or lectures 2011 - part 2

8/13/2019 Or Lectures 2011 - Part 2

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Prof. M. Hamdy El wany

1 s t Year P roduc t ion Eng ineer ingDepar tmen t

2 0 11

Operations Research1


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Lecture 5

Integer Programming2


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Introduction

An integer programming problem (IP) is anLP in which some or all of its variables arerequired to be non-negative integers.

Many real-life situations may beformulated as IPs.

Mainly 3 classes depending on variabletype:

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Pure Integer Programming

An IP in which all variables are required tobe integers is called a pure integerprogramming problem. For example,

Max z = 3x1 + 2x2 s.t x 1 + x2 6

x1, x2 0, x1, x2 are integers

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Mixed Integer Programming

An IP in which only some of the variablesare required to be integers is called amixed integer programming problem. Forexample:

Max z = 3x1 + 2x2 s.t x 1 + x2 6

x1, x2 0, x1 is an integer

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0 1 IP or Binary IP

An IP problem in which all the variables mustequal 0 or 1 is called a 0 1 IP or Binary IP .The following is an example of a 0 1 IP:

Max z = x1 - x2s.t. x 1 + 2x2 2

2x1 - x2 1x1, x2 = 0 or 1

0 1 IPs occur in surprisingly many situations.

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Example 5.17

Stockco is considering four investments.Stockco has only $14,000 for investment.Stockco wants to maximize Net Income (NI)obtained.

Investment 1 2 3 4

Initial Cost 5,000 7,000 4,000 3,000

NI 16,000 22,000 12,000 8,000


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How Does it Differ?9 max z = 21x 1 + 11x2

s.t. 7x 1 + 4x2 13 x1, x2 0 x1, x2 integer


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How Does it Differ? (Contd)

Feasible is not a convex set.Feasible region is the set of points:

S = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1)}.

If the feasible region for a pure IPs LP relaxationis bounded, then the feasible region for the IPwill consist of a finite number of points.Could be solved by enumeration: The optimalsolution here is x 1 = 0, x2 = 3 with a value of z =33.

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Solving this Simple Example11

Unfortunately, IPs are usually much harderto solve than LPs.

Could be solved by:Enumeration of all possible solutions.Graphical Solution.

With the help of LP Relaxation.The LP obtained by omitting all integer or 0 1constraints on variables.


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Solving this Simple Example (Contd)

Suppose that a naive analyst suggests thefollowing approach for solving an IP:

First solve the LP relaxation;

Then round off (to the nearest integer) each variablethat is required to be an integer and that assumes afractional value in the optimal solution to the LPrelaxation.Applying this approach here, we first find the optimalsolution to the LP relaxation: x 1 = 13/7, x 2 = 0.

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Solving this Simple Example (Contd)

Rounding this solution yields the solution x 1 = 2,x2 = 0 as a possible optimal solution. But it ishowever infeasible.

Even if we round x 1 downward (yielding thecandidate solution x 1 = 1, x2 = 0),

For some IPs, it can even turn out that everyroundoff of the optimal solution to the LPrelaxation is infeasible.

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LP Relaxation

The concept of LP Relaxation of an integerprogramming problem plays a key role in thesolution of IPs.DEFINITION: The LP obtained by omitting allinteger or 0 1 constraints on variables is calledthe LP relaxation of the IP.

This means that the feasible region for any IPmust be contained in the feasible region for thecorresponding LP relaxation.

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LP Relaxation (Contd)

For any IP that is a max problem, this impliesthat;Optimal z- value for LP relaxation optimal z -

value for IP

If you solve the LP relaxation of a pure IP andobtain a solution in which all variables areintegers, then the optimal solution to the LPrelaxation is also the optimal solution to the IP.

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The Branch-and-Bound Method

In practice, most IPs are solved byusing the technique of branch-and-

bound.Branch-and bound methods find theoptimal solution to an IP by efficientlyenumerating the points in a sub-problems feasible region.

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The Branch-and-Bound Method18


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Unfortunately, the optimal solution to the LPrelaxation is z = 165/4, x 1 = 15/4, x 2 = 9/4 (seeFigure).

We know that (optimal z-value for IP) (optimalz-value for LP relaxation). This implies that theoptimal z-value for the IP cannot exceed 165/4.

Thus, the optimal z-value for the LP relaxation isan upper bound for Telfas profit.

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Our next step is to partition the feasible regionfor the LP relaxation in an attempt to find outmore about the location of the IPs optimal

solution.We arbitrarily choose a variable that is fractionalin the optimal solution to the LP relaxation say,

x1.

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Now observe that every point in the feasibleregion for the IP must have either x 1 3 or x 1 4. (Why cant a feasible solution to the IP have 3

< x1 < 4?)With this in mind, we branch on the variablex1 and create the following two additionalsubproblems:

Subproblem 2: Subproblem 1 + Constraint x 1 4. Subproblem 3: Subproblem 1 + Constraint x 1 3.

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Observe that neither subproblem 2 norsubproblem 3 includes any points with x 1 = 15/4.This means that the optimal solution to the LPrelaxation cannot recur when we solvesubproblem 2 or subproblem 3.

