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Make to Stock (MTS) vs. Make to Order (MTO). Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety inventory protects against stockouts due to variability of arrival time and processing time. Inventory also permits economies of scale.

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make to stock mts vs make to order mto
Make to Stock (MTS) vs. Make to Order (MTO)

Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety inventory protects against stockouts due to variability of arrival time and processing time. Inventory also permits economies of scale.

Make-to-order (MTO) operations. Each order is specific, cannot be stored in advance. Ex. banks, restaurants, retail checkout counters, airline reservation, hospitals , repair shops, call centres. Production systems also try to follow Dell Computer model. We needs to maintain sufficient capacity to deal with uncertainty in both arrival and processing time. Safety Capacity vs. Safety Inventory.

a call centre
A Call Centre

The Call Centre Process

Sales Reps

Processing

Calls

(Service Process)

Incoming Calls

(Customer Arrivals)

Answered Calls

(Customer Departures)

Calls

on Hold

(Service Inventory)

Blocked Calls

(Due to busy signal)

Abandoned Calls

(Due to long waits)

Calls In Process

(Due to long waits)

characteristics of waiting lines queuing systems
Characteristics of Waiting Lines (Queuing Systems)
  • The time of the arrival of an order is not known ahead of time. It is a random variable with estimated mean and standard deviation.
    • The time of the next telephone call is not known.
  • The service time is not known (precisely) ahead of time. It is a random variable with estimated mean and standard deviation.
    • The time a customers spends on the web page of amazon.com is not precisely known.
    • The time a customer spends speaking with the teller in the bank is unknown
article the psychology of waiting lines
Article: The Psychology of Waiting Lines
  • Unoccupied time feels longer than occupied time.
  • Pre-process waits feels longer than in-process waits.
  • Anxiety makes waits seem longer.
  • Uncertain waits are longer than known, finite waits.
  • Unexplained waits are longer than explained waits.
  • Unfair waits are longer than equitable waits.
  • The more valuable the service, the longer I will wait.
  • Solo waiting feels longer than group waiting.
characteristics of queuing systems
Characteristics of Queuing Systems
  • Variability in arrival time and service time leads to
    • Idleness of resources
    • Waiting time of customers (orders) to be processed
  • We are interested in evaluating two measures:
    • Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor).
    • Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).
  • Let us first look at the Servers or Processors
average processing time tp average processing rate rp
AVERAGE Processing Time TpAVERAGE Processing Rate Rp

Tp: Processing time.

Tpunits of time. Ex. on average it takes 5 minutes to serve a customer.

Rp: processing rate.

Rpflow units are handled per unit of time.

If Tp is 5 minutes. Compute Rp.

Rp= 1/5 per minute, or 60/5 = 12 per hour.

more than one server c servers
More than One Server; c Servers

Tp: processing time.

Rp: processing rate.

What is the relationship between Rp and Tp?

If we have one resource  Rp= 1/Tp

What is the relationship between Rp and Tp when we have more than one resource; We have c recourses

Rp= c/Tp

Each customer always spends Tp unites of time with the server

average processing rate of c servers
Average Processing Rate of c Servers

Tp= 5 minutes. Processing time is 5 minute. Each customer on average is with the server for 5 minutes.

c = 3, we have three servers.

Processing rate of each server is 1/5 customers per minute, or 12 customer per hour.

Rp is the processing rate of all three servers.

Rp = c/Tp

Rp= 3/5 customers/minute, or 36 customers/hour.

inter arrival time ta and arrival rate ra
Inter-arrival Time (Ta) and Arrival Rate (Ra)

Ta:customer inter-arrival time.

On average each 10 minutes one customer arrives.

Ra:customer arrival (inflow) rate.

What is the relationship between Ta and Ra

Ta = every ten minutes one customer arrives

How many customers in a minute? 1/10;Ra= 1/Ta= 1/10

Ra = 1/10 customers per min; 6 customers per hour

Ra= 1/Ta

throughput min ri rp
Throughput = Min (Ri,Rp)

Ra MUST ALWAYS <= Rp.

We will show later that even Ra=Rp is not possible.

Incoming rate must be less than processing rate.

Throughput = Flow Rate R = Min (Ra, Rp) .

Stable Process = Ra< RpR = Ra

Safety Capacity Rs = Rp – Ra

buffer waiting line and processors servers
Buffer (waiting line) and Processors (Servers)

What is the waiting time in the servers (processors)?

