CS代考 Operations Management

Operations Management

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• Assignment 3 (Due Sunday Feb 13, 11:59pm – UTORMAT)
• Case 1 (Due Sunday Feb 27, 11:59pm – Quercus)
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• Case 2 (Due April 8th, 11:59 pm – Quercus)

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! Inventory can build up (If Input Rate > Capacity Rate) ! Inventory Build-up diagram
– Shows graph of inventory versus time
– Build-up Rate = Input Rate – Capacity Rate
– Using the inventory build up diagram we can compute average inventory
Setting: Predictable Variability, Short Run Analysis (therefore, ok to have Input Rate > Capacity Rate), Variable rates ok

Little’s Law: Pay attention to the Units
Inventory = Throughput Rate x Waiting Time I= RxT
All units should match:
• Inventory (flow units)
• Throughput rate (flow units / unit time) • Waiting time (time)
• This equation can be re-arranged to solve for other quantities depending on the question!

Little’s Law:
Coffee Example
… … … … …
Average Waiting Time (T) [hr]
! Time to make one coffee: 30 seconds
What is the throughput rate in minutes? 2 customers/minute
! 60 customers in the system (Inventory) Waiting time:
Waiting Time = Inventory / Throughput Rate Waiting Time = 60 customers / 2 (customer/min)
Coffee Shop

Inventory Buildup: Cranberry Example
During harvesting season, a processing factory works around the clock.
• Farmers deliver their loads of cranberries from 12am to 12pm (last truck
arrives at 11 am) at a constant rate of 2 tons/hour.
• The fruits are dumped into a big storage bin and processed at a rate of 1 ton/hour.
• Draw an Inventory buildup diagram.
(Assume the flow unit is cranberries and that the they arrive at the
station at a constant pace all day).
• What is the average inventory of cranberries in the factory?

Inventory Buildup: Cranberry Example
Average Inventory in the System
(Area Under the Curve) / (Time) = (12*(24)/2)/24=6 tons
Build-up rate=
2-1=1 ton/hour
Build-up rate= 0-1=-1 ton/hour
12:00am 12:00pm

Inventory Buildup: Cranberry Example Continued
Suppose the storage bin has room to hold only 6 tons of cranberries. Once this space is filled, the farmers’ trucks must wait to dump their contents
• Notice there are now two buffers:
• Cranberrybuffer • Truck buffer
Questions:
• What is the average inventory of cranberries
• In the bin?
• In the trucks?
• In both?

Inventory Buildup: Cranberry Example
Average Inventory in the System (Area Under the Curve) / (Time)
Avg Inv (truck)= (12*6/2)/24=1.5 Tons
Average Inv (bin)= ((12+24)*6/2)/24=4.5 Tons
Storage bin
12:00am 12:00pm

Inventory Buildup: Cranberry Example Continued
Now let us change the flow unit to a “truck”
• Assumptions:
• Each truck carries 1 ton of cranberries, i.e., two trucks arrive every hour
between 12:00am to 12:00pm.
• The storage bin has a capacity of 6 tons.
• At the start of every hour, the processor takes 1 ton of cranberries from the storage bin.
• Draw an Inventory buildup diagram of trucks.
• At what time will the trucks likely start to wait to unload?
• What is the “average” inventory of trucks waiting?

Inventory Buildup: Cranberry Example Continued
7 6 5 4 3 2 1 0
Inventory of trucks
Avg Inv (Trucks)= (1+2+3+4+5+6+5+4+3+2+1)*1/24=1.5 trucks
Max truck inv: 6 trucks
At 6AM the first truck starts
waiting to unload.
0:00 1:00 2:00 3:00 4:00 5:00 6:00 7:00 8:00 9:00
10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00 18:00 19:00 20:00 21:00 22:00 23:00
Number of trucks waiting

Little’s Law: When the Input Rate Fluctuates
• How long does a truck wait on average?
• We can apply Little’s Law:
• Average inventory = 1.5 trucks
• (Average) throughput rate = 24 trucks/24 hours = 1 truck/hr • Therefore, by Little’s Law
Waiting Time= Inventory/ Throughput rate
What is the waiting time for each truck?
T =1.5/ 1= 1.5 hours

