ (44) (the upper boud o χ i the give iterval), thus χ< (1 μ) + i the give iterval This implies that SW(T ) > SW(R) I case (), we compare SW(T )= (μ 1)(χ()+(μ 1)(+)) () vs SW(R) = ɛ (χ+μ 1) χ(χ(+) ɛ()) IthisitervalSW(T) >SW(R) give all the bouds o the parameters I case (3), we compare SW(T )= (μ 1)(χ()+(μ 1)(+)) () vs SW(R) = (μ 1)(ɛ()+(μ 1)(+)) () This compariso is quite easy Sice χ<ɛi this iterval ad sice the coefficiet of χ ad ɛ i SW(T )adsw(r) ( (μ 1) <0 ) is egative, SW(T ) >SW(R) I the iterval of 1 μ <χ<1 μ, The optimal solutio for the tax model is c =1adτ = 1+()χ+μ ad SW(T )= (1 χ μ) The expressios ad the iterval bouds have complicated structures which makes this proof tedious Hece, the required computatios for the proof have bee performed i Mathematica The Mathematica code is available from the authors 11
For the rate model there are two feasible optima depedig o the value of χ: (1) whe χ ɛ() +, c =1adSW(R) = (χ+μ 1) (+) ad () whe ɛ() + <χ< ɛ(1 μ)( 1) ɛ() (1 μ)(+), c (0, 1) ad SW(R) = ɛ () (χ+μ 1) χ(χ(+) ɛ()) I case (1), we compare SW(T )= (1 χ μ) vs SW(R) = (χ+μ 1) (+) () This implies that SW(T ) >SW(R) (1 χ μ) (χ + μ 1) ( +) ( +1) (45) 1 ( +) ( +1) (46) ( +1) ( + ) (47) 1 > 0 (48) I case (), we compare SW(T )= (1 χ μ) vs SW(R) = ɛ (χ+μ 1) χ(χ(+) ɛ()) This compariso iterval ca be writte as max( 1 μ, ɛ() + ) <χ<mi(ɛ, 1 ɛ(1 μ)( 1) μ, ɛ() (1 μ)(+) ) If we solve SW(T ) SW(R) = 0, we fid 4 roots: χ 1 = ɛ (49) χ,3 = 1 μ (50) χ 4 = ɛ + (51) χ 1 ad χ,3 are already equal to the upper bouds of the compariso iterval Hece, they are out of the compariso iterval χ 4 = ɛ + < ɛ() +, hece this root is also out of the iterval Thus, there are o roots i the compariso iterval This implies that SW(T ) SW(R) ever hits 0 ad the sig of the differece does ot chage i the compariso iterval To determie the sig of SW(T ) SW(R) throughout the iterval, it is sufficiet to determie the sig at a arbitrary poit of the iterval Cosider the poit {ɛ = 301/4096,μ = 13/16, = } i the iterval For this data istace the value iterval for χ ca be computed as (1/16, 301/4096) We select 9/18 (1/16, 301/4096) from this iterval ad at this particular poit SW(T ) SW(R) = 89375/3488435 > 0 This shows that SW(T ) >SW(R) i the give iterval Whe (+)(1 μ) ɛ: This case is very similar to the secod case (ie, χ<ɛ< (+)(1 μ) ) with respect to the compariso itervals ad the optimal SW values i the relevat itervals The oly differece is that there exists oly oe feasible optima for the rate model Both i the iterval of 0 <χ 1 μ model is c = 1, ad the optimal SW(R) = (χ+μ 1) (+) () 1 μ ad <χ<1 μ, the optimal solutio for the rate 1
Similar to the previous case, the optimal social welfare for the tax model is SW(T )= (μ 1)(χ()+(μ 1)(+)) () whe 0 <χ 1 μ (1 χ μ),adsw(t)= whe 1 μ <χ Hece, the comparisos give i the secod case hold for this case as well ad we ca coclude that SW(T ) >SW(R) Give this aalysis, we ca coclude that SW(T )=SW(R) oly whe ɛ 1 μ ad SW(T ) >SW(R) otherwise as show i Table EC4 ad ɛ χ, To show the secod part of the corollary ote that as the degree of competitio () icreases, the upper boud ( 1 μ ) o the equality regio decreases ( 1 μ ) < 0 (5) This implies that as icreases the equality regio gets smaller while the regio where the tax model domiates elarges I other words, the social plaer prefers the tax model i a wider rage of ɛ values Hece, icreased competitio () reiforces the domiace of the tax model from the perspective of the social plaer Maufacturer ad Cosumer Perspectives We compare the two polices with respect to the total maufacturer profit ad the evirometal impact The comparisos with respect to the cosumer surplus parallel those for the total maufacturer profit i this settig because the cosumer surplus is always equal to a costat ( ) times the total maufacturer profit Hece, we do ot make a separate compariso for the cosumer surplus However, we ote that this is ot the case whe operatig cost exteralities are take ito accout, as illustrated i 31 Corollary If 1 μ (+)(1 μ) <ɛ< () ad ɛ χ, theπ M (R) > Π M (T ) ad Π C (R) > Π C (T ), ie, the rate model is preferred by the maufacturers ad the cosumers Otherwise, Π M (R) Π M (T ) ad Π C (R) Π C (T ), ie, the tax model is at least as good as the rate model (see Table EC5) Furthermore, with more itese competitio () the rate model is preferred by the maufacturers ad the cosumers for lower ɛ values ɛ 1 μ ɛ χ 1 μ (+)(1 μ) <ɛ< () χ<ɛ< (+)(1 μ) (+)(1 μ) () ɛ<1 μ χ< χ χ χ 1 μ (+)(1 μ) ɛ 1 μ <χ<1 μ 0 <χ<1 μ Π M (T )=Π M (R) Π M (T ) < Π M (R) Π M (T ) Π M (R) Π M (T )=Π M (R) Π M (T ) > Π M (R) Table EC5: Compariso of the total maufacturer profit betwee the two models Proof Similar to the social welfare compariso, we ca evaluate the two models i terms of the maufacturer profit uder three cases; whe ɛ χ, χ<ɛ< (+)(1 μ) ad (+)(1 μ) ɛ 13