Throughput?

Flow time T = Ti+ Tp

Inventory I = Ii + Ip

Ti: waiting time in the inflow buffer

Ii: number of customers in the inflow buffer

utilization is always less than 1
Utilization is Always Less than 1

U = Utilization

U =inflow rate / processing rate

U = throughout / process capacity

U = R/ Rp < 1

Safety Capacity = Rp– R

For example , R = 6 per hour, processing time for a single server is 6 min  Rp= 12 per hour,

U = R/ Rp = 6/12 = 0.5

Safety Capacity = Rp– R = 12-6 = 6

given the utilization how many flow units are in the processor s
Given the Utilization, How Many Flow Units are in the Processor(s)

Given a single server. And a utilization of U= 0.5

How many flow units are in the server ?

  • U = 0.5 means
  • 50% of timethere is 1 flow unit in the server
  • 50% of time there is 0 flow unit in the server
  • 0.5  1 + 0.5  0 = 0.5
  • Average Inventory in the server is equal to utilization
  • Ip= 1U = U
given the utilization how many flow units are in the processor s1
Given the Utilization, How Many Flow Units are in the Processor(s)
  • U = 0.5 means
  • 50% of timethere is 1 flow unit in each server
  • 50% of time there is 0 flow unit in each server
  • 0.5  1 + 0.5  0 = 0.5 flow unit in each server
  • Average Inventory in the server is equal to utilization timesthe number of servers Ip= 2U = cU
  • Given 2 servers. And a utilization of U = 0.5
  • How many flow units are in the servers ?
what we have learned without looking for any formula
What We Have Learned Without Looking for any Formula

Processing time: Tp, Ex. Tp = 5 minutes

Number of servers: c, Ex. c=3

Tp is also waiting time in the server, no mater one server or c servers. Tp in this example is always 5 min.

Processing rate Rp= c/Tp. Ex. Rp =3/5 per min; 36/hr

Utilization: U. Ex. U = 0.8 in our example

Number of the flow units in all servers, Ip = cU

In our example, Ip = 3  0.8 = 2.4

Can we compute R?  TR = I

Tp R = cU R = cU/Tp

5 R = 2.4  R = 0.48 flow units per minute or 28.8 / hr

We learned it without looking at any formula

characteristics of queuing systems1
Characteristics of Queuing Systems
  • Variability in arrival time and service time leads to
    • Idleness of resources
    • Waiting time of customers (orders) to be processed
  • We are interested in evaluating two measures:
    • Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor).
    • Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).
utilization and variability
Utilization and Variability
    • Two key drivers of process performance are Utilization and Variability.
  • The higher the utilization the longer the waiting line.
  • High capacity utilization U= R/Rp or low safety capacity Rs =R –Rp, due to
    • High inflow rate R
    • Low processing rate Rp = c / Tp, which may be due to small-scale c and/or slow speed 1 /Tp
  • The higher the variability, the longer the waiting line.
drivers of process performance
Drivers of Process Performance
  • Variability in the interarrival time and processing time can be measured using standard deviation. Higher standard deviation means greater variability.
  • Coefficient of Variation: the ratio of the standard deviation of interarrival time (or processing time) to the mean.
    • Ca = coefficient of variation for interarrival times
    • Cp = coefficient of variation for processing times
now let s look at the rest of the system the little s law applies everywhere
Now Let’s Look at the Rest of the System; The Little’s Law Applies Everywhere

R

  • Flow time T Ti+ Tp
  • Inventory I = Ii + Ip

Ip= R  Tp

I = R T

Ii = R Ti

R = I/T = Ii/Ti = Ip/Tp

We have already learned

Rp = c/Tp

U= R/Rp = Ip/c

operational performance measures
Operational Performance Measures

Flow time T = Ti+ Tp

Inventory I = Ii + Ip

Ti: waiting time in the inflow buffer = ?

Ii: number of customers waiting in the inflow buffer =?

  • A set of excel sheet are provided at the end of this lecture.
  • We relay on the approximation formula, given shortly, and our understanding of the waiting line logics.
  • We also provide an exact table for a special case.
the queue length approximation formula
The Queue Length APPROXIMATION Formula
  • U= R/Rp,whereRp = c / Tp
  • Caand Cp are the Coefficients of Variation
  • Standard Deviation/Mean of the inter-arrival or processing times (assumed independent)

Utilization effect U-part

Variability effect

V-part

factors affecting queue length
Factors affecting Queue Length

This part captures the capacity utilization effect. It shows that queue length increases rapidly as U approaches 1.

