Solving Optimization Problems with Diseconomies of Scale via

Solving Optimization Problems with Diseconomies of Scale via

Solving Optimization Problems with Diseconomies of Scale via
Decoupling
Konstantin Makarychev – Microsoft Research
Maxim Sviridenko – Yahoo Labs
Simons Symposium on Approximation Algorithms, February 26, 2014
What is a diseconomy of scale?
Cost of Resources
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
What is a diseconomy of scale?
Cost of Resources
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
What is a diseconomy of scale?
Cost of Energy used for Computing
8
7
Energy consumption grows as 𝑥 𝑞 as a function of speed 𝑥.
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Example: Energy Efficient Routing
Given:
• graph 𝐺
• set of demand pairs (𝑠𝑖 , 𝑡𝑖 , 𝑑𝑖 )
Goal:
• route 𝑑𝑖 units of unsplittable flow
from 𝑠𝑖 to 𝑡𝑖
So as to minimize energy:
𝑞
𝑐𝑒 𝑥𝑒
𝑒∈𝐸
Example: Energy Efficient Routing
Given:
• graph 𝐺
• set of demand pairs (𝑠𝑖 , 𝑡𝑖 , 𝑑𝑖 )
Goal:
• route 𝑑𝑖 units of unsplittable flow
from 𝑠𝑖 to 𝑡𝑖
So as to minimize energy:
𝑞
𝑐𝑒 𝑥𝑒
𝑒∈𝐸
Example: Energy Efficient Routing
Given:
• graph 𝐺
• set of demand pairs (𝑠𝑖 , 𝑡𝑖 , 𝑑𝑖 )
Goal:
• route 𝑑𝑖 units of unsplittable flow
from 𝑠𝑖 to 𝑡𝑖
So as to minimize energy:
𝑞
𝑐𝑒 𝑥𝑒
𝑒∈𝐸
= 5 × 1𝑞 + 2 × 2𝑞
What’s known?
 Arbitrary 𝑑𝑖 :
Andrews, Fernández Anta, Zhang, Zhao – 𝑂(#𝑑𝑒𝑚𝑎𝑛𝑑𝑠 + log 𝑑𝑚𝑎𝑥 )
approximation
 All 𝑑𝑖 are the same:
 Andrews, Fernández Anta, Zhang, Zhao – 𝑂(1) approximation
𝑞
 Bampis, Kononov, Letsios, Lucarelli, Sviridenko – 𝑃 𝑞 approximation
 We give 𝑃
𝑞
𝑞
approximation for the general case
Graph of
𝑃
𝑞
𝑞
2.5
2
1.5
1
0.5
0
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
LP Relaxation
 𝒫𝑖 is the set of 𝑠𝑖 → 𝑡𝑖 paths
 variable 𝜆𝑝 for all 𝑝 ∈ 𝒫𝑖
constraints:
• 𝑝∈𝒫𝑖 𝜆𝑝 = 𝑑𝑖 ∀𝑖
• 𝜆𝑝 ∈ [0, 𝑑𝑖 ]
Let Λ be the set of feasible 𝜆’s.
𝜆𝑝
LP Relaxation
 𝒫𝑖 is the set of 𝑠𝑖 → 𝑡𝑖 paths
 variable 𝜆𝑝 for all 𝑝 ∈ 𝒫𝑖
Minimize
𝑒 𝑐𝑒
𝑝:𝑒∈𝑝 𝜆𝑝
𝑞2
Subject to 𝜆 ∈ Λ
constraints:
𝜆𝑝 = 1 𝑛
• 𝑝∈𝒫𝑖 𝜆𝑝 = 𝑑𝑖 ∀𝑖
• 𝜆𝑝 ∈ [0, 𝑑𝑖 ]
Let Λ be the set of feasible 𝜆’s.
