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The efficiency of resource allocationmechanisms for budget-constrained users
Ioannis Caragiannis Alexandros A. Voudouris
University of Patras
A resource allocation problem
Question: How can we distribute a single divisible resourceamong a set of self-interested users with budget constraints?
Mechanism(auction)
A resource allocation problem
Question: How can we distribute a single divisible resourceamong a set of self-interested users with budget constraints?
Mechanism(auction)
A resource allocation problem
Question: How can we distribute a single divisible resourceamong a set of self-interested users with budget constraints?
Mechanism(auction)
user signal
𝑠𝑖
A resource allocation problem
Question: How can we distribute a single divisible resourceamong a set of self-interested users with budget constraints?
Mechanism(auction)
user signal allocation, payment
𝑠𝑖 𝑔𝑖 𝐬 , 𝑝𝑖 𝐬
Example: pay-your-signal mechanisms
• Each user pays her own signal: pi(s) = si
• Kelly mechanism [Kelly 1997, Kelly et al. 1998]:
gi(s) = si∑j sj
• SHmechanism [Sanghavi and Hajek 2004]:
gi(s) = si
maxℓ sℓ
∫ 1
0
∏j =i
(1 − sj
maxℓ sℓt
)dt
Example: pay-your-signal mechanisms
• Each user pays her own signal: pi(s) = si
• Kelly mechanism [Kelly 1997, Kelly et al. 1998]:
gi(s) = si∑j sj
• SHmechanism [Sanghavi and Hajek 2004]:
gi(s) = si
maxℓ sℓ
∫ 1
0
∏j =i
(1 − sj
maxℓ sℓt
)dt
Example: pay-your-signal mechanisms
• Each user pays her own signal: pi(s) = si
• Kelly mechanism [Kelly 1997, Kelly et al. 1998]:
gi(s) = si∑j sj
• SHmechanism [Sanghavi and Hajek 2004]:
gi(s) = si
maxℓ sℓ
∫ 1
0
∏j =i
(1 − sj
maxℓ sℓt
)dt
User characteristics and selfishness
Each user i has• a valuation function vi : [0, 1] → R≥0 that represents her value forfractions of the resource (increasing, differentiable, concave)
• a budget ci ∈ R≥0 ∪ {+∞} that restricts the payments she can afford
Strategic behavior• Users are utility-maximizers• Select si in order to maximize utility: ui(s) = vi(gi(s)) − pi(s)
• if pi(s) > ci then ui(s) = −∞
Equilibrium• A signal vector s for which all players simultaneously maximize theirpersonal utilities
User characteristics and selfishness
Each user i has• a valuation function vi : [0, 1] → R≥0 that represents her value forfractions of the resource (increasing, differentiable, concave)
• a budget ci ∈ R≥0 ∪ {+∞} that restricts the payments she can afford
Strategic behavior• Users are utility-maximizers• Select si in order to maximize utility: ui(s) = vi(gi(s)) − pi(s)
• if pi(s) > ci then ui(s) = −∞
Equilibrium• A signal vector s for which all players simultaneously maximize theirpersonal utilities
User characteristics and selfishness
Each user i has• a valuation function vi : [0, 1] → R≥0 that represents her value forfractions of the resource (increasing, differentiable, concave)
• a budget ci ∈ R≥0 ∪ {+∞} that restricts the payments she can afford
Strategic behavior• Users are utility-maximizers• Select si in order to maximize utility: ui(s) = vi(gi(s)) − pi(s)
• if pi(s) > ci then ui(s) = −∞
Equilibrium• A signal vector s for which all players simultaneously maximize theirpersonal utilities
User characteristics and selfishness
Each user i has• a valuation function vi : [0, 1] → R≥0 that represents her value forfractions of the resource (increasing, differentiable, concave)
• a budget ci ∈ R≥0 ∪ {+∞} that restricts the payments she can afford
Strategic behavior• Users are utility-maximizers• Select si in order to maximize utility: ui(s) = vi(gi(s)) − pi(s)
• if pi(s) > ci then ui(s) = −∞