Whe ɛ χ: I the iterval of ɛ 1 μ, The optimal solutio for tax model is c =0adτ = 0 ad the optimal total maufacturer profit is Π M (T )= (μ 1) () Similarly, the optimal solutio for the rate model is c = 0 ad the optimal Π M (R) = (μ 1) () Hece, it is obvious that Π M (T )=Π M (R) ithisiterval I the iterval of ɛ> 1 μ, The optimal solutio for the tax model is c =0adτ = 1+()ɛ+μ,adΠ M (T )= (1 ɛ μ) For the rate model there are two feasible optima depedig o the value of ɛ: (1) whe ɛ (1 μ)χ ), c =0adΠ M(R) = (μ 1),()whe (1 μ)χ () ) < ɛ<χ( (+) ), optimal c (0, 1) ad Π M(R) = ɛ (χ+μ 1) (χ(+) ɛ()) I case (1) we compare Π M (T )= (1 ɛ μ) vs Π M (R) = (μ 1) () The compar- ( ) iso iterval ca be writte as 1 μ <ɛ<mi χ, (1 μ)χ ) χ> (1 μ)χ ), hece the tightest compariso iterval ca be summarized as 1 μ (1 μ)χ <ɛ< ) If we solve we fid roots: Π M (T ) Π M (R) = 0 (53) (1 ɛ μ) (μ 1) ( +1) = 0 (54) ɛ 1 = 1 μ +1 (μ 1)( +1) ɛ = +1 (55) (56) ɛ 1 is out of the compariso iterval because 1 μ <ɛi this iterval To see that ɛ is also out of the iterval, we compare ɛ with (1 μ)χ ) (the upper boud of the iterval): (μ 1)( +1) (1 μ)χ + ( +1 1 μ + χ +1 ) (57) ( + 1)(1 μ + χ) χ( + ) (58) χ + χ +( + 1)(1 μ) χ +χ (59) χ χ ( + 1)(1 μ) (60) χ( 1) ( + 1)(1 μ) (61) 14
Sice 1adμ<1 by assumptio, the right-had side of (61) is (-) Thus, the iequality holds with (>) which meas that ɛ is greater tha the upper boud ad out of the compariso iterval Hece, o roots exist i the compariso iterval ad this implies that Π M (T ) Π M (R) ever hits 0 ad the sig of the differece does ot chage withi the compariso iterval To determie the sig of Π M (T ) Π M (R), it is sufficiet to determie the sig at a arbitrary poit i the iterval Cosider the parameter istace { χ = 11 3,μ= 1 4,ɛ = 1 4,=9} At this parameter istace, the value iterval for ɛ is (3/40, 363/1400) We choose ɛ =1/4 (3/40, 363/1400) ad compute Π M (T ) Π M (R) = 39/14400 < 0 which implies that Π M (T ) < Π M (R) throughout this compariso iterval I case () we compare Π M (T )= (1 ɛ μ) vs Π M (R) = ɛ (χ+μ 1) Takig (χ(+) ɛ()) ito accout all the upper ad the lower bouds o ɛ the tightest compariso iterval ca be writte as (1 μ)χ ) <ɛ χ (for detailed aalysis please see Case () o page 10 which refers to the same compariso iterval) If we solve (1 ɛ μ) Π M (T ) Π M (R) = 0 (6) ɛ (χ + μ 1) (χ( +) ɛ( +1)) = 0 (63) we fid 4 roots ɛ 1 = ((μ 1)(3+) χ) +8χ(μ 1)()(+)+χ (μ 1)(3+) ɛ = ((μ 1)(3+) χ) +8χ(μ 1)()(+) χ+(μ 1)(3+) 4() (64) 4() (65) ɛ 3 = ( + )(1 μ) ( +1) (66) ɛ 4 = χ (67) ɛ < (1 μ)χ ) <ɛ 4 = χ<ɛ 1, hece the roots ɛ 1, ɛ,adɛ 4 are out of the compariso iterval 3 Olyɛ 3 ca be located i the compariso iterval ad split the compariso iterval ito Hece, ɛ 3 istheolyrootatwhichπ M (T ) Π M (R) hits 0 ad chages sig i the compariso iterval The to determie the sig of Π M (T ) Π M (R) (thus the greater Π M )whe (1 μ)χ ) <ɛ<ɛ 3 ad ɛ 3 < ɛ χ, we take two arbitrary poits i each subiterval Cosider the parameter istace {μ =1/4,χ =3/5, =6} for which the value iterval for ɛ i the first subiterval ca be writte as (8/1, 9/1) ad i the secod oe as (9/1, 3/5] We choose ɛ =17/4 from the first subiterval ad compute Π M (T ) Π M (R) = 3 The compariso of the roots withi the iterval bouds is complicated ad tedious Hece, the proof that these roots are outside the compariso iterval has bee performed i Mathematica The Mathematica code is available from the authors 15
8569/894348 < 0 From thesecod subitervalwe chooseɛ = 5/4 ad calculate Π M (T ) Π M (R) = 341/13604 > 0 This aalysis proves that Π M (T ) < Π M (R) whe (1 μ)χ ) <ɛ<ɛ 3 (68) Π M (T ) > Π M (R) whe ɛ 3 <ɛ χ (69) Π M (T )=Π M (R) whe ɛ = ɛ 3 (70) Give these results ad the results of Case (1), we ca coclude that Π M (T ) < Π M (R) whe 1 μ (+)(1 μ) () <ɛ () (71) Π M (T ) > Π M (R) whe (+)(1 μ) () <ɛ<1 μ (7) Π M (T )=Π M (R) whe ɛ = (+)(1 μ) () (73) Note that at the poit ɛ = (+)(1 μ) (), the maufacturer profits are equal uder the two models For the ease of presetatio we do ot show the equality poit separately i Table EC5 ad combie it with the secod case as: whe (+)(1 μ) () ɛ<1 μ, Π M (T ) Π M (R) Whe χ<ɛ< (+)(1 μ) : I the iterval of 0 <χ 1 μ, The optimal solutio for the tax model is c =1adτ =0adΠ M (T )= (μ 1) () For the rate model depedig o the value of χ there are 3 feasible optima: (1) whe χ ()ɛ (+), c =1adΠ M(R) = (χ+μ 1),()whe ɛ() + <χ< χ = () ɛ(1 μ)( 1) ɛ() (1 μ)(+), c (0, 1) ad Π M (R) = χ χ, c =0adΠ M (R) = (μ 1) () I case (1) we compare Π M (T )= (μ 1) () (μ 1) ( +1) ɛ (χ+μ 1) (χ(+) ɛ()), (3) fially, whe vs Π M (R) = (χ+μ 1) () (χ + μ 1) ( +1) (74) (1 μ) > (1 χ μ) (75) Sice we assume μ + χ<1 (see page ), it is obvious that the left-had side of (75) is greater tha the right-had side Hece, Π M (T ) > Π M (R) 16
