Load balance
Course: Price of Anarchy
Professor: Michal Feldman
Student: Nave Frost
23/04/2014
Tight Bounds for Worst-Case Equilibria, Artur Czumaj and Berthold V¨ocking
Model
• 𝑚 independent parallel links with speeds
𝑠1 , 𝑠2 , … , 𝑠𝑚 .
• 𝑛 tasks with weights 𝑤1 , 𝑤2 , … , 𝑤𝑛 .
• Goal: allocate tasks to links to minimize the
maximum load of the links in the system.
Definitions
• Given a combination (𝑗1 , 𝑗2 , … , 𝑗𝑛 ) ∈ [𝑚]𝑛 of
pure strategies, one for each task:
• The load of link 𝑗 is: 𝑗𝑘 =𝑗 𝑤𝑠 𝑘.
𝑗
𝑤𝑘
𝑗𝑘 =𝑗𝑖 𝑠𝑗
𝑖
• The cost for task 𝑖 is:
• 𝑂𝑝𝑡 ≔
min
(𝑗1 ,𝑗2 ,…,𝑗𝑛
)∈[𝑚]𝑛
• Observation: 𝑂𝑝𝑡 ≥
( max (
𝑗∈[𝑚]
max 𝑤𝑖
𝑖
max 𝑠𝑗
𝑗
.
𝑖:𝑗𝑖 =𝑗
𝑤𝑖
𝑠𝑗
))
Mixed strategies
• Let 𝑝𝑖,𝑗 denote the probability that an task 𝑖
∈ 𝑛 sends the entire weight 𝑤𝑖 on link 𝑗.
• Let 𝑙𝑗 denote the expected load on link j.
1
𝑙𝑗 ≔ 𝑠 𝑖∈ 𝑛 𝑤𝑖𝑝𝑖,𝑗.
𝑗
• Let 𝑐𝑖,𝑗 denote the expected cost for task 𝑖 on
link j.
𝑤𝑡 𝑝𝑡,𝑗
𝑤𝑖
𝑤𝑖
𝑐𝑖,𝑗 ≔
+
= 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
𝑠𝑗
𝑠𝑗
𝑠𝑗
𝑡≠𝑖
Nash equilibrium
• Probabilities 𝑝𝑖,𝑗 define Nash equilibrium if
𝑝𝑖,𝑗 > 0 implies 𝑐𝑖,𝑗 ≤ 𝑐𝑖,𝑞 for every 𝑞 ∈ [𝑚].
Price of anarchy
• 𝑝𝑖,𝑗 - probabilities that define Nash equilibrium.
Define 𝐶𝑗 to be random variable indicating the
load of link 𝑗.
• 𝐶𝑗 = 𝑠1
𝑗
• 𝐽𝑖,𝑗
•
𝑛
𝑖=1 𝑤𝑖 𝐽𝑖,𝑗
1, 𝑤. 𝑝. 𝑝𝑖,𝑗
=
0,
𝑒𝑙𝑠𝑒
𝑛
E[𝐶𝑗 ] = 𝐸
𝑖=1 𝑤𝑖 𝐽𝑖,𝑗 =
1
1
𝑛
𝑛
𝑤
𝐸[𝐽
]
=
𝑤𝑝
𝑖,𝑗
𝑠𝑗 𝑖=1 𝑖
𝑠𝑗 𝑖=1 𝑖 𝑖,𝑗
1
𝑠𝑗
= 𝑙𝑗
Price of anarchy (2)
• Let 𝑐 denote the maximum expected load over
all links, that is:
𝑐 = max 𝑙𝑗 = max 𝐸[𝐶𝑗 ]
𝑗∈[𝑚]
𝑗∈[𝑚]
• Let the social cost 𝐶 denote the expected
maximum load over all links, that is:
𝐶 = 𝐸[ max 𝐶𝑗 ]
𝑗∈ 𝑚
• Note that 𝑐 ≤ 𝐶 and possibly 𝑐 ≪ 𝐶.
Price of anarchy (3)
• The price of anarchy defined as :
𝐶
𝑅 = max 𝑂𝑝𝑡
𝑝𝑖,𝑗
Claim 1
• Claim 1:
In Nash equilibrium, if 𝑝𝑖,𝑞 > 0 for certain task 𝑖 ∈ 𝑛 and link 𝑞 ∈ 𝑚 , then
for every link j ∈ [m]
𝑙𝑗 + 𝑤𝑠 𝑖 ≥ 𝑙𝑞
𝑗
In particular, if 𝑙𝑗 + 1 < 𝑙𝑞 then 𝑤𝑖 > 𝑠𝑗 .
