a)
VCG mechanism:
outcomes:
x ∗ ( v ) = a r g m a x x ( ∑ v i ( x ) ) x^*(v)=arg\ max_x(\sum v_i(x)) x ∗ ( v ) = a r g ma x x ( ∑ v i ( x ))
x ∗ ( v − i ) = a r g m a x x ( ∑ j ≠ i v j ( x ) ) x^*(v_{-i})=arg\ max_x(\sum_{j\neq i} v_j(x)) x ∗ ( v − i ) = a r g ma x x ( ∑ j = i v j ( x ))
transfers:
Agent i receives
t i ( v ) = ∑ j ≠ i v j ( x ∗ ( v ) ) − ∑ j ≠ i v j ( x ∗ ( v − i ) ) t_i(v)=\sum_{j\neq i}v_j(x^*(v))-\sum_{j\neq i}v_j(x^*(v_{-i})) t i ( v ) = ∑ j = i v j ( x ∗ ( v )) − ∑ j = i v j ( x ∗ ( v − i ))
for a A size of the TV:
x is root of equation v A ( x ) = a A x − x 3 / 36 v_A(x) = a_Ax − x^3/36 v A ( x ) = a A x − x 3 /36
for a B size of the TV:
x is root of equation v B ( x ) = a B x − x 3 / 36 v_B(x) = a_Bx − x^3/36 v B ( x ) = a B x − x 3 /36
for a C size of the TV:
x is root of equation v C ( x ) = a C x − x 3 / 36 v_C(x) = a_Cx − x^3/36 v C ( x ) = a C x − x 3 /36
b)
for a A = 10 :
v A ( x ) = 10 x − x 3 / 36 v_A(x) = 10x − x^3/36 v A ( x ) = 10 x − x 3 /36
for a B = 40 :
v B ( x ) = 40 x − x 3 / 36 v_B(x) = 40x − x^3/36 v B ( x ) = 40 x − x 3 /36
for a C = 50 :
v C ( x ) = 50 x − x 3 / 36 v_C(x) = 50x − x^3/36 v C ( x ) = 50 x − x 3 /36
∑ v i ( x ) = 100 x − x 3 / 12 \sum v_i(x)=100x-x^3/12 ∑ v i ( x ) = 100 x − x 3 /12
( ∑ v i ( x ) ) ′ = 100 − x 2 / 4 = 0 (\sum v_i(x))'=100-x^2/4=0 ( ∑ v i ( x ) ) ′ = 100 − x 2 /4 = 0
x = 20 x=20 x = 20
outcome that maximizes the sum of values:
x ∗ ( v ) = 20 x^*(v)=20 x ∗ ( v ) = 20
for Alice(A):
∑ j ≠ i v j ( x ) = 90 x − x 3 / 18 \sum_{j\neq i} v_j(x)=90x-x^3/18 ∑ j = i v j ( x ) = 90 x − x 3 /18
∑ j ≠ i v j ( x ) = 90 − x 2 / 6 = 0 \sum_{j\neq i} v_j(x)=90-x^2/6=0 ∑ j = i v j ( x ) = 90 − x 2 /6 = 0
x ∗ ( v − A ) = 540 = 23.24 x^*(v_{-A})=\sqrt{540}=23.24 x ∗ ( v − A ) = 540 = 23.24
Alice receives:
t A = 90 ⋅ 20 − 2 0 3 / 18 − ( 90 ⋅ 23.24 − 23.2 4 3 / 18 ) = − 38.72 t_A=90\cdot20-20^3/18-(90\cdot23.24-23.24^3/18)=-38.72 t A = 90 ⋅ 20 − 2 0 3 /18 − ( 90 ⋅ 23.24 − 23.2 4 3 /18 ) = − 38.72
for Bob(B):
∑ j ≠ i v j ( x ) = 60 x − x 3 / 18 \sum_{j\neq i} v_j(x)=60x-x^3/18 ∑ j = i v j ( x ) = 60 x − x 3 /18
∑ j ≠ i v j ( x ) = 60 − x 2 / 6 = 0 \sum_{j\neq i} v_j(x)=60-x^2/6=0 ∑ j = i v j ( x ) = 60 − x 2 /6 = 0
x ∗ ( v − B ) = 360 = 18.97 x^*(v_{-B})=\sqrt{360}=18.97 x ∗ ( v − B ) = 360 = 18.97
Bob receives:
t B = 60 ⋅ 20 − 2 0 3 / 18 − ( 60 ⋅ 18.97 − 18.9 7 3 / 18 ) = − 3.39 t_B=60\cdot20-20^3/18-(60\cdot18.97-18.97^3/18)=-3.39 t B = 60 ⋅ 20 − 2 0 3 /18 − ( 60 ⋅ 18.97 − 18.9 7 3 /18 ) = − 3.39
for Charlie(C):
∑ j ≠ i v j ( x ) = 50 x − x 3 / 18 \sum_{j\neq i} v_j(x)=50x-x^3/18 ∑ j = i v j ( x ) = 50 x − x 3 /18
∑ j ≠ i v j ( x ) = 50 − x 2 / 6 = 0 \sum_{j\neq i} v_j(x)=50-x^2/6=0 ∑ j = i v j ( x ) = 50 − x 2 /6 = 0
x ∗ ( v − A ) = 300 = 17.32 x^*(v_{-A})=\sqrt{300}=17.32 x ∗ ( v − A ) = 300 = 17.32
Charlie receives:
t C = 50 ⋅ 20 − 2 0 3 / 18 − ( 50 ⋅ 17.32 − 17.3 2 3 / 18 ) = − 21.79 t_C=50\cdot20-20^3/18-(50\cdot17.32-17.32^3/18)=-21.79 t C = 50 ⋅ 20 − 2 0 3 /18 − ( 50 ⋅ 17.32 − 17.3 2 3 /18 ) = − 21.79
So, Alice has to pay 38.72, Bob has to pay 3.39, Charlie has to pay 21.79
Comments