cost = 2𝑛 ×
1
𝑛2
= 2/n
LP Relaxation – Local Distributions of Paths
Minimize
Let
 𝐹𝑞,𝑒 = min 𝔼𝑌 𝑒 ∼𝒟𝑒
𝒟𝑒
𝑒 𝑐𝑒 𝐹𝑞,𝑒 (𝜆) subject to
𝑒
𝑌
𝑖 𝑖
𝑞
,
where
• 𝒟𝑒 is a distribution over r.v. {𝑌𝑖𝑒 };
• 𝑌𝑖𝑒 is the amount of flow 𝑠𝑖 → 𝑡𝑖 going via 𝑒
• 𝔼 𝑌𝑖𝑒 = 𝑝∈𝒫𝑖 :𝑒∈𝑝 𝜆𝑝
• 𝑌𝑖𝑒 ∈ {0, 𝑑𝑖 }
𝜆∈Λ
Rounding Algorithm
LP:
Minimize
𝑒 𝑐𝑒 𝐹𝑞,𝑒 (𝜆) subject
where 𝐹𝑞,𝑒 = min 𝔼𝑌 𝑒 ∼𝒟𝑒
𝒟𝑒
to 𝜆 ∈ Λ
𝑞
𝑒
𝑝:𝑒∈𝑝 𝑌𝑝
.
Algorithm A:
For each 𝑖, pick a random path 𝑝: 𝑠𝑖 → 𝑡𝑖 with probability 𝜆𝑝 /𝑑𝑖 .
Pick paths 𝑝 independently for all 𝑖.
Analysis
Algorithm A:
For each 𝑖, pick a random path 𝑝: 𝑠𝑖 → 𝑡𝑖 with probability 𝜆𝑝 /𝑑𝑖 .
Pick paths 𝑝 independently for all 𝑖.
Analysis:
Compare LP and ALG edge by edge. For every edge 𝑒:
 Let 𝑋𝑖 = amount of flow from 𝑠𝑖 to 𝑡𝑖 routed via 𝑒 by A.
 All 𝑋𝑖 are independent.
 𝑎𝑙𝑔. 𝑐𝑜𝑠𝑡 𝑒 = 𝑐𝑒 𝑖 𝑋𝑖𝑒 𝑞 .
Analysis
We need to compare
 𝔼[𝑎𝑙𝑔. 𝑐𝑜𝑠𝑡 𝑒 ] = 𝔼[𝑐𝑒
 𝐿𝑃. 𝑐𝑜𝑠𝑡 𝑒
= 𝔼[𝑐𝑒
𝑒 𝑞
𝑋
]
𝑖 𝑖
𝑒 𝑞
𝑌
]
𝑖 𝑖
 Each 𝑋𝑖𝑒 has the same distribution as 𝑌𝑖 .
 All 𝑋𝑖𝑒 are independent.
 But 𝑌𝑖𝑒 are not independent 
Decoupling Inequality (de la Peña‘90)
If
◦ 𝑌1 , … , 𝑌𝑛 — jointly distributed nonnegative r.v.’s;
◦ 𝑋1 , … , 𝑋𝑛 — independent nonnegative r.v.;
◦ Each 𝑋𝑖 has the same distribution as 𝑌𝑖 ;
Then
𝑋1 + ⋯ + 𝑋𝑛
𝑞
for some absolute constant 𝐶𝑞 .
≤ 𝐶𝑞 ⋅ 𝑌1 + ⋯ + 𝑌𝑛
𝑞
Decoupling Inequality (de la Peña‘90)
If
◦ 𝑌1 , … , 𝑌𝑛 — jointly distributed nonnegative r.v.’s;
◦ 𝑋1 , … , 𝑋𝑛 — independent nonnegative r.v.;
◦ Each 𝑋𝑖 has the same distribution as 𝑌𝑖 ;
Then
𝑋1 + ⋯ + 𝑋𝑛
𝑞
≤ 𝐶𝑞 ⋅ 𝑌1 + ⋯ + 𝑌𝑛
De la Peña, Ibragimov, Sharakhmetov ‘03:
𝑞
𝐶𝑞
𝑞
𝐶𝑞
𝑞
≤ 2 for 𝑞 ∈ (1,2];
≤
𝑃
𝑞
𝑞
for 𝑞 ≥ 2.