• Proof:
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
𝑐𝑖,𝑞 = 𝑙𝑞 + 1 −
𝑤𝑖
𝑤𝑖
≤
𝑙
+
𝑗
𝑠𝑗
𝑠𝑗
𝑖 ≥ 𝑙
𝑝𝑖,𝑞 𝑤
𝑞
𝑠𝑞
Since 𝑝𝑖,𝑞 > 0, the
definition of Nash equilibria implies that 𝑐𝑖,𝑞 ≤ 𝑐𝑖,𝑗 and
𝑤𝑖
hence, 𝑙𝑞 ≤ 𝑙𝑗 + .
𝑠𝑗
• The second part of the claim follows trivially from the first one.
Identical links
• In the case of identical links all speeds are
identical, i.e., for all 𝑗1 , 𝑗2 ∈ 𝑚
𝑠𝑗1 = 𝑠𝑗2
Identical links (2)
• Theorem:
For 𝑚 identical links the price of anarchy is
𝑂
log 𝑚
log log 𝑚
.
• Proof:
Let us first re-scale the weights and speeds in the
problem and assume, without loss of generality,
that for all j ∈ 𝑚 : 𝑠𝑗 = 1 and 𝑂𝑝𝑡 = 1.
Identical links (3)
• Since 𝑂𝑝𝑡 = 1 and we observed that 𝑂𝑝𝑡
max 𝑤𝑖
≥ 𝑖
, then ∀i ∈ 𝑛 : 𝑤𝑖 ≤ 1.
max 𝑠𝑗
𝑗
• Since 𝑂𝑝𝑡 = 1 there must exist a 𝑗 such that
𝑙𝑗 ≤ 1.
• Since 𝑝𝑖,𝑗 define Nash equilibrium it implies
that, according to claim 1:
𝑐 = max 𝑙𝑗 ≤ 2
𝑗∈[𝑚]
Identical links (4)
• We will estimate the load 𝐶𝑗 of any link 𝑗 ∈ [𝑚] by applying to 𝐶𝑗 =
𝑛
𝑖=1 𝑤𝑖 𝐽𝑖,𝑗 a standard concentration inequality due to Hoeffding
bound.
• Hoeffding bound: Let 𝑋1 , … , 𝑋𝑁 be independent random variables
with values in the interval [0, 𝑧] for some 𝑧 > 0, and let 𝑡𝑋 =
𝑛
𝑖=1 𝑋𝑖 ,
then for any 𝑡 it holds that Pr 𝑋 ≥ 𝑡 ≤
• Pr 𝐶𝑗 ≥ 𝑡 = Pr
𝑒𝑙𝑗 𝑡
𝑡
≤
2𝑒 𝑡
𝑡
𝑛
𝑖=1 𝑤𝑖 𝐽𝑖,𝑗
≥𝑡 ≤
For any 𝑡 > 0.
𝑒𝐸
𝑛
𝑖=1 𝑤𝑖 𝐽𝑖,𝑗
𝑡
𝑒𝐸 𝑋
𝑡
𝑡
=
𝑧
𝑒𝐸 𝐶𝑗
𝑡
𝑡
=
Identical links (5)
2𝑒 𝑡
We saw that Pr 𝐶𝑗 ≥ 𝑡 ≤
For any 𝑡 > 0.
𝑡
3 ln 𝑚
Therefore, if we pick 𝑡 =
we will get the following:
ln ln 𝑚
∞
C = 𝐸 max 𝐶𝑗 ≤ t+
𝑗∈ 𝑚
∞
𝑚·𝜏 2𝑒 𝜏
≤ t+
𝜏=𝑡
Pr ∃𝑗 ∈ 𝑚 : 𝐶𝑗 ≥ 𝜏 ·𝜏
𝜏=𝑡
𝜏
∞
2−𝜏
≤ t+
𝜏=𝑡
3 ln 𝑚
≤𝑡+1=
+1
ln ln 𝑚
Upper bound – Speed varies
• Lemma 1:
The maximum expected load 𝑐 satisfies
c = 𝑂𝑝𝑡·𝑂 min
log 𝑚
𝑠1
, log
log log 𝑚
𝑠𝑚
where it is assumed that the speeds satisfy 𝑠1 ≥ ⋯ ≥ 𝑠𝑚 .
• Lemma 2:
The social cost 𝐶 satisfies
log 𝑚
C = 𝑂𝑝𝑡·𝑂
+1
𝑂𝑝𝑡· log 𝑚
log
𝑐
Lemma 1 - Proof
• 𝑝𝑖,𝑗 define a Nash equilibrium.
• Without loss of generality, assume 𝑠1 ≥ ⋯
≥ 𝑠𝑚 .
• Scale weights of such that 𝑂𝑝𝑡 = 1.
• Will show:
c=O
log 𝑚
log log 𝑚
and c = O log
𝑠1
𝑠𝑚
c=O
log 𝑚
log log 𝑚
– Proof
Proof:
For 𝑘 ≥ 1, define 𝑗𝑘 to be the smallest index in
{0,1, … , 𝑚} such that 𝑙𝑗𝑘 +1 < 𝑘 or, if no such index
exists, 𝑗𝑘 = 𝑚.
