chapter 8 operations scheduling. scheduling problems in operations job shop scheduling personnel...

Chapter 8

Operations Scheduling

Scheduling Problems in Operations

• Job Shop Scheduling

• Personnel Scheduling

• Facilities Scheduling

• Vehicle Scheduling and Routing

• Project Management

• Dynamic versus Static Scheduling

The Hierarchy of Production Decisions

Characteristics of the Job Shop Scheduling Problem

• Job Arrival Pattern

• Number and Variety of Machines

• Number and Skill Level of Workers

• Flow Patterns

• Evaluation of Alternative Rules

Objectives in Job Shop Scheduling

• Meet due dates

• Minimize work-in-process (WIP) inventory

• Minimize average flow time

• Maximize machine/worker utilization

• Reduce set-up times for changeovers

• Minimize direct production and labor costs(note: that these objectives can be conflicting)

Terminology• Flow shop: shop design in which machines are arranged in series

A Pure Flow Shop

• In general flow shop a job may skip a particular machine

Machine 1 Machine 2 Machine 3 Machine 4

Input parts

Finished Products

Machine 1 Machine 2 Machine 3 Machine 4

Input parts

Finished Products

Input parts Input parts Input parts

Terminology• Job shop: the sequencing of jobs through machines

– A job shop does not have the same restriction on workflow as a flow shop. In a job shop, jobs can be processed on machines in any order

– Usual job shop contains m machines and n jobs to be processed

– Each job requires m operations (one on each machine) in a specific order, but the order can be different for each job

– Real job shops might not require to use all m machines and yet may have to visit some machines more than once

– Workflow is not unidirectional in a job shop

One Machine in a Job Shop

Machine i

Finished Products

Jobs arriving from WIP

Input parts

Jobs leaving as WIP

Terminology• Parallel processing vs. sequential processing: parallel

processing means that the machines are identical

– In practice, there are often multiple copies of the same machine

– A job arriving at a work center can be scheduled on any one of a number machines more flexibility, complicating the scheduling problem further

– A factory might have multiple “identical machines”, purchased from the same manufacturer, that produce parts with higher quality on one machine than on any other

• Schedule: provides the order in which jobs are to be executed, and projects start time for each job at each work center

• Sequence: lists the order in which jobs are to be done

Terminology: Performance Measures

• Average WIP level: ….(is exactly what it sounds like)

• Flowtime: The amount of time a job spends from the moment it is ready for processing until its completion, and includes any waiting time prior to processing– Average WIP level is directly related to the time jobs spend in the

shop (flowtime)

• Makespan: The total time for all jobs to finish processing– For a single machine problem, the makespan is the same

regardless of the schedule, assuming we do not allow any idle time between jobs

Terminology: Performance Measures

Performances that have to do with each job’s due date• Lateness: The amount of time a job is past its due date

– Lateness is a negative number if a job is early

• Earliness: The amount of time a job a early

• Tardiness: Equals to zero if job is on time or early, and equals to lateness if the job is late

Measures of the cost of production:

Machine utilization and labor utilization are primary measures of shop utilization

Deterministic Scheduling of a Single Machine: Priority Sequencing Rules

• Random: Choose the next job at random. Do not use it!

• FCFS: First Come First Served. Jobs processed in the order they arrive to the shop. Viewed as a “fair” rule.

• SPT: Shortest Processing Time. Jobs with the shortest processing time are scheduled first. Popular method to determine the next homework assignment by many students.

Deterministic Scheduling of a Single Machine: Priority Sequencing Rules

• SWPT: Shortest Weighted Processing Time. A weight is assigned to each job based on the job’s value (holding cost) or on its cost of delay

• EDD: Earliest Due Date. Jobs are sequenced according to their due dates.

• CR: Critical Ratio. Compute the ratio of processing time of the job and remaining time until the due date. Schedule the job with the largest CR value next, however, if the job is late, the ration will be negative, or the denominator will be zero, and this job should be given highest priority

(Processing time remaining until completion) / (Due Date – Current Time)

424

1212

5

1865

63

871

Due Date, dj, (day)

Processing Time, pj, in Days

Job

424

1212

5

1865

63

871

Due Date, dj, (day)


