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1. Solve the following L P problem through SIMPLEX method

Maximize Z = 5x1 + 6x2

Subject to

2x1 + 3x2 3000

5x1 + 7x2 1000

x1 + x2 500

x1 and x2 0

Introducing slack variables, x3 0, x4 0, x5 0, so that the constraints now become equations of

the form:

2x1 + 3x2 + x3 = 3000

5x1 + 7x2 + x4 = 1000

x1 + x2 + x5 = 500

Also, z 5x1 6x2 = 0

Now, constructing Table 1,

x1 x2 x3 x4 x5 z

2 3 1 0 0 0 3000

5 7 0 1 0 0 1000

1 1 0 0 1 0 500

_________________________________________

-5 -6 0 0 0 1 0

The pivot element should now be determined. For this, first the pivot column is selected which will

be the one with the most negative element in the objective row which is the fourth row.

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Here, -6 is the most negative element.

Now, divide the constant column, which is the last column after the vertical line, with the

corresponding values from the pivot column and select the smallest non-negative element. The row

with this smallest non-negative element is the pivot row.

The values from dividing the constant column with the pivot column are:

3000 / 3

1000 / 7

500 / 1

Here, 1000 / 7 is the smallest non-negative value. The pivot element will be the element at the

intersection of the pivot row and pivot column.

So from Table 1, the pivot element is 7.

x1 x2 x3 x4 x5 z

2 3 1 0 0 0 3000

5 7 0 1 0 0 1000

1 1 0 0 1 0 500

_________________________________________

-5 -6 0 0 0 1 0

Now the pivot column in Table 1 is made into a Unit Column by converting the Pivot Element to 1

and the other elements to 0.

R2 = R2 / 7

R1 = R1 3R2

R3 = R3 R2

R4 = R4 + 6R2

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Now constructing, Table 2,

x1 x2 x3 x4 x5 z

-1/7 0 1 -3/7 0 0 18000/7

5/7 1 0 1/7 0 0 1000/7

2/7 0 0 -1/7 1 0 2500/7

_________________________________________

-5/7 0 0 6/7 0 1 6000/7

Since there is a negative value in the objective row, the process should be carried out again by

finding the pivot element.

Here, -5/7 is the most negative element in the objective row and dividing the constant column with

the pivot column gives the following values:

-18000

200

1250

Here, 200 is the smallest non-negative value and thereby, from Table 2, the pivot element is 5/7.

x1 x2 x3 x4 x5 z

-1/7 0 1 -3/7 0 0 18000/7

5/7 1 0 1/7 0 0 1000/7

2/7 0 0 -1/7 1 0 2500/7

_________________________________________

-5/7 0 0 6/7 0 1 6000/7

Now the pivot column in Table 2 is made into a Unit Column by converting the Pivot Element to 1

and the other elements to 0.

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R2 = R2 / (5/7)

R1 = R1 + (1/7)R2

R3 = R3 (2/7)R2

R4 = R4 + (5/7)R2

Now, constructing Table 3,

x1 x2 x3 x4 x5 z

0 1/5 1 -2/5 0 0 2600

1 7/5 0 1/5 0 0 200

0 -2/5 0 -1/5 1 0 300

_________________________________________

0 1 0 1 0 1 1000

Now all the values in the objective row are non-negative.

Therefore, maximum value of the objective function,

z = 1000

when, x1 = 200

x2 = 0

x3 = 2600

x4 = 0

x5 = 300

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2. Write the Dual of the Problem 1 and give the economic interpretation of Dual Variables.

Problem 1: Maximize z = 5x1 + 6x2

Subject to:

2x1 + 3x2 3000

5x1 + 7x2 1000

x1 + x2 500

x1 and x2 0

The above problem is the primal problem in the standard form, which means that the primal

problem:

Considers maximization of the objective function.

Has less than or equal to type constraints.

Has non-negative constraints on the variables.

Since the above primal problem is in the standard form, the dual problem can be formulated by

using the following rules:

The number of constraints on the primal problem is equal to the number of dual variables.

The number of constraints in the dual problem is equal to the number of variables in the primal

problem.

The primal problem is a maximization problem while the dual problem is a minimization

problem.