Observe:
• for every 𝑘 ≥ 1 with 0 < 𝑗𝑘 , all links j < 𝑗𝑘 have
load at least 𝑘.
• for every 𝑘 ≥ 1 with 𝑗𝑘 < 𝑚, link 𝑗𝑘 + 1 has load
less than 𝑘.
c=O
log 𝑚
log log 𝑚
– Proof (2)
• Let 𝑐 ∗ = 𝑐 − 1 .
• Will show that 𝑗1 ≥ 𝑐 ∗ !
• Hence, (recall that 𝑗1 ≤ 𝑚)
𝑐≤Γ
−1
𝑚 +1=
log 𝑚
log log 𝑚
1+𝑜 1
• Gamma function: Γ N + 1 = N! and Γ −1 𝑁 is
the inverse.
Known:
log 𝑁
−1
Γ
𝑁 =
(1 + 𝑜(1))
log log 𝑁
c=O
•
log 𝑚
log log 𝑚
– Proof (3)
Claim: 𝑗𝑐 ∗ > 0 and hence 𝑙1 ≥ 𝑐 ∗ .
• Proof:
For the purpose of contradiction, assume:
𝑙1 < 𝑐 ∗ ≤ 𝑐 − 1.
Let 𝑞 denote the link with the maximum expected load.
𝑙𝑞 = 𝑐 > 𝑙1 + 1
If there exist a task 𝑖 with weight 𝑤𝑖 ≤ 𝑠1 :
𝑖 ≤ 𝑙 + 𝑤𝑖 ≤ 𝑙 + 1 < 𝑙
𝑐𝑖,1 = 𝑙1 + 1 − 𝑝𝑖,1 𝑤
1
1
𝑞
𝑠1
𝑠1
which contradicts to Claim 1.
Hence, there is no task 𝑖 with weight 𝑤𝑖 ≤ 𝑠1 and since:
𝑂𝑝𝑡 ≥
max 𝑤𝑖
𝑖
max 𝑠𝑗
𝑗
It leads to contradiction.
=
max 𝑤𝑖
𝑖
𝑠1
>1
c=O
log 𝑚
log log 𝑚
– Proof (4)
• Claim: For 𝑘 ≥ 1, 𝑗𝑘 ≥ (𝑘 + 1)𝑗𝑘+1 .
• Proof:
Let 𝑇 be the set of tasks in the system that have
positive probability on at least one of the links
in {1, . . . , 𝑗𝑘+1 }.
Fix an optimal allocation strategy 𝑂𝑃𝑇𝑆𝑡𝑟. We
distinguish between two different ways of how
𝑂𝑃𝑇𝑆𝑡𝑟 might allocate the tasks in 𝑇 to the links.
c=O
log 𝑚
log log 𝑚
– Proof (5)
• Case 1:
For the purpose of contradiction, assume 𝑂𝑃𝑇𝑆𝑡𝑟 allocates at
least one of the tasks in 𝑇 to a link 𝑞, 𝑞 > 𝑗𝑘 .
𝑊𝑇 denote the minimum weight of the tasks in 𝑇.
For all j ∈ 1, . . . , 𝑗𝑘+1 :
𝑙𝑗 ≥ 𝑘 + 1
𝑙𝑗𝑘 +1 < 𝑘
Therefore, according to claim 1 𝑊𝑇 > 𝑠𝑗𝑘 +1 .
But this implies that allocating a task from 𝑇 to link 𝑞 gives
𝑊𝑇
𝑊𝑇
cost at least
≥
> 1.
𝑠𝑞
𝑠𝑗𝑘 +1
This means 𝑂𝑝𝑡 > 1 and hence the contradiction.
c=O
log 𝑚
log log 𝑚
– Proof (6)
• Case 2:
Now let us assume 𝑂𝑃𝑇𝑆𝑡𝑟 allocates all tasks in 𝑇 to the links {1, … , 𝑗𝑘 }.
We will show that this implies 𝑗𝑘 ≥ (𝑘 + 1)𝑗𝑘+1 .
𝑊ΣT denote the sum of the weights of the tasks in 𝑇.
𝑗𝑘+1
𝑊ΣT =
𝑤𝑖 ≥
𝑖∈𝑇
𝑤𝑖
𝑖∈𝑇
𝑗𝑘+1
𝑗𝑘+1 𝑛
𝑝𝑖,𝑗 =
𝑗=1
𝑤𝑖 𝑝𝑖,𝑗 =
𝑗=1 𝑖∈𝑇
𝑗𝑘+1
𝑤𝑖 𝑝𝑖,𝑗 =
𝑗=1 𝑖=1
𝑗𝑘+1
𝑙𝑗 𝑠𝑗 ≥ (𝑘 + 1)
𝑗=1
On the other hand:
𝑗𝑘
𝑊ΣT ≤
𝑠𝑗
𝑗=1
Hence:
𝑗𝑘
𝑗𝑘+1
𝑠𝑗 ≥ (𝑘 + 1)
𝑗=1
𝑠𝑗
𝑗=1
Since the sequence of link speeds is non-increasing, this implies that 𝑗𝑘 ≥ (𝑘 + 1)𝑗𝑘+1 .
