©the mcgraw-hill companies, inc. 2008mcgraw-hill/irwin chapter 8 timeliness: scheduling and project...

©The McGraw-Hill Companies, Inc. 2008McGraw-Hill/Irwin

Chapter 8

Timeliness: Scheduling and Project Management

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Learning Objectives

• Explain the effect time has on each profitability measure.• Describe the impact of feedback delay on quality.• Explain the common time reduction strategies.• Construct a Gantt chart.• Sequence orders using the traditional sequencing rules.• Describe the physical features of queues and describe how they affect

queue performance.• Compute the probability of x arrivals per unit time for a queue.• Describe the psychological approaches to managing perception of queue

length.• Construct a network diagram for a project.• Identify the critical path for a project using CPM calculations• Calculate the likelihood of completing a project in a specified time.• Complete the calculations necessary to effectively crash a project.

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• Time is money– It affects customer perceived value– It affects Net Income – It affects Return on Assets

Operations Management Framework

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Time and Customer Value

• Speed and flexibility– Faster processes allow greater flexibility

• Deliver more options in the same amount of time– Customers will only wait so long.

• Start later, closer to demand– Better predictions of what will be needed

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Time Really Is Money

• Time affects profitability through net income– Time-based value attributes - Increasing net sales

• Response time

• Delivery dependability

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Time Really Is Money

• Managing timing of payments to make money (managing “float”)– Example. . . .Travel Agents

• You pay right away (on a credit card)– They pay in 30 days (business billing cycle)

They deep the interest

– Example . . . . .Insurance companies– Collect only as much as (or less than) they will eventually pay in claims

– Invest money while they are waiting to pay out claims

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The Effect of Time on ROA

• The cash-to-cash cycle

• Not just production time. . .– Waiting for supplies to arrive -- Opportunity cost of sales

– Waiting to sell product -- Delay of return on investment in inventory and production

– Waiting to collect from customer -- Financing the customer’s purchase

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Suppose the transaction in exhibit 8.1 netted a $200 return on total expenditures of $1000 (a 20% return). What would happen to ROA if the cash-to-cash cycle went from 17 weeks to 13 weeks?

With a 17 week cash-to-cash cycle, there are approximately three transactions per year for a return of $600 (60%).

Reducing the cash-to-cash cycle time to 13 weeks results in the $1000 being turned over four times instead of three. $800 will now be generated. The yield shifts from 60% to 80%.

A 23.5% reduction in time results in a 33% increase in return financial return

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• Shorter delivery lead-times from suppliers mean lower inventory requirements

Weekly DeliveryAverage Inventory (Cases) = 400 casesAverage Inventory (Investment) = $106,000

Daily DeliveryAverage Inventory (Cases) = 80 casesAverage Inventory (Investment) = $21,200

Example from Chapter 2• Deliveries of 800 cases weekly (all are consumed)• Cases cost $265, Sell for $300• Weekly Net Income is $28,000

ROA = $28,000/$106,000 = 26.42%

ROA = $28,000/$21,200 = 132.08%

As inventory is reduced to 1/5, ROA increases five-fold

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Feedback Delay. . . Another benefit of time reduction.

• Elimination of causes of quality problems depends on timely information.

• The greater the elapsed time between producing a defect and discovering it, the less likely the defects cause will be found.

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General Time Reduction Strategies

• Two general approaches:– Compressing time of activities

– Making tasks concurrent instead of sequential

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Insert exhibit 8.3

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Insert exhibit 8.4

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Insert exhibit 8.5

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Scheduling

• Definition: Determination of when something is to be done and the tasks and activities required to do it– Scheduling aids in on-time completion

• Direct link to value perceived by customers

– Scheduling improves the utilization of the firm’s resources• Direct link to productivity

• Forward scheduling– When start date is known and completion needs determination

• Backward scheduling– When completion date is known and start date needs determination

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Scheduling

• Gantt Charts– The layout of steps within a process

– The layout of multiple tasks that utilize a single resource.

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Traditional Job Shop Sequencing

• Sequencing rules for scheduling multiple jobs at a single resource– First Come, First Served

– Earliest Due Date

– Shortest Processing Time

– Critical Ratio• Time Remaining/effort remaining

• C.R. < 1 means there is no way to complete the job on time

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Example 8.1 Job Sequencing Using EDD

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Example 8.1 Job Sequencing Using EDD

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Example 8.2 Job Sequencing Using SPT

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Example 8.3 Job Sequencing Using Critical Ratio

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Traditional Job Shop Sequencing

• Most traditional sequencing rules assume a single operation.

• Slack Time Remaining– Slack = Days to due date - Total processing time (all operations)

– Slack per remaining operation = Slack / Number of remaining operations

Exhibit 8.12: Slack per Remaining Operations

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People Are Not Products:People Expect To Be Treated “Fairly”

• Queue configuration: The physical design of the lines and servers in a queuing system.

• Server: A resource that is able to complete the process or service that customers or jobs wait in queue for.