The profit coefficients of the primal problem appear on the right hand side of the

constraints of the dual problem.

The primal problem has less than or equal to type constraints while the dual problem has

greater than or equal to type constraints.

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The coefficients of the constraints of the primal problem which appear from left to right are

placed from top to bottom in the constraints of the dual problem and vice versa.

Therefore, the corresponding dual problem of the primal problem 1, is,

Minimise 3000w1 + 1000w2 +500w3

Subject to:

2w1 + 5w2 + w3 5

3w1 + 7w2 + w3 6

w1 0, w2 0, w3 0

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3. Write the generalized form of transportation problem as LP Problem. Also explain how to

solve the degeneracy in transportation problem.

GENERALIZED FORM OF TRANSPORTATION PROBLEM

In order to obtain the generalized form of transportation problem, we shall make use of an

example. A manufacturer operates three factories and dispatches his products to four different

dealers. The table shows the total units that each factory can supply, the total units required at each

dealer and the cost of transportation of one unit from each factory to each dealer.

DEALERS

FACTORY 1 2 3 4 SUPPLY

1 2 2 2 4 1000

2 4 6 4 3 700

3 3 2 1 0 900

REQUIREMENT 900 800 500 400

The total units that factories 1, 2 and 3 can supply are 1000 units, 700 units and 900 units, denoted

by a1, a2 and a3, respectively.

The total units required at dealers 1, 2, 3 and 4 are 900 units, 800 units, 500 units and 400 units,

denoted by b1, b2, b3 and b4, respectively.

The cost of transportation of one unit from Factory 1 to Dealer 1 is 2, from Factory 1 to Dealer 2 is

2 and so on.

A transportation problem is formulated as an LP problem using variables with two subscripts.

Let,

X11 = Amount to be transported from Factory 1 to Dealer 1.

X12 = Amount to be transported from Factory 1 to Dealer 2, and so on until,

X34 = Amount to be transported from Factory 3 to Dealer 4

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Let,

C11 = Cost of Transportation of one unit from Factory 1 to Dealer 1, and so on until,

C34 = Cost of Transportation of one unit from Factory 3 to Dealer 4

Then the transportation problem can be formulated as:

Minimise C11X11 + C12X12 = ..C34X34

Subject to,

X11 + X12 + X13 + x14 = a1

X21 + X22 + X23 + X24 = a2

X31 + X32 + X33 + X34 = a3

X11 + X21 + X31 = b1

X12 + X22 + X32 = b2

X13 + X23 + X33 = b3

X14 + X24 + X34 = b4

X11 0, X12 0,.X34 0

The generalized form of the above transportation problem in Linear Programming problem can be

written as:

m nMinimise CijXij (where, m = number of factories, n = number of dealers)

i=1 j=1

n

Subject to Xij = ai, where i = 1 to mj=1

m

Xij = bj, where j = 1 to ni=1

Xij 0, where i = 1 to m, j = 1 to n

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DEGENERACY IN TRANSPORTATION PROBLEM

If a basic feasible solution of a transportation problem with m origins and n destinations has fewer

than m + n 1 positive Xij (occupied cells) the problem is said to be a degenerate transportation

problem.

Degeneracy can occur at two stages:

At the initial solution.

During the testing of the optimum solution.

A degenerate basic feasible solution in a transportation problem exists if and only if some partial

sum of availabilitys (row) is equal to a partial sum of requirements (column).

We shall make use of the previous example to solve the degeneracy in transportation problem. A

manufacturer operates three factories and dispatches his products to four different dealers. The

table shows the total units that each factory can supply, the total units required at each dealer and

the cost of transportation of one unit from each factory to each dealer.

DEALERS


1 2 2 2 4 1000

2 4 6 4 3 700

3 3 2 1 0 900


Here, S1 = 1000, S2 = 700, S3 = 900R1 = 900, R2 = 800, R3 = 500, R4 = 400

Since R3 + R4 = S3 so the given problem is a degeneracy problem.

Now we will solve the transportation problem by Matrix Minimum Method.

To resolve degeneracy, we make use of an artificial quantity, d.

The quantity, d, is so small that it does not affect the supply and demand constraints.

