Load balance

Course: Price of AnarchyProfessor: Michal Feldman

Student: Nave Frost23/04/2014

Tight Bounds for Worst-Case Equilibria, Artur Czumaj and Berthold V¨ocking

Model

• independent parallel links with speeds .• tasks with weights .

• Goal: allocate tasks to links to minimize the maximum load of the links in the system.

Definitions

• Given a combination of pure strategies, one for each task:

• The load of link is: .• The cost for task is: .

• Observation:

Mixed strategies

• Let denote the probability that an task sends the entire weight on link .

• Let denote the expected load on link . .

• Let denote the expected cost for task on link .

Nash equilibrium

• Probabilities define Nash equilibrium if implies for every .

Price of anarchy

• - probabilities that define Nash equilibrium. Define to be random variable indicating the load of link .

• E[

Price of anarchy (2)

• Let denote the maximum expected load over all links, that is:

• Let the social cost denote the expected maximum load over all links, that is:

• Note that and possibly .

Price of anarchy (3)

• The price of anarchy defined as :

Claim 1

• Claim 1: In Nash equilibrium, if for certain task and link , then for every link j [m]∈

In particular, if then .

• Proof:

Since , the definition of Nash equilibria implies that and hence, .• 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

Identical links (2)

• Theorem: For identical links the price of anarchy is .

• Proof: Let us first re-scale the weights and speeds in the problem and assume, without loss of generality, that for all and .

Identical links (3)

• Since and we observed that , then .• Since there must exist a such that

. • Since define Nash equilibrium it implies that,

according to claim 1:

Identical links (4)

• We will estimate the load of any link by applying to a standard concentration inequality due to Hoeffding bound.

• Hoeffding bound: Let be independent random variables with values in the interval for some , and let , then for any it holds that

• For any .

Identical links (5)

We saw that For any . Therefore, if we pick we will get the following:

Upper bound – Speed varies

• Lemma 1: The maximum expected load satisfies

where it is assumed that the speeds satisfy .

• Lemma 2: The social cost satisfies

Lemma 1 - Proof

• define a Nash equilibrium. • Without loss of generality, assume . • Scale weights of such that . • Will show:

and

– Proof

Proof: For , define to be the smallest index in such that or, if no such index exists, .Observe:• for every with , all links have load at least .• for every with , link has load less than .

– Proof (2)

• Let . • Will show that • Hence, (recall that )

• Gamma function: and is the inverse. Known:

– Proof (3)• Claim: and hence .

• Proof: For the purpose of contradiction, assume:

. Let denote the link with the maximum expected load.

If there exist a task with weight :

which contradicts to Claim 1.Hence, there is no task with weight and since:

It leads to contradiction.

– Proof (4)

• Claim: For , .

• Proof: Let be the set of tasks in the system that have positive probability on at least one of the linksin . Fix an optimal allocation strategy . We distinguish between two different ways of how might allocate the tasks in to the links.

– Proof (5)

• Case 1: For the purpose of contradiction, assume allocates at least one of the tasks in to a link . denote the minimum weight of the tasks in . For all

Therefore, according to claim 1 .But this implies that allocating a task from to link gives cost at least .This means and hence the contradiction.

– Proof (6)

• Case 2: Now let us assume allocates all tasks in to the links .We will show that this implies . denote the sum of the weights of the tasks in .

On the other hand:

Hence:

Since the sequence of link speeds is non-increasing, this implies that .

– Proof (7)

We combine both proven claims: • Claim: and hence .• Claim: For , .And get:

As wanted.

Hence, .

– Proof

• Proof:We will claim that the speeds of the links increase in a geometric fashion.

– Proof (2)

• Proof:Fix an optimal strategy . For every link :

Hence, There is task and link with that allocate to link in .Note that:

Therefore, there exists a link and a task of weight with .

– Proof (3)

On the one hand:

On the other hand:

Since we assume the system is in a Nash equilibrium, and the cost of task on link cannot be larger than the cost of task on link

Clearly, this inequality implies that:

Hence, the claim is shown.