I case (), we compare Π M (T )= (μ 1) () vs Π M (R) = ɛ (χ+μ 1) (χ(+) ɛ()) Sice χ< (μ 1) ɛ (χ + μ 1) ( +1) (χ( +) ɛ( +1)) (76) (1 μ) ɛ(1 χ μ) ( +1) (ɛ( +1) χ( +)) (77) (1 μ)(ɛ( +1) χ( +)) ɛ(1 χ μ)( + 1) (78) ɛ(1 μ)( +1) χ(1 μ)( +) ɛ(1 μ)( +1) ɛχ( + 1) (79) Π M (T ) > Π M (R) ɛ(1 μ)( +1)+ɛχ( +1) χ(1 μ)( + ) (80) ɛ > χ(1 μ)( +) (1 μ + χ)( +1) (81) ɛ(1 μ)( 1) ɛ() (1 μ)(+) i this iterval, (81) holds with (>) which implies that I case (3), Π M (T )=Π M (R) = (μ 1) () I the iterval of 1 μ <χ<1 μ, The optimal solutio for the tax model is c =1adτ = 1+()χ+μ ad Π M (T )= (1 χ μ) For the rate model depedig o the value of χ there are two feasible optima: (1) whe χ ()ɛ (+), c =1adΠ M(R) = (χ+μ 1) ad () whe ɛ() () + <χ< ɛ(1 μ)( 1) ɛ() (1 μ)(+), c (0, 1) ad Π M(R) = ɛ (χ+μ 1) (χ(+) ɛ()) I case (1), sice 1 >,Π () M (T ) > Π M (R) I case, we compare Π M (T )= (1 χ μ) vs Π M (R) = ɛ (χ+μ 1) (χ(+) ɛ()) (1 χ μ) ɛ (χ + μ 1) (χ( +) ɛ( +1)) (8) (χ( +) ɛ( +1)) ɛ (83) ɛ( +1) χ( +) ɛ (84) ɛ( +) > χ( + ) (85) Sice ɛ>χi this iterval, (85) holds with (>) which implies that Π M (T ) > Π M (R) Whe (+)(1 μ) ɛ: As discussed i the social welfare compariso, this case is very similar to the secod case (ie, χ<ɛ< (+)(1 μ) ) The oly differece is that there exists oly oe feasible optima for the rate model: Both i the iterval of 0 <χ 1 μ 1 μ ad <χ<1 μ, the optimal solutio for the rate model is c = 1 ad optimal Π M (R) = (χ+μ 1) () 17
ɛ 1 μ ɛ χ 1 μ ()ɛ <ɛ ɛ ɛ <ɛ<1 μ χ (+) ()ɛ 1 μ (+) <χ χ<ɛ< (+)(1 μ) 1 μ ()ɛ <χ (+) ()ɛ (+) <χ<1 μ (+)(1 μ) ɛ Π E (T )=Π E (R) Π E (T ) Π E (R) Π E (T ) < Π E (R) Π E (T )=Π E (R) Π E (T ) > Π E (R) Π E (T )=Π E (R) Π E (T ) > Π E (R) Π E (T )=Π E (R) Table EC6: Compariso of evirometal beefits betwee the two models Hece, the compariso results from the secod case whe optimal c =1fortherate model (Π M (T ) > Π M (R)) are valid i this case as well Give this aalysis, we ca coclude that the compariso of the two models with respect to the maufacturer profit ca be summarized as i Table EC5 To see how the maufacturers prefereces are affected by the itesity of the competitio () ad prove the secod part of Corollary, ote that the lower ad the upper bouds o ɛ i the iterval where the rate model is preferred ( 1 μ (+)(1 μ) ad () ) decrease as icreases ( 1 μ ) ( (+)(1 μ) () ) = 1 μ ( +1) < 0 (86) = μ 1 ( +1) < 0 (87) Thus, the regio where the rate model is preferred shifts to the left, eg, lower ɛ values 3 Evirometal Perspective Corollary 3 There exists a ɛ, such that if ɛ <ɛ<χ,theπ E (T ) < Π E (R), ie, evirometal beefits are higher uder the rate model (see Table EC6) Furthermore, with more itese competitio (), the eviromet is better off with the rate model eve at lower ɛ levels Proof Similar to the social welfare ad the total maufacturer profit comparisos, we compare the two models with respect to the evirometal impact uder three cases; Whe ɛ χ: I the iterval of ɛ 1 μ, zero take-back is optimal for both models ad Π E(T )= Π E (R) = ɛ(1 μ) I the iterval of 1 μ <ɛ, Π E (T )= ɛ(1 ɛ μ) There are two feasible optima for the rate model: (1) whe ɛ (1 μ)χ ), optimal c =0adΠ E (R) = ɛ(1 μ) ad () whe (1 μ)χ optimal c (0, 1) ad Π E (R) = ɛ (χ+μ 1) (ɛ+ɛ χ(+)) χ(χ(+) ɛ()) ) <ɛ<χ( + ), 18
I case (1) we compare Π E (T )= ɛ(1 ɛ μ) vsπ E (R) = ɛ(1 μ) ɛ(1 μ) ɛ(1 ɛ μ) (88) +1 ( + 1)(1 μ)+( +1)ɛ (1 μ) (89) ( +1)ɛ 1 μ (90) ɛ > 1 μ (91) +1 We already kow that (91) is true due to the boud o ɛ i the iterval This shows that Π E (T ) > Π E (R) ithisiterval I case () we compare Π E (T )= ɛ(1 ɛ μ) vs = ɛ (χ+μ 1) (ɛ+ɛ χ(+)) χ(χ(+) ɛ()) The tightest compariso iterval ca be summarized as (1 μ)χ ) < ɛ χ (please see Case () o page 10 which refers to the same compariso iterval) This compariso is very complicated ad tedious Hece, we oly provide a sketch of the proof here We solve Π E (T ) Π E (R) =0forɛ adwefid4rootsoeof which is χ χ is already the upper boud of the iterval We also show that oly oe of the other 3 roots is i the compariso iterval 4 Let s call this root as ɛ, the we show that whe ɛ< ɛ, Π E (T ) > Π E (R), whe ɛ> ɛ, Π E (T ) < Π E (R), ad whe ɛ = ɛ, Π E (T )=Π E (R) Note