𝑠𝑗
𝑗=1
c=O
log 𝑚
log log 𝑚
– Proof (7)
We combine both proven claims:
• Claim: 𝑗𝑐 ∗ > 0 and hence 𝑙1 ≥ 𝑐 ∗ .
• Claim: For 𝑘 ≥ 1, 𝑗𝑘 ≥ (𝑘 + 1)𝑗𝑘+1 .
And get:
𝑗1 ≥ 𝑐 ∗ ! 𝑗𝑐 ∗ ≥ 𝑐 ∗ !
As wanted.
Hence, 𝑐 = O
log 𝑚
log log 𝑚
.
c = O log
• c = O log
𝑠1
𝑠𝑚
– Proof
𝑠1
𝑠𝑚
• Proof:
We will claim that the speeds of the links
𝑗1 , 𝑗2 , … increase in a geometric fashion.
c = O log
𝑠1
𝑠𝑚
– Proof (2)
• Claim For 1 ≤ 𝑘 ≤ 𝑐 − 3, 𝑠𝑗𝑘+2 +1 ≥ 2𝑠𝑗𝑘 +1 .
• Proof:
Fix an optimal strategy 𝑂𝑃𝑇𝑆𝑡𝑟.
For every link 𝑗 ′ ≤ 𝑗𝑘+2 :
𝑙𝑗 ′ ≥ 𝑘 + 2 > 1 = 𝑂𝑝𝑡
Hence, There is task 𝑖 and link 𝑗 ∈ {1, … , 𝑗𝑘+2 } with 𝑝𝑖,𝑗 > 0
that 𝑂𝑃𝑇𝑆𝑡𝑟 allocate to link in {𝑗𝑘+2 + 1, … , 𝑚}.
Note that:
𝑤𝑖 ≤ 𝑠𝑗𝑘+2 +1
Therefore, there exists a link j ∈ {1, … , 𝑗𝑘+2 } and a task 𝑖 of
weight 𝑤𝑖 ≤ 𝑠𝑗𝑘+2+1 with 𝑝𝑖,𝑗 > 0.
c = O log
𝑠1
𝑠𝑚
On the one hand:
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
On the other hand:
𝑐𝑖,𝑗𝑘 +1 = 𝑙𝑗𝑘 +1 + 1 − 𝑝𝑖,𝑗𝑘 +1
𝑤𝑖
– Proof (3)
𝑤𝑖
≥ 𝑙𝑗 ≥ 𝑘 + 2
𝑠𝑗
≤ 𝑙𝑗𝑘 +1 +
𝑤𝑖
<𝑘+
𝑤𝑖
𝑠𝑗𝑘 +1
𝑠𝑗𝑘 +1
𝑠𝑗𝑘+1
Since we assume the system is in a Nash equilibrium, and 𝑝𝑖,𝑗 > 0 the
cost of task 𝑖 on link 𝑗 cannot be larger than the cost of task 𝑖 on link 𝑗𝑘
𝑠𝑗𝑘+2 +1
𝑤𝑖
𝑘+2≤𝑘+
≤𝑘+
𝑠𝑗𝑘 +1
𝑠𝑗𝑘+1
Clearly, this inequality implies that:
2𝑠𝑗𝑘+1 ≤ 𝑠𝑗𝑘+2 +1
Hence, the claim is shown.
c = O log
𝑠1
𝑠𝑚
– Proof (4)
The previous claim says that in a Nash
equilibrium the speeds increase geometrically
with the expected load.
This implies that:
𝑠𝑚 ≤ 𝑠𝑗1+1 ≤
Thus,
𝑐−2
−
2 2
· 𝑠𝑗𝑐−1+1 ≤
𝑐−2
−
2 2
𝑠1
c ≤ 2 log
+ O(1)
𝑠𝑚
· 𝑠1
Lemma 2 - Proof
We wish to show that 𝐶 satisfies:
log 𝑚
C = 𝑂𝑝𝑡·𝑂
+1
𝑂𝑝𝑡· log 𝑚
log
𝑐
Our goal is to show, for every link j ∈ [𝑚], it is
unlikely that 𝐶𝑗 deviates much from its
expectation.
For this purpose, we will use a Hoeffding bound.