Little’s Law: Organ Example
Patients waiting for an organ transplant are placed on a list till a suitable organ is available.
Patients matched to donated organs
Patients in need Patients leaving the list
of a transplant hopefully with a successful transplant

Little’s Law: Organ Example
! On average:
Ø 300 people are waiting for an organ transplant. Ø Patients stay on the list for 3 years.
! Assume no patients die during the wait!
! Question:
Ø Number of transplants performed per year?

Little’s Law: Organ Example
Average Queue Length = 300 Patients
Patients matched to donated organs
Average Wait Time = 3 years
Inventory = Throughput Rate x Waiting Time
R=100 Transplants per year

Insights from Little’s Law
Key Relationship:
! Throughput rate, waiting time and inventory are
intimately related:
Depending on the situation, a manager can influence any
one of these measures by controlling the other two: – Once two are chosen, the third is determined.

Little’s Law:
Shipping Containers
! You are managing the construction of a new container terminal at the Port of Vancouver. You expect that containers will spend about 2 days waiting to be shipped, and you have promised customers to “process” 2000 containers/day.
Containers to be shipped
Containers shipped

Little’s Law:
Shipping Containers
! On average, your container storage yard can hold 3000 containers.
! Question:
Ø Is your yard big enough? Ø Justify!
Avg Inv= 2000 x 2 = 4000 containers

Unpredictable Variability
• Unpredictable variability refers to “unknowable” changes in input and/or capacity rates
• Supply of pumpkins changes each year due to crop yield • Exact demand for pumpkins each day

Basic Questions
• What are the effects of variability on processes • In particular, how does variability affect
Average Throughput Rate
Average Inventory
• Iftheeffectsarenegative,howcanwedealwithit? 23
Average Flow Time

Consider a process with no variability
(1 person/min)
Throughput Rate?
• Assume that all customers are identical
• Customers arrive exactly 1 minute apart
• The service time is exactly 1 minute for all the customers
Service time
(exactly 1 min)

Effect of Input Variability (no buffer)
Random Input
0, 1, 2 customers/min (with equal probability)
Throughput Rate?
Service time
(exactly 1 min)
• Assume that customers who find the ATM busy do not wait 1234567

Effect of Input Variability (no buffer)
• When a process faces input variability, and a buffer
cannot be built, some input may get lost
• Input variability can reduce the throughput
• Lower throughput means
• Lost customers; lost revenue • Customer dissatisfaction
• Less utilization of resources

Dealing with Variability
• When the arrival rate of customers is unpredictable, what could you do to increase throughput?
Add Buffer
Increase Capacity
(e.g., Add another ATM;
Decrease the time it takes the ATM to serve a customer)

Dealing with Variability
• When the arrival rate of customers is unpredictable and contains variability, what could you do to increase throughput?
• Add a buffer before the process.
• Staffing, equipment, physical space, etc. • Decrease the variability via information.

Effect of Input Variability (with buffer)
Random Input
0, 1, 2 customers/min (with equal probability)
Throughput Rate?
Waiting time Service time (exactly 1 min)
• Now assume that customers wait We can build-up an inventory buffer

Effect of Input Variability (with buffer)
Random Input
0, 1, 2 customers/min (with equal probability)
Throughput Rate?
Waiting time Service time (exactly 1 min)

Variability (with buffer)
• If we can build-up inventory (i.e., insert a buffer into the process) variability leads to
• An increase in the average inventory in the process
• An increase in the average flow time
• We are not immediately losing customers due to abandonment (although they may still be unhappy)
• Fewer customers may be unhappy
• More utilization of resources
• Little’s Law still holds

Quantifying Variability
• So far, we focused on qualitative effect of variability • Without buffer, input may get lost and throughput may
• With buffer, queue may build up, flow time may increase
• How long is the queue on average?
• How long does a customer have to wait?
• We would like to quantify average inventory, average waiting time, and average throughput rate.