that Π E (T )=Π E (R) alsowheɛ = χ Whe χ<ɛ< (+)(1 μ) I the iterval of 0 <χ 1 μ, Perfect take-back (c = 1) is optimal for the tax model ad Π E (T )=0 For the rate model, there are three feasible optima depedig o the value of χ: (1) whe χ ()ɛ (+), perfect take-back is optimal for the rate model too ad Π E (R) =0,()whe ()ɛ + <χ< ɛ(1 μ)( 1) ɛ() (1 μ)(+), partial take-back (c (0, 1)) is optimal ad Π E (R) = ɛ (χ+μ 1) (ɛ+ɛ χ(+)) ɛ(1 μ)( 1),(3)whe χ(χ(+) ɛ()) ɛ() (1 μ)(+) χ, zero-take (c = 0) is optimal ad Π E (R) = ɛ(1 μ) Oly i case (1), Π E (T )=Π E (R) By costructio Π E 0 (see page 3), hece i cases () ad (3) Π E (T ) > Π E (R) I the iterval of 1 μ <χ<1 μ, Agai for the tax model the optimal c =1adΠ E (T )=0 For the rate model, there are two feasible optima: (1) whe χ ()ɛ (+),Π E(R) =0 ad () whe ()ɛ + <χ< ɛ(1 μ)( 1) ɛ() (1 μ)(+),π E(R) = ɛ (χ+μ 1) (ɛ+ɛ χ(+)) χ(χ(+) ɛ()) 4 Three of the four roots are too log ad difficult to preset i the text We used Mathematica to fid the roots ad determie whether they are i the compariso iterval usig umerical aalysis The Mathematica codes are available from the authors 19
I case (1) Π E (T )=Π E (R) =0adicase()Π E (T ) > Π E (R) Whe (+)(1 μ) ɛ, c = 1 is still optimal for the tax model ad Π E (T )=0 Ithis iterval Π E (R) = 0 as well Hece, Π E (T )=Π E (R) throughout this iterval Give this aalysis, the compariso of the two models with respect to the evirometal impact ca be summarized as i Table EC6 To prove the last part of the corollary we examie how the lower boud of the rate model domiatig iterval i Table EC6 chages as icreases This lower boud is ɛ ɛ decreases as icreases I other words, the eviromet is better off with the rate model at lower ɛ values uder more itese competitio Due to the very complicated structure of ɛ, weusedumerical aalysis to show this result 3 The Impact of Operatig ad Moitorig Costs I this sectio, we ivestigate the impact of two importat implemetatio related exteralities, amely the operatig costs ad the moitorig costs, ad their impacts o differet stakeholders policy prefereces 31 Operatig Costs This additioal cost results i slight differeces i our model Each maufacturer i has a additioal cost term i the objective fuctio: max q i Π Mi (k) =q i (p μ) q i ρ k (τ,c) Ω k (c) (9) where 0, if k=t; Ω k (c) = (93) αc, if k=r The social plaer s objective is also slightly modified; the last compoet of the total social welfare becomes: qi φ k (τ,c) = (τ cχ) αc, if k=t; (94) 0, if k=r I the tax model the social plaer icurs the operatig cost I the rate model, the total operatig cost remais the same (αc ) as i the tax model ad each maufacturer icurs a equal fractio of the operatig cost, ie, αc to assure a take-back rate of c 0
311 Social Plaer Perspective Propositio 3 (Rate Model with Operatig Costs) There is a uique solutio to the social plaer s problem uder the rate model with the operatig cost The optimal take-back rate decreases i α ad icreases i Proof The profit fuctio of maufacturer i is writte as; max Π Mi = (p μ)q i cχq i αc (95) q i where p = 1 q i (96) Sice the operatig cost is idepedet of the quatity, the optimal quatity for each maufacturer i=1 ad the price remai the same as before ad writte as; Give the optimal quatity, q i = 1 μ cχ +1 ad p = 1+(cχ + μ) +1 Π M = i=1 Π Mi = (cχ + μ 1) αc ( +1) ( +1) Π C = (cχ + μ 1) ( +1) ɛ(c 1)(cχ + μ 1) Π E = +1 The problem of the social plaer is writte as; max c SW = (cχ + μ 1)(ɛ( +1)+(μ 1)( +)+c(χ( +) ɛ( + 1))) αc ( +1) ( +1) (97) st 0 c 1 (98) 0 τ 1 μ (99) SW c SW c = = (c(+)χ +((1 c)ɛ()+(μ 1)(+))χ ɛ(μ 1)()) αc() () (100) χ(χ( +) ɛ( +1)) α( +1) ( +1) (101) We check the sig of the secod order derivative to determie whether SW is cocave, covex or liear i c χ(χ( +) ɛ( +1)) α( +1) ( +1) =0 ɛ = χ ( +) α( +1) χ( +1) (10) 1
Whe ɛ> χ (+) α() χ(), SW c < 0adSW is cocave i c From the first order coditios, c = (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) α() +χ(ɛ() χ(+)) if ɛ χ(χ+μ 1)(+) α() (χ+μ 1)(), c 1 ad the optimal c (c )issetto1 if (1 μ)χ χ(χ+μ 1)(+) α() ) <ɛ< (χ+μ 1)(),0<c<1ad c = (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) α() +χ(ɛ() χ(+)) if χ (+) α() χ() <ɛ (1 μ)χ ), c 0adc =0 Whe ɛ< χ (+) α() χ(), SW > 0adSW is covex i c To fid the optimal solutio c we compare the objective value at two boudary solutios; c =1adc =0 SW c=1 = (χ + μ 1) ( +) α( +1) (μ 1)(ɛ( +1)+(μ 1)( +)) ( +1) ad SW c=0 = ( +1) If we solve (χ + μ 1) ( +) α( +1) ( +1) = ɛ 1 = (μ 1)(ɛ( +1)+(μ 1)( +)) ( +1) (103) χ( +)(χ +μ ) α( +1) (μ 1)( +1) (104) The upper boud o ɛ for covexity