• Phase: A distinct step in a process that requires a separate queue.• Jockeying: When customers switch lines hoping to move faster.• Queue Discipline: The rules that management enforces to

determine the next customer served in a queue.• Calling Population: the population of arriving customers or

orders.

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The Queue Arrival Process: Determining the Probability of x arrivals in a time period

• Queue arrivals follow a Poisson distribution.

• P(x) = the probability of x arrivals in a time period

• x = the number of arrivals per unit time

• λ = the average arrival rate in a certain time increment

• e = 2.7283 (the base of the natural logarithms)

( )!

xep x

x

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Managing Queue Arrivals

• A Call center receives 36 calls per hour, Poisson distributed. What is the probability of receiving 41 calls?

• λ = 36• x = 41• e = 2.7283 (the base of the natural logarithms)

36 4136( )

41!

4.46%

ep x

( )!

xep x

x

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People Are Not Products:Treating People “Fairly”is Important

• First Come, First Served seems the most fair

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People Are Not Products:Serving Customers in Queue

• Psychological features of queues• Make the wait seem less be . . .

– Keeping customers busy– Keeping customers informed– Treating customers fairly– Starting the service as soon as possible– Exceeding the customer’s expectations

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People Are Not Products:Queue Alternatives

• Avoid Queues: Developing Alternate Procedures– Express check-in and check-out (hotels, rental cars)– Pay bills by mail and online, not in person

• Parking tickets paid online

– Automatic Checking Account Debits• Phone,Utilities, Credit Cards, Mortgages

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Project Management

• A project is a set of activities aimed at meeting a goal, with a defined beginning and end.

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Project Management with Certain Time Estimates

• Summary of steps:– Determine activities that need to be accomplished– Determine precedence relationships and completion

times– Construct network diagram– Determine the critical path– Determine early start and late start schedules

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• Create Network Diagram– Based on order of precedence

among activities

Project Management: Scheduling Projects with Certain Time Estimates

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Calculation of the Critical Path

• Network approach helps calculate project duration

• A “path” is a sequence of activities that begins at the start of the project and goes to the end of the project– 1,2,3,5,6,7,8

– 1,2,4,6,7,8

– The “critical path” is the path that takes the longest to complete• and thus determines the minimum duration of the project

1 2

4

3 5

6 7 8

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Exhibit 8.17: Project Detail

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• Critical Path– The path that takes

the longest to complete

1 2

4

3 56 7 8

4 weeks 10 weeks

4 weeks

2 weeks 3 weeks

16 weeks 4 weeks 1 week


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• Critical Path– The path that takes the

longest to complete

C.P. = 40 weeks


1 2

4

3 56 7 8

4 weeks 10 weeks

4 weeks

2 weeks 3 weeks


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• It is possible for multiple Critical Paths to exist

• Suppose new information suggested that Activity 4 would take 5 weeks instead of 4. . .

X5


1 2

4

3 56 7 8

4 weeks 10 weeks

4 weeks

2 weeks 3 weeks


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X5


1 2

4

3 56 7 8

4 weeks 10 weeks

5 weeks

2 weeks 3 weeks


There are now two critical paths, each one is 40 weeks.

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• The Critical Path may also shift if non-critical activity is lengthened or Critical Path activity is shortened

• Suppose another update indicates

that it will actually take 6 weeks for Activity 4

The Critical Path becomes 41 days.

4 weeks

1 2

4

3 5

6 7 8

10 weeks

6 weeks

2 weeks 3 weeks


X6


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Determining Slack

• Slack - The amount of time an activity on a non-critical path can be delayed without affecting the duration of the project (i.e., without putting it on the critical path)– Four values are used

• Early start - Earliest an activity can start (based on prerequisites)

• Early finish - Earliest it can finish (based on prerequisites & duration)

• Late start - Latest an activity can start and not delay the project

• Late finish - Latest an activity can finish and not delay the project

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Calculating Early Start (ES) and Early Finish (EF)

• The process moves from left to right in network

The ES for the 1st activity is usually zero

The EF equals the ES plus activity duration

ES is the latest of the EF times

of an activity’s predecessors


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• This process moves from right to left in the network

Calculating Late Start (LS) and Late Finish (LF)

LF for the last activity equals the EF for the last activity

LS equals LF minus the activity durationLF is the earliest

of the LS times of an activity’s followers

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Calculating Slack

• Slack - The amount of time an activity on a non-critical path can be delayed without affecting the duration of the project (i.e., without putting it on the critical path)

• Can be computed by either:

Late Start - Early Start

or

Late Finish - Early Finish

• Activities that have zero slack form the critical path

Exhibit 8.21: ES, EF, LS, LF, and Slack Values

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Project Scheduling When the Times of Activities are Uncertain

• Summary of steps:– Determine the activities that need to be accomplished

– Determine the precedence relationships and completion times

– Construct the network diagram

– Determine the critical path

– Determine the early start and late start schedules

– Calculate the variances for the activity times

– Calculate the probability of completing by the desired due date

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Project Scheduling with Time Uncertainty

• The “Heuristic” approach to dealing with timing uncertainty– Based on understanding of individual activities as conforming to a “beta” distribution

• Take three time estimates– Optimistic - What is the (realistic) fastest we can get an activity done?

