BPMS

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Overview

• Processes and Flows – Important Concepts– Throughput

– WIP

– Cycle Time

– Little’s Formula

• Cycle Time Analysis

• Managing Cycle Time and Capacity– Cycle time reduction

– Increasing Process Capacity

• Theory of Constraints

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• A process = A set of activities that transforms inputs to outputs

• Two main methods for processing jobs1. Discrete – Identifiable products or services

Examples: Cars, cell phones, clothes etc.

2. Continuous – Products and services not in identifiable distinct units Examples: Gasoline, electricity, paper etc.

• Three main types of flow structures1. Divergent – Several outputs derived from one input

Example: Dairy and oil products

2. Convergent – Several inputs put together to one output Example: Car manufacturing, general assembly lines

3. Linear – One input gives one output Example: Hospital treatment

Processes and Flows – Concepts

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• Inflow and Outflow rates typically vary over time– IN(t) = Arrival/Inflow rate of jobs at time t– OUT(t) = Departure/Outflow rate of finished jobs

at time t– IN = Average inflow rate over time– OUT = Average outflow rate over time

• A stable system must have IN=OUT= = the process flow rate = process throughput

Process Throughput

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1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

t

Jobs IN(t)

OUT(t)

Process Inflow and Outflow vary over time

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• All jobs that have entered the process but not yet left it

• A long lasting trend in manufacturing has been to lower WIP by reducing batch sizes– The JIT philosophy

– Forces reduction in set up times and set up costs

• WIP = Average work in process over time

• WIP(t) = Work in process at time t– WIP(t) increases when IN(t)>OUT(t)

– WIP(t) decreases when IN(t)<OUT(t)

Work-In-Process

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The WIP Level Varies With Process Inflow and Outflow

t1 t2 t3

WIP(t)

WIP

OUT(t) > IN(t)

IN(t) > OUT(t)

OUT(t) = IN(t)

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• The difference between a job’s departure time and its arrival time = cycle time– One of the most important attributes of a process– Also referred to as throughput time

• The cycle time includes both value adding and non-value adding activity times– Processing time– Inspection time– Transportation time– Storage time– Waiting time

• Cycle time is a powerful tool for identifying process improvement potential

Process Cycle Time

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• States a fundamental and very general relationship between the average: WIP, Throughput (= ) and Cycle time (CT)– The cycle time refers to the time the job spends in the system or process

• Implications, everything else equal– Shorter cycle time lower WIP

– If increases to keep WIP at current levels CT must be reduced

• A related measure is (inventory) turnover ratio– Indicates how often the WIP is entirely replaced by a new set of jobs

Little’s Formula (Due to J.D.C. Little (1961))

Little’s Formula: WIP = ·CT

Turnover ratio = 1/CT

Little’s Law

λ•= rate at which units arrive

•quantity/time-unit or #/time-unit;

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Little’s Law

• L = inventory/quantity/number in the system

• (eg. WIP: Work-In-Process, customers);

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Little’s Law

• W = time a unit spends in the system = throughput time

• (eg. hours)

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WIP=·CT

Little’s LawThe average waiting time and the average number of items waiting for..a service in a service system are important measurements for a manager. Little's Law Relates these two metrics via the average rate of arrivals to the system. This fundamental law has found numerous uses in operations management and managerial decision making

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Example

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Example inventory management

• Inventory Management

• L = average inventory;

• W = average time in inventory;

• λ = average throughput rate.

• The quantity 1/CT = λ/WIP is often referred to as the turnover ratio.

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Inventory Management

• A fast food restaurant processes on the average 5000 lbs. of hamburger per week. The observed inventory level of raw meat, over a long period of time, averages 2500 lbs.

• L = 2500 lbs., λ = 5000 lbs./week;• W = L/λ = 2500/5000 = 1/2 week is the average

time spent by a pound of meat in inventory; • 1/W = 2 times per week is the inventory turnover

ratio.

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Example Services Management

• L = average number of customers;

• W = average customer’s delay;

• λ = average customers’ throughput rate.

A restaurant processes on average 1500 customers per day (=15 hours). On average, there are 50 customers waiting to place an order, waiting for an order to arrive or eating.

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• λ = 1500 customers/day = 100 customers/hour;

• L = 50 customers;

• W = L/λ = 50/100 = 1/2 hours, average time in the restaurant.

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• Out of the 50 customers, 40 customers on the average are eating.