is χ (+) α() χ() SW c=1 <SW c=0 iff ɛ<ɛ 1 (105) SW c=1 >SW c=0 iff ɛ>ɛ 1 (106) SW c=1 = SW c=0 iff ɛ>ɛ 1 (107) χ ( +) α( +1) χ( +)(χ +μ ) α( +1) χ( +1) (μ 1)( +1) (108) χ ( +) α( +1) α( +1) + χ( + )((1 μ) χ) χ (1 μ) (109) χ ( + )(1 μ) α( +1) (1 μ) χα( +1) + χ ( + )((1 μ) χ) (110) χ ( +)(χ (1 μ)) < ( +1) α(χ +1 μ) (111) Note that (χ (1 μ) < 0 i the left-had side of (111) ad the right-had side is positive, hece (111) holds with (<) Give ɛ< χ (+) α() χ() (covexity coditio), i this iterval SW c=0 >SW c=1 ad the optimal c (c )is0 Whe ɛ = χ (+) α() χ(), SW c c =0 =0adSW is liear i c SW c < 0atthisɛ value Hece, Give this aalysis, we ca coclude that there is a uique solutio to the social plaer s problem with the operatig cost uder the rate model, as summarized i Table EC7
0 <ɛ (1 μ)χ ) (1 μ)χ c 0 q Π M Π E (1 μ) χ(χ+μ 1)(+) α() (χ+μ 1)() χ(χ+μ 1)(+) α() ) <ɛ< (χ+μ 1)() (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) α() +χ(ɛ() χ(+)) 1 (χɛ(χ+μ 1)+α(μ 1)()) χ(χ(+) ɛ()) α() (1 χ μ) (μ 1) ((χɛ(χ+μ 1)+α(μ 1)()) α(ɛ(μ 1)() χ(ɛ()+(μ 1)(+))) ) (χ+μ 1) α() () (α() +χ(ɛ() χ(+))) () ɛ(μ 1) ɛ(χɛ(χ+μ 1)+α(μ 1)())(α() +(χ+μ 1)(ɛ+ɛ χ(+))) (α() +χ(ɛ() χ(+))) 0 (μ 1) Π (χɛ(χ+μ 1)+α(μ 1)()) (χ+μ 1) C () (α() +χ(ɛ() χ(+))) () SW (μ 1)(ɛ()+(μ 1)(+)) (ɛ (χ+μ 1) +α(μ 1)(ɛ()+(μ 1)(+))) (χ+μ 1) (+) α() () 4α() χ(χ(+) ɛ()) () Table EC7: The optimal solutio for the rate model with the operatig cost ɛ To prove the secod part of the propositio, first ote that the lower boud o ɛ i the perfect-take back regio is icreasig i α ( χ(χ+μ 1)(+) α() (χ+μ 1)() α ) ( +1) = (χ + μ 1) > 0 (11) This meas that as α icreases, the social plaer prefers perfect take-back at higher ɛ values Moreover, the optimal take-back rate i the partial take-back regio is decreasig i α; ( ) (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) α() +χ(ɛ() χ(+)) = () (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) < 0 (113) α (α() +χ(ɛ() χ(+))) Note that the partial derivative above is less tha 0 because ɛ> (1 μ)χ ) i the partial takeback regio Give these observatios, we ca coclude that as α icreases, the optimal take-back rate decreases The lower ad the upper bouds o ɛ decrease as icreases: ( ) (1 μ)χ ) ( χ(χ+μ 1)(+) α() (χ+μ 1)() ) = = χ(μ 1) (χ μ +1)( +1) < 0 (114) α (χ + μ 1) χ ( +1) < 0 (115) This implies that as icreases, the social plaer prefers perfect take-back at lower ɛ values Also the partial take-back rate icreases as icreases: ( ) (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()) α() +χ(ɛ() χ(+)) = α()(χ(ɛ+ɛ+μ ) ɛ(μ 1)()) χ ɛ(χ+μ 1) > 0 (α() +χ(ɛ() χ(+))) (116) Hece, we ca coclude that the optimal take-back rate icreases as icreases To esure that the maufacturer profit is always positive i the solutios, we assume α < (χ+μ 1) () 3
For c =0Π M = (μ 1) () > 0 for all α For c =1Π M = (χ+μ 1) α() () > 0ifα< (χ+μ 1) () For c = (χ(ɛ()+(μ 1)(+)) ɛ(μ 1)()), α() +χ(ɛ() χ(+)) Π M = ((χɛ(χ+μ 1)+α(μ 1)()) α(ɛ(μ 1)() χ(ɛ()+(μ 1)(+))) ) (α() +χ(ɛ() χ(+))) > 0ifα< (χ+μ 1) () Propositio 4 (Tax Model with Operatig Costs) There is a uique solutio to the social plaer s objective uder the tax model with the operatig cost Whe α>0, a take-back cost lower tha the evirometal cost (ie, χ<ɛ) o loger guaratees perfect take-back Whe α>0, perfect take-back is oly possible at relatively lower χ ad higher ɛ values Moreover, as the operatig cost icreases, the optimal take-back rate decreases ad the optimal tax icreases Proof The objective of each maufacturer i is writte as; max q i Π Mi = (p μ τ)q i (117) where p = 1 q i (118) The maufacturer i s objective fuctio is ot affected by the operatig cost ad the optimal quatity (qi ) remais the same as before ad writte as; i=1 qi = 1 μ τ +1 Give the optimal quatity, ad p = 1+(μ + τ) +1 Π M = Π Mi = i=1 (μ + τ 1) ( +1) Π C = (μ + τ 1) ( +1) ɛ(1 c)(1 μ τ) Π E = +1 The problem of the social plaer is writte as; max c,τ SW = (μ + τ 1)(c(χ ɛ)( +1)+ɛ( +1)+(μ 1)( +) τ) ( +1) αc (119) st 0 c 1 (10) 0 τ 1 μ (11) 4