Equilibrium• A signal vector s for which all players simultaneously maximize theirpersonal utilities
Efficiency without budgets
• The social welfare of an allocation d in game G is
SW(d, G) =∑
i
vi(di)
• Price of anarchy of mechanism M ,
PoA(M) = supG∈M
sups∈EQ(G)
maxd SW(d, G)SW(g(s), G)
Efficiency without budgets
• PoA(Kelly) = 4/3 [Johari and Tsitsiklis, 2004]• PoA(SH) ≈ 8/7 [Sanghavi and Hajek, 2004]
• There exist mechanisms that achieve full efficiency: MBmechanisms[Maheswaran and Basar, 2006]
gi(s) = si∑j sj
pi(s) = si
∑j =i
sj
Efficiency without budgets
• PoA(Kelly) = 4/3 [Johari and Tsitsiklis, 2004]• PoA(SH) ≈ 8/7 [Sanghavi and Hajek, 2004]• There exist mechanisms that achieve full efficiency: MBmechanisms[Maheswaran and Basar, 2006]
gi(s) = si∑j sj
pi(s) = si
∑j =i
sj
Efficiency with budgets
• The price of anarchy may be arbitrarily large
• Example: a player with high value, but really low budget
0 1
v(x)
fraction
value
Efficiency with budgets
• The price of anarchy may be arbitrarily large• Example: a player with high value, but really low budget
0 1
v(x)
fraction
value
Efficiency with budgets
• The price of anarchy may be arbitrarily large• Example: a player with high value, but really low budget
0 EQ 1
v(EQ)
v(x)
fraction
value
Efficiency with budgets
• The price of anarchy may be arbitrarily large• Example: a player with high value, but really low budget
0 EQ OPT 1
v(EQ)
v(OPT)
v(x)
fraction
value
Efficiency with budgets
• Alternative: liquid welfare [Dobzinski and Paes Leme, 2014]• The liquid welfare of an allocation d in game G is
LW(d, G) =∑
i
min{vi(di), ci}
• Liquid price of anarchy of mechanism M ,
LPoA(M) = supG∈M
sups∈EQ(G)
maxd LW(d, G)LW(g(s), G)
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞
• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)
• wlog d1 ≤ d2 ⇔ d1 ≤ 12 , d2 ≥ 1
2• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)
• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well
• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 32
LPoA(any n-player mechanism M ) ≥ 2 − 1n
• Proof for two players: lower bound of 3/2
• Game G: v1(x) = v2(x) = x and c1 = c2 = +∞• Equilibrium: s = (s1, s2) with allocation d = (d1, d2)• wlog d1 ≤ d2 ⇔ d1 ≤ 1
2 , d2 ≥ 12
• Same valuations, no budget constraints ⇒ LPoA(G) = 1
• Game G: v1(x) = x, c1 = +∞ and v2(x) = x + d2, c2 = d2
• u1(y, s2) = v1(g1(y, s2)) − p1(y, s2) = u1(y, s2)• u2(s1, y) = v2(g2(s1, y)) − p2(s1, y) = u2(s1, y) + d2
• s is an equilibrium of G as well• LW(d, G) = d1 + d2 = 1 vs. OPT ≥ 1 + d2 ⇒ LPoA(G) ≥ 3
2
Characterization of worst-case games
For every signal vector s, define the following game:
• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) ≤ sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
Characterization of worst-case games
For every signal vector s, define the following game:• player 1 has c1 = +∞ and linear valuation v1(x) = λ1(s)x, with
λ1(s) =
(∂g1(y, s−1)
∂y
∣∣∣∣y=s1
)−1
· ∂p1(y, s−1)∂y
∣∣∣∣y=s1
• all other players i ≥ 2 have budgets ci = pi(s) and affine valuationvi(x) = λi(s)x + ci
• Player 1 contributes her value to the liquid welfare, and all otherscontribute their budgets
LPoA(M) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Concave allocation + convex payment
LPoA bounds (overview)
n players• LPoA(Kelly) = 2 (best possible for large number of players)• LPoA(SH) = 3
Two players• LPoA(E2-PYS) = 1.792 (best possible among all pay-your-signalmechanisms with concave allocation functions)
• LPoA(E2-SR) ≤ 1.529 (almost optimal over all two-playermechanisms)
LPoA bounds (overview)
n players• LPoA(Kelly) = 2 (best possible for large number of players)• LPoA(SH) = 3