Lemma 2 – Proof (2)
• Claim For every task 𝑖 and link 𝑗 with 𝑝𝑖,𝑗
1
∈ 0, 4 , 𝑤𝑖 ≤ 12·𝑠𝑗 ·𝑂𝑝𝑡
• Proof:
We extend the definition of 𝑗𝑘 to hold for
arbitrary 𝑂𝑝𝑡 in natural way: for 𝑘 ≥ 1, we
define 𝑗𝑘 as the smallest index in {1, … , 𝑚} such
that 𝑙𝑗𝑘 +1 < 𝑘·𝑂𝑝𝑡, or, 𝑗𝑘 = 𝑚 if no such index
exists.
Lemma 2 – Proof (3)
Let’s look at link 𝑗 ∈ {𝑗𝑘 + 1, … , 𝑗𝑘−1 } for some 𝑘 ≤ 𝑐 − 3.
On the one hand:
𝑖 ≥ 𝑘 − 1 ·𝑂𝑝𝑡 + 3𝑤𝑖
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗 𝑤
𝑠
4𝑠
𝑗
On the other hand:
𝑐𝑖,𝑗𝑘+2 +1 = 𝑙𝑗𝑘+2 +1 + 1 − 𝑝𝑖,𝑗𝑘+2 +1
𝑤𝑖
≤ 𝑘 + 2 ·𝑂𝑝𝑡 + 2𝑠
𝑗
𝑤𝑖
𝑠𝑗𝑘+2 +1
≤ 𝑙𝑗𝑘+2 +1 + 𝑠
𝑤𝑖
𝑗𝑘+2 +1
𝑗
Since we assume the system is in a Nash equilibrium, and 𝑝𝑖,𝑗 > 0 the
cost of task 𝑖 on link 𝑗 cannot be larger than the cost of task 𝑖 on link
𝑗𝑘+2 + 1.
𝑖 ≤ 𝑘 + 2 ·𝑂𝑝𝑡 + 𝑤𝑖 . Which imply :
Hence, 𝑘 − 1 ·𝑂𝑝𝑡 + 3𝑤
4𝑠
2𝑠
𝑗
𝑗
𝑤𝑖 ≤ 12·𝑂𝑝𝑡 ·𝑠𝑗
Lemma 2 – Proof (4)
It remains to investigate the case 𝑗 ≤ 𝑗𝑘 , where 𝑘 = 𝑐 − 3 .
On the one hand:
𝑤𝑖
𝑐𝑖,1 = 𝑙1 + 1 − 𝑝𝑖,1
≤ 𝑐·𝑂𝑝𝑡 + 𝑂𝑝𝑡 = 𝑐 + 1 ·𝑂𝑝𝑡
𝑠1
On the other hand:
𝑤𝑖
3𝑤𝑖
3𝑤𝑖
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
≥ 𝑘·𝑂𝑝𝑡 +
≥ 𝑐 − 4 ·𝑂𝑝𝑡 +
𝑠𝑗
4𝑠𝑗
4𝑠𝑗
Since we assume the system is in a Nash equilibrium, and 𝑝𝑖,𝑗 > 0 the cost of
task 𝑖 on link 𝑗 cannot be larger than the cost of task 𝑖 on link 1.
Hence, 𝑐 − 4 ·𝑂𝑝𝑡 +
This proves the claim.
3𝑤𝑖
4𝑠𝑗
≤ 𝑐 + 1 ·𝑂𝑝𝑡. Which imply :
20
𝑤𝑖 ≤
·𝑂𝑝𝑡·𝑠𝑗 < 12·𝑂𝑝𝑡·𝑠𝑗
3
Lemma 2 – Proof (5)
Now let us focus on a single link 𝑗 ∈ [𝑚].
We apply the claim in order to show that it is unlikely that 𝐶𝑗 deviates
much from its expectation.
Denote:
𝑇𝑗,1 - the set of tasks with 𝑝𝑖,𝑗 ∈ 0, 14 .
𝑇𝑗,2 - the set of tasks with 𝑝𝑖,𝑗 ∈ 14, 1 .
𝐶𝑗,1 - random variable that describe the cost on link 𝑗 only counting
tasks in 𝑇𝑗,1 .
𝐶𝑗,1 - random variable that describe the cost on link 𝑗 only counting
tasks in 𝑇𝑗,2 .
Since only tasks with 𝑝𝑖,𝑗 > 0 can be allocated to link 𝑗:
𝐶𝑗 = 𝐶𝑗,1 + 𝐶𝑗,2
Lemma 2 – Proof (6)
Let us consider only the tasks in 𝑇𝑗,1 .