• We first learn how to represent variability.
• We then discuss some measures of variability that we will need.
• We then look at a single server Queuing model.
• We first introduce a few standard notations
• We then introduce PK formula to calculate average inventory as a function of utilization and variability measure (and by using Little’s law average waiting time).

Representing variability:
Short Review on Probability (1)
Discrete Random Variable and Probability
Throw a dice; the number you get is a discrete random variable:
1, w.p. 1/6
2, w.p. 1/6
3, w.p. 1/6
4, w.p. 1/6
5, w.p. 1/6
6, w.p. 1/6
P{X = 2} = 1/6
Probability
(Probability mass)
Cumulative probability
P{X ≤ 2} = P{X=1 or X=2} = 1/3 P{X ≤ 2.1} = P{X=1 or X=2} = 1/3
P(X≤x) 123456
1/6 1/3 1/2
P{X ≤ x} is a function of x, called the cumulative distribution function (CDF) 34

Representing variability:
Short Review on Probability (2)
Continuous Random Variable and Probability
The time between two customers’ arrival times is a
continuous random variable
Probability density
Cumulative distribution function (CDF)
òx P(X≤x)= 0 f(s)ds

Variability Measures
Source of variability:
Variability from inflow of flow units.
Measures of Variability
Variability in processing time.
For a given random variable X with mean μ where mean denotes the expected value of
a random variable., E[X] we define
Variance V[X]: is a measure of variability which is equal to the expected value of the square of deviations of the random variable around its mean, i.e., Var[X] = E[(X μ)2]. Also sometimes referred to as X2 or 2.
Standard deviation denoted by : A measure of variability, which is equal to square root of vpariance. It shows how spread out the instances of the random variable. X = Var[X].
Coecient of Variation CV Standard deviation divided by the mean, i.e.,
CV = μ . It is a relative measure of uncertainty in a random variable. Since both standard deviation and mean have the same measurement unit, CV is a unitless measure.
Based on these measures of variability we introduce a simple formula to calculate the average inventory.

Single Server Queuing Model
• As we will see, average inventory is a function of
• Average input rate
• Average capacity
• Utilization of the server
• The variation of the interarrival times
• More precisely the coefficient of variation of interarrival time
• The variation of the service time
• More precisely the coefficient of variation of service time