– Pessimistic - What is the (realistic) worst case scenario for delay?

– Most likely - What is our “most likely” estimate?

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• Calculate the “expected time” for the activity

Where:

t = Expected Time

o = Optimistic estimate

m = Most likely estimate

p = Worst-case (pessimistic) estimate

4

6

o m pt

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3 4(4) 54.00

6t

8 4(10) 1210.00

6t

Estimates Activity # Optimistic Most Likely Pessimistic

1 3 4 5 2 8 10 12 3 1 2 4

Activity #1 weeks

Activity #3 weeks

Activity #2 weeks

1 4(2) 42.17

6t

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1 2

4

3 5

6 7 8

4 weeks 10 weeks

4.17 weeks

2.17 weeks 3 weeks

16.17 weeks 4.17 weeks 1.17 weeks

C.P. = 40.68 weeks

How do we interpret this estimate???

There is a 50% chance we will finish faster, and a 50% chance that we will finish slower than our estimate

We assume that finish times across the entire project are distributed normally.

Using Variation in Time Estimates to Assess Duration Probabilities

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• So, if the “expected” duration (t) of the project is 40.68 weeks, how do we determine the probability of finishing it in 39 weeks (or less)?– Take advantage of the estimates of t and make an estimate of 2… – … draw on knowledge of statistics (especially Normal distributions),… – … and use the “standard normal probability” table along with the following equation:

Using Variation in Time Estimates to Access Duration Probabilities

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Where:

Z = Number of standard deviations D is from Te

t = Expected Time

D = Project duration I am thinking about

Note that this is , not 2 ,We have to take the square root of the “variance” to get the standard

deviation

D tZ

Using Variation in Time Estimates to Determine Duration Probabilities

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• We recognize that there is variation around our estimates– We are estimating “most likely” or “expected” not “exact”– Commonly assume to be a “Beta” distribution

• Calculating variance of the activity’s duration estimate:

22

6

op


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Estimates Activity # Optimistic Most Likely Pessimistic

1 3 4 5 2 8 10 12 3 1 2 4

Activity #1 weeks11.06

352

2


8122

2


142

2


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• Calculate Critical Path using t

• Accumulate the variances of the individual activities

• Apply formula to estimate project duration probabilities

Where:

Z = Number of standard deviations D is from t

t = Expected Time

D = Activity duration I am thinking about

D tZ

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Exhibit 8.22: Expected Values and Variances of Time Estimates

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1 2

4

3 5

6 7 8

4 weeks 10 weeks

4.17 weeks

2.17 weeks 3 weeks



What is the probability of finishing in 39 weeks or less?


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1 2

4

3 5

6 7 8

4 weeks 10 weeks

4.17 weeks

2.17 weeks 3 weeks


C.P. = 40.68 weeksSum of Variances = 2.56 weeks


What is the probability of finishing in 39 weeks or less?


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• Look up the cumulative probability of the activity being completed before -1.05 standard deviations (Appendix B, near the center)

• There is a 14.686% chance of finishing in 39 weeks or less.

Where:

Z = Number of standard deviations D is from t

t = Expected Time

D = Activity duration I am thinking about

05.16.1

68.4039

Z

D tZ

USE:

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Crashing Projects

• A methodical approach to reducing project duration– Focus on the time of activities on the critical path– Looking for greatest improvement with least cost

• Additional labor, machinery• Overtime and temporary employees• Premiums paid to outside contractors for early delivery

• Steps– Create network– Identify critical path– Identify costs of reducing each activity on path– Reduce most cost effective activity– Look for critical path changes

• Beware of multiple critical paths

– Crash next activity

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Crashing Projects: Create the Network

Exhibit 8.24: Network to Crash

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A8

B7

D10

E12

C7

F8

G9

H5

Crashing Projects: Identify the Critical Path

C.P. = 35 days

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Crashing Projects:Identify Costs of Crashing Each Activity

Exhibit 8.25: Crash Time and Costs

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A8

B7

D10

E12

C7

F8

G9

H5

Crashing Projects:Reduce Most Cost Effective Activity

6

X

Critical path starts out at 35 days

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A8

B7

D10

E12

C7

F8

G9

H5

Crashing Projects:Look for Critical Path Changes

6

X

Critical path now= 35 days

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A8

B7

D10

E12

C7

F8

G9

H5

Crashing Projects:Look for Critical Path Changes

6

X

There are now 2 critical paths, so reducing the project length will require shortening each

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Crashing Projects:Crash Next Activity

Both C.P. Path 1 Only Path 2 Only

Exhibit 8.25: Crash Time and Costs

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Crashing Projects: Summary

Solution Exhibit 8.26: Crashing Summary

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Caveats

• Time estimates are frequently wrong• Early finishes are absorbed• Cushions get wasted• Resources aren’t always available when needed

©the mcgraw-hill companies, inc. 2008mcgraw-hill/irwin chapter 8 timeliness: scheduling and project...

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money time

effect time

effect of time

cash cycle time

production time

time results

time customers

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