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From Figure, we see that every point in the feasibleregion for the Telfa IP is included in the feasibleregion for subproblem 2 or subproblem 3.

Also, the feasible regions for subproblems 2 and 3have no points in common.Because subproblems 2 and 3 were created by

adding constraints involving x 1, we say thatsubproblems 2 and 3 were created by branching onx1.

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Telfa Subproblems 1 and 2 Solved26


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A display of all subproblems that have been createdis called a tree.Each subproblem is referred to as a node of thetree, and each line connecting two nodes of thetree is called an arc.The constraints associated with any node of thetree are the constraints for the LP relaxation plusthe constraints associated with the arcs leadingfrom subproblem 1 to the node.The label t indicates the chronological order inwhich the subproblems are solved.

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The optimal solution to subproblem 2 did notyield an all-integer solution, so we choose to usesubproblem 2 to create two new subproblems.

We choose a fractional valued variable in theoptimal solution to subproblem 2 and thenbranch on that variable.

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Because x 2 is the only fractional variable in theoptimal solution to subproblem 2, we branch onx2. We partition the feasible region for

subproblem 2 into those points having x 2 2and x 2 1.This creates the following two subproblems:

Subproblem 4 = Subproblem 1 + x 1 4 and x 2 2= Subproblem 2 + x 2 2.

Subproblem 5 = Subproblem 1 + x 1 4 and x 2 1= Subproblem 2 + x 2 1.

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Feasible Regions Subproblems 4, 530


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Feasible Regions Subproblems 4, 532


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Subproblems 1, 2, 4, and 5 Solved33


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Feasible Regions Subproblems 6,734


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After Six Subproblems35


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Solve Sub-problem 336


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Final Branch-and-Bound Tree37


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Fathoming

If it is unnecessary to branch on a subproblem,then it is fathomed. The following threesituations result in a subproblem being

fathomed:The subproblem is infeasible;The subproblem yields an optimal solution in which

all variables have integer values; andThe optimal z-value for the subproblem does notexceed (in a max problem) the current LB.

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Another Example39


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Solving MIP Problems40

Modify the method described by branching onlyon variables that are required to be integers.Optimum solution needs integer values for onlyinteger variables.For Example

Max z = 2X1 + X2

s.t. 5X1+ 2X2 8 X1 + X2 3 X1, X2 0, X1 integer


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Branch-and-Bound tree for Mixed IP41


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Solving BIP Problems42


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Lecture 6

Multi-Objective Linear

Programming43


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Introduction

Consider the impact of multiple, and conflictingobjectivesMathematical representation for a multi-objective programming problem:

Minimize (or maximize) set of objectives:

Subject to set of constraints

= 1 , 2

( ) = 0 ( ) 0

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Introduction

Multiple-objective techniques :Weighting or utility methods.Ranking or prioritizing methods.

Efficient solution (or generating) methods.

Goal programming approach (a ranking orprioritizing method) is the technique understudy.

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Introduction

Reasons for goal programming:The model development is relatively simple andstraightforward.Minor modifications may be employed so as toencompass the alternative approaches.The method of solution is quite simple and is, in fact, justa refinement to the two phase Simplex method.The goal programming models, and variations, havealready found extensive implementation in actualproblems since the early 1950s.The model and its assumptions seem consistent withtypical real-world problems.

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Terminology

Efficient solutionResults form the model and it is called with that namebecause it may not be optimum with respect to all theconflicting objectives of the problem.

ObjectiveAn objective is a relatively general statement that reflectsthe desires of the decision maker.

Aspiration level

Is a specific value associated with a desired or acceptablelevel of achievement of an objective.Used to measure the achievement of an objective andgenerally to anchor the objective to reality.

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Terminology Cont.

GoalAn objective in conjunction with an aspirationlevel is termed a goal.

Goal deviation

The difference between what we accomplishand what we aspire to is the deviation fromour goal.

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Goal formulation

Goal Type Goal Programming Form Deviation Variables to Be Minimized

+ =

+

=

= + = + f i (x)= mathematical representation of objective i as a function of

the decision variables .b i = value of the aspiration level associated with objective i. i =positive deviation. i =negative deviation

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Goal formulation (Contd)

ExampleLet us assume that the profit of a firm may beexpressed as a linear function of two variables:

If our aspiration level is to obtain at least $1,000 ofprofit per time period, the goal maybe written as:

Adding negative and positive deviation variables

1 should be minimized.

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+ 72

=

5 1 + 7 2 1000 5 1 + 7 2 + = 1000

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The Achievement Function

The achievement function answers the question of:How good the solution is?

Some of the measures used to evaluate the goodness of a solutioninclude:

How well does it minimize the sum of weighted goal deviations?How well does it minimize some polynomial (or other nonlinear)form of the goal deviations?How well does it minimize the maximum (worst) goal deviation?

How well does it lexicographically minimize an ordered (i.e.ranked or prioritized) set of goal deviations?Various combinations of the above.