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Degeneracy can be avoided if we ensure that no partial sum of S i (supply) and Rj (requirement) are

the same. We set up a new problem where:

Si = Si + d i = 1, 2, ....., m

Rj = Rj

Rn = Rn + md

DEALERS


1 2900 2100+d 2 4 1000 + d

2 4 6700d 42d 3 700 + d

3 3 2 15002d 0400+3d 900 + d

REQUIREMENT 900 800 500 400 + 3d

Substituting d = 0.

DEALERS


1 2900 2100 2 4 1000

2 4 6700 40 3 700

3 3 2 1500 0400 900


Initial basic feasible solution:

2 * 900 + 2 * 100 + 6 * 700 + 4 * 0 + 1 * 500 + 0 * 400 = 6700

Now degeneracy has been removed.

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4. Arrivals at an STD booth is considered poisson with an average time of 10 minutes

between them. Length of phone call is assumed exponential with mean 3 minutes. Determine

a. The probability that a person arriving at the booth will have to wait.

b. Average Length of queue that form time to time

c. Average utilization of STD booth.

Average arrival time = 1 = 10 minutes = 1 hour

6

Therefore, = 6 / hour

Average service time = 1 = 3 minutes = 1 hour

20

Therefore, = 20 / hour

Here, < , so a steady state solution exists.

a. The probability that a person arriving at the booth will have to wait

Average waiting time for a person arriving at the booth,

Wq = = 6

( - ) 20 (20 - 6)

= 6 x 1 hours

20 14

= 1.28 minutes

Therefore, the average waiting time for a person arriving at the booth is 1.28 minutes

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b. Average length of queue that form time to time

Expected number of customers in the queue,

Lq = 2 = 62

( - ) 20 (20 - 6)

= 36

20 x 14

= 9 / 70

Therefore, Lq = 9 / 70

c. Average utilization of STD booth

Average time spent in the booth,

Ls = 1

-

= 1

20 - 6

= 1 hour

14

= 4.28 minutes

Therefore the average utilization of the STD booth is 4.28 minutes

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5. Write short notes on:

i) Dynamic Programming

ii) Integer Programming

iii) Pure and Mixed Strategy

iv) Queue Parameters.

i. DYNAMIC PROGRAMMING

Dynamic programming is an optimization approach that transforms a complex problem into a

sequence of simpler problems. The main characteristic of dynamic programming is the multistage

nature of the optimization procedure and provides a general framework for analyzing many

problem types. Within this framework a variety of optimization techniques can be employed to

solve particular aspects of a more general formulation. Dynamic programming solves problems by

combining the solutions into sub-problems and can be used when a problem can be divided into

sub-problems and when these sub-problems are not independent. There are three important

characteristics of dynamic programming problems explained below.

Stages

The essential feature of the dynamic programming approach is the structuring of optimization

problems into multiple stages, which are solved sequentially one stage at a time. Although each

one-stage problem is solved as an ordinary optimization problem, its solution helps to define the

characteristics of the next one-stage problem in the sequence. Often, the stages represent different

time periods in the problems planning horizon but also, sometimes the stages do not have time

implications.

States

Associated with each stage of the optimization problem are the states of the process. The states

reflect the information required to fully assess the consequences that the current decision has upon

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future actions. The specification of the states of the system is perhaps the most critical design

parameter of the dynamic programming model. There are no set rules for doing this but the

essential properties that should motivate the selection of states are that the states should convey

enough information to make future decisions without regard to how the process reached the current

state and the number of state variables should be small, since the computational effort associated

with the dynamic programming approach is prohibitively expensive when there are more than two,

or possibly three, state variables involved in the model formulation.

Principle of Optimality

In dynamic programming terminology, each point where decisions are made is usually called a

stage of the decision-making process. At any stage, knowledge of the intersection currently

available is needed to be able to make subsequent decisions. Information that summarizes the

knowledge required about the problem in order to make the current decisions, such as the

intersection at a particular stage, is called a state of the decision-making process. On these terms,

the solution to the minimum delay problem involves what is referred to as the principle of

optimality.

Any optimal policy has the property that, whatever the current state and decision, the remaining

decisions must constitute an optimal policy with regard to the state resulting from the current

decision. To make this principle more concrete, the optimal-value function can be defined in the

context of the minimum delay problem.