SW is joitly cocave i c ad τ if det(hessia) = ( α (χ ɛ) ) ( +1) > 0 χ ɛ<χ+ α or χ α<ɛ χ (1) The Lagragia is L(τ, c, λ, γ) = (μ+τ 1)(c(χ ɛ)()+ɛ()+(μ 1)(+) τ) () αc +λ 1 c+λ (1 c)+γ 1 τ +γ (1 μ τ) (13) L τ L c = γ 1( +1) γ ( +1) + (ɛ + ɛ + μ + c(χ ɛ)( +1) τ 1) ( +1) (14) = λ 1 αc λ + (χ ɛ)(μ + τ 1) +1 (15) As log as the cocavity coditios are satisfied; the first order coditios, the complemetary slackess, ad the dual ad primal feasibility coditios are the ecessary ad the sufficiet coditios for the optimality The complemetary slackess coditios are: λ 1 c =0,λ (1 c) =0, γ 1 τ =0,γ (1 μ τ) =0whereλ 1, λ, γ 1 ad γ are Lagrage multipliers The dual ad the primal feasibility coditios are 0 c 1, 0 τ 1 μ, λ i 0adγ i 0fori =1, By simultaeously solvig the first order coditios ad the complemetary slackess equatios we obtai the followig solutio sets: { } (ɛ + μ 1) S1 = λ 1 =0,λ =0,γ 1 =0,γ =,τ =1 μ, c =0 +1 { S = λ 1 =(ɛ χ)(ɛ + μ 1),λ =0,γ 1 =0,γ =0,τ = ɛ + ɛ + μ 1 },c=0 { } (χ ɛ)(μ 1) (ɛ + ɛ + μ 1) S3 = λ 1 =,λ =0,γ 1 = +1 ( +1),γ =0,τ =0,c=0 { } S4 = λ 1 =0,λ =0,γ 1 = ((μ 1)(χ ɛ) +α(ɛ+ɛ+μ 1)),γ α() =0,τ =0,c= (χ ɛ)(μ 1) α() { } (χ + μ 1) S5 = λ 1 =0,λ = α, γ 1 =0,γ =,τ =1 μ, c =1 +1 { S6 = λ 1 =0,λ =(χ ɛ)(χ + μ 1) α, γ 1 =0,γ =0,τ = χ + χ + μ 1 },c=1 { } S7 = λ 1 =0,λ = (χ ɛ)(μ 1) α(),γ 1 = (χ+χ+μ 1),γ () =0,τ =0,c=1 { } S8 = λ 1 =0,λ =0,γ 1 =0,γ =0,τ = (μ 1)(χ ɛ) +α(ɛ+ɛ+μ 1),c= (χ ɛ)(ɛ+μ 1) (α (χ ɛ) ) α (χ ɛ) To see that these 8 solutios give the complete list of solutio sets from simultaeous solutio of the first order coditios ad the complemetary slackess coditios, ote that due to the complemetary slackess coditios the Lagrage Multipliers (λ 1, λ, γ 1 ad γ ) ca either take a positive or zero value i the optimal solutio Hece, there are 16 cadidate solutio sets However, a solutio with both λ 1 > 0adλ > 0 is ot possible because for λ 1 > 0 c should be 0 5
while for λ > 0 c should be 1 Similarly, a solutio with both γ 1 > 0adγ > 0 is ot possible because for γ 1 > 0 τ should be 0 while for γ > 0 τ should be 1 μ Hece, we elimiate 7 of the 16 cadidate solutios Moreover, {λ 1 > 0, λ =0,γ 1 =0,γ > 0, c =0,τ =1 μ} is ot also a possible solutio because whe λ =0,c =0,adτ =1 μ due to (15) λ 1 should be 0 Hece, we get 8 solutio sets as give above To determie the optimal solutio i differet parameter itervals, we check the dual ad the primal feasibility coditios The iterval i which each solutio satisfies the feasibility coditios is give below S1: whe ɛ 1 μ (required to esure γ 0) S: whe ɛ<1 μ (required to esure τ<1 μad λ 1 0), ɛ χ (to esure λ 1 0), ad ɛ> 1 μ (to esure τ>0) S3: whe ɛ χ (required to esure λ 1 0) ad ɛ 1 μ (to esure γ 1 0) S4: whe ɛ > χ (required to esure c > 0), χ > ɛ α() (1 μ) α(μ 1)(ɛ+ɛ+μ 1) χ (μ 1) + ɛ (to esure γ 1 0) (to esure c < 1) ad S5: ever feasible sice λ = α <0 α >0 S6: whe χ> 1 μ (required to esure τ>0) ad χ ɛ α 1 χ μ (to esure λ 0) S7: whe χ 1 μ (required to esure γ 1 0) ad χ ɛ α() (1 μ) (to esure λ 0) α(μ 1)(ɛ+ɛ+μ 1) S8: wheɛ > χad ɛ < 1 μ (requiredto esurec > 0), ɛ+ <χ<ɛ+ α(ɛ+ɛ+μ 1) α(μ 1)(ɛ+ɛ+μ 1) (to esure τ>0) ad (to esure c<1) (1 μ+ɛ 8α+(ɛ+μ 1) ) <χ< We check the order of the bouds o χ ad ɛ give for the solutios (μ 1) (1 μ+ɛ+ ) 8α+(ɛ+μ 1) For S1, S, ad S3, the itervals where they are feasible, ad thus optimal, are clear They are immediately placed i Table EC8 α(μ 1)(ɛ+ɛ+μ 1) For S4, the upper boud o χ, χ oc = (μ 1) + ɛ, is ot a real umber whe ɛ< 1 μ 1 μ Hece, this boud is oly icluded i the solutio table whe <ɛ ( For S6, the upper boud o χ is ɛ α 1 χ μ From S8 we kow that if χ< 1 μ+ɛ 8α+(ɛ+μ 1) ) (let call this χ oc3 ), c>1 ad the optimal c (c ) is set to 1; i other words S6 becomes the optimal solutio χ oc3 <ɛ solutio tables α 1 χ μ, hece the upper boud for S6 is take as χoc3 i the 6
For S7, there are upper bouds o χ: ɛ< (μ 1) +α() (μ 1)() 1 μ ad ɛ α() (1 μ) (let s call this χoc1 ) Whe (let s call this ɛ oc1 ), χ oc1 < 1 μ, hece i this case χoc1 is icluded as the lower boud i the optimal solutio table Whe ɛ>ɛ oc1, o the other had, 1 μ is lower ad it is used i the solutio table For S8, there are two upper bouds ( ɛ + α(ɛ+ɛ+μ 1) (1 μ+ɛ+ ) 8α+(ɛ+μ 1) ) ad α(μ 1)(ɛ+ɛ+μ 1) ad two lower bouds (χ oc ad χ oc3 )oχ χ oc >χ oc3 whe ɛ<ɛ oc1 (16) χ oc <χ oc3 whe ɛ>ɛ oc1 (17) χ caot be greater tha either of the upper bouds give ɛ<1 μ ad χ<ɛ Give the above aalysis, the optimal solutio of the model for differet realizatios of ɛ ad χ ca be summarized as i Table EC8 SW is ot joitly cocave i τ ad c whe ɛ χ α or χ + α ɛ This implies that the optimal solutio will occur at some boudary of the feasible regio Hece, to fid the optimal solutio, we fid ad compare the boudary solutios, which are the cadidate maximizers of this case To fid the boudary solutios we cosidered the boudary values of each variable (ie, c =0adc =1,adτ =0adτ =1 μ), ad at each boudary value we optimized over the other variable to fid its value i the correspodig solutio This allows us to determie the 8 cadidate maximizers ad their feasible itervals as follows; (1) P1 c =0adτ = ()ɛ (1 μ) : Feasible whe 1 μ <ɛ<1 μ ad SW P 1 = 1 (ɛ + μ 1) () P c =0adτ =0adSW P = (μ 1)(ɛ()+(μ 