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The Achievement Function (Contd)

We shall measure the achievement in terms of thelexicographic minimization of an ordered set of goaldeviations, wherein within each set of goals at a

particular rank, weights may be used.Lexicographical minimum

Given an ordered array a of nonnegative elements a ks,the solution given by a (1) is preferred to a (2) if ak(1) < ak(2) and all higher-order elements (i.e. a 1, a2, ak-1) areequal. If no other solution is preferred to a , then a is thelexicographic minimum.

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Model Construction

The steps in the construction of the model aresummarized then as:

Step 1: Formulate the baseline model.

Step 2: Specify aspiration levels for each and everyobjective.Step 3: Include negative and positive deviationvariables for each and every goal and constraint.

Step 4: Rank the goals in terms of importance.Priority 1 is always reserved for the rigid constraints.Step 5: Establish the achievement function.

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Model Construction (Contd)

ExampleA small cosmetics firm until recently made only nailpolish remover. However, when an employeeaccidentally dropped a jar of peanut butter into theremover, it was found that the resultant mixturecould temporarily remove facial wrinkles.Thus, the firm now produces two product: Wipe out(the nail polish remover), and Ageless (The wrinkle

remover). The new, improved formulae for eachproduct require differing amounts of two basechemicals (whose names are a company secret) asshown in the coming.

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Produc t Prof i t per

Gallon

Pounds o f

Chemical A

Required per

Gallon

Pounds o f

Chemical B

Required per

Gallon

Pounds o f

Peanu t Better

per Gallon

Ageless

wrinkle

remover

80 4 4 1

Wipe out nai l

po l i sh

remover

100 5 2 0

A m o u n t s

available per

day (pounds )

- 80 48 6

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The daily availability of these two chemicalscannot be exceeded, as there is only onesupplier, who is producing at maximal capacity.

Note that, although the daily supply of peanutbutter is unlimited, the owner does not want topurchase more than 6 pounds per day, so as to

keep secret the fact that this ingredient is goinginto production.

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When the company president was questioned, thefollowing facts were determined:The daily availability of chemicals A and B cannot, in anyway, be exceeded. That is, these are the rigid constraints.

The firm would like to maintain daily profits at a levelabove $800.The daily amount of peanut butter ordered should bekept under 6 pounds (that way, it may be disguised asshipments to the company cafeteria).The total number of gallons produced per day of bothproducts should be minimized so as to simplify shippingand handling.

First, we start by constructing the baseline model:

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d l ( d)


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Decision variables: X1 = gallons per day of Ageless producedX2 = gallons per day of Wipe out produced

Goals and constraints: 4X1 + 5X2 < 80 (1)4X1 + 2X2 < 48 (2)80X1 + 100X2 > 800 (3)

X1 < 6 (4)Minimize X1 + X2 (5)X1, X2 > 0

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The only relation for which we need an aspiration level is(5). We shall assume that the aspired level is 7 gallons perday.Now we can form our linear goal programming model as:

Find X1 and X2 so as to lexicographically minimizea = {(1 + 2), (3), (4), (5)}Subject to

4X1 + 5X2 + 1 - 1 = 804X1 + 2X2 + 2 - 2 = 4880X1 + 100X2 + 3 3 = 800 X1 + 4 - 4 = 6 X1 + X2 + 5 - 5 = 7X , , > 0

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Methods of solution

There are 2 basic approaches to thesolution of the linear goal programmingmodel with an achievement function thatis to be lexicographically minimized.

Graphical methodSequential Linear Goal Programming (SLGP)

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G hi l M h d


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Graphical Method

Find X1 and X2 so as to lexicographicallyminimize

a = {(1 + 2), (3), (4), (5)}

Subject to,G1: 4X1 + 5X2 + 1 - 1 = 80G2: 4X1 + 2X2 + 2 - 2 = 48G3: 80X1 + 100X2 + 3 3 = 800G4: X1 + 4 - 4 = 6G5: X1 + X2 + 5 - 5 = 7X, , > 0

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G hi l M h d (C d)


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Graphical Method (Contd)

G1: 4X 1 + 5X 2 + 1 - 1 = 80

G2: 4X 1 + 2X 2 + 2 - 2 = 48

G3: 80X 1 + 100X 2 + 3 3 = 800G4: X 1 + 4 - 4 = 6

G5: X 1 + X2 + 5 - 5 = 7

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G hi l M h d (C d)


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G1: 4X 1 + 5X 2 + 1 - 1 = 80

G2: 4X 1 + 2X 2 + 2 - 2 = 48

G3: 80X 1 + 100X 2 + 3 3 = 800G4: X 1 + 4 - 4 = 6

G5: X 1 + X2 + 5 - 5 = 7

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G hi l M h d (C d)


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G1: 4X 1 + 5X 2 + 1 - 1 = 80

G2: 4X 1 + 2X 2 + 2 - 2 = 48

G3: 80X 1 + 100X 2 + 3 3 = 800G4: X 1 + 4 - 4 = 6

G5: X 1 + X2 + 5 - 5 = 7

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G hi l M th d (C td)


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G1: 4X 1 + 5X 2 + 1 - 1 = 80

G2: 4X 1 + 2X 2 + 2 - 2 = 48

G3: 80X 1 + 100X 2 + 3 3 = 800G4: X 1 + 4 - 4 = 6

G5: X 1 + X2 + 5 - 5 = 7

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Sequential Linear Goal Programming (SLGP)

Lexicographically minimizea = {( 1 + 2), ( 3), ( 4), ( 1 + 1.5 2)}

Subject to,X1 + 1 - 1 = 30 X2 + 2 - 2 = 15

8X1 + 12X2 +3 3 = 1000 X1 + 2X2 +4 4 = 40X, , > 0.