JobFCFS Example

Job j pj Dj Cj Fj Lj Ej Tj

1 7 8 7 7 -1 1 0

2 1 12 8 8 -4 4 0

3 5 6 13 13 7 0 7

4 2 4 15 15 11 0 11

5 6 18 21 21 3 0 3

Average 12.8 3.2 1 4.2

Max 21 11 4 11

Processing Due date Completion Flowtime Lateness Earliness Tardiness

Flowtime: The amount of time a job spends from the moment it is ready for processing until its completion, and includes any waiting time prior to processingEarliness: The amount of time a job a early

SPT Example


2 1 12 1 1 -11 11 0

4 2 4 3 3 -1 1 0

3 5 6 8 8 2 0 2

5 6 18 14 14 -4 4 0

1 7 8 21 21 13 0 13

Average 9.4 -0.2 3.2 3

Max 21 13 11 13


Shortest Processing Time

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

is optimal for minimizing Average and Total flowtime Average waiting time Average and Total lateness

SWPT Example

Job j Pj Dj wj pj//wj Cj Fj wjFj Lj Ej Tj

4 2 4 5 0.4 2 2 10 -2 2 0

2 1 12 2 0.5 3 3 6 -9 9 0

5 6 18 4 1.5 9 9 36 -9 9 0

3 5 6 3 1.67 14 14 42 8 0 8

1 7 8 2 3.5 21 21 42 13 0 13

Ave 9.8 27.2 0.2 4 4.2

Max 21 42 13 9 13

Shortest Weighted Processing Time-total weighted down time

-sequencing 424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

n

jjjw FwF

1

n

n

wp

wp

wp

2

2

1

1

Processing Due date Completion Flowtime Lateness Earliness TardinessWeights

EDD Example

Earliest Due DateEarliest Due Date

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job


4 2 4 2 2 -2 2 0

3 5 6 7 7 1 0 1

1 7 8 14 14 6 0 6

2 1 12 15 15 3 0 3

5 6 18 21 21 3 0 3

Average 11.8 2.2 0.4 2.6

Max 21 6 2 6


CR Example

Critical Ratio:Critical Ratio:

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

Job j pj Dj CRj

1 7 8 0.875

2 1 12 0.083

3 5 6 0.833

4 2 4 0.500

5 6 18 0.333

TimeCurrent -Date Due

completion until remaining timeProcessing

Job j pj Dj Dj-CT CRj

2 1 12 5 0.200

3 5 6 -1 -5.000

4 2 4 -3 -0.667

5 6 18 11 0.545

Subtract Current Time

Schedule jobs 1 4 3 2 5

CR Example (cont)

Critical Ratio:Critical Ratio:

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

TimeCurrent -Date Due

completion until remaining timeProcessing


1 7 8 7 7 -1 1 0

4 2 4 9 9 5 0 5

3 5 6 14 14 8 0 8

2 1 12 15 15 3 0 3

5 6 18 21 21 3 0 3

Average 13.2 3.6 0.2 3.8

Max 21 8 1 8


Comparing Methods

Method Fj Lj Ej Tj

FCFS Ave 12.8 3.2 1 4.2

Max 21 11 4 11

SPT Ave 9.4 -0.2 3.2 3

Max 21 13 11 13

SWPT Ave 9.8 0.2 4 4.2

Max 21 13 9 13

EDD Ave 11.8 2.2 0.4 2.6

Max 21 6 2 6

CR Ave 13.2 3.6 0.2 3.8

Max 21 8 1 8

Results for Single Machine Sequencing

• The rule that minimizes the mean flow time of all jobs is SPT.

• The following criteria are equivalent: – Mean flow time– Mean waiting time.– Mean lateness

• Moore’s algorithm minimizes number of tardy jobs

• Lawler’s algorithm minimizes the maximum flow time subject

to precedence constraints.

EDD Example

Earliest Due DateEarliest Due Date

424

1212

5

1865

63

871

Due Date, dj, (day)


Job

424

1212

5

1865

63

871

Due Date, dj, (day)


Job


4 2 4 2 2 -2 2 0

3 5 6 7 7 1 0 1

1 7 8 14 14 6 0 6

2 1 12 15 15 3 0 3

5 6 18 21 21 3 0 3

Average 11.8 2.2 0.4 2.6

Max 21 6 2 6


Minimizing the Number of Tardy JobsMorre Algorithm - minimizes number of tardy jobs

Step 1: Sequence the jobs according to EDD rule and initially put all jobs in set V

Step 2: Find the first tardy job in set V {say it is job [k] in the sequence}. If there are no tardy jobs in the set V, stop; the sequence is optimal