  • This part captures the variability effect. The queue length increases as the variability in interarrival and processing times increases.
  • Even if the processing capacity is not fully utilized, whenever there is variability in arrival or in processing times, queues will build up and customers will have to wait.
utilization variability delay curve
Utilization – Variability - Delay Curve

Average

Time in System T

Variability

Increases

Tp

100%

U

Utilization (U)

lessons learned
Lessons Learned
  • If inter-arrival and processing times are constant, queues will build up if and only if the arrival rate is greater than the processing rate.
  • If there is (unsynchronized) variability in inter-arrival and/or processing times, queues will build up even if the average arrival rate is less than the average processing rate.
  • If variability in interarrival and processing times can be synchronized (correlated), queues and waiting times will be reduced.
terminology and classification of waiting lines
Terminology and Classification of Waiting Lines

Terminology: The characteristics of a queuing system is captured by five parameters; Arrival pattern, Service pattern, Number of server, Restriction on queue capacity, The queue discipline.

  • M/M/1; Poisson arrival rate, Exponential service times, one server, No capacity limit.
  • M/G/12/23; Poisson arrival rate, General service times, 12 servers, Queue capacity is 23
example coefficient of variation of interarrival time
Example; Coefficient of Variation of Interarrival Time

A sample of 10 observations on Interarrival times in minutes  10,10,2,10,1,3,7,9, 2, 6 minutes.

=AVERAGE ()  Avg. interarrival time = 6 min.

Ra= 1/6 arrivals / min.

=STDEV()  Std. Deviation = 3.94

Ca = 3.94/6 = 0.66

C2a = (0.66)2 =0.4312

example coefficient of variation of processing time
Example: Coefficient of Variation of Processing Time

A sample of 10 observations on Processing times in minutes  7,1,7 2,8,7,4,8,5, 1 minutes.

Tp= 5 minutes

Rp = 1/5 processes/min.

Std. Deviation = 2.83

Cp = 2.83/5 = 0.57

C2p = (0.57)2 = 0.3204

example utilization and safety capacity
Example: Utilization and Safety Capacity

Ra=1/6 < RP =1/5  R = Ra

U= R/ RP = (1/6)/(1/5) = 0.83

With c = 1, the average number of passengers in queue is as follows:

Ii = [(0.832)/(1-0.83)] ×[(0.662+0.572)/2] = 1.56

On average 1.56 passengers waiting in line, even though safety capacity is Rs= RP - Ra= 1/5 - 1/6 = 1/30 passenger per minute, or 2 per hour.

example other performance measures
Example: Other Performance Measures

Waiting time in the line?

Ti=Ii/R = (1.56)(6) = 9.4 minutes.

Waiting time in the system?

T = Ti+Tp

Since Tp= 5  T = Ti+ Tp= 14.4 minutes

Total number of passengers in the process is:

I = RT = (1/6) (14.4) = 2.4

Alternatively, 1.56 in the buffer. How many in the process?

I = 1.56 + 0.83 = 2. 39

example now suppose we have two servers
Example: Now suppose we have two servers.

Compute R, Rp and U: Ta= 6 min, Tp = 5 min, c=2

R = Ra= 1/6 per minute

Processing rate for one processor 1/5 for 2 processors

Rp = 2/5

U = R/Rp = (1/6)/(2/5) = 5/12 = 0.417

On average 0.076 passengers waiting in line.

safety capacity is Rs= RP - Ra= 2/5 - 1/6 = 7/30 passenger per minute, or 14 passengers per hour

other performance measures for two servers
Other Performance Measures for Two Servers

Other performance measures:

Ti=Ii/R = (0.076)(6) = 0.46 minutes

Compute T?

T = Ti+Tp

Since TP = 5  T = Ti + Tp= 0.46+5 = 5.46 minutes

Total number of passengers in the process is: I=

0.08 in the buffer and 0.417 in the process.

