[SOLVED] CS代考 AM-11 AM, and by appointment on zoom

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Operations Management
Lecture 5 slides
After class slides does not include the review
we cover at the beginning

Copyright By PowCoder代写加微信 assignmentchef

of the lectures and only includes the new materials. The additional
Materials(answers to questions,
are added to after class slides in pink)

! My lectures will continue to be on zoon until February 27th.
! Office hours: Fridays, 9 AM-11 AM, and by appointment on zoom
https://utoronto.zoom.us/j/88361623271 (Links to an external site.) passcode: 604020
! Grades for Assignment 1 has been posted (feedbacks on utorsubmit, grades on Quercus).
! Individual Assignments: Assignment #3 (weight 2%) is available on utorsubmit and will be due on 2021/02/13, 11:59 PM. After you download your assignment from utorsubmit follow the instructions provided on the html file (the assignment) carefully. You should upload back the generated text file on utorsubmit (it is explained on the html file of the assignment).The assignments should be submitted on utorsubmit. If you have any questions regarding how to submit your assignments, please contact:
– For windows users, please use firefox to access utorsubmit
! Case(group) work is available on Quercus under modules, case 1.
– All students should also submit a teamwork cover page, posted in the same module.
– Case 1 is due on Sunday Feb. 27th,11:59PM.
– Mahsa has posted the instructions on your case submission. For case assignments, submission should be through
Quercus. If you have any questions regarding your case submission, please email:

! There will be no classes or office hours during the reading week
! There will be no classes during the midterm week
! There will be extra office hour next Wednesday Feb 16th, 10AM-11AM on zoom (same zoom link as for the regular office hour)
! Midterm Exam: Feb 18, 7pm-9pm, )
– Ifyouhaveacademicortime-zoneconflictsyoushouldhave
emailed me.
– MakeupMidtermExam:Feb17th,9AM-11AM.Theexamwillbe available only to students with conflicts (The students who informed us will receive an email from us with more information). If you have a conflict and did not let us know, you should email us as soon as possible.
Read all slides, summary slides posted on Quercus, assignments,
sample problems, and sample past midterms (Class 1

Midterm Exam Info
! The midterm exam (as well as the make-up midterm exam) will be online through Quercus.
! Please make sure you are ready at least 10 minutes prior to the start of the exam.
! You will need a (1) computer, (2) calculator, (3) ruler, and (4) either a tablet (for writing down your answers) or pencil and paper.
! The exam is designed to only require 1 hour and 45 minutes of your time. We have included an additional 15 minutes for you to (1) scan your answers if you decide to use pencil and paper with either a scanner or a camera (it is your responsibility to make sure your answers are legible in the photos) and (2) upload your (scanned) answers to Quercus.
! Please make sure to upload your answers to Quercus by the end of the exam. If you fail to do so, the instructor reserves the right to refuse your answers.
! The final answers entered on Quercus should match your derivations on the paper.

Today’s Lecture
Review of Basic Queueing & the OM Triangle
– The Pollaczek-Khinchin (PK) Formula for average queue length and average system times
– Inventory, Capacity and Information
Queueing Models
– Single server queues: M/M/1, M/D/1, D/D/1
– Queues with multiple servers
– Pooling versus separate buffers

Example: Hospital
Suppose you are the system administrator of the emergency department for Toronto East General Hospital.

Review: Example: Hospital Point (a) – Extra Capacity
! To run an emergency room there is a high level of variability. Why?
– Inter-arrival and service times are random
! Assume expected demand for service is 2 patients/hour
! If you choose to position hospital at point (a)
– What do you need to do to achieve a utilization of 50%?
» Patients will not have to wait very long and there will be very few customers waiting in the buffer areas.
» This is indicated by a low-vertical position on the y axis.
4 patients/hour

Review: Example: Hospital Point (b) – High Inventory
! What happens if you can sustain a utilization of 0.9? – You can get by with much less capacity while still
accommodating the same demand of 2 patients/hour
! More capacity means more equipment, a bigger facility and more doctors and nurses (Costly!)
– This position may only appear to be much better
! You must tolerate a high level of inventory = many
patients waiting in the buffer areas
! You must consider whether this high inventory level (and long waiting time) is acceptable

Review: Example: Hospital Point (c) – Low Variability
! What happens if you reduce the unpredictable variability to an extremely low level (0.01)?
– Better information (or knowledge) is, more often than not, a key factor in reducing variability
Associate lower variability with better information!