𝐶𝑗,1 = 𝑠1
𝑤𝑖 𝐽𝑖,𝑗
𝑗
𝑖∈𝑇𝑗,1
• Remainder :
Hoeffding bound: Let 𝑋1 , … , 𝑋𝑁 be independent random variables with values
in the interval [0, 𝑧] for some 𝑧 > 0, and let 𝑋 = 𝑛𝑖=1 𝑋𝑖 , then for any 𝑡 it
𝑒𝐸 𝑋 𝑡 𝑧
holds that Pr 𝑋 ≥ 𝑡 ≤ (
)
𝑡
According to the previous claim:
𝑖 ≤ 12·𝑂𝑝𝑡
max 𝑤
𝑠
𝑖∈𝑇𝑗,1 𝑗
·
𝛼 𝑂𝑝𝑡
Pr 𝐶𝑗,1 ≥ 𝛼·𝑂𝑝𝑡 ≤
For any 𝛼 > 1.
𝑒𝐸 𝐶𝑗,1
𝛼·𝑂𝑝𝑡
·
12 𝑂𝑝𝑡
≤
𝑒𝑐
𝛼·𝑂𝑝𝑡
𝛼
12
Lemma 2 – Proof (7)
Let us consider only the tasks in 𝑇𝑗,2 .
𝐶𝑗,2 = 𝑠1
𝑤𝑖 𝐽𝑖,𝑗 ≤ 𝑠1
𝑗
𝑖∈𝑇𝑗,2
𝐸 𝐶𝑗,2 = 𝐸
1
𝑠𝑗
𝑖∈𝑇𝑗,2
𝑤𝑖 𝐽𝑖,𝑗 = 𝑠1
𝑤𝑖 𝐸 𝐽𝑖,𝑗 > 4𝑠1
𝑗
𝑖∈𝑇𝑗,2
𝑤𝑖
𝑗
𝑖∈𝑇𝑗,2
𝑖∈𝑇𝑗,2
Hence, we get:
𝐶𝑗,2 < 4𝐸 𝐶𝑗,2 ≤ 4𝑐
Hence, for every 𝛼 > 1
𝑒𝑐
Pr 𝐶𝑗 ≥ 4𝑐 + 𝛼·𝑂𝑝𝑡 ≤
𝛼·𝑂𝑝𝑡
𝑤𝑖
𝑗
𝛼
12
Lemma 2 – Proof (8)
Denote ℒ = max 𝐶𝑗.
𝑗∈[𝑚]
𝑒𝑐
Pr ℒ ≥ 4𝑐 + 𝛼·𝑂𝑝𝑡 ≤ 𝑚
𝛼·𝑂𝑝𝑡
Since of union bound.
𝛼
12
Lemma 2 – Proof (9)
Recall that 𝐶 is defined to be 𝐸 ℒ .
Hence, for every 𝜆 > 1, we can estimate 𝐶 as follows.
∞
𝐶 = 𝐸 ℒ ≤ 4𝑐 + λ·𝑂𝑝𝑡 + 𝑂𝑝𝑡·
Pr ℒ ≥ 4𝑐 + 𝜆 + 𝑡 ·𝑂𝑝𝑡 𝑑𝑡
0
∞
≤ 4𝑐 + λ·𝑂𝑝𝑡 + 𝑂𝑝𝑡·𝑚·
0
𝑒𝑐
≤ 4𝑐 + λ·𝑂𝑝𝑡 + 𝑂𝑝𝑡·𝑚·
𝜆·𝑂𝑝𝑡
Notice that for λ ≥
21𝑒𝑐
𝑂𝑝𝑡
we have
∞
𝑒𝑐
0
𝜆·𝑂𝑝𝑡
𝑡
12
𝜆+𝑡
𝑒𝑐
𝜆 + 𝑡 ·𝑂𝑝𝑡
𝜆
∞
12
𝑑𝑡 =
·
0
𝑑𝑡
·
𝑡
𝑒𝑐
𝜆·𝑂𝑝𝑡
12
ln
12
𝜆 𝑂𝑝𝑡
𝑒𝑐
12
𝑑𝑡
≤ 4 and therefore we obtain:
𝑒𝑐
𝐶 ≤ 4𝑐 + λ·𝑂𝑝𝑡 + 4·𝑂𝑝𝑡·𝑚·
𝜆·𝑂𝑝𝑡
𝜆
12
Lemma 2 – Proof (10)
For λ ≥
𝑐
(2Γ −1
𝑂𝑝𝑡
(𝑒·𝑚
·
12 𝑂𝑝𝑡
𝑐
)) then:
𝑒𝑐
𝜆·𝑂𝑝𝑡
To prove this inequality, let us set:
𝜆0 =
𝑐
1+Γ
𝑂𝑝𝑡
𝜆
12
≤
−1
1
𝑚
(𝑒·𝑚
12·𝑂𝑝𝑡
𝑐)
Observe that for any integer 𝑘 ≥ 1
(𝑘 𝑒)𝑘 ≥ 𝑘 − 1 !