Decoupling Inequality (de la Peña‘90)
If
◦ 𝑌1 , … , 𝑌𝑛 — jointly distributed nonnegative r.v.’s;
◦ 𝑋1 , … , 𝑋𝑛 — independent nonnegative r.v.;
◦ Each 𝑋𝑖 has the same distribution as 𝑌𝑖 ;
Then
𝑋1 + ⋯ + 𝑋𝑛
We show that
𝑞
𝐶𝑞
= 𝑃
𝑞
𝑞,
𝑞
≤ 𝐶𝑞 ⋅ 𝑌1 + ⋯ + 𝑌𝑛
𝑞
where 𝑃 is a Poisson r.v. with parameter 1.
What is the Poisson distribution?
 𝑃𝜆 is Poisson with parameter 𝜆 ⇒ ℙ 𝑃 = 𝑘 =
𝜆𝑘 𝑒 −𝜆
.
𝑘!
 𝑃𝜆 = number of “chosen” points in the Poisson process:
𝑁 points; each chosen w.p. 𝜆/𝑁
What is the Poisson distribution?
 𝑃𝜆 is Poisson with parameter 𝜆 ⇒ ℙ 𝑃 = 𝑘 =
𝜆𝑘 𝑒 −𝜆
.
𝑘!
 𝑃𝜆 = number of “chosen” points in the Poisson process:
𝑁 points; each chosen w.p. 𝜆/𝑁
Decoupling Inequality (de la Peña‘90)
If
◦ 𝑌1 , … , 𝑌𝑛 — jointly distributed nonnegative r.v.’s;
◦ 𝑋1 , … , 𝑋𝑛 — independent nonnegative r.v.;
◦ Each 𝑋𝑖 has the same distribution as 𝑌𝑖 ;
Then
𝑋1 + ⋯ + 𝑋𝑛
We show that
𝑞
𝐶𝑞
= 𝑃
𝑞
𝑞,
𝑞
≤ 𝐶𝑞 ⋅ 𝑌1 + ⋯ + 𝑌𝑛
𝑞
where 𝑃 is a Poisson r.v. with parameter 1.
Better Language?
Yes – Convex Stochastic Order
Convex Order
Def: 𝑋 ≤𝑐𝑥 𝑌 if for every convex 𝜑: 𝔼𝜑 𝑋 ≤ 𝔼𝜑(𝑌).
Basic Properties:





≤𝑐𝑥 is a property of distribution: 𝑋 =𝑠𝑡 𝑍, 𝑋 ≤𝑐𝑥 𝑌 ⟹ 𝑍 ≤𝑐𝑥 𝑌
If 𝑋 ≤𝑐𝑥 𝑌, 𝑌 ≤𝑐𝑥 𝑍 ⟹ 𝑋 ≤𝑐𝑥 𝑍
If 𝑋 ≤𝑐𝑥 𝑌, then 𝛼𝑋 + 𝛽 ≤𝑐𝑥 𝛼𝑌 + 𝛽
If 𝑋 ≤𝑐𝑥 𝑌, then 𝔼𝑋 = 𝔼𝑌
We can verify “≤” only for 𝜑 with 𝜑 0 = 0
Example: Random Walk
𝑋 ≤𝑐𝑥 𝑌
8
𝑋 = 𝑊𝜏1
7
𝜏1 ≤ 𝜏2
6
𝑌 = 𝑊𝜏2
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Example: 𝐵 ≤𝑐𝑥 𝑃
 𝐵 – Bernoulli r.v. with parameter 𝑝;
 𝑃 – Poisson r.v. with parameter 𝑝
 Then, 𝐵 ≤𝑐𝑥 𝑃
Proof: 𝜑 – convex with 𝜑 0 = 0
10
𝜑 𝑃 ≥𝑙 𝑃
𝜑 𝐵 = 𝑙(𝐵)
9
8
𝜑 𝑡
7
6
5
4
𝔼𝜑 𝑃 ≥ 𝔼𝑙 𝑃 = 𝑙 𝔼𝑃
= 𝑙 𝔼𝐵 = 𝔼𝑙 𝐵 = 𝔼𝜑(𝐵)
ℓ 𝑡
3
2
1
0
0
0.5
1
1.5
2
2.5
3
Lemma I
 If 𝑋, 𝑌, 𝑍 are independent r.v’s and 𝑋 ≤𝑐𝑥 𝑌,
 Then Z + 𝑋 ≤𝑐𝑥 𝑍 + 𝑌.