• Then

• λ = 100, L = 50 − 40 = 10 customers at the service counter;

• W = L/λ = 10/100 hours = 6 minutes average wait at the counter.

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Examples

• Insurance: An insurance company processes 10,000 claims per year. The average processing time of a claim is 3 weeks. Assuming 50 weeks per year, we have

• λ = 10,000 claims/year = 200 claims/week;

• W = 3 weeks;

• L = λW = 3 × 200 = 600 claims backlog on the average

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Examples

• A company sells 300M$ worth of finished• goods per year. The average amount of accounts

receivable is 45M$.• λ = 300M$/year;• L = 45M$;• W = L/λ = 45/300 = 0.15 years = 1.8 months.• So it takes, on average, 1.8 months from the time a

customer is billed until the time• payment is received

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Example in IT

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Example in IT

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Example1-solution

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Example 2

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Solution Example2

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Note

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• Let pi be the ith packet into the queue• Let Ni = # of pkts already in the queue when pi arrives• Let Ti = time spent by pi in the system: includes

– time sitting in queue– time it takes processor to process pi

• If K = ∞ (unlimited queue size) then

lim E[Ni] = lim E[Ti] i∞ i∞

Holds for any distribution of , μ (which means for any distribution of Ti as well)!!

Note

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Examples

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• People arrive at a bank at an avg. rate of 5/min. They spend an average of 20 min in the bank. What is the average # of people in the bank at any time?

• To keep the average # of people under 50, how much time should be spent by customers on average in the bank?

=5, T=20, E[N] = E[T] = 5(20) = 100

=5, E[N] < 50, E[T] = E[N] / < 50 / 5 = 10

Examples

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Example- 3 scenarios

• At the supermarket a checkout operator has on average 4 customers and customers arrive every 2 minutes. Therefore customers on average will be in line for 8 minutes.

• A restaurant holds about 60 people, and the average person will be in there about 2 hours, so they're entering at the rate of about 30 people an hour. The queue for the restaurant has 30 people in it, so that means I'll wait about an hour for a table.

• A financial services organisation receives on average 160 enquiries per day about its products and services. If it takes around 30 minutes to process each enquiry and management want to ensure each enquiry is responded to on the same day its received, how many people are needed to process the enquiries?

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Problem Throughput Time Work in Process Throughput Rate

1 ? minutes/customer 4 customers ½ customer/minute

2 2 hours/customer 60 customers ? customers/hour

3 ½ hour/enquiry ? people @ 1 per

enquiry

20 enquiries/hour

• The trick to using Little’s Law is to ensure you are working with the same time interval for the throughput time and throughput rate i.e. minutes, hours, days, weeks, etc. Otherwise your answers will be incorrect. (This is where most people get Little’s law wrong).

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• In the first problem, we want to find the Throughput Time at the checkout, which is 4 customers x ½ customer / minute = 8 minutes. Note that I converted all figures to minutes.

• In the second problem, we want to find the Throughput Rate, which is 60 customers / 2 hours per customer = 30 customers / hour. If the queue has 30 people in it then, based on this throughput rate, we’ll have to wait for an hour before getting a seat in the restaurant.

• In the third problem, we need to find the Work in Process, which will equate to the required number of people if we consider that each enquiry will require one person to handle it. We firstly convert the figures to an hourly basis. We then calculate the Work in Process which is 20 enquiries per hour x ½ hour per enquiry = 10 enquiries in process at any point in time, i.e. We need 10 people to handle the enquiries 34


• A fast food hamburger restaurant uses 3,500 kilograms of hamburger mince each week. The manager of the restaurant wants to ensure that the meat is always fresh i.e. the meat should be no more than two days old on average when used. How much hamburger mince should be kept in the refrigerator as inventory?

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Exercise-1

• Law:

• Throughput = 3,500 kilograms per week (= 500 kilograms per day)

• Average flow time = 2 days

• Average inventory = Throughput x Average flow time

• = 500 x 2 = 1,000 kilograms

• (Note that the variables are all in the same time frame i.e. days)

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Exercise1 -Solutions

Excercise2

• The Acme Australia insurance company processes 12,000 insurance claims per year. At any point in time, there are 480 insurance claims in head office in various phases of processing. Assuming that the office works 50 weeks per year, find out the average processing time of an insurance claim in days.