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SLGP E l L l 1


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SLGP Example Level 1

We begin by formulating the modelassociated with priority level 1 only.

Minimize a 1 = {(1 + 2)}Subject to,

X1 + 1 - 1 = 30 X2 + 2 - 2 = 15

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SLGP E l L l 1 (C td)


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SLGP Example Level 1 (Contd)

CN 0 0 1 1

CB V X1 X2 1 2 XB

0 1 1 0 -1 0 30

0 2 0 1 0 -1 15

0 0 -1 -1 0

Initial tableau for priority level 1.

X1 = 0 1 = 30 2 = 0

X2 = 0 2 =15 1 = 0

a 1*= 1 + 2 = 0

This initial tableau isalso optimal, and thusthe solution to the firstlevel is:

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CN 0 0 0

CB V X1 X2 3 XB

0 1 1 0 0 300 2 0 1 -1 15

1 3 8 12 -1 1000

8 12 -1 1000

Initial tableau for priority level 2

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V X1 2 3 XB

1 1 0 0 30X2 0 1 0 15

3 8 -12 -1 820

8 -12 -1 820

Second tableau for priority level 2

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V X1 2 3 XB

X1 1 0 0 30

X2 0 1 0 15

3 - 8 -12 -1 580

- 8 -12 -1 580

Third and optimum tableau for priority level 2.

X1 = 30 1 = 0 3 = 0

X2 = 15 2 =0 3 = 580

a 2*= 3 = 580

The solution is, thus, forlevel 2:

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Note that the indicator row elements under thenon-basic variables 1, 2, and 3 are all negative inthis optimal tableau.They should degrade the previous solution for level2.Column dropping rule: Any non-basic variable thathas a negative indicator row value in the optimal

tableau may be dropped (and its correspondingcolumn in the tableau dropped) from the problem,as the introduction of such a variable woulddegrade the solution.

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SLGP Example Level 3 (Cont d)

Thus, 1, 2, and 3 may all be dropped, givingrise to the following formulation for level 3:

Minimize a = 4

Subject to,X1 = 30 X2 = 15

8X1 + 12X2 +3 = 1000 X1 + 2X2 +4 4 = 40 3 = 580

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Moving to priority level 4, note that,a4 = 1 + 1.5 2

But 1 and 2 have already been dropped (i.e.

set to zero) in the problem. Thus, it isunnecessary to formulate or solve the model forlevel 4 as the solution is fixed:

X1 *= 30X2 * = 15

a 4* = (0, 580, 20, 0)

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Introduction


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Introduction

Inventory is one of the most expensive assets ofmany companies, representing as much as 40%of total invested capital.

A firm can reduce costs by reducing on-handinventory levels.Production may stop and customers become

dissatisfied when an item is out of stock

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Introduction Cont


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Introduction Cont.

All organizations have some type of inventoryplanning and control systemBankHospital

Government agenciesSchools

In inventory managementDecide whether to produce goods or to purchase them.

Forecast demand.Determine the inventory necessary to service thatdemand

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Functions of Inventory


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Functions of Inventory

To provide a stock of goods to meet anticipatedcustomer demand and provide a selection ofgoods.

To decouple suppliers from production andproduction from distribution.To take advantage of quantity discounts,

because purchases in larger quantities mayreduce the cost of goods or delivery.

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Functions of Inventory Cont


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Functions of Inventory Cont.

To hedge against inflation and upward pricechanges.To protect against delivery variation due to

weather, supplier shortages, quality problems,or improper deliveries.To permit operations to continue smooth/v with

the use of work -in-process inventory

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Types of Inventory


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Types of Inventory

To accommodate the functions of inventory,firms maintain four types of inventories:

Raw material inventory

Materials that purchased but not processedWork-in-process inventory

Components or raw material that has undergone somechange but is not completed

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ABC Analysis


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ABC Analysis

Also called Pareto analysisIt discuss how inventory items can be classifiedThe Pareto principle states that there are acritical few and trivial many. The idea is to establish inventory policies thatfocus resources on the few critical inventory notthe many trivial ones.

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ABC Analysis Cont


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ABC Analysis Cont.

Class A items are those on which the annual dollarvolume is high.Represent about 15% of the inventory.Represent 70% to 80% of the total dollar usage.

Class B items inventory items of medium annualdollar volume.Represent about 30% of inventory.Represent 15% to 25% of the total value

Those with low dollar volume are Class CRepresent only 5% of the annual dollarRepresent about 55% of the total inventory items.

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ABC Analysis Cont


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ABC Analysis Cont.87

Inventory Models


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Inventory Models

Inventory cost elementsHolding Costs

Are the costs associated with holding or carrying inventoryover time. (Space, lighting, ..)