Step 3: Select the job with largest processing time among first k jobs. Place this job in set U. Go to step 2

Comments:

1. Placing a job in set U means that it will be tardy and will occupy a position in sequence after all non-tardy jobs

2. Tardy jobs may be schedules in any order because the performance measure is the number of tardy jobs

Example Moore AlgorithmJob j pj Dj Cj Fj Lj Ej Tj

4 2 4 2 2 -2 2 0

3 5 6 7 7 1 0 1

1 7 8 14 14 6 0 6

2 1 12 15 15 3 0 3

5 6 18 21 21 3 0 3

Average 11.8 2.2 0.4 2.6

Max 21 6 2 6

4 2 4 2 2 -2 2 0

1 7 8 9 9 1 0 1

2 1 12 10 10 -2 2 0

5 6 18 16 16 -2 2 0

3 5 6 21 21 15 0 15

4 2 4 2 2 -2 2 0

2 1 12 3 3 -9 9 0

5 6 18 9 9 -9 9 0

3 5 6 14 14 8 0 8

1 7 8 21 21 13 0 13

Average 9.8 0.2 4 4.2

Max 21 13 9 13

Iteration 1

Iteration 2

Iteration 3

Alt. solution: 1— 4 — 2 — 5 — 3

Lawler’s Algorithm

minimizes the maximum flow time subject to precedence constraints. Goal: Scheduling a set of simultaneously arriving tasks on

one machine with precedence constraints to minimize maximum lateness (tardiness).

Precedence constraints occur when certain jobs must be completed before other jobs can begin.

Algorithm:

Tasks are scheduled in reverse order: job to be completed last is scheduled first.

At each step selection is made from the jobs that are not required to precede any other unscheduled job.

Select a job that achieves iiVi

dF

min

Lawler’s Example:Processing for all jobs is 1 day

1

2 3

4 5 6

D1=2

D2=5 D3=4

D4=3 D5=5 D6=6

One machine Ffinal = 1+1+1+1+1+1 = 61) Select from jobs {4,5,6} such that gives

min{6-3, 6-5, 6-6}=0 job 6 is a last job2) Recalculate F: F = 6-1= 5 Select from jobs {3,4,5} such that gives min{5-4, 5-3, 5-5}=0 order x-x-x-x-5-6

3) Recalculate F: F = 5-1= 4 Select from jobs {3,4} such that gives min{4-4, 4-3}=0 order x-x-x-3-5-64) Recalculate F: F = 4-1= 3 Select from jobs set {4} order x-x-4-3-5-6

5) Recalculate F: F = 3-1= 2 Select from jobs set {2} order x-2-4-3-5-6

6) Recalculate F: F = 2-1= 1 Select from jobs set {1} order 1-2-4-3-5-6

Lawler’s Example:


1 1 2 1 1 -1 1 0

2 1 5 2 2 -3 3 0

4 1 3 3 3 0 0 0

3 1 4 4 4 0 0 0

5 1 5 5 5 0 0 0

6 1 6 6 6 0 0 0

Average 3.5 -0.67 0.67 0

Max 6 0 3 0


Processing for all jobs is 1 day

1

2 3

4 5 6

D1=2

D2=5 D3=4

D4=3 D5=5 D6=6

Production is done in next order:1 – 2 – 4 – 3 – 5 – 6

Lawler’s algorithm minimizes the maximum flow time subject to precedence constraints

Gantt Charts

Pictorial representation of a schedule is called Gantt Chart

The purpose of the chart is to graphically display the state of each machine at all times

Horizontal axis – timeVertical axis – machines 1, 2, …, m

11 2

2

3 5 6 11

Machine 1Machine 2

Time (days)

Question: Is it an optimal schedule? Are there any precedence constrains?

Processing Job 1 Job 2

Machine 1 3 5

Machine 2 2 1


Machine 1 3/1 5/2

Machine 2 2/2 1/1

Gantt Charts


Machine 1 3 5

Machine 2 2 1


Machine 1 3/1 5/2

Machine 2 2/2 1/1

11 2

2

3 5 6 11

Machine 1Machine 2

Time (days)8

112

2

3 5 6 11

Machine 1Machine 2

Time (days)8

11 2

2

3 5 6 11

Machine 1Machine 2

Time (days)8Question: How to determine THE optimal solution?

What makes scheduling problem more difficult?

