MS&E246:Lecture9SequentialbargainingRamesh JohariNashbargainingsolutionRecall Nash’s approach to bargaining:The planner is given the set of achievable payoffs and status quo point.Implicitly:The process of bargaining does not matter.DynamicsofbargainingIn this lecture:We use a dynamic game of perfect information to model the process of bargaining.AninterferencemodelRecall the interference model:•Two devices• Device 1 given channel afraction q of the time• For efficiency:When device n has control,it transmits at full power PAninterferencemodel• When timesharing is used,the set of Pareto efficient payoffs becomes:{ (Π1, Π2) : Π1= qR1, Π2= (1 - q) R2}• We now assume the devices bargain through a sequence of alternating offers.Alternatingoffers•At time 0:•Stage 0A:Device 1 proposes a choice of q(denoted q1)•Stage 0B:Device 2 decides to accept or rejectdevice 1’s offerTwoperiodmodelAssumption 1:If device 2 rejects at stage 0B,then predetermined choice Q ∈ [0, 1]is implemented at time 1DiscountingAssumption 2:Devices care about delay: Any payoff received by device i at time kis discounted by δik.0 < δi< 1: discount factor of device iGametree0A0B1q1ACCEPTREJECTDevice 1 makesinitial offerDevice 2 acceptsor rejectsΠ1= q1R1Π2= (1 - q1)R2Π1= δ1QR1Π2= δ2(1 - Q)R2Gametree•This is a dynamic game of perfect information.• We solve it using backward induction.Backwardinduction1. Given q1, at Stage 0B:• Device 2 rejects if:δ2(1 - Q) R2> (1 - q1) R2Backwardinduction1. Given q1, at Stage 0B:• Device 2 rejects (s2(q1) = R) if:q1> 1 - δ2(1 - Q)• Device 2 accepts (s2(q1) = A) if:q1< 1 - δ2(1 - Q)• Device 2 is indifferent(s2(q1) ∈ { A, R }) ifq1= 1 - δ2(1 - Q)Backwardinduction2. At Stage 0A:• Device 1 maximizes Π1(q1, s2(q1))over offers ( 0 ≤ q1≤ 1 )• Claim: Maximum value of Π1is(1 - δ2( 1 - Q)) R1Backwardinduction2. At Stage 0A:• Claim: Maximum value of Π1isΠ1MAX= ( 1 - δ2( 1 - Q) ) R1• Proof:(a) Maximum is achievable:If q1increases to 1 - δ2( 1 - Q),then Π1increases to Π1MAXBackwardinduction2. At Stage 0A:• Claim: Maximum value of Π1isΠ1MAX= ( 1 - δ2( 1 - Q) ) R1• Proof:(b) If q1>1 -δ2( 1 - Q), then Π1 < Π1MAX:Device 2 rejects ⇒ Π1= δ1QR1Backwardinduction2. At Stage 0A:• Claim: Maximum value of Π1isΠ1MAX= ( 1 - δ2( 1 - Q) ) R1• Proof:(b) If q1>1 -δ2( 1 - Q), then Π1 < Π1MAX:But note that: δ1Q + δ2(1 - Q) < 1Backwardinduction2. At Stage 0A:• Claim: Maximum value of Π1isΠ1MAX= ( 1 - δ2( 1 - Q) ) R1• Proof:(b) If q1>1 -δ2( 1 - Q), then Π1 < Π1MAX:But note that: δ1Q < 1 - δ2(1 - Q)Backwardinduction2. At Stage 0A:• Claim: Maximum value of Π1isΠ1MAX= ( 1 - δ2( 1 - Q) ) R1• Proof:(b) If q1>1 -δ2( 1 - Q), then Π1 < Π1MAX:So δ1QR1< ( 1 - δ2(1 - Q) ) R1Backwardinduction2. At Stage 0A:• Best responses for device 1 :All choices of q1that achieve Π1MAXThe only possibility:q1* = 1 - δ2(1 - Q)Backwardinduction2. At Stage 0A:•If s2(q1*) = reject,no best response exists for device 1!•If s2(q1*) = accept,best response for device 1 is q1= q1*UniqueSPNEWhat is the unique SPNE?