Ordering/Setup CostsInclude costs of supplies, forms, order processing, clericalsupport and so forth. (Phone call, transportation cost,.)

Is a part of what is called setup costs.Setup cost is the cost to prepare a machine or pr formanufacturing an order.

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Inventory Models Cont


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Inventory Models Cont.

Two main types of inventory models:Deterministic Inventory Models

Basic economic order quantity (EOQ) model.

Production order quantity model.Quantity discount model.

Probabilistic Models with Constant Lead Time

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EOQ Model


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EOQ Model

Basic economic order quantity (EOQ) modelIs one of the oldest and most commonly knowninventory-control techniques.Model assumptions

Demand is known, constant, and independent.Lead time that is, the time between placement and receipt ofthe order is known and constant.Receipt of inventory is instantaneous and complete.Quantity discounts are not possible.The only variable costs are the cost of setting up or placing anorder and the cost of holding or storing inventory over time .Stock-outs (shortages) can be completely avoided if orders areplaced at the right time.

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EOQ Model Cont.


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EOQ Model Cont.

The objective of most inventory models is tominimize total costs

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EOQ Model Cont.


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EOQ Model Cont.

Q = Number of pieces per orderQ* = Optimum number of pieces per order (EOQ)D = Annual demand in units for the inventory item

S = Setup or ordering cost for each orderH = Holding or carrying cost per unit per yearL = Lead time

D = Slope (units/day )

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EOQ Model Cont.


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EOQ Model Cont.

Annual setup cost

Annual holding cost

Annual Purchase cost

The Total Annual cost

EOQ

=

=2

= = +

2+

= 2

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EOQ Model Cont.


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EOQ Model Cont.

Reorder Points (ROP)It answer when to order question

=

=

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EOQ Model Cont.


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EOQ Model Cont.

ExampleCompany markets painless hypodermic needlesto hospitals would like to reduce its inventory

cost by determining the optimal number ofhypodermic needles to obtain per order. Theannual demand is 1000 units; the setup or

ordering cost is $10 per order: and the holdingcost per unit per year is $.50.

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EOQ Model Cont.


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OQ ode Co t.

r units/orde2000.5

)2(1000)(10H

2DSQ*

= =

=

= =

= 1000/2200 = 5

= 250/5 = 50 = (1000/200) (5) + (200/2) (0.50) = 50 + 50 = $ 100

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EOQ Model Cont.


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Q

Notice that we ignored the purchase price here incalculating the total annual costs as they areconstant for any ordering policy, except for the caseof quantity discount models which we will discusslater onIf we assume that the annual demand and the priceper hypodermic needle are known values (forexample, 1,000 hypodermics per year at P = $10)and annual cost should include purchase cost,

= 50 + 50 + (1000 10) = $ 10,100

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POQ Model


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Q

Production Order Quantity (POQ) ModelConsider that the firm may receive its inventory overa period of time

This model is applicable under two situations: When inventory continuously flows or builds up over aperiod of time after an order has beenWhen units are produced and sold simultaneously.

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POQ Model Cont.


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Q100

POQ Model Cont.


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Q

Q = Number of pieces per order (run size)H = Holding cost per unit per yearp = Daily production rated = Daily demand rate, or usage ratet = Length of the production run in days

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POQ Model Cont.


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Q

Annual inventory holding cost

Average Inventory Level

Maximum inventory level

= = /2

= 1

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POQ Model Cont.


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Q

The Total Annual cost

Optimum run size

= PD)pd

H(12Q

SQD

++

=

2

1

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POQ Model Cont.


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Q

Annual Demand = D = 1000 units per yearSetup Cost = $10/setupHolding Cost = H = $0.50 per unit per yearDaily production rate = p = 8 units per dayDaily Demand rate = 4 units per day

= 21=

2 1000 (10)0.50 148

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Quantity Discount Model Cont.


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y

The equation for total annual inventory cost:Total cost = Setup cost + Holding cost + Product cost

Where:Q = Quantity orderedD = Annual demand in unitsS = Ordering or setup cost per order or per setupP = Price per unitH = Holding cost per unit per year

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y

Quantity Discount ModelsSteps

Step 1: For each discount, calculate a value for optimal ordersize Q*.

Step 2: For any discount, if the order quantity is too low toqualify for the discount, adjust the order quantity upward tothe lowest quantity that will qualify the discount.Step 3: Using the total cost equation, compute a total costfor every Q* determined in steps 1 and 2.Step 4: Select the Q* that has the lowest total cost, ascomputed in step 3.

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y

ExampleA Discount Store stocks toy race cars. Recently, the storehas been given a quantity discount schedule for thesecars (The one shown in the previous table). The normal

cost for the toy race cars is $5.00. For orders between1,000 and 1,999 units, cost drops to $4.80; for orders of2,000 or more units, the unit cost is only $4.75.Furthermore, ordering cost is $49.00 per order. Annualdemand is 5,000 race cars, and inventory carrying charge,as a percentage of cost, labeled I, is 20% or 0.2. Whatorder quantity will minimize the total inventory cost?