Vn(Sn) = Optimal value or minimum delay over the current and subsequent stages or intersections,

given that we are in state Sn in a particular intersection with n stages to go.

A recursive relationship for computing the optimal value function by recognizing that, at each

stage, the decision in a particular state is determined simply by choosing the minimum total delay.

If we number the states at each stage as Sn = 1, which is the bottom intersection, up to Sn = 6 which

is the top intersection, then

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Vn (Sn) = Min { tn (Sn) + Vn 1(Sn 1)} , (1)

subject to:

Sn 1 = Sn + 1 if we choose up and n even,

Sn 1 = Sn 1 if we choose down and n odd,

Sn 1 = Sn otherwise,

Where, tn (Sn) is the delay time in intersection Sn at stage n.

ii. INTEGER PROGRAMMING

An integer programming problem is a mathematical optimization orfeasibility program in which

some or all of the variables are restricted to be integers. When formulating linearprograms it is

often found that, certain variables should have been regarded as taking integer values but, for the

sake of convenience, they are allowed to take fractional values reasoning that the variables were

likely to be so large that any fractional part could be neglected. While this is acceptable in some

situations, in many cases it is not, and in such cases we must find a numeric solution in which the

variables take integer values.

Problems in which this is the case are called integer programs and the subject of solving such

programs is called integer programming. Integer programming occurs frequently because many

decisions are essentially discrete in that one or more options must be chosen from a finite set of

alternatives. Problems in which some variables can take only integer values and some variables can

take fractional values are called mixed-integer programs.

Branch-and-Bound

Branch-and-bound is essentially a strategy of divide and conquer. The idea is to partition the

feasible region into more manageable subdivisions and then, if required, to further partition the

subdivisions. In general, there are a number of ways to divide the feasible region, and as a

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consequence there are a number of branch-and-bound algorithms. An integer linear program is a

linear program further constrained by the integrality restrictions. Thus, in a maximization problem,

the value of the objective function, at the linear-program optimum, will always be an upper bound

on the optimal integer-programming objective. In addition, any integer feasible point is always a

lower bound on the optimal linear-program objective value.

The idea of branch-and-bound is to utilize these observations to systematically subdivide the linear

programming feasible region and make assessments of the integer-programming problem based

upon these subdivisions.

Implicit Enumeration

A special branch-and-bound procedure can be given for integer programs with only binary

variables. The algorithm has the advantage that it requires no linear programming solutions. One

way to solve such problems is complete enumeration. List all possible binary combinations of the

variables and select the best such point that is feasible. The approach works very well on a small

problem, where there are only a few potential 01 combinations for the variables.

In general, though, an n-variable problem contains 2n 01 combinations; for large values of n, the

exhaustive approach is prohibitive. Instead, one might implicitly consider every binary

combination, just as every integer point was implicitly considered, but not necessarily evaluated,

for the general problem via branch-and-bound. Here, we adopt the opposite tactic of always

maintaining the 01 restrictions, but ignoring the linear inequalities. The idea is to utilize a branch-

and-bound (or subdivision) process to fix some of the variables at 0 or 1. The variables remaining

to be specified are called free variables.

Cutting Planes

The cutting plane algorithm solves integer programs by modifying linear-programming solutions

until the integer solution is obtained. It does not partition the feasible region into subdivisions, as

in branch-and-bound approaches, but instead works with a single linear program, which it refines

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by adding new constraints. The new constraints successively reduce the feasible region until an

integer optimal solution is found. In practice, the branch-and-bound procedures almost always

outperform the cutting-plane algorithm. Nevertheless, the algorithm has been important to the

evolution of integer programming.

In addition, even though the algorithm generally is considered to be very inefficient, it has

provided insights into integer programming that have led to other, more efficient, algorithms.

iii. PURE STRATEGY AND MIXED STRATEGY

A Pure Strategy in game theory, also known as Nash equilibrium is a fundamental concept in the

theory of games and the most widely used method of predicting the outcome of a strategic

interaction in the social sciences. A game in strategic or normal form consists of the following

three elements:

A set of players.

A set of actions or pure-strategies available to each player.

A payoff or utility function for each player.