Lecture 5:
1) Queue Representation 2) Multiple Servers

Attributes of a Queueing System
! Distribution of time between arrivals – Constant, Uniform, Exponential, etc.
! Distribution of service time
– Constant, Uniform, Exponential, etc.
! Number of servers (one/multi-server)
! Number of queues/buffers (i.e., waiting areas)
! Maximum queue length (buffer size) – “capacity”, e.g., waiting room area
! Service & Routing Policies
– First-Come, First-Served (FCFS), Priorities, Reservations, Service Time, Network Queueing Models

Representing Single Queues
Single queue is represented by
(inter-arr. distribution, service distribution, #of resources(servers))
Number of servers (c)
queue represents the queue length in a system with a single server where interarrival times have a general (meaning arbitrary) distribution and service times have another general distribution.
(A / S / c / K / N / D ) Queue
(A): Inter-arrival times distribution
Example: if interarrival times is from Exponential distribution and service time is also from exponential distribution and we have only one server we represent it as
M/ 1 queue
(S):Service time distribution
M: exponential distribution
D: deterministic (not variable) G: general distribution

Review: Justification of Exponential Assumptions
! In many situations, the exponential distribution assumption is a good approximation of what really happens
– Such an arrival process is called “Poisson process”
» Number of customers arriving per time unit is Poisson distributed
! Easy to analyze because coefficient of variation = 1 for the exponential distributionààà CX=s{X}/E{X}=1
! Recall the PK formula:
r2 C2 +C2 as

Single-Server Queues: M/M/1
The Simplest ‘Stochastic’ Queue
– The first “M” indicates the inter-arrival times are exponentially
distributed Þ Ca = 1
– The second “M” indicates the service times are exponentially
distributed Þ Cs = 1
– The last “1” indicates one single server
! For M/M/1 queue, the P-K formula is exact (=, not »):
Iq = ρ2 = λ2 1-ρ μ(μ-λ)
! Average waiting time in queue: Tq = Iq / l (Little’s Law)
! Self-test: I=Iq +Is =Iq +l/μ T=Tq+Ts=Iq /l+1/μ

Single-Server Queues: M/M/1
The Simplest ‘Stochastic’ Queue
Iq = ρ2 = λ2 1-ρ μ(μ-λ)
Average Arrival Rate
6 person/hour
Average Service
Time (per person) 5 min
μ= 1/5 person/min= 12 person/hour
! Ave. Number in Queue Iq = 36/(12*(12-6)) cust = 0.5 cust
! Ave. Waiting Time in Queue Tq = 0.5/6 hour = 5 min
! Ave.TimespentinSystem T=5min+5min=10min
! Ave. Number in System I = 0.5 cust + 6/12 cust = 1 customer

Single-Server Queues: M/D/1
The Simplest ‘Semi-Stochastic’ Queue
Assumptions: (A/S/c/K/N/D)
– Inter-arrival times (A) follow an Exponential distribution – M » M stands for memoryless
» Arrival process follows a Poisson Distribution
– Service times (S) are constant (i.e., Deterministic) – D » There is no service variability!
– Number of servers (c) – 1
! Other technical assumptions:
» There is a single buffer that serves the entire queue.
» There is no limit on the length the buffer can grow to (K).
» The population the queue can service is unlimited (N).
» Service discipline (D) is First Come – First Served (FCFS).
» All units that arrive enter the queue (no balking)
» Any unit entering the system stays in the queue till served (no reneging)
» All units arrive independently of each other (no batching or correlation). 16