A Single Server Process (a few notations)
Long-run average input rate
Process Boundary
A queue forms in a buffer
Average processing time by one server
(Average) Customer inter-arrival time
Long-run average processing rate of a single server
A single phase service system is stable whenever l < μ • We are focusing on long-run averages, • Ignoring the predictable variability that may be occurring in the short run. Single-Server Process Assumption: l < μ Arrival rate: l persons/min (average input rate) On average, 1 person arrives every E{a} unit of time. Service rate: μ persons/min (average capacity rate) Average throughput rate l persons/min Thus, l = 1 / E{a} ... On average, 1 person can be served every E{s} unit of time. Thus, μ = 1 / E{s} Inter-arrival times: Service times: a1a2 a3 a4a5 a6 a7... s1 s2 s3 s4 s5 s6 s7 ATM Example (adjusting to notations) Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. • Throughput rate. • Utilization • Average interarrival time . • Average service time. ATM Example (adjusting to notations) Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Calculate • Throughput rate. Average Input Rate = l = 40 People per hour Average Capacity Rate = μ = 45 People per hour Average Throughput Rate = min {input rate, capacity} = min{40, 45} = 40 People per hour ATM Example (adjusting to notations) Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Calculate • Utilization ** Remember that the Utilization is the time that ATM is busy divided by the time that is available. r = !"#$%&"'%( )*(+ = l = 01 ,*'*-.(/ μ 02 ATM Example (adjusting to notations) Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Calculate • Average interarrival time . E[a]=3 = 3 *60=1.5minutes l 01 ATM Example (adjusting to notations) Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Calculate • Average service time. E[s]=3 = 3 *60=0minutes μ02 4 What are we trying to quantify? (a few notations) Avg inventory: I Avg queue length Iq Avg number of people being served: Is Little’s Law holds Iq = lTq Throughput rate = l Waiting time Tq Service time Ts T System Characteristics Performance Measures Average queue length Average number of customers being served Average number of customers in the process Utilization (In a stable system, r = l/μ < 100%) Average waiting time (in queue) Is I=Iq+Is Average time spent at the server Average flow time (in process) Average Inventory Assumption: l ≤ μ Arrival rate: l persons/min (average input rate) Service rate: μ persons/min (average capacity rate) Average throughput rate l persons/min • Average number of persons in the system: I = Iq + Is • Question: Is=??? (Express Is in terms of l and μ) Answer: Is=l/μ Pollaczek-Khinchin (PK) Formula I @ ρ 2 ́ C 2a + C 2s What is the relationship among variability, qinventory (queue length) and utilization? 1-ρ 2 “=” for special cases “»” in general Average queue length (excl. the one in service) (Long run) Average utilization = Average Throughput / Average Capacity = l / μ Ca = s{a}/E{a} Coefficient of variation (CV) of inter-arrival times Cs = s{s}/E{s} Coefficient of variation (CV) of service times Performance Measure: Average queue length: Pollaczek-Khinchin (PK) Formula Ca= coefficient of variation of inter-arrival time = s{a}/E{a} average queue length (excl. the one in service) C 2a + C 2s 2 coefficient of variation of service times = s{s}/E{s} average utilization= average throughput / average capacity = l /μ ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. What are the coefficients of variation of interarrival time and service time? Calculate Utilization. Calculate the average queue length. Calculate the average waiting time. ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. 1. What is the coefficients of variation of interarrival time and service time? Coefficients of Variation of Interarrival Time: E(a)=! = ! *60=1.5minutes l "# " % 𝐶! = ".$ = & ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. 2. What is the coefficients of variation of service time? E(s)= ! = ! *60=" minutes μ"$ % !" " 𝐶! = # = # ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. 3. Calculate Utilization. Ρ = l = !" = $ μ !# % ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. 4. Calculate the average queue length. Iq @ ρ2 ́C2a+C2s 1-ρ 2 !" ⌃ ( #$ ⌃ ( * %& ⌃ ( 𝐼&=!)!* ( " ATM Example Suppose on average 40 people arrive at the ATM per hour. The ATM can serve on average 45 people per hour. Suppose the variance of interarrival times is 1 minute squared, and the standard deviation of service time is 1/3 minute. 5. Calculate the average waiting time. 𝑇 = #! = #! hours " l $% Another Example • Customers arrive at rate 4/hour, and mean service time is 10 minutes • Assume that standard deviation of inter-arrival times equals 5 minutes, and the standard deviation of service time equals 3 minutes • What is the average size of the queue? What is the average time that a flow unit spends in the queue? l = 4 E[a] =1/ 4 hour μ=6 E[s]=1/6hour r=lμ=46=23 s[a]=1/12hour Ca =s[a]=1/12=1 E[a] 1/4 3 s[s]=1/20hour I @ ρ2 ́C2+C2=(23)2 ́(13)2+(310)2 Cs =s[s]=1/20= 3 E[s] 1/6 10 1-ρ 2 13 2 T=I l=I 4 qqq PK Formula and OM Triangle ρ2 C2 +C2 l l C2 +C2 1-ρ 2 μμ-l 2 I@ ́a s= ́ ́a s μ= Capacity Rate l = Input Rate INFORMATION Variability Impact of Utilization (ρ = l/μ) Impact on Queue Length (Inventory) I @ ρ2 ́C2+C2 as Impact on Waiting Time (Flow Time) T = I l Little’s Law qq Queue Length Waiting Time Utilization ρ 0% 100% Impact of Variability Iq @ ρ2 ́ 1-ρ Queue length or waiting time Increasing variability Utilization r C 2a + C 2s 2 程序代写 CS代考加微信: powcoder QQ: 1823890830 Email: powcoder@163.com

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