I = 0.076 + 2(0.417) = 0.91

processing time waiting time long waiting line
Processing Time, Waiting Time; Long Waiting Line

Ra =R= 10/hour

Tp= 5 minutes

Interarrival time Poisson

Service time exponential

Ra =

10/hour

Increasing Capacity

Polling

more servers specialization
More Servers + Specialization

Ra/2 =

10/hour

Ra/2 =

10/hour

effect of pooling
Effect of Pooling

Ra =R= 10/hour

Tp= 5 minutes

Interarrival time Poisson

Service time exponential

Ra/2

Server 1

Queue 1

Ra

Ra/2

Server 2

Queue 2

Server 1

Ra

Queue

Server 2

effect of pooling 2m m 1
Effect of Pooling : 2M/M/1

Ra/2

Server 1

Ra/2 = R= 5/hour

Tp= 5 minutes

c = 1  Rp = 12 / hour

U= 5/12 = 0.417

Queue 1

Ra

Ra/2

Server 2

Queue 2

comparison of 2m m 1 with m m 2
Comparison of 2M/M/1 with M/M/2

Ra/2

Server 1

Queue 1

Ra

Ra/2

Server 2

Queue 2

Server 1

Ra

Queue

Server 2

effect of pooling m m 2
Effect of Pooling: M/M/2

Ra=R= 10/hour

Tp= 5 minutes

c = 2

Rp = 24 /hour

U= 10/24

U= 0.417 AS BEFORE for each processor

Server 1

Ra

Queue

Server 2

effect of pooling1
Effect of Pooling
  • Under Design A,
    • We have Ra = 10/2 = 5 per hour, and Tp = 5 minutes, c =1, we arrive at a total flow time of 8.6 minutes
  • Under Design B,
    • We have Ra =10 per hour, Tp= 5 minutes, c=2, we arrive at a total flow time of 6.2 minutes
  • So why is Design B better than A?
    • Design A the waiting time of customer is dependent on the processing time of those ahead in the queue
    • Design B, the waiting time of customer is partially dependent on each preceding customer’s processing time
    • Combining queues reduces variability and leads to reduce waiting times
performance measures
Performance Measures
  • Financial Performance Measures
    • Sales: Throughput Rate
    • Cost: Capacity utilization, Number in queue / in system
    • Customer service: Waiting Time in queue /in system
  • Performance Improvement Levers
    • Decrease variability in customer inter-arrival and processing times.
    • Decrease capacity utilization.
    • Synchronize available capacity with demand.
1 variability reduction levers
1. Variability Reduction Levers
  • Customers arrival are hard to control
    • Scheduling, reservations, appointments, etc….
  • Variability in processing time
    • Increased training and standardization processes
    • Lower employee turnover rate  more experienced work force
    • Limit product variety, increase commonality of parts
2 capacity utilization levers
2. Capacity Utilization Levers
  • If the capacity utilization can be decreased, there will also be a decrease in delays and queues.
  • Since U=R/Rp, to decrease capacity utilization there are two options
    • Manage Arrivals: Decrease inflow rate Ra
    • Manage Capacity: Increase processing rate Rp
  • Managing Arrivals
    • Better scheduling, price differentials, alternative services
  • Managing Capacity
    • Increase scale of the process (the number of servers)
    • Increase speed of the process (lower processing time)
3 synchronizing capacity with demand
3. Synchronizing Capacity with Demand
  • Capacity Adjustment Strategies
    • Personnel shifts, cross training, flexible resources
    • Workforce planning & season variability
    • Synchronizing of inputs and outputs, Better scheduling
assignment 1 m m 1 performance evaluation
Assignment 1: M/M/1 Performance Evaluation

Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution.

  • On average how many customers are in the waiting line?
  • How long a customer stays in the line?
  • How long a customer stays in the processor (with the server)?
  • On average how many customers are with the server?
  • On average how many customers are in the system?
  • On average how long a customer stay in the system?
assignment 2 m m 1 performance evaluation
Assignment 2: M/M/1 Performance Evaluation
  • What if the arrival rate is 11 per hour?
assignment 3 m g c
Assignment 3: M/G/c

A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute.

  • On average how many customers are in the waiting line?
  • How long a customer stays in the line?
  • How long a customer stays in the processors (with the servers)?
  • On average how many customers are with the servers?
  • On average how many customers are in the system ?
  • On average how long a customer stay in the system ?
assignment 4 m m c example
Assignment 4: M/M/c Example

A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)?

assignment 5 m d c example
Assignment 5: M/D/c Example
  • Suppose the service time is a constant
  • What is the answer of the previous question?