Single-Server Queues: M/D/1
The Simplest ‘Semi-Stochastic’ Queue
– The “M” indicates the inter-arrival times are exponentially
distributed Þ Ca = 1
– The “D” indicates the service times are a constant Þ Cs = 0 – The “1” indicates single server
! First come first served (FCFS)
! For M/D/1 queue, the P-K formula gives us:
I=ρ2æ1ö= λ2
q 1 – ρ çè 2 ÷ø 2 μ ( μ – λ )
! Average waiting time in queue: Tq = Iq / l (Little’s Law)
! Self-test: I=Iq +Is =Iq +l/μ T=Tq+Ts=Iq /l+1/μ

Single-Server Queues: M/D/1
The Simplest ‘Semi-Stochastic’ Queue
ρ2 æ1ö 1-ρçè2÷ø
Average Arrival Rate 6 person/hour
2μ(μ-λ) Service Time (per person)
5 min and is NOT RANDOM
! l=6cust/hour,1/μ=5minÞμ=12cust/hour
! Ave. Number in Queue, Iq = 36/(2*12(12-6)) cust = 0.25 cust
! Ave. Waiting Time in Queue, Tq = 0.25/6 hour = 2.5 min
! Ave.TimespentinSystem, T=2.5min+5min=7.5min
! Ave. Number in System, I = 0.25 cust + 6/12 cust = 0.75 cust

Single-Server Queues: D/D/1
The Simplest ‘Non-Stochastic’ Queue
Assumptions: (A/S/c/K/N/D)
– Inter-arrival times (A) are constant (i.e., Deterministic) – D
» There is no arrival variability!
– Service times (S) are constant (i.e., Deterministic) – D » There is no service variability!
– Number of servers (c) – 1
! Other technical assumptions:
» There is a single buffer that serves the entire queue.
» There is no limit on the length the buffer can grow to (K).
» The population the queue can service is unlimited (N).
» Service discipline (D) is First Come – First Served (FCFS).
» All units that arrive enter the queue (no balking)
» Any unit entering the system stays in the queue till served (no reneging)
» All units arrive independently of each other (no batching or correlation). 19

Single-Server Queues: D/D/1
The Simplest ‘Non-Stochastic’ Queue
l<μ Arrival Rate(person/min)Inter-arrival timeService Rate (persons/min)Throughp ut?Service Time (min)Service time Waiting Time ? Single-Server Queues: D/D/1The Simplest ‘Non-Stochastic’ Queuel>μ Arrival Rate
(person/min)
Inter-arrival time
Service Rate (persons/min)
Throughp ut?
Service Time (min)
Customers waiting:
Waiting Time ? , the queue size “blows-up” and the average
waiting time in the long run goes to infinity and average number in queue also goes to infinity.
Service time

Single-Server Queues: D/D/1
The Simplest ‘Non-Stochastic’ Queue
! This is queueing under a Deterministic scenario – There is no uncertainty or variability in the process.
– l and μ are constants (not even average rates!)
! Customers in queue (Iq) and time spent in queue (Tq) – If l < μ, each customer (job) is processed before the next arrival.The average waiting time and the average queue size is 0.– If l > μ, the queue size “blows-up” and the average waiting time in the long run goes to infinity and average number in queue also goes to infinity.

Today’s outline
Queue Analysis:
! We learned so far:
– Single resources: » PK Formula
» OM Triangle
! We learn today:
– Multiple Identical Resources (e.g., multiple ATMs): » Updated PK formula for multiple resource
» Pooling versus separate queues

Multiple Servers
Separate Queue
Pooled Queue

Multi-Server Queueing Models c servers
Arrival rate (average input rate)
Average throughput l=1/E[a] persons/min
c servers,
capacity utilization
l=1/E[a] persons/min inter-arrival time
distribution a Ca = s[a]/E[a]
τ = l / ( c x μ) Service rate (per server)
Weassumethat: l CS代考 加微信: assignmentchef QQ: 1823890830 Email: [email protected]

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[SOLVED] CS代考 AM-11 AM, and by appointment on zoom
30 $