The payoff functions represent each players preferences over action profiles, where an action

profile is simply a list of actions, one for each player. A pure strategy Nash equilibrium is an action

profile with the property that no single player can obtain a higher payoff by deviating unilaterally

from this profile. This concept can best be understood by looking at some examples.

Consider first a game involving two players, each of whom has two available actions, called A and

B. If the players choose different actions, they each get a payoff of 0. If they both choose A, they

each get 2, and if they both choose B, they each get 1. This coordination game may be represented

as follows, where player 1 chooses a row, player 2 chooses a column, and the resulting payoffs are

listed in parentheses, with the first component corresponding to player 1s payoff. The action

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profile (B,B) is an equilibrium, since a unilateral deviation to A by any one player would result in a

lower payoff for the deviating player. Similarly, the action profile (A,A) is also an equilibrium.

Mixed Strategy is another concept in game theory. A game in strategic form does not always have

a Nash equilibrium in which each player deterministically chooses one of his strategies. However,

players may instead randomly select from among these pure strategies with certain probabilities.

Suppose a consumer purchases a license for a software package, agreeing to certain restrictions on

its use. The consumer has an incentive to violate these rules. The vendor would like to verify that

the consumer is abiding by the agreement, but doing so requires inspections which are costly. If the

vendor does inspect and catches the consumer cheating, the vendor can demand a large penalty

payment for the noncompliance. The standard outcome, defining the reference payoff zero to both

vendor (player 1) and consumer (player 2), is that the vendor chooses Dont Inspect and the

consumer chooses to Comply. Without inspection, the consumer prefers to Cheat since that gives a

positive payoff, with resulting negative payoff to the vendor. The vendor may also decide to

Inspect and if the consumer complies, inspection leaves the consumers payoff unchanged, while

the vendor incurs a cost resulting in a negative payoff. If the consumer cheats, however, inspection

will result in a heavy penalty and still create a certain amount of hassle for player 1.

In all cases, player 1 would strongly prefer if player 2 complied, but this is outside of player 1s

control. However, the vendor prefers to inspect if the consumer cheats. If the vendor always

preferred Dont Inspect, then this would be a dominating strategy and be part of a unique

equilibrium where the consumer cheats. If any of the players settles on a deterministic choice, like

Dont Inspect by player 1, the best response of the other player would be unique, here Cheat by

player 2, to which the original choice would not be a best response. The strategies in Nash

equilibrium must be best responses to each other, so in this game this fails to hold for any pure

strategy combination.

iv. QUEUEING THEORY AND ITS PARAMETERS

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The main object of Queueing Theory is to develop formulae, expressions, or algorithms

for performance metrics, such as the average number of entities in a queue, mean time spent in the

system, resource availability, probability of rejection, and the like. The results from queueing

theory can directly be used to solve design and capacity planning problems, such as determining

the number of servers, an optimum queueing discipline, and schedule for service, number

of queues, system architecture, and the like. Besides making such strategic design decisions,

queueing theory can also be used for tactical as well as operational designs and controls. The

queueing parameters can be listed as:

Arrival pattern/process

The arrival process or input process to a queueing system is often measured in terms of the average

number of arrivals per unit of time or by the average time between successive arrivals. Customers

may arrive for service individually or in groups. Single arrivals are illustrated by customers

visiting a beautician, students reaching at a library counter, and so on. On the other hand, families

visiting restaurants, ship discharging cargo at a dock are examples of bulk, or batch, arrivals.

Customers can also arrive in the system at regular or irregular times, or they might arrive in a

random way. The queueing models wherein customers arrival times are known with certainty are

categorized as deterministic models and are easier to handle. On the other hand, a substantialmajority of the queueing models are based on the premise that the customers enter the system

stochastically, at random points in time. With random arrivals, the number of customers reaching

the system per unit might be described by a probability distribution. Generally, the queueing

models are based on the assumption that arrival pattern follows Poisson distribution.