11 2

2

3 5 6 11

Machine 1Machine 2

Time (days)

ExampleProcessing time / machine

number

Job Operation 1

Operation 2

Operation 3

Release date

Due date

1 4/1 3/2 2/3 0 16

2 1/2 4/1 4/3 0 14

3 3/3 2/2 3/1 0 10

4 3/2 3/3 1/1 0 8

M 1

M 3

M 2

Find a solution!

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

3 24132 4

312 4

Completed by141113 (late)10 (late)

Deterministic Scheduling with Multiple Machines

• For the case of m machines and n jobs, there are n! distinct sequenced on each machine (permutations), so (n!)m is the total number of possible schedules

• For m = 3 and n = 4, total number of possible schedules is 243=13,824

• Assume that each job must be processed in the order– First on machine 1, then machine 2….

• The optimal solution for scheduling n jobs on two machines to minimize the total flow time is always a permutation schedule

– Assume flow shop: in each job operations have to be done on both machines

– Permutation schedule is when jobs are done in the same order on both machines

– This is the basis for Johnson’s algorithm

Example

Jobs

Machines

1 2 3 4 Total time

1 5 4 3 2 14

2 2 5 2 6 15

Is it optimal?

MetalFrame makes 4 different types of metal door frames.Preparing the hinge upright is a two-step operation.

Natural schedule:

7 14 16 22

12

3 41

243

If idle time for machine 2 is equal to zero, then we have found an optimal solution

Deterministic Scheduling with Multiple Machines: Johnson’s Rule

• Name Machine 1 = A, Machine 2 = B,

then ai = processing time for job i on A

and bi = processing time for job i on B

• Johnson’s Rule says that job i precedes job j in the optimal sequence if

Algorithm:

• Step 1: Record the values of ai and bj in two columns

• Step 2: Find the smallest remaining value in two columns. If this value in column a, schedule this job in the first open position in the sequence; if this value in column b, schedule this job in the last open position in the sequence; Cross off each job as it is scheduled

ijji baba ,min ,min

Example (cont)Jobs

Machines 1 2 3 4 Total time

1 5 4 3 2 14

2 2 5 2 6 15

Johnson’s schedule:4 – x – x – x

job A B

1 5 2

2 4 5

3 3 2

4 2 6

4 – x – x – 3

4 – x – 1 – 3

4 – 2 – 1 – 3

Natural schedule:

7 14 17 22

12

341

24 3Johnson’s schedule:

Is it optimal?

7 14 16 22

12

3 41

243

Results for Multiple Machines

• For three machines, a permutation schedule is still optimal if we restrict attention to total flow time only (not necessarily the case for average flow time).

• Under some circumstances, the two machine algorithm can be used to solve the three machine case:– Label the machines A, B and C

– or

– Redefine Ai’= Ai + Bi and Bi’= Bi + Ci

• When scheduling two jobs on m machines, the problem can be solved by graphical means.

iforBA ii ,maxmin iforBC ii ,maxmin

Sequencing Theory: The Two-Job Flow Shop Problem

Assume that two jobs are to be processed through m machines. Each job must be processed by the machines in a particular order, but the sequences for the two jobs need not be the same

Graphical procedure developed by Akers (1956):

– Draw a Cartesian coordinate system with the processing times corresponding to the first job on the horizontal axis and the processing times corresponding to the second job on the vertical axis (keeping order)

– Block out areas corresponding to each machine at the intersection of the intervals marked for that machine on the two axes

– Determine a path from the origin to the end of the final block that does not intersect any of the blocks and that minimizes the vertical movement. Movement is allowed only in three directions: horizontal, vertical, and 45-degree diagonal. The path with minimum vertical distance corresponds to the optimal solution

Example 8.7 (in the book)

A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps must be sanded,

lacquered, and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The

times required for the three operations for each of the two shipments are

The order of operations is the same for both jobs: A B C

Job 1 Job2Operation Time Operation Time

Sanding (A)

3 A 2

Lacquering (B)

4 B 5

Polishing (C)

5 C 3

Minimizing the flow time is equivalent to finding the path from the origin to the upper right point F (for this problem it is art the end of block C) that maximizes the diagonal movement and therefore minimizes either the horizontal or the vertical movement.

12 +(3)=15

10 + (6)=16

Job 1 Job2

Time

Time

A 3 A 2

B 4 B 5

C 5 C 3

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Example

Job 1 Job2

Order & Operation

Time Order & Operation

Time

B 3 A 2

D 4 D 5

C 2 B 4

A 5 C 3

B D C A

A

D

B

C

A

C

D

B

F

14+4=1814+4=18

14+2+2=18

7 11 15 18

BA

D CB

ACDJ1

J2

BA

D CB

ACDJ1

J2

Schematic of a Typical Assembly LineThe problem of balancing an assembly line is a classic engineering problem

• A set of n distinct tasks that must be completed on each item• The time required to complete task i is a known constant ti

• The goal is to organize the tasks into groups, with each group of tasks being performed at a single workstation• The amount of time allotted to each workstation is determined in advance (C = cycle time), based on the desired rate of production of the assembly line

• Assembly line balancing is traditionally thought of as a facilities design and layout problem

• There are a variety of factors that contribute to the difficulty of the problem– Precedence constrains: some tasks may have to be

completed in a particular sequence– Zoning restriction: Some tasks cannot be performed at

the same workstation

• Let t1, t2, …, tn be the time required to complete the respective tasks

Assembly Line Balancing

• The total work content (time) associated with the production of an item, say T, is given by


n

iitT

1

• For a cycle time of C, the minimum number of workstations possible is [T/C], where the brackets indicate that the value of T/C is to be rounded to the next larger integer

• Ranked positional weight technique: the method places a weight on each task based on the total time required by all of the succeeding tasks. Tasks are assigned sequentially to stations based on these weights

Assembly Line BalancingExample 8.11

The Final assembly of Noname personal computers, a generic mail-order PC clone, requires a total of 12 tasks. The assembly is done at the Lubbock, Texas, plant using various components imported from the Far East. The network representation of this particular problem is given in the following figure.


PreconditionThe job times and precedence relationships for

this problem are summarized in the table below.

Task Immediate Predecessors Time1 _ 122 1 63 2 64 2 25 2 26 2 127 3, 4 78 7 59 5 1

10 9, 6 411 8, 10 612 11 7

Assembly Line Balancing: Helgeson and Birnie Heuristic (1961)

Ranked positional weight technique

The solution precedence requires determining the positional weight of each task.

The positional weight of task i is defined as the time required to perform task i plus the times required to perform all tasks having task i as a predecessor.

t3 + t7 + t8 + t11 + t12 = 31Task Time Positional

Weight

1 12 70

2 6 58

3 6 31

4 2 27

5 2 20

6 12 29

7 7 25

8 5 18

9 1 18

10 4 17

11 6 1312 7 7

The ranking: 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12

ti = 70, and the production rate is a unit per 15 minutes;The minimum number of workstations = [70 / 15] = 5

Assembly Line Balancing: Helgeson and Birnie Heuristic (1961)

The ranking: 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12

Task Immediate Predecessors

Time

1 _ 12

2 1 63 2 64 2 25 2 26 2 127 3, 4 78 7 59 5 1

10 9, 6 411 8, 10 612 11 7

Station 1 2 3 4 5 6

Tasks 1 2, 3, 4 5, 6, 9 7, 8 10, 11 12

Processing time 12 14 15 12 10 7

Idle time 3 1 0 3 5 8

C=15

T2=6

C=15

Station 1 2 3 4 5 6

Tasks 1 2,3,4 5,6,9 7,8 10,11 12


Idle time 3 1 0 3 5 8

Cycle Time=15

T1=12

T2=6 T3=6 T4=2

T5=2 T6=12 T9=1

T5=2

T8=5T7=7 T10=4

T10=4 T11=6 T12=7

T12=7

15Evaluate the balancing results by the efficiency ti/NC

The efficiencies for C=15 is 77.7%, C=16 is 87.5%, and C=13 is 89.7% is the best one

Helgeson and Birnie Heuristic (1961)

C=15

Station 1 2 3 4 5 6

Tasks 1 2,3,4 5,6,9 7,8 10,11 12


Idle time 3 1 0 3 5 8


Station 1 2 3 4 5

Tasks 1 2,3,4,5

6,9 7,8,10 11,12

Idle time 4 0 3 0 3

C=16Increasing the cycle time from 15 to 16, the total idle time has been cut down from 20 min/units to 10 improvement in balancing rate.