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Step 1

1 = 2 = 2 5000 490.2 5.00 = 700 / 2 = 2 = 2 5000 490.2 4.80 = 714 / 3 = 2 = 2 5000 490.2 4.75 = 718 /

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Step 2

Step 3

Step 4

Q1* = 700Q2* = 1000 adjustedQ3* = 2000 adjusted

TC1 = $ 25,700TC2 = $ 24,725TC3 = $ 24,822.50

Q2* = 1000 units.

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Probabilistic model


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Probabilistic models with constant lead timeApply when productdemand is not knownbut can be specified bymeans of a probabilitydistribution

= +

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Probabilistic model Cont.


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ExampleRivera Optical has determined that its reorderpoint for eyeglass frames is 50 (d X L) units. Itscarrying cost per frame per year is $5, and stock-out (or lost sale) cost is $ 40 per frame. Thestore has experienced the following probabilitydistribution for inventory demand during the

reorder period. The optimum number of ordersper year is six.

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Number of Units Probability

30 0.2

40 0.250 0.3

60 0.2

70 0.1

1.0

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SafetyStock

AdditionalHolding Cost

Stock-out CostTotalCost

20 (20)*($5) = $100 $0 $ 100

10 (10)*($5) = $ 50 (10)*(0.1)*($40)*(6) = $240 $ 290

0 $ 0 (10)*(0.2)*($40)*(6) +(20)*(0.1)*($ 40)*(6) =$ 960

$ 960

The safety stock with the lowest total cost is 20frames. Therefore, the new reorder point = 50 + 20= 70 frames.

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ExampleA regional Hospital stocks a code blueresuscitation kit that has a normally distributeddemand during the reorder period. The mean

(average) demand during the reorder period is 350kits, and the standard deviation is 10 kits. Thehospital administrator wants to follow a policy thatresults in stock-outs occurring only 5% of the time.

What is the appropriate value of Z?How much safety stock should the hospital maintain?What reorder point should be used?

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The following figure may help you visualize theexample:

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(c) The reorder point =Expected demand during lead time + SS = 350 + 16.5= 366.5 pr 367 kits.

Now, assumes that both an estimate of expected demand

during lead times and its standard deviation are available.When data on lead time demand is not at hand, theseformulas cannot be applied and we need to determine if:

(a) demand is variable and lead time is constant; or

(b) only lead time is variable; or(c) both demand and lead time are variable. For each of thesesituations, a different formula as follows:

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(a) If only the demand d is variable, then

(b) If only lead time is variable, then

(c) If both are variable, then

= + , = =

= ( ) + ( )

=

+ 2 + 2 2

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Lecture 8

Queuing Theory121


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Introduction


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Queuing analysis is a probabilistic form of analysis,not a deterministic technique. Thus, the results ofqueuing analysis, referred to as operatingcharacteristics, are probabilistic.These operating statistics (such as the average timea person must wait in line to be served) are used bythe manager of the operation containing the queue

to make decisions.

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Waiting Line Analysis


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Operating characteristicsaverage values for characteristics that describe theperformance of a waiting line system

Queue

A single waiting lineWaiting line system consists of

ArrivalsServers

Waiting line structures

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Elements of a Waiting Line (cont.)


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17-127

Queue disciplineOrder in which customers are servedFirst come, first served is most common

Length can be infinite or finiteInfinite is most commonFinite is limited by some physical

Basic Waiting Line Structures


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Channels are the number of parallel serversSingle channelMultiple channels

Phases denote number of sequential servers the customer mustgo through

Single phaseMultiple phases

Steady stateA constant, average value for performance characteristics that system willreach after a long time

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Essential Features of Queuing Systems


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DepartureQueue

discipline

Arrival

process Queueconfiguration

Serviceprocess

Renege

Balk

Callingpopulation

No futureneed for

service

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Arrival Process


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Static Dynamic

AppointmentsPriceAccept/Reject BalkingReneging

Randomarrivals withconstant rate

Random arrivalrate varyingwith time

Facility-controlled

Customer-exercised

control

Arrivalprocess

130

Queue Configuration


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Multiple Queue Single queue

Take a NumberEnter

3 4

8

2

6 10

1211

5

79

131

Queue Discipline


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Queuediscipline

Static(FCFS rule)

Dynamic

selectionbased on status

of queue

Selection basedon individual

customerattributes

Number ofcustomers

waitingRound robin Priority Preemptive

Processing timeof customers

(SPT rule)

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Service Facility Arrangements


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Service facility Server arrangement

Parking lot Self-serveCafeteria Servers in series

Toll booths Servers in parallelSupermarket Self-serve, first stage; parallel servers, second

stageHospital Many service centers in parallel and series, not all

used by each patient

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Kendall Notation


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SD=Service DisciplineFCFS,FCLS

Defaults B= , K= SD=FCFS

M/M/1 = M/M/1/ / /FCFS

135

Multi-server queue


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136

Multiple Single-Server Queues


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137

Operating Characteristics


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NOTATION OPERATINGCHARACTERISTIC

L Average number of customers in the system(waiting and being served)

Lq Average number of customers in the waiting lineW Average time a customer spends in the system

(waiting and being served)

W q Average time a customer spends waiting in line

P 0 Probability of no (zero) customers in the system

P n Probability of n customers in the system

Utilization rate; the proportion of time the system is in use

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Queuing System Design Objective