Proof:
𝔼𝜑 𝑍 + 𝑋 = 𝔼 𝔼 𝜑 𝑍 + 𝑋
𝑍] ≤𝑐𝑥 𝔼 𝔼 𝜑 𝑍 + 𝑌 𝑍] = 𝔼𝜑 𝑍 + 𝑌
Since for any fixed 𝑧, 𝔼𝜑 𝑧 + 𝑋 ≤𝑐𝑥 𝔼𝜑 𝑧 + 𝑌 .
Lemma II
 𝑋1 , … , 𝑋𝑛 are independent r.v.’s; 𝑌1 , … , 𝑌𝑛 are independent r.v.’s;
 𝑋𝑖 ≤𝑐𝑥 𝑌𝑖 for all 𝑖
 Then 𝑋1 + ⋯ + 𝑋𝑛 ≤𝑐𝑥 𝑌1 + ⋯ + 𝑌𝑛
Proof:
𝑋1 + 𝑋2 + 𝑋3 + 𝑋4 ≤𝑐𝑥
𝑋1 + 𝑋2 + 𝑋3 + 𝑌4 ≤𝑐𝑥
𝑋1 + 𝑋2 + 𝑌3 + 𝑌4 ≤𝑐𝑥
𝑋1 + 𝑌2 + 𝑌3 + 𝑌4 ≤𝑐𝑥
𝑌1 + 𝑌2 + 𝑌3 + 𝑌4
Decoupling Inequality
If
◦ 𝑌1 , … , 𝑌𝑛 — jointly distributed nonnegative r.v.’s;
◦ 𝑋1 , … , 𝑋𝑛 — independent nonnegative r.v.;
◦ Each 𝑋𝑖 has the same distribution as 𝑌𝑖 ;
Then
𝑋1 + ⋯ + 𝑋𝑛 ≤𝑐𝑥 𝑃 ⋅ (𝑌1 + ⋯ + 𝑌𝑛 )
where 𝑃 is a Poisson r.v. with parameter 1 independent of 𝑌𝑖 ’s.
Proof of de la Peña’s Inequality


𝑞
𝑋1 + ⋯ + 𝑋𝑛 𝑞 = 𝔼 𝑋1 + ⋯ + 𝑋𝑛 𝑞
𝑞
𝑌1 + ⋯ + 𝑌𝑛 𝑞 = 𝔼 𝑌1 + ⋯ + 𝑌𝑛 𝑞
𝔼 𝑋1 + ⋯ + 𝑋𝑛
𝑞
≤ 𝔼 𝑃 ⋅ 𝑌1 + ⋯ + 𝑌𝑛
𝑞
= 𝔼 𝑃𝑞 𝔼 𝑌1 + ⋯ + 𝑌𝑛
𝑞
Why Poisson?
 𝑋1 , … , 𝑋𝑛 : independent Bernoulli r.v.’s with parameter 1 𝑛;
 𝑌1 , … , 𝑌𝑛 : Pick a random 𝑖 ∈ [𝑛], let 𝑌𝑖 = 1; 𝑌𝑗 = 0 for 𝑗 ≠ 𝑖;
 Then
𝑋1 + ⋯ + 𝑋𝑛 ≈ 𝑃
𝑌1 + ⋯ + 𝑌𝑛 = 1
So
𝑋1 + ⋯ + 𝑋𝑛 ≈ 𝑃(𝑌1 + ⋯ + 𝑌𝑛 )
Special Case of the Theorem
 𝑋1 , … , 𝑋𝑛 are independent Bernoulli random variables with 𝔼 𝑋𝑖 = 1/𝑛;
 𝑌1 , … , 𝑌𝑛 are Bernoulli random variables s.t.
𝑖 𝑌𝑖
= 1 (always)
 Then
𝑌1 + ⋯ + 𝑌𝑛 ≤𝑐𝑥 𝑋1 + ⋯ + 𝑋𝑛 ≤𝑐𝑥 𝑃 𝑌1 + ⋯ + 𝑌𝑛
1 ≤𝑐𝑥 𝑋1 + ⋯ + 𝑋𝑛 ≤𝑐𝑥 𝑃
Special Case of the Theorem
 𝑋1 , … , 𝑋𝑛 are independent Bernoulli random variables;
 𝑌1 , … , 𝑌𝑛 are Bernoulli random variables s.t.