The production rate has to be reduced from one unit/15 minutes to one unit/16minute;

Station 1 2 3 4 5 6

Tasks 1 2,3 6 4,5,7,9 8,10 11,12

Idle time 1 1 1 1 4 0

C=13

C=15

Station 1 2 3 4 5 6

Tasks 1 2,3,4 5,6,9 7,8 10,11 12


Idle time 3 1 0 3 5 8


Station 1 2 3 4 5

Tasks 1 2,3,4,5

6,9 7,8,10 11,12

Idle time 4 0 3 0 3

C=16

Station 1 2 3 4 5 6

Tasks 1 2,3 6 4,5,7,9 8,10 11,12

Idle time 1 1 1 1 4 0

C=13

13 minutes appear to be the minimum cycle time with six station balance.

Increasing the number of stations from 5 to 6 results in a great improvement in production rate;

Stochastic Scheduling: Static Case

• Single machine case: Suppose that processing times are random variables. If the objective is to minimize average weighted flow time, jobs are sequenced according to expected weighted SPT. That is, if job times are t1, t2, . . ., and the respective weights are u1, u2, . . . then job i precedes job i+1 if

E(ti)/ui < E(ti+1)/ui+1

• Multiple Machines: Requires the assumption that the distribution of job times is exponential, (memoryless property). Assume parallel processing of n jobs on two machines. Then the optimal sequence is to to schedule the jobs according to LEPT (longest expected processing time first).

• Johnsons algorithm for scheduling n jobs on two machines in the deterministic case has a natural extension to the stochastic case as long as the job times are exponentially distributed.

Stochastic Scheduling: Queueing TheoryA typical queueing process

• “The basic phenomenon of queueing arises whenever a shared facility needs to be accessed for service by a large number of jobs or customers.” (Bose)

• “The study of the waiting times, lengths, and other properties of queues.” (Mathworld)

Applications:

Telecommunications Health services Traffic control Predicting computer performanceAirport traffic, airline ticket sales Layout of manufacturing systemsDetermining the sequence of computer operations

Service FacilityCustomers arriving Served customers leaving

Discouraged customers leaving

Examples of Queueing Theory

http://www.bsbpa.umkc.edu/classes/ashley/Chaptr14/sld006.htm

Stochastic Scheduling: Dynamic Analysis

• View network as collections of queues – FIFO data-structures

• Queuing theory provides probabilistic analysis of these queues

• Typical operating characteristics of interest include: – Lq = Average number of units in line waiting for service– L = Average number of units in the system (in line waiting for service

and being serviced)– Wq = Average time a unit spends in line waiting for service– W = Average time a unit spends in the system– Pw = Probability that an arriving unit has to wait for service– Pn = Probability of having exactly n units in the system– P0 = Probability of having no units in the system (idle time)– U = Utilization factor, % of time that all servers are busy

Characteristics of Queueing Processes

• Arrival pattern of customers

• Service pattern of servers

• Queue discipline

• System capacity

• Number of service channels

• Number of service stages


• Arrival pattern of customers– Probability distribution describing the times between

successive customer arrivals• Time independent Stationary arrival patterns• Time dependent Non-stationary

– Batch or Bulk customer arrivals• Probability distribution describing the size of the batch

– Customers behavior while waiting• Wait no matter how long the queue becomes• If the queue is too long, customer may choose not to enter into the

system• Enter, wait, and choose to leave without being serviced• If there is more than one waiting line, customer may switch “jockey”


• Arrival pattern of customers

• Service pattern of servers– Single or Batch– May depend on the number of customers waiting state dependent– Stationary or Non-stationary

• Queue discipline– Manner in with customers are selected to service– First Come First Served (FCFS)– Last Come First Served (LCLS)– Random Selection for Service (RSS)– Priority Schemes

• Preemptive case• Non-preemptive case


• Arrival pattern of customers• Service pattern of servers• Queue discipline• System capacity

– Finite queueing situations = Limiting amount of waiting room

• Number of service channels– Single-channel system– Multi-channel system, generally assumed that parallel channels

operate independently of each other

• Number of service stages

Notation Used in Queueing Processes

Full notation: A / B / X / Y / Z Shorthand: A / B / X

A – indicates the interarrival-time distribution Assumes: Y is infinity,

B – the probability distribution for service time Z = FCFS

X – number of parallel service channels

Y – the restriction on system capacity

Z – the queue discipline (FCFS)Symbol = Explanation

A

B

M = Exponential, D = Deterministic, Ek = Erlang type

Hk = Mixture of k exponentials, PH = Phase type, G = General

X

Y

1, 2, ... , infinity

1, 2, ... , infinity

Z FCFS, LCLS, RSS, PR = priority, GD = general discipline

Queueing Processes: Little’s Formulas

One of the most powerful relationships in queueing theory was developed by John D.C. Little in the early 1960s.

