Economic of Negotiations  Assignment  Question 1            Ua(Xa)= Xa^α                                             Xa+Xb=1  Ub(Xb)=Xb                                             (da,db) = (0,0)        Ub(Xb)= 1-Xa     Ub(Xb)=1-Ua(Xa)^(1/α)         -g´(Ua)=(Ub-db)/(Ua-da)               (1/α)*Ua^(1/α-1)=(1-Ua^█(1/α@   ))/Ua             (1/α)Ua^(1/α   )=1-Ua^(1/α)        (1/α) Ua^(1/α)+Ua^(1/α   )=1        (1+α) Ua^(1/α)=α     Ua=(α/(1+α))^α     NBS is:    Xa=(α/(1+α))    ;    Xb=1-(α/(1+α)) ;   Ua=(α/(1+α))^α; Ub=1- (α/(1+α))          α can be interpreted as how much somebody A values the  saloon in terms of utility. If he doesnt value the  bar (if he doesnt  alike(p) the cake) then α is 0, and he  fares 0 from the cake. If he really likes the cake, so α = 1, and he gets ½ of the cake.                Question 2    The  direction of i   mplementing the discount factors in NBS is by giving or  fetching bargaining  forcefulness depending if a player is more or less patient. Its necessary to use the asymmetric NBS to  turn this question.

    Ua(Xa)= Xa                                    Xa+Xb=1           Ub(Xb)=Xb                                 ( da,db)= (0,0)         Ub(Xb)=1-Ua(Xa)           -g´(Ua)=γ/(1-γ)    (Ub-db)/(Ua-da)             1= γ/(1-γ)   (1-Ua)/Ua     Ua= γ                         NBS is   Ua= γ ; Ub=1- γ ; Xa= γ   ; Xb=1- γ        b)    If they apply Rubenstein model then:                            1-Xa= δbXb   ;   1-Xb= δaXa                           By    solving the  dodge:                       X!   ^* a=(1-δb)/(1-δaδb) ; X^* b=(1-δa)/(1-δaδb)                          equipoise Strategies    participant A  invariably  plead X^* a=(1-δb)/(1-δaδb) and accepts any offer Xb≤(1-δa)/(1-δaδb)   Player B always offer X^* b=(1-δa)/(1-δaδb) and accepts any offer Xa≤(1-δb)/(1-δaδb)    Equilibrium payoffs...If you want to get a full essay, order it on our website: 
OrderEssay.netIf you want to get a full information about our service, visit our page: How it works.  
 
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.