Speed of service / service mechanism

In a queueing system, the speed with which service is provided can be expressed in either of two

ways, as service rate and as service time. The service rate describes the number of customers

serviced during particular time period. The service time indicates the amount of time needed to

service a customer. Service rates and times are reciprocals of each other and either of them is

sufficient to indicate the capacity of the facility. Thus, if a cashier can attend, on average, to 10

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customers in an hour, the service rate would be expressed as 10 customers per hour and service

time would be equal to 6 minutes per customer. Generally, however, we consider the service time

only. If these service times are known exactly, the problem can be handled easily. But, as generally

happens, if these are different and not known with certainty, we have to consider the distribution of

the service times in order to analyze the queueing system. Generally, the queueing models are

based on the assumption based on the assumption that service times are exponentially distributed.

Queue discipline

It refers to the manner by which customers are selected for service when a queue has formed. The

most common discipline that can be observed in everyday life is first come, first served order, or

first in-first out (FIFO), as it sometimes called some other in common usage are last in-last out

(LIFO), which is applicable to many inventory systems when there is no obsolescence in stored

units as it is easier to reach the nearest items which are the last in, and a variety of priority

schemes, where customers are given priorities upon entering the system, the ones with higher

priorities to be selected for service ahead of those with lower priorities, regardless of their time of

arrival to the system.

Number of servers

The number of servers or service channels refers to the number of parallel service stations which

can service customers simultaneously. There can be single servers or multiple server queueing

system. The two multichannel systems differ in that the first has a single queue, while the second

allows a queue for each channel. A barber shop with many chairs is an Example of the first type of

multichannel system, while number of queues before the railway counters is an example of the

second type of multichannel system.

System capacity

In some queueing processes there is a physical limitation to the amount of waiting room, so that

when the line reaches a certain length, no further customers are allowed to enter, until space

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becomes available by a service completion. These are referred to as finite queueing situation; that

is; there is a finite limit to the maximum queue size.

Size of the population

The source of customers for a queueing system can be infinite or finite. For example, all people of

a city or state (and others) could be the potential customers at a super bazaar. The number of

people being very large, it can be taken to the infinite. On the other hand, there are many situations

in business and industrial conditions where we cannot consider the population to be infinite-it is

finite. Thus, the ten machines in the factory requiring repairs and maintenance by the maintenance

crew would exemplify finite population.

REFERENCES

Dynamic Programming, [Online].

Available at: http://www.cs.berkeley.edu/~vazirani/algorithms/chap6.pdf

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John, R., A Dynamic Programming, [Online].

Available at: http://gemini.econ.umd.edu/jrust/research/papers/dp.pdf

Integer Programming, [Online].

Available at: http://web.mit.edu/15.053/www/AMP-Chapter-09.pdf

Wiesemann, W.Integer Programming, [Online].

Available at: http://www.doc.ic.ac.uk/~br/berc/IPlecture1.pdf

Mixed Strategy, [Online].

Available at: http://www.columbia.edu/~rs328/MixedStrategy.pdf

Nash Equilibrium, [Online].

Available at: http://www.columbia.edu/~rs328/NashEquilibrium.pdf

Game Theory, [Online],

Available at: http://www.cdam.lse.ac.uk/Reports/Files/cdam-2001-09.pdf

Queueing Theory, [Online],

Available at: http://www.win.tue.nl/~iadan/queueing.pdf
http://gemini.econ.umd.edu/jrust/research/papers/dp.pdfhttp://web.mit.edu/15.053/www/AMP-Chapter-09.pdfhttp://www.doc.ic.ac.uk/~br/berc/IPlecture1.pdfhttp://www.columbia.edu/~rs328/MixedStrategy.pdfhttp://www.columbia.edu/~rs328/NashEquilibrium.pdfhttp://www.cdam.lse.ac.uk/Reports/Files/cdam-2001-09.pdfhttp://www.win.tue.nl/~iadan/queueing.pdfhttp://gemini.econ.umd.edu/jrust/research/papers/dp.pdfhttp://web.mit.edu/15.053/www/AMP-Chapter-09.pdfhttp://www.doc.ic.ac.uk/~br/berc/IPlecture1.pdfhttp://www.columbia.edu/~rs328/MixedStrategy.pdfhttp://www.columbia.edu/~rs328/NashEquilibrium.pdfhttp://www.cdam.lse.ac.uk/Reports/Files/cdam-2001-09.pdfhttp://www.win.tue.nl/~iadan/queueing.pdf

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