𝑖 𝑌𝑖
= 1 (always)
 𝛼1 , … , 𝛼𝑛 are non-negative numbers
 Then
𝛼1 𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 ≤𝑐𝑥 𝛼1 𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≤𝑐𝑥 𝑃 𝛼1 𝑌1 + ⋯ +𝛼𝑛 𝑌𝑛
General Case
We compare

𝑖 𝑌𝑖

𝑦∈𝒴 𝐵𝑦

𝑦∈𝒴

𝑖 𝑋𝑖
=
𝑦∈𝒴 𝜒
⋅
𝑌=𝑦 ⋅
𝒴
𝑖 𝑦𝑖
𝑖 𝑦𝑖
𝑖𝑦
𝐵
𝑦
𝑖
𝑖
𝐵𝑦 —Bernoulli: ℙ 𝐵𝑦 = 1 = ℙ 𝑌 = 𝑦
𝑦 = (𝑦1 , … , 𝑦𝑛 )
General Case
Let 𝒴 be the support of 𝑌 = (𝑌1 , … , 𝑌𝑛 ). Then
𝑌𝑖 =
𝑦∈𝒴 𝜒 𝑌 = 𝑦 ⋅ 𝑦𝑖 ≤𝑐𝑥
where 𝐵𝑦𝑖 is a Bernoulli with parameter ℙ 𝑌 = 𝑦 .
𝑖 ⋅𝑦
𝐵
𝑖
𝑦∈𝒴 𝑦
General Case
Let 𝒴 be the support of 𝑌 = (𝑌1 , … , 𝑌𝑛 ). Then
𝑋𝑖 =𝑠𝑡 𝑌𝑖 =
𝑦∈𝒴 𝜒 𝑌 = 𝑦 ⋅ 𝑦𝑖 ≤𝑐𝑥
where 𝐵𝑦𝑖 is a Bernoulli with parameter ℙ 𝑌 = 𝑦 .
𝑖 ⋅𝑦
𝐵
𝑖
𝑦∈𝒴 𝑦
General Case
Let 𝒴 be the support of 𝑌 = (𝑌1 , … , 𝑌𝑛 ). Then
𝑋𝑖 =𝑠𝑡 𝑌𝑖 =
𝑦∈𝒴 𝜒 𝑌 = 𝑦 ⋅ 𝑦𝑖 ≤𝑐𝑥
𝑖 ⋅𝑦
𝐵
𝑖
𝑦∈𝒴 𝑦
where 𝐵𝑦𝑖 is a Bernoulli with parameter ℙ 𝑌 = 𝑦 . Thus,
𝑖 𝑋𝑖 ≤𝑐𝑥
𝑖
𝑖 ⋅𝑦 =
𝐵
𝑖
𝑦∈𝒴 𝑦
𝑦∈𝒴(
𝑖 ⋅𝑦 )
𝐵
𝑖
𝑖 𝑦
Now,
𝑦∈𝒴 ( 𝑖 𝑦𝑖 )
𝐵𝑦 ≤𝑐𝑥 𝑃 ⋅
𝑦∈𝒴( 𝑖 𝑦𝑖 )
𝜒 𝑌=𝑦 =𝑃
𝑖 𝑌𝑖
Lemma:
𝑖 𝑦𝑖 𝐵
𝑖
≤
𝑖 𝑦𝑖 𝐵
Rescale 𝑦 s.t. 𝑦𝑖 = 1.
Then,
𝔼𝜑
𝑖
𝑦
𝐵
𝑖 𝑖
≤
𝑖
𝑦
𝔼
𝜑
𝐵
𝑖 𝑖
=
𝑖 𝑦𝑖
𝔼𝜑 𝐵
=𝔼𝜑 𝐵
Special Case of the Theorem
 𝑋1 , … , 𝑋𝑛 are independent Bernoulli random variables;
 𝑌1 , … , 𝑌𝑛 are Bernoulli random variables s.t.