– Proof (4)

The previous claim says that in a Nash equilibrium the speeds increase geometrically with the expected load. This implies that:

Thus,

Lemma 2 - Proof

We wish to show that satisfies:

Our goal is to show, for every link , it is unlikely that deviates much from its expectation. For this purpose, we will use a Hoeffding bound.

Lemma 2 – Proof (2)

• For every task and link with , • Proof: We extend the definition of to hold for arbitrary in natural way: for , we define as the smallest index in such that , or, if no such index exists.

Lemma 2 – Proof (3)

Let’s look at link for some .On the one hand:

On the other hand:

Since we assume the system is in a Nash equilibrium, and the cost of task on link cannot be larger than the cost of task on link . Hence, . Which imply :

Lemma 2 – Proof (4)

It remains to investigate the case , where .On the one hand:

On the other hand:

Since we assume the system is in a Nash equilibrium, and the cost of task on link cannot be larger than the cost of task on link .Hence, . Which imply :

This proves the claim.

Lemma 2 – Proof (5)

Now let us focus on a single link . We apply the claim in order to show that it is unlikely that deviates much from its expectation. Denote: - the set of tasks with . - the set of tasks with . - random variable that describe the cost on link only counting tasks in . - random variable that describe the cost on link only counting tasks in .

Since only tasks with can be allocated to link :

Lemma 2 – Proof (6)

Let us consider only the tasks in .

• Remainder :Hoeffding bound: Let be independent random variables with values in the interval for some , and let , then for any it holds that

According to the previous claim:

For any .

Lemma 2 – Proof (7)

Let us consider only the tasks in .

Hence, we get:

Hence, for every

Lemma 2 – Proof (8)

Denote .

Since of union bound.

Lemma 2 – Proof (9)

Recall that is defined to be .Hence, for every , we can estimate as follows.

Notice that for we have and therefore we obtain:

Lemma 2 – Proof (10)

For then:

To prove this inequality, let us set:

Observe that for any integer

Therefore

Since ensures that:

We can conclude that for any we have:

Lemma 2 – Proof (11)

Therefore, if we set:

We obtain:

Upper bound - Proof

In Lemma 2 we saw

If then

If then by Lemma 1

Hence,

Lower boundLet’s assume without loss of generality that is an integer.We consider groups of links .And groups of tasks .For all

For all

For all

Will consider a pure strategy in which for each it assigns exactly tasks from with probability to be allocated to .

Lower bound – (2)

In our construction can be chosen to be any positive integer that satisfies:

Note that:

Thus, our analysis can be carried over for all satisfying:

And hence for all satisfying:

Lower bound – (3)

• Strategy satisfies the following properties:1. The maximum load is .2. The social optimum is .3. The system is in Nash equilibrium.

• Proof: 1. For all the load .

Lower bound – (4)2. First we will observe that:

Let’s consider a strategy in which for for each it assigns a single task from with probability to be allocated to .Notice that:

For all

For all

The cost of every link in the system is at most .

Lower bound – (5)

3. Let us take any task that is allocated to link , for .Lets see that it is not beneficial for to move to link .

Since for any non-negative and , none of the tasks allocated to 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 to obtain a mixed strategy for which:

For , will allocate tasks in the same manner did.For all and , will set .

Lower bound – Mixed strategy – (2)

• Strategy satisfies the following properties:1. The maximum load is .2. The social optimum is .3. The system is in Nash equilibrium.4. The social cost is

• Proof: 1. The maximum load is the same as for strategy .

Lower bound – Mixed strategy – (3)

. The value of opt is unaffected by the modification of the probabilities.

. We have to show that for it is not beneficial to move to link .For

We can see that the costs for task on link slightly increased.

The cost for task on link remains the same as before.Since for any non-negative and , none of the tasks allocated to has an incentive to migrate to another link.Therefore, the system is in a Nash equilibrium.

Lower bound – Mixed strategy – (4)

. To observe this property, we notice that the allocation of the tasks in to the links in corresponds to the allocation problem of throwing balls uniformly at random into bins. In expectation, it is known that the expected maximum occupancy in this allocation problem is:

Since in our case. We get:

Since for and , , this bound on the maximum occupancy directly implies a lower bound on the social cost.