Two players• LPoA(E2-PYS) = 1.792 (best possible among all pay-your-signalmechanisms with concave allocation functions)
• LPoA(E2-SR) ≤ 1.529 (almost optimal over all two-playermechanisms)
LPoA bounds (overview)
n players• LPoA(Kelly) = 2 (best possible for large number of players)• LPoA(SH) = 3
Two players• LPoA(E2-PYS) = 1.792 (best possible among all pay-your-signalmechanisms with concave allocation functions)
• LPoA(E2-SR) ≤ 1.529 (almost optimal over all two-playermechanisms)
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
LPoA(Kelly) = 2
LPoA(Kelly) = sups
∑i≥2 pi(s) + λ1(s)∑
i≥2 pi(s) + λ1(s) g1(s)
• Pay-your-signal:∑
i≥2 pi(s) =∑
i≥2 si =: C
• For player 1: g1(s) = s1s1+C
g1(y, s−1) = yy+C ⇒ ∂g1(y,s−1)
∂y = C(y+C)2
p1(y, s−1) = y ⇒ ∂p1(y,s−1)∂y = 1
λ1(s) = (s1 + C)2
C
LPoA(Kelly) = sups1,C
C + (s1 + C)2/C
C + (s1 + C)s1/C= 2
Two players: the design of E2-PYS
• s1 ≤ s2, z = s1s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS• s1 ≤ s2, z = s1
s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS• s1 ≤ s2, z = s1
s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS• s1 ≤ s2, z = s1
s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS• s1 ≤ s2, z = s1
s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS• s1 ≤ s2, z = s1
s2
• Pay-your-signal:p1(s) = s1 p2(s) = s2
• Allocation:g1(s) = f(z) g2(s) = 1 − f(z)
• Compute λ1(s):
g1(y, s2) = f(
ys2
)⇒ ∂g1(y,s2)
∂y = 1s2
f ′(
ys2
)p1(y, s2) = y ⇒ ∂p1(y,s2)
∂y = 1
λ1(s) = s2
f ′(z)
• Characterization:
p2(s) + λ1(s)p2(s) + λ1(s)g1(s)
=s2 + s2
f ′(z)
s2 + s2f ′(z) f(z)
= f ′(z) + 1f ′(z) + f(z)
Two players: the design of E2-PYS
• Require:
f ′(z) + 1f ′(z) + f(z)
= β ⇔ f ′(z) + β
β − 1f(z) = 1
β − 1
• Solve the differential equation:
f(z) = 1β
− 1β
exp(
− β
β − 1z
)• Definition of allocation function:
gi(s) =
1β − 1
β exp(
− ββ−1
si
s3−i
)si ≤ s3−i
β−1β + 1
β exp(
− ββ−1
s3−i
si
)si > s3−i
Two players: the design of E2-PYS
• Require:
f ′(z) + 1f ′(z) + f(z)
= β ⇔ f ′(z) + β
β − 1f(z) = 1
β − 1
• Solve the differential equation:
f(z) = 1β
− 1β
exp(
− β
β − 1z
)
• Definition of allocation function:
gi(s) =
1β − 1
β exp(
− ββ−1
si
s3−i
)si ≤ s3−i
β−1β + 1
β exp(
− ββ−1
s3−i
si
)si > s3−i
Two players: the design of E2-PYS
• Require:
f ′(z) + 1f ′(z) + f(z)
= β ⇔ f ′(z) + β
β − 1f(z) = 1
β − 1
• Solve the differential equation:
f(z) = 1β
− 1β
exp(
− β
β − 1z
)• Definition of allocation function:
gi(s) =
1β − 1
β exp(
− ββ−1
si
s3−i
)si ≤ s3−i
β−1β + 1
β exp(
− ββ−1
s3−i
si
)si > s3−i
LPoA(E2-PYS) = 1.792
• For s1 ≤ s2, by definition, LPoA(E2-PYS) = β
• For s1 > s2, we can show that LPoA(E2-PYS) ≤ β
• By anonymity
f(1) = 12
⇔ 1β
− 1β
exp(
− β
β − 1
)= 1
2⇔ β ≈ 1.792
LPoA(E2-PYS) = 1.792
• For s1 ≤ s2, by definition, LPoA(E2-PYS) = β
• For s1 > s2, we can show that LPoA(E2-PYS) ≤ β
• By anonymity
f(1) = 12
⇔ 1β
− 1β
exp(
− β
β − 1
)= 1
2⇔ β ≈ 1.792
LPoA(E2-PYS) = 1.792
• For s1 ≤ s2, by definition, LPoA(E2-PYS) = β
• For s1 > s2, we can show that LPoA(E2-PYS) ≤ β
• By anonymity
f(1) = 12
⇔ 1β
− 1β
exp(
− β
β − 1
)= 1
2⇔ β ≈ 1.792
The design of E2-SR
• Change the payment function: p1(s) = s1s2
and p2(s) = s2s1
• Follow the same reasoning as with E2-PYS
Open problems
• Is there a simple mechanism that achieves the 2 − 1n bound?
• LPoA bounds over more general equilibrium concepts?• Other resource allocation problems with budget constraints?• Other efficiency benchmarks?
Thank you!
Open problems
• Is there a simple mechanism that achieves the 2 − 1n bound?
• LPoA bounds over more general equilibrium concepts?• Other resource allocation problems with budget constraints?• Other efficiency benchmarks?
Thank you!