Therefore
𝜆0 ·𝑂𝑝𝑡
𝑒·𝑐
𝜆0 ·𝑂𝑝𝑡
𝑐
≥
𝜆0 ·𝑂𝑝𝑡
− 1 ! ·𝑒 −1
𝑐
Since 𝜆0 ensures that:
𝜆0 ·𝑂𝑝𝑡
12·𝑂𝑝𝑡
𝑐
− 1 ! ·𝑒 −1 ≥ 𝑚
𝑐
We can conclude that for any 𝜆 ≥ 𝜆0 we have:
𝑐
12·𝑂𝑝𝑡
𝜆0
𝜆0 ·𝑂𝑝𝑡
𝜆
𝑐
12
12
𝜆·𝑂𝑝𝑡
𝜆0 ·𝑂𝑝𝑡
𝜆0 ·𝑂𝑝𝑡
12·𝑂𝑝𝑡
𝑐
≥
=
≥ 𝑚
𝑒·𝑐
𝑒·𝑐
𝑒·𝑐
𝑐
12·𝑂𝑝𝑡
=𝑚
Lemma 2 – Proof (11)
Therefore, if we set:
𝜆 = max
21𝑒𝑐
𝑐
1,
,
𝑂𝑝𝑡 𝑂𝑝𝑡
2Γ
−1
𝑒·𝑚
·
12 𝑂𝑝𝑡
𝑐
=𝑂 1
𝑐
+
𝑂𝑝𝑡
Upper bound - Proof
In Lemma 2 we saw
𝑐
log 𝑚
𝐶 = 𝑂𝑝𝑡·𝑂 1 +
+
𝑂𝑝𝑡 log 𝑂𝑝𝑡· log
𝑚
𝑐
If 𝑐 ≤ 𝑂𝑝𝑡 then
C = 𝑂𝑝𝑡·𝑂
log 𝑚
+1
𝑂𝑝𝑡· log 𝑚
log
𝑐
If 𝑐 > 𝑂𝑝𝑡 then by Lemma 1
𝑐
log 𝑚
log 𝑚
=𝑂
=𝑂
𝑂𝑝𝑡
log log 𝑚
log 𝑂𝑝𝑡· 𝑐log 𝑚
Hence,
𝑐
log 𝑚
log 𝑚
C = 𝑂𝑝𝑡·𝑂 1 +
+
= 𝑂𝑝𝑡·𝑂
+1
𝑂𝑝𝑡 log 𝑂𝑝𝑡· log 𝑚
log 𝑂𝑝𝑡· log 𝑚
𝑐
𝑐
Lower bound
Let’s assume without loss of generality that 𝑚 is an integer.
We consider 𝐾 + 1 groups of links 𝐸0 , 𝐸1 , … , 𝐸𝐾 .
And 𝐾 + 1 groups of tasks 𝑇0 , 𝑇1 , … , 𝑇𝐾 .
For all 0 ≤ 𝑘 ≤ 𝐾
𝐾!
𝐸𝑘 = · 𝑚
𝑘!
𝑇𝑘 = 𝑘· 𝐸𝑘
For all 𝑗 ∈ 𝐸𝑘
For all 𝑖 ∈ 𝑇𝑘
𝑠𝑗 = 2𝑘
𝑤𝑖 = 2𝑘
Will consider a pure strategy 𝑆 in which for each 𝑗 ∈ 𝐸𝑘 it assigns exactly 𝑘 tasks from
𝑇𝑘 with probability 1 to be allocated to 𝑗.
Lower bound – (2)
In our construction 𝐾 can be chosen to be any positive integer that satisfies:
𝐾
𝐸𝑘 ≤ 𝑚
𝑘=0
Note that:
𝐾
𝐾
𝐸𝑘 =
𝑘=0
𝑘=0
𝑚·𝐾!
= 𝑚·𝐾! ·
𝑘!
𝐾
𝑘=0
1
≤ 𝑚·𝐾! ·𝑒
𝑘!
Thus, our analysis can be carried over for all 𝐾 satisfying:
𝑚·𝐾! ·𝑒 ≤ 𝑚
And hence for all 𝐾 satisfying:
𝑚
log 𝑚
𝐾 ≤ Γ −1
−1=𝛩
𝑒
log log 𝑚
Lower bound – (3)
• Claim Strategy 𝑆 satisfies the following
properties:
1. The maximum load is c = 𝐾.
2. The social optimum is 1 ≤ 𝑂𝑝𝑡 ≤ 2.
3. The system is in Nash equilibrium.
• Proof:
1. For all 𝑗 ∈ 𝐸𝑘 the load 𝐶𝑗 =
𝑘 ·2 𝑘
𝑠𝑗
=
𝑘·2𝑘
2𝑘
= 𝑘.
Lower bound – (4)
2. First we will observe that:
max 𝑤𝑖
2𝑘
𝑂𝑝𝑡 ≥
= 𝑘=1
max 𝑠𝑗
2
𝑖
𝑗
Let’s consider a strategy in which for 0 ≤ 𝑘 < 𝐾 for each 𝑗 ∈ 𝐸𝑘 it assigns a single task from 𝑇𝑘+1 with
probability 1 to be allocated to 𝑗.