𝑖 𝑌𝑖
= 1 (always)
 𝛼1 , … , 𝛼𝑛 are non-negative numbers
 Then
𝛼1 𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 ≤𝑐𝑥 𝛼1 𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≤𝑐𝑥 𝑃 𝛼1 𝑌1 + ⋯ +𝛼𝑛 𝑌𝑛
Proof: 𝛼1𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 ≤𝑐𝑥 𝛼1𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛
 Consider a convex function 𝜑 with 𝜑 0 = 0
 𝜑 is superadditive: 𝜑 𝑎 + 𝑏 ≥ 𝜑 𝑎 + 𝜑(𝑏)
 𝜑 𝛼1 𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≥ 𝜑 𝛼1 𝑋1 + ⋯ + 𝜑 𝛼𝑛 𝑋𝑛
𝔼𝜑
𝔼𝜑
𝑖 𝛼𝑖 𝑌𝑖
=
𝑖 𝛼𝑖
𝑋𝑖 ≥
𝑖 𝔼𝜑
𝑖 𝜑(𝛼𝑖 )ℙ(𝑌𝑖
𝛼𝑖 𝑋𝑖
= 1) =
𝑖 𝔼𝜑
𝛼𝑖 𝑌𝑖
Proof: 𝛼1𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≤𝑐𝑥 𝑃(𝛼1𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 )
 Let 𝑃𝑖 be a Poisson r.v. with parameter ℙ(𝑋𝑖 = 1), then
𝛼1 𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≤𝑐𝑥 𝛼1 𝑃1 + ⋯ + 𝛼𝑛 𝑃𝑛
 𝑃 = 𝑃1 + ⋯ 𝑃𝑛 is a Poisson r.v. with parameter 1.
Proof: 𝛼1𝑃1 + ⋯ + 𝛼𝑛 𝑃𝑛 ≤𝑐𝑥 𝑃(𝛼1𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 )
 Let 𝑃𝑖 be a Poisson r.v. with parameter ℙ(𝑋𝑖 = 1), then
𝛼1 𝑋1 + ⋯ + 𝛼𝑛 𝑋𝑛 ≤𝑐𝑥 𝛼1 𝑃1 + ⋯ + 𝛼𝑛 𝑃𝑛
 𝑃 = 𝑃1 + ⋯ 𝑃𝑛 is a Poisson r.v. with parameter 1.
Proof: 𝛼1𝑃1 + ⋯ + 𝛼𝑛 𝑃𝑛 ≤𝑐𝑥 𝑃(𝛼1𝑌1 + ⋯ + 𝛼𝑛 𝑌𝑛 )
 Let 𝑃 = 𝑃1 + ⋯ + 𝑃𝑛 . Condition on 𝑃 = 𝑘 > 0:
𝔼 𝜑
𝑖 𝛼𝑖 𝑃𝑖 ) 𝑃 = 𝑘 = 𝔼 𝜑(
≥𝔼
𝑃𝑖
𝑖 𝑘
𝑃𝑖
𝑖 𝑘 𝛼𝑖 𝑘) |𝑃
⋅ 𝜑 𝛼𝑖 𝑘 |𝑃 = 𝑘
=
𝑖 𝜑(𝛼𝑖 𝑘) 𝔼
𝑃𝑖
|𝑃
𝑘
=
𝑖 𝜑(𝛼𝑖 𝑘) 𝔼
𝑃𝑖 =
=𝔼 𝜑 𝑃
=𝑘
𝑖 𝛼𝑖 𝑌𝑖 )
=𝑘
𝑖 𝜑(𝛼𝑖 𝑘) 𝔼
𝑃=𝑘
𝑌𝑖
Conclusion
 Introduced a new framework for problems with “diseconomies of scale”:
 Energy minimization
 𝐿𝑝 -minimization
 Found the exact constant in de la Peña’s inequality.
 Generalized it to other convex and concave functions.
𝑋1 + ⋯ + 𝑋𝑛 ≤𝑐𝑥 𝑃 ⋅ (𝑌1 + ⋯ + 𝑌𝑛 )
Thank you!