•Must give strategies for both players!UniqueSPNEWhat is the unique SPNE?• Device 1:At Stage 0A, offer q1= q1*• Device 2:At Stage 0B,accept if q1≤ q1*, reject if q1> q1*PayoffsatuniqueSPNE• So the offer of device 1 isaccepted immediately by device 2.• Device 1 gets: Π1= ( 1 - δ2(1 - Q)) R1• Device 2 gets: Π2= δ2(1 - q0) R2InfinitehorizonMore realistic model:Devices alternate offers indefinitely.For simplicity: assume δ1= δ2= δFinitehorizon0A 0Bq10ACCEPTREJECTΠ1= q10R1Π2= (1 - q10)R21Device 1Device 2Infinitehorizon0A 0Bq10ACCEPTREJECTΠ1= q10R1Π2= (1 - q10)R2Π1= δq21R1Π2= δ(1 - q21)R21A 1BACCEPTREJECTq212A. . .Device 1 Device 1Device 2 Device 2Device 1Infinitehorizon:formalmodel• Device 1 offers q1kat stage kA, for k even• Device 2 offers q2kat stage kA, for k odd• Device 2 accepts/rejects stage kA offerat stage kB, for k even• Device 1 accepts/rejects stage kA offerat stage kB, for k oddInfinitehorizon:formalmodel• Payoffs:Π1= Π2= 0 if no offer ever accepted(similar to status quo in NBS)Infinitehorizon:formalmodel• Payoffs:If offer made at stage kA by player iaccepted at stage kB:Π1= δkqikR1Π2= δk(1 - qik) R2Infinitehorizon• Can’t use backward induction!•Use stationarity:Subgame rooted at 1A isthe same as the original game,with roles of 1 and 2 reversed.SPNEDefine V and v:VR1= highest time 0 payoff to device 1 among all SPNEvR1= lowest time 0 payoff to device 1 among all SPNESPNEThen if device 2 rejects at 0B:VR2= highest time 1 payoff to device 2among all SPNEvR2= lowest time 1 payoff to device 2among all SPNESPNE:Twoinequalities• vR1≥ (1 - δ V ) R1At Stage 0B:Device 2 will accept any q10<1 -δ VSo at Stage 0A:Device 1 must earn at least (1 - δ V ) R1SPNE:Twoinequalities• VR1≤ (1 - δ v) R1If offer q10is accepted at stage 0B,device 2 must get a timeshareof at least δ v⇒ q10≤ 1 - δ vSPNE:Twoinequalities• VR1≤ (1 - δ v) R1If offer q10is rejected at stage 0B,device 1 earns at most δ (1 - v) R1since device 2 earns at least δ vR2⇒ Π1≤ δ (1 - v) R1≤ (1 - δ v) R1Combininginequalities• v ≤ V• v ≥ 1 - δ V• V ≤ 1 - δ vCombininginequalities• v ≤ V• v + δ V ≥ 1• V + δ v ≤ 1So: V + δ v ≤ v + δ V⇒ (1 - δ) V ≤ (1 - δ) v⇒ V = vUniqueSPNE•So V = 1 - δ V ⇒• SPNE strategies for device 1:At Stage kA , k even:Offer q1k= 1 - δ VAt Stage kB, k odd:Accept if q2k≥ δ VUniqueSPNE•So V = 1 - δ V ⇒• SPNE strategies for device 2:At Stage kA , k odd:Offer q2k= δ VAt Stage kB, k even:Accept if q1k≤ 1 - δ VUniqueSPNE:PayoffsStage 0A offer by device 1 isaccepted in Stage 0B by device 2.Infinitehorizon:Discussion• Outcome is efficient:No “lost utility” due to discounting• Stationary SPNE strategies:Actions do not depend on time k• First mover advantage:Π1SPNE> Π2SPNEShorteningtimeperiodsShorten each time step to length t < 1 …… Same as changing discount factor to δt As t → 0, note that ΠiSPNE→ Ri/2.Nash bargaining solution!IngeneralIf δ1≠δ2: Find SPNE using two period model:Note that Q must be SPNE payoff when device 2 offers firstCan show (for an appropriate limit) that