Formulas:

and ,

where λ is an average rate of customers entering the system, and

W is an expected time customer will spend in the system

WL qq WL

1

2

3

Number of customersin system

Time, tt1 t2 t3 t4 t5 t6 t7 T

Poisson Process & Exponential Distribution

• M: stands for "Markovian", implying exponential distribution for service times or inter-arrival times, that carries the memoryless property – past state of the system does not help to predict next arrival /

departure

n+1nn-1

Calculating Expected System Measures for M/M/1

The utilization rate: ρ = λ / μ

P0 = 1 – ρ

Pi = ρi(1 – ρ), for i = 1, 2, 3,…

these formulas hold only if <

210

CHARACTERISTIC SYMBOL FORMULA

Utilization ρ λ / μ Exp. No. in System L λ / (μ – λ) = ρ / (1-ρ)Exp. No. in Queue Lq λ2/ μ(μ – λ) = ρ2 / (1-ρ)Exp. Waiting Time W=L/ λ 1 / (μ – λ) = ρ / λ(1-ρ)Exp. Time in Queue Wq=Lq/ λ λ / μ(μ – λ) = ρ2 / λ(1-ρ)Prob. System is Empty P0 1 – (λ / μ) = 1 - ρ

Calculating Expected System Measures for M/M/m

http://www.ece.msstate.edu/~hu/courses/spring03/notes/note4.ppt


Assumption

- m servers

- all servers have the same service rate μ

- single queue for access to the servers

- arrival rate λn = λ

- departure rate

…

,1,,

1,,2,1,0,

mmnm

mnnn

m-1 m m+1210

λ λ λ λ λ λ

mμ

λ

mμmμ(m-1)μ3μ2μμ


1

11

00 1

!!

m

m

n

mP

mm

n

n

02

1

1!P

m

mL

mm

q

mLL q qq LW 1 qWW

m-1 m m+1210

λ λ λ λ λ λ

mμ

λ

mμmμ(m-1)μ3μ2μμ

…

Example• Unisex hair salon runs on a first-come, first-served basis.

Customers seem to arrive according to a Poisson process with mean arrival rate of 5/hr. Because of Ms. H.R. Cutt’s excellent reputation, customers are always willing to wait. Average service time of 10 min is exponentially distributed.

• Calculate the average number of customers in the shop and the average number of customers waiting for a haircut.

• Calculate the percentage of time an arrival can walk right in without having to wait at all.

• The waiting room has only 4 seats. What is the probability that a customer upon arrival rill have to stand?

• Calculate average system waiting time, and the line delay.

Other Systems

M/M/1/K - system with a capacity Kλeff = effective arrival rate

M/D/1; M/G/1; M/G/∞

Assignment: download the QTS add-in for Excel software to check the homework problems answers

http://www.geocities.com/qtsplus/http://www.geocities.com/qtsplus/DownloadInstructions.htm#DOWNLOAD_INSTRUCTIONSDownloadInstructions.htm#DOWNLOAD_INSTRUCTIONS

Homework Assignment

• Read Ch. 8 (8.1 – 8.10)

• Read Supplement Two (S2.1 - S2.13)

• 8.4, 8.5, 8.7, 8.12, 8.15,

• 8.18, 8.23, 8.25, 8. 27, 8.28

References

• Presentation by McGraw-Hill/Irwin

• Presentation by Professor JIANG Zhibin, Department of Industrial Engineering & Management, Shanghai Jiao Tong University

• “Production & Operations Analysis” by S.Nahmias

• “Production: Planning, Control, and Integration” by Sipper and Bulfin Jr.

• “Inventory Management and Production Planning and Scheduling” by Silver, Pyke and Peterson

• “Fundamentals of Queueing Theory” by Cross and Harris

• http://www.geocities.com/qtsplus/DownloadInstructions.htm#DOWNLOAD_INSTRUCTIONS QTS analysis for Excel

http://www.geocities.com/qtsplus/DownloadInstructions.htm#DOWNLOAD_INSTRUCTIONS

http://www.geocities.com/qtsplus/DownloadInstructions.htm#DOWNLOAD_INSTRUCTIONS

chapter 8 operations scheduling. scheduling problems in operations job shop scheduling personnel...

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