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Exercise 2-solution

• Throughput time = Work In Process / Throughput Rate

•

• Work in Process = 480 claims

• Throughput Rate = 12,000 claims per year = 12,000/50 claims per week

• = 240 claims per week

• Throughput Time = 480 / 240 = 2 weeks

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Summary

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• The task of calculating the average cycle time for an entire process or process segment– Assumes that the average activity times for all involved activities

are available

• In the simplest case a process consists of a sequence of activities on a single path– The average cycle time is just the sum of the average activity times

involved

• … but in general we must be able to account for – Rework

– Multiple paths

– Parallel activities

Cycle Time Analysis

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• Many processes include control or inspection points where

if the job does not conform it will be sent back for rework – The rework will directly affect the average cycle time!

• Definitions– T = sum of activity times in the rework loop

– r = percentage of jobs requiring rework (rejection rate)

• Assuming a job is never reworked more than once

• Assuming a reworked job is no different than a regular job i.e rework more than once

Rework

CT = (1+r)T

CT = T/(1-r)

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Example – Rework effects on the average cycle time

• Consider a process consisting of – Three activities, A, B & C taking on average 10 min. each– One inspection activity (I) taking 4 minutes to complete.– X% of the jobs are rejected at inspection and sent for rework

What is the average cycle time?a) If no jobs are rejected and sent for rework.b) If 25% of the jobs need rework but never more than once.c) If 25% of the jobs need rework but reworked jobs are no different in

quality than ordinary jobs.

0.75

0.25

A(10)

B(10)

C(10)

I(4)

If no rejection

•CT= 10+10+10+4= 34 minutes

If rejection is 25% for once rework

•CT=10+(1+.25)X(10+10+4)=40 minutes

If rejection is 25% and is more than once

•CT=10+(10+10+4)/ (1-0.25)= 42 minutes

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Example – Rework effects on the average cycle time

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• It is common that there are alternative routes through the process– For example: jobs can be split in “fast track”and normal jobs

• Assume that m different paths originate from a decision point– pi = The probability that a job is routed to path i – Ti = The time to go down path i

Multiple Paths

CT = p1T1+p2T2+…+pmTm=

m

1iiiTp

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Example – Processes with Multiple Paths

• Consider a process segment consisting of 3 activities A, B & C with activity times 10,15 & 20 minutes respectively

• On average 20% of the jobs are routed via B and 80% go straight to activity C.

What is the average cycle time? =10+0.2X15+0.8X20 =10+3+16=29 minutes

0.8

0.2

A(10)

B(15)

C(20)

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A(5)

Easy return

B(30)

C(60)

D(6)

yes

0.30

0.70

No

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Solution

CT= 5+0.30X30+0.70X60+6= 62 minutes

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• If two activities related to the same job are done in parallel the contribution to the cycle time for the job is the maximum of the two activity times.

• Assuming– M process segments in parallel

– Ti = Average process time for process segment i to be completed

Processes with Parallel Activities

CTparallel = Max{T1, T2,…, TM}

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• Consider a process segment with 5 activities A, B, C, D & E with average activity times: 12, 14, 20, 18 & 15 minutes

What is the average cycle time for the process segment? CT= 12+max(14,20,18)+15= 47

Example – Cycle Time Analysis of Parallel Activities

A(12)

B(14)

C(20)

D(18)

E(15)

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• Measured as the percentage of the total cycle time spent on value adding activities.

• Theoretical Cycle Time = the cycle time which we would have if only value adding activities were performed– That is if the activity times, which include waiting times, are

replaced by the processing times

Cycle Time Efficiency

Cycle Time Efficiency = CT

TimeCyclelTheoretica

Example cycle time analysis

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For cycle time efficiency

• CT is calculated

• Theoretical Cycle time or process time is to be calculated

• For process time – activity time – waiting time.

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Cycle time and capacity analysis provide valuable information about process performance

– Helps identify problems

– Increases process understanding

– Useful for assessing the effect of design changes

• Ways of reducing cycle times through process redesign1. Eliminate activities

2. Reduce waiting and processing time

3. Eliminate rework

4. Perform activities in parallel

5. Move processing time to activities not on the critical path

6. Reduce setup times and enable batch size reduction

Cycle time Reduction

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• Two fundamental ways of increasing process capacity1. Add resource capacity at the bottleneck

– Additional equipment, labor or overtime

– Automation

2. Reduce bottleneck workload

– Process redesign Shifting activities from the bottleneck to other resources Reducing activity time for bottleneck jobs

• When the goal is to reduce cycle time and increase capacity careful attention must be given to– The resource availability

– The assignment of activities to resources

Increasing Process Capacity

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BPMS