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Cost Relationship in Waiting Line Analysis

Service capacity Optimum

Cost of service

capacity

customer

waiting Cost

Total cost C o s t

Totalcost

Customerwaiting cost

Capacitycost

= +

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Waiting Time vs. Service System Utilization


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System Utilization

A v e r a g e n u m

b e r o r t i m e w a i t i n g i n

l i n e

0 100%

140


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Single-Server Model Assumptions142


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Poisson arrival rateExponential service timesFirst-come, first-served queue discipline

Infinite queue lengthInfinite calling population

= mean arrival rate = mean service rate

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Waiting Line Performance Measures for M/M/1143


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Average queue time, W q = * W = / ( - ) Average time in system, W = 1 / -

Average queue length, Lq = * L = 2 / ( - )

Average number of customers in system, L = / -

Average system utilization rate, /

Probability of idle service facility, P 0 1 - 1 /Probability of exactly n customers in system, P n (1 - ) n (1 / ) ( / ) n

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M/M/1 Example145


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A) What is the average utilization of the employee?

= 25 cust / hr

=1 customer

2 mins (1hr / 60 mins) = 30 cust / hr

= = 25 cust / hr 30 cust / hr

= .8333

145

M/M/1 Example146


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B) What is the average number of customers in line?

C) What is the average number of customers in the system?

4.167=25)-30(30

(25) =

)-( =

22

q L

5=25)-(30

25 =

- =

L

Note: L= L q+ 4.167 + 0.833

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M/M/1 Example148


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F) What is the probability that exactly two cars are inthe system?

n))(-(1=Pn

.1157=)3025

)(3025

-(1=P2

2

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Waiting Lines - Summary149


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Queuing System Design RulesTradeoff between capacity cost and cost of waitingIncreasing average utilization ( = / ) may causevery long waiting times

Model Characteristics

Arrival process distributionService process distribution and discipline

Number of serversQueue capacity/length

Model PresentedM/M/1

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Constant Service Times150


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Constant service timesoccur with machineryand automatedequipmentConstant service timesare a special case of thesingle-server model withundefined service times

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Constant Service Times151


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P 0 = 1 -Probability that no customersare in system

Average number of

customers in system

L = L q +

Average number ofcustomers in queue Lq =

2 2 ( - )

151

Constant Service Times (Contd) 152


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=Probability that the

server is busy

Average time customerspends in the systemW = W q +

1

Average time customerspends in queue

W q = L q

152

M/C/1 Example153


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Assume a drive-up window at a fast foodrestaurant. Customers arrive at the rate of 25 perhour. The employee can serve one customer everytwo minutes. Assume variable arrivals and aconstant service rate.Determine:

A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in thesystem?D) What is the average waiting time in line?

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M/C/1 Example154


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A) What is the average utilization of the employee?

.8333=cust/hr 30cust/hr 25 ==

cust/hr 30=mins)(1hr/60mins2customer 1

=

cust/hr 25=

154

M/C/1 Example

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B) What is the average number of customers in line?

C) What is the average number of customers in the system?

D) What is the average waiting time in line?

2.0833=25)-(2)(30)(30

(25) =

)-(2 =

22

Lq

mins5=hrs.08333=25

08333.2 =

=

LqW q

mins7=hrs.116667= 30/hr

1 +hrs.08333=

1

+=

WqWs

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Finite Queue Length156


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A physical limit exists on length of queueM = maximum number in queueService rate does not have to exceed arrival rate( ) to obtain steady-state conditions

156

Finite Queue Length

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P 0 =Probability that nocustomers are in system

1 - /

1 - ( / ) M + 1

Probability of exactly n customers in system P n = ( P 0 ) for n M

n

L = -Average number ofcustomers in system

/

1 - /

( M + 1)( / ) M + 1

1 - ( / ) M + 1

Finite Queue Length158


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Let P M = probability a customer will not join system

Average time customerspends in system

W = L

(1 - P M )

L q = L -(1- P M )Average number of

customers in queue


W q = W -1


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Finite Queue Length Solution

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Probability that nocars are in thesystem

P 0 = = = 0.381 - 20/30

1 - (20/30) 5

1 - /

1 - ( / ) M + 1

P n = ( P 0 ) = (0.38) = 0.076Probability ofexactly 4 cars inthe system

2030

4

n = M

L = - = 1.24Average numberof cars in thesystem

/ 1 - /

( M + 1)( / ) M + 1

1 - ( / ) M + 1

Finite Queue Length Solution

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Average time a carspends in the system

W = = 0.067 hr L

(1 - P M )

L q = L - = 0.62(1- P M )Average number of

cars in the queue

Average time a carspends in the queue

W q = W - = 0.033 hr1


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Finite Calling Population163


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W q = L q

( N - L )Average time customerspends in queue

L = L q + (1 - P 0 )Average number ofcustomers in system

W = W q +Average time customerspends in system

1


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Finite Calling Population Solution165


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Probability that notrucks are in the system P

0 = 0.652

Average number oftrucks in the queue

L q = 0.169

Average number oftrucks in system

L = 0.169 + (1 - 0.652) = .520

Average time truck

spends in queue

W q = 1.74 days

Average time truckspends in system

W = 5.33 days

Multiple-Server Model

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Two or more independent servers serve a single waitingline. Poisson arrivals, exponential service, infinite callingpopulation

s >

Computing P0 can be time-consuming.Tables can used to find P0 for selected values of and s.