Notice that:
𝐾!
𝐸𝑘 = · 𝑚
𝑘!
𝐾!
𝐾!
𝑇𝑘+1 = (𝑘+1)· 𝐸𝑘+1 = (𝑘+1)·
· 𝑚= · 𝑚
𝑘+1 !
𝑘!
For all 𝑗 ∈ 𝐸𝑘
𝑠𝑗 = 2𝑘
For all 𝑖 ∈ 𝑇𝑘+1
𝑤𝑖 = 2𝑘+1
The cost of every link in the system is at most 2.
𝑂𝑝𝑡 ≤ 2
Lower bound – (5)
3. Let us take any task 𝑖 that is allocated to link 𝑗 ∈ 𝐸𝑘 , for 𝑘
≥ 1.
Lets see that it is not beneficial for 𝑖 to move to link 𝑟 ∈ 𝐸𝑡 .
𝑤𝑖
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
= 𝑙𝑗 = 𝑘
𝑠𝑗
𝑤𝑖
2𝑘
𝑐𝑖,𝑟 = 𝑙𝑟 + 1 − 𝑝𝑖,𝑟
= 𝑙𝑟 + 𝑡 = 𝑡 + 2𝑘−𝑡
𝑠𝑟
2
Since 𝑡 + 2𝑘−𝑡 ≥ 𝑘 for any non-negative 𝑡 and 𝑘, none of the
tasks allocated to 𝑗 has an incentive to migrate to another link.
Therefore, the system is in a Nash equilibrium.
Lower bound – Mixed strategy
We will slightly modify the pure strategy 𝑆 to
obtain a mixed strategy 𝑆′ for which:
log 𝑚
𝐶 = 𝑐·𝛩
log(log(𝑚) 𝑐)
For 0 ≤ 𝑘 < 𝐾, 𝑆′ will allocate tasks in the same
manner 𝑆 did.
1
For all 𝑖 ∈ 𝑇𝐾 and 𝑗 ∈ 𝐸𝐾 , 𝑆′ will set 𝑝𝑖,𝑗 = 𝑚.
Lower bound – Mixed strategy – (2)
• Claim Strategy 𝑆′ satisfies the following properties:
1. The maximum load is c = 𝐾.
2. The social optimum is 1 ≤ 𝑂𝑝𝑡 ≤ 2.
3. The system is in Nash equilibrium.
4. The social cost is 𝐶 = Ω
log 𝑚
log(log(𝑚) 𝐾)
• Proof:
1. The maximum load 𝑐 is the same as for strategy 𝑆′.
Lower bound – Mixed strategy – (3)
𝟐. The value of opt is unaffected by the modification of the probabilities.
𝟑. We have to show that for 𝑖 ∈ 𝑇𝐾 it is not beneficial to move to link 𝑟 ∈ 𝐸𝑡 .
For 𝑗 ∈ 𝐸𝐾
𝑤𝑖
1 2𝐾
1
𝑐𝑖,𝑗 = 𝑙𝑗 + 1 − 𝑝𝑖,𝑗
=𝐾+ 1−
=
𝐾
+
1
−
𝑠𝑗
𝑚 2𝐾
𝑚
We can see that the costs for task 𝑖 on link 𝑗 slightly increased.
𝑤𝑖
2𝐾
𝑐𝑖,𝑟 = 𝑙𝑟 + 1 − 𝑝𝑖,𝑟
= 𝑙𝑟 + 𝑡 = 𝑡 + 2𝐾−𝑡
𝑠𝑟
2
The cost for task 𝑖 on link 𝑟 ∈ 𝐸𝑡 remains the same as before.
1
Since 𝑡 + 2𝐾−𝑡 ≥ 𝐾 + 1 ≥ 𝐾 + 1 −
for any non-negative 𝑡 and 𝐾, none
𝑚
of the tasks allocated to 𝑗 has an incentive to migrate to another link.
Therefore, the system is in a Nash equilibrium.
Lower bound – Mixed strategy – (4)
𝟒. To observe this property, we notice that the allocation of the tasks
in 𝑖 ∈ 𝑇𝐾 to the links in 𝐸𝐾 corresponds to the allocation problem of
throwing 𝑚·𝐾 balls uniformly at random into 𝑚 bins.
In expectation, it is known that the expected maximum occupancy in
this allocation problem is:
log 𝑚
𝛩 𝐾+
log(log(𝑚) 𝐾)
Since 𝐾 = 𝛩
log 𝑚
log log 𝑚
in our case. We get:
log 𝑚
𝛩
log(log(𝑚) 𝐾)
Since for 𝑖 ∈ 𝑇𝐾 and j ∈ 𝐸𝐾 , 𝑤𝑖 = 𝑠𝑗 , this bound on the maximum
occupancy directly implies a lower bound on the social cost.