Question #262163

Three friends Alice(A), Bob(B), and Charlie(C) are planning to buy a TV.

TV’s are available in all sizes (size is the diagonal length of the display in

inches). Each friend i has a private valuation of the form

vi(x) = aix − x^3/36

for a TV of size x, where i ∈ {A, B, C}

(a)

Use the VCG mechanism to decide which size of the TV should be

bought. Describe the size as a function of aA, aB, aC.

(b)

Compute the payments for each friend under the VCG mechanism in

the situation where Alice has aA = 10, Bob has aB = 40, and Charlie

has aC = 50.


1
Expert's answer
2021-11-17T15:40:12-0500

a)

VCG mechanism:

outcomes:

x(v)=arg maxx(vi(x))x^*(v)=arg\ max_x(\sum v_i(x))

x(vi)=arg maxx(jivj(x))x^*(v_{-i})=arg\ max_x(\sum_{j\neq i} v_j(x))

transfers: 

Agent i receives

ti(v)=jivj(x(v))jivj(x(vi))t_i(v)=\sum_{j\neq i}v_j(x^*(v))-\sum_{j\neq i}v_j(x^*(v_{-i}))


for aA size of the TV:

x is root of equation vA(x)=aAxx3/36v_A(x) = a_Ax − x^3/36

for aB size of the TV:

x is root of equation vB(x)=aBxx3/36v_B(x) = a_Bx − x^3/36

for aC size of the TV:

x is root of equation vC(x)=aCxx3/36v_C(x) = a_Cx − x^3/36


b)

for aA = 10 :

vA(x)=10xx3/36v_A(x) = 10x − x^3/36

for aB = 40 :

vB(x)=40xx3/36v_B(x) = 40x − x^3/36


for aC = 50 :

vC(x)=50xx3/36v_C(x) = 50x − x^3/36


vi(x)=100xx3/12\sum v_i(x)=100x-x^3/12


(vi(x))=100x2/4=0(\sum v_i(x))'=100-x^2/4=0

x=20x=20

outcome that maximizes the sum of values:

x(v)=20x^*(v)=20


for Alice(A):

jivj(x)=90xx3/18\sum_{j\neq i} v_j(x)=90x-x^3/18

jivj(x)=90x2/6=0\sum_{j\neq i} v_j(x)=90-x^2/6=0

x(vA)=540=23.24x^*(v_{-A})=\sqrt{540}=23.24

Alice receives:

tA=9020203/18(9023.2423.243/18)=38.72t_A=90\cdot20-20^3/18-(90\cdot23.24-23.24^3/18)=-38.72


for Bob(B):

jivj(x)=60xx3/18\sum_{j\neq i} v_j(x)=60x-x^3/18

jivj(x)=60x2/6=0\sum_{j\neq i} v_j(x)=60-x^2/6=0

x(vB)=360=18.97x^*(v_{-B})=\sqrt{360}=18.97

Bob receives:

tB=6020203/18(6018.9718.973/18)=3.39t_B=60\cdot20-20^3/18-(60\cdot18.97-18.97^3/18)=-3.39


for Charlie(C):

jivj(x)=50xx3/18\sum_{j\neq i} v_j(x)=50x-x^3/18

jivj(x)=50x2/6=0\sum_{j\neq i} v_j(x)=50-x^2/6=0

x(vA)=300=17.32x^*(v_{-A})=\sqrt{300}=17.32

Charlie receives:

tC=5020203/18(5017.3217.323/18)=21.79t_C=50\cdot20-20^3/18-(50\cdot17.32-17.32^3/18)=-21.79


So, Alice has to pay 38.72, Bob has to pay 3.39, Charlie has to pay 21.79


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