P 0 = 1

1s!

ss

s -

n = s -1

n =0

1n !

n

+


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Probability of exactlyn customers in the

system

P n =

P 0 , for n >s

1s ! s n-s

n

P 0, for n > s

1

n !

n

Probability an arrivingcustomer must wait

P w = P 01s !

ss -

s

Average number ofcustomers in system

L = P 0 +( / ) s

( s - 1)!( s - ) 2


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W = L Average time customerspends in system

= / s Utilization factor


W q = W - =1 L q

L q = L

-

Average number of

customers in queue

Multichannel (M/M/S) Model Characteristics169


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Type: Multichannel systemInput source : Infinite; no balks, no renegingArrival distribution : PoissonQueue : Unlimited; multiple servers

Queue discipline : FIFO (FCFS)Service distribution : Negative exponentialRelationship : Independent service & arrival

Service rates > arrival rate

Multiple-Service Model Example170


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Student Health Service Waiting Room = 10 students per hour = 4 students per hour per service representative

s = 3 representativess = (3)(4) = 12

Multiple-Service Model Solution171


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P 0 = 0.045Probability no students are in the system

Number of students in the service area L = 6

L q = L - / = 3.5Number of students waiting tobe served

Average time students will wait inline W q = L q / = 0.35 hrs

Probability that a student must wait P w = 0.703

Waiting time in the service area W = L / = 0.60

Multiple-Service Model Solution172


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Add a 4th server to improve serviceRecompute operating characteristicsP0 = 0.073 prob of no studentsL = 3.0 students

W = 0.30 hour, 18 min in serviceLq = 0.5 students waitingW q = 0.05 hours, 3 min waiting, versus 21 earlierP

w = 0.31 prob that a student must wait


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lec ture 9

Basic Simulation Modeling173

What is Simulation?174


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It refers to a broad collection of methods andapplications to mimic the behavior of realsystems.

Examples of Modeled Systems175


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A manufacturing plant with machines, people.Transport devices, conveyor belts and storagesystems.A bank with different kinds of customers, servers,and facilities like teller windows, automated tellermachines (ATMs), and safety deposit boxes.A distribution network of plants, warehouses, and

transportation links.A supermarket with inventory control, checkout,and customer service.

Ways of Studying a System176


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System

Experiment with the Actual System

Experiment with aModel of the System

PhysicalModel

MathematicalModel

Simulation AnalyticalSolution


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When NOT to use Simulation?178


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If the problem can be solved analytically.If the problem is easier to be solved bydirect experiments.If simulation costs more that it saves.If resources or time is not available.

If systems behavior is too complex or cant be defined.

Simulation Advantages179


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Experimenting with the real system is too expensive or

sometimes impossible.

Experiments are made without disrupting the real world

system.Studying a certain parameter over a long period of time.

Better understanding of systems.

Easy manipulation of models.

Bottle-neck analysis.


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Types of Systems181


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Discrete-system state variable.


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Systems Components 183


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Entity: object of interest in the system.Attribute: property of an entity.Activity: represents a time period of specifiedlength.The state of a system : collection of variablesnecessary to describe the system at any time,relative to the objectives of the study.

An event: instantaneous occurrence that maychange the state of the system.

Examples of System s and Components184


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System Entities Attributes Activities Events State Variables

Production Machines

Speed;capacity;

breakdown

rate

Welding;stamping

BreakdownStatus of machines(busy, idle or down)

Inventory Warehouse Capacity Withdrawing DemandLevels of inventory;

backloggeddemands

Banking CustomersCheckingaccountbalance

Makingdeposits

Arrival;departure

Number of busytellers; queue length


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Simulation Methods186


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First simulation used to be performed byhand, then evolved to various types ofcomputer simulation packages, as follows:

1. By Hand2. Programming in General-Purpose Languages3. Simulation Languages4. High-Level Simulators


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Example of Simulation by Hand (Contd) 188


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Manual Simulation Outputs:Customer

Number (n)Arrival

Time (2)ServiceTime (3)

DepartureTime (4)

Time in Queue= 4 n-1 - 2 n

Time in Bank= 4 n -2 n

1 3.2 3.8 7.0 0.0 3.8

2 10.9 3.5 14.4 0.0 3.53 13.2 4.2 18.6 1.2 5.4

4 14.8 3.1 21.7 3.8 6.9

5 17.7 2.4 24.1 4.0 6.4

6 19.8 4.3 28.4 4.3 8.6

7 21.5 2.7 31.1 6.9 9.68 26.3 2.1 33.2 4.8 6.9

9 32.1 2.5 35.7 1.1 3.6

10 36.6 3.4 40.0 0.0 3.4



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Event-oriented description of bank teller simulation.



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Graphic portrayal of bank teller simulation.


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or lectures 2011 - part 2

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