1)(+)) () (3) P3 c =1adτ =0adSW P 3 = (μ 1)(χ()+(μ 1)(+)) α() () (4) P4 c = (ɛ χ)(1 μ) α() ad τ = 0: Feasible whe χ oc1 = ɛ α() (1 μ) SW P 4 = (μ 1)((μ 1)(χ ɛ) +α(ɛ()+(μ 1)(+))) 4α() <χ<ɛad (5) P5 c =1adτ = ()χ (1 μ) : Feasible whe 1 μ <χad SW P 5 = 1 ( (χ + μ 1) α ) (6) P6, P7 ad P8 c =0,c (0, 1) or c =1adτ =1 μ: Whe τ =1 μ, there is o productio (q = 0) Hece, irrespective of the value of csw P 6 = 0 That is why we preset these 3 solutios as a sigle solutio ad refer to it as P6 i the rest of the paper Whe ɛ χ α ad ɛ 1 μ : Due to the feasibility itervals of the cadidate maximizers (please see above), P4 caot be feasible; the feasibility coditio for P4 (χ <ɛ) doesot hold i this iterval P1 caot be feasible either because 1 μ <ɛis required for the 7
feasibility of this poit whereas ɛ 1 μ i this iterval P6 caot also be a maximizer, because as log as ɛ<1 μthe other solutios, which always provide a positive SW, always domiate P6, which provides a zero SW P, P3, ad P5 are the feasible cadidate maximizers i this iterval, sice there is o feasibility coditio for P ad P3 ad the feasibility coditio of P5 ( 1 μ <χ) does ot coflict with the bouds of this iterval First we compare SW P vs SW P 3 (μ 1)(ɛ( +1)+(μ 1)( +)) (μ 1)(χ( +1)+(μ 1)( +)) α( +1) ( +1) ( +1) (18) α( +1) (μ 1)(χ( +1) ɛ( + 1)) (19) α( +1) > (μ 1)(χ ɛ) (130) The right-had side of (130) is egative sice χ>ɛi this iterval, hece SW P >SW P 3 Compariso betwee SW P ad SW P 5 is more complicated We solve (μ 1)(ɛ( +1)+(μ 1)( +)) ( +1) = 1 ( (χ + μ 1) α ) (131) ( ) ɛ = () (µ 1) () α+χ(χ+μ ) (μ 1) is the root of the equality (131) ad whe ɛ is lower tha the root, SW P >SW P 5 This root is greater tha the upper boud o ɛ ( 1 μ ) i the give iterval Hece, ɛ ca ever be greater tha the root ad we ca coclude that SW P >SW P 5 i the give iterval Give that the optimal solutio i this iterval is c =0adτ =0 Whe ɛ χ α ad ɛ> 1 μ : Due to the feasibility coditios (please see page 7), P4 caot be feasible The feasibility coditio for P4 (χ <ɛ) does ot hold i this iterval P6 caot be a cadidate maximizer due to the same reaso discussed i the previous iterval Recall that there is o feasibility coditio for P ad P3 ad the feasibility coditios for P1 ( 1 μ <ɛ<1 μ) ad P5 ( 1 μ <χ) do ot coflict with the bouds of the iterval Hece, i this iterval the feasible cadidate maximizers are P1, P, P3, ad P5 I the previous iterval we already showed that SW P >SW P 3 This result also holds here because χ>ɛis still valid It is also easy to show that SW P 1 is greater tha SW P Ideed this result is obvious with o eed for further aalysis Give c = 0, solutio τ = 0 (P) will always be domiated by τ = ()ɛ (1 μ) (P1) which is a local maximizer if both of these poits are feasible i a give iterval Still, to be complete, we prove this observatio aalytically below 1 (ɛ + μ 1) (μ 1)(ɛ( +1)+(μ 1)( +)) ( +1) (13) ( +1) ((μ 1) + ɛ +ɛ(μ 1)) ɛ( +1)(μ 1) + (μ 1) ( + ) (133) (μ 1) ɛ(μ 1)( + 1) ( +1) ɛ (134) (μ 1) ɛ( + 1)(1 μ) ( +1) ɛ (135) (1 μ ( +1)ɛ) > 0 (136) Hece, SW P 1 >SW P The we compare SW P 1 ad SW P 5 Siceɛ<χad α>0, SW P 1 >SW P 5 8
Hece, SW P 1 is the greatest social welfare amout i this iterval ad the optimal solutio is c =0 ad τ = ()ɛ (1 μ) We examie χ + α ɛ iterval i two parts First, we examie χ + α ɛ<1 μ where the optimal quatity is always positive ad P6 is ever optimal: whe ɛ 1 μ, P6 is domiated by P ad whe ɛ> 1 μ, it is domiated by P1 The, we examie 1 μ ɛ part where P6 is also a cadidate for the optimal solutio Before startig to compare the cadidate maximizers i these ew itervals, we should ote oe importat issue about the order of the feasibility coditios for the cadidate maximizers I the two previous itervals, P4 was ot feasible However, whe χ + α ɛ, P4 ca also be feasible because its feasibility coditio, χ<ɛ, holds i this case Whe P4 is feasible, due to the feasibility coditios of P4 (ie, χ oc1 <χ) ad P5 (ie, 1 μ <χ), we observe two lower bouds o χ These bouds should be ordered i order to arrage the compariso itervals properly ad make the maximizer comparisos i accurate itervals We compared χ oc1 ad 1 μ i the cocave case while examiig S7 o page 7 We stated that whe ɛ<ɛ oc1, whe ɛ>ɛ oc1, χ oc1 < 1 μ χ oc1 > 1 μ We use the same result i the iterval of χ + α ɛ to specify the compariso itervals for the cadidate maximizers as below Whe χ + α ɛ<mi(ɛ oc1, 1 μ), χ oc1 < 1 μ, hece we arrage the compariso itervals as follows: if χ χ oc1 : Cosiderig the feasibility coditios (see page 7) oly P1, P, ad P3 are feasible P4 ad P5 are ot feasible sice their feasibility coditios, χ oc1 <χad 1 μ <χrespectively, do ot hold i this iterval From (13)-(136) we kow that SW P 1 >SW P The we compare SW P 1 ad SW P 3 We solve SW P 1 SW P 3 = 0 ad check whether the root(s) are i this iterval SW P 1 SW P 3 = 0 (137) 1 (ɛ + μ (μ 1)(χ( +1)+(μ 1)( +)) α( 1) +1) = ( +1) (138) ( ( )) ( +1) α + 1 (μ 1) + ɛ(ɛ +μ ) () χ 1 = (139) (μ 1) SW P 1 >SW P 3 oly whe χ is greater tha the above root However, the root is greater tha χ oc1 (the upper boud o χ i this iterval), thus χ ca ever be greater tha the root give this upper boud 5 ThisimpliesthatSW P 1 <SW P 3 throughout 5 The compariso of the root with χ oc1 is complicated ad tedious Hece, the required computatios for the proof have bee performed i Mathematica The Mathematica code is available from the authors 9
this iterval Hece, SW P 3 is greater tha the others ad c =1adτ =0isthe optimal solutio i this iterval if χ oc1 <χ 1 μ : P1, P, P3, ad P4 are feasible P5 is ot feasible sice its feasibility coditio ( 1 μ <χ)doesothold We already kow that SW P 1 >SW P (see (13)-(136)) It is also easy to see that SW P 4 >SW P 3 Wheτ =0,solutioc = 1 (P3) is always domiated by the solutio c = (ɛ χ)(1 μ) α() (P4) which is a local maximizer Hece, whe both poits are feasible SW P 4 is always greater tha SW P 3 The we compare SW P 1 with SW P 4 to fid the optimal poit i this iterval If we solve SW P 1 SW P 4 = 0 we fid the roots: α(μ 1) χ 1 = ɛ (ɛ + ɛ + μ 1) (μ 1) (140) α(μ 1) χ = ɛ + (ɛ + ɛ + μ 1) (μ 1) (141) Note that i this iterval χ ɛ α (please see 9) χ >ɛ Hece, χ is always smaller tha χ i this iterval Furthermore, give the bouds o other parameters, χ is always lower tha χ 1 Wheχ oc1 <χ<χ 1, SW P 1 <SW P 4 Hece, the optimal solutio i this iterval is c = (ɛ χ)(1 μ) α() ad τ =0 if 1 μ <χ: Not possible Whe ɛ oc1 <ɛ<1 μ, χ oc1 > 1 μ, hece we arrage the compariso itervals as follows: if χ 1 μ : Cosiderig the feasibility coditios (see page 7) feasible cadidates ca be listed as P1, P, ad P3 P4 ad P5 are ot feasible sice their feasibility coditios, χ oc1 <χad 1 μ <χrespectively, do ot hold i this iterval We already kow that SW P is domiated by SW P 1 (see (13)-(136)) The we compare SW P 1 ad SW P 3 We kow the root of SW P 1 SW P 3 = 0 from (139) ad we also kow that SW P 1 <SW P 3 whe χ is smaller tha this root This root is always greater tha the upper boud ( 1 μ )oχ i this iterval Hece, we ca coclude that SW P 3 is the greatest social welfare ad the optimal solutio is c =1adτ =0 throughout the give iterval if 1 μ <χ: Feasible cadidates are P1, P, P3, ad P5 P4 is ot feasible sice its feasibility coditio, χ oc1 <χdoes ot hold i this iterval P is domiated by P 1 as we kow from (13)-(136) The we cosider SW P 1 vs SW P 5 Whe we solve 30
SW P 1 SW P 5 = 0, we fid two roots χ 1 = 1 μ + α +(ɛ + μ 1) (14) χ = 1 μ α +(ɛ + μ 1) (143) Obviously χ 1 is out of the iterval because it is greater tha 1 μ Give the bouds o the other parameters, χ ca ever be greater tha χ i this iterval throughout the iterval χ<χ ad uder this coditio SW P 1 <SW P 5 Hece, Fially, we compare SW P 3 vs SW P 5 We fid the roots χ 1, = 1 μ whe we solve SW P 3 SW P 5 = 0 Obviously, χ caot be smaller tha 1 μ i this iterval Whe 1 μ <χ, SW P 3 <SW P 5 Hece, c =1adτ = ()χ (1 μ) i this iterval χ oc1 <χ: Not possible is the optimal solutio I the previous cases we have examied ɛ<1 μ case where P6 (o productio case) is ever optimal Now, we look at 1 μ ɛ case where o productio is also a cadidate for the optimal solutio The feasible itervals ad the order of the bouds o χ ad ɛ are same as i the ɛ<1 μ case However, i this case, the optimal solutios foud i the previous aalysis may be domiated by P6 Hece, we compare the optimal solutio of each iterval with P6 below First we compare P6 with P Recall that SW P 6 = 0 Hece, we solve SW P =0adget the root ɛ = (1 μ)(+) () SW P 6 >SW P (ie, P is domiated by P6) if ɛ> (1 μ)(+) () This root is smaller tha 1 μ Hece, whe ɛ>1 μ, ɛ is always greater tha (1 μ)(+) () ad P is always domiated by P6 For that reaso we will ot show P amog the cadidate maximizers i the followig comparisos TocompareP6withP3,wesolveSW P 3 =0adgettherootχ = Let χ oc5 = P6) ( ) () α (μ 1) (+) () (μ 1) ( ) () α (μ 1) (+) () (μ 1) Wheχ oc5 χ, SW P 6 >SW P 3 (ie, P3 is domiated by TocompareP6withP4,wesolveSW P 4 = 0 We fid two roots α(μ 1)(ɛ( +1)+(μ 1)( +)) χ 1 = ɛ (μ 1) α(μ 1)(ɛ( +1)+(μ 1)( +)) χ = ɛ + (μ 1) (144) (145) Sice μ 1 < 0 by assumptio (please see page ), ɛ<χ 1 I this compariso iterval χ<1 μ<ɛ, hece χ 1 is out of the compariso iterval which implies that the oly root 31