1 aggregate planning. 2 master scheduling material requirements planning order scheduling weekly...

1

Aggregate Planning

2

Master scheduling

Material requirements planning

Order schedulingWeekly workforce andcustomer scheduling

Daily workforce and customer scheduling

Process planning

Strategic capacity planning

Sales and operations (aggregate) planning

Longrange

Intermediaterange

Shortrange

ManufacturingServices

Sales plan Aggregate operations plan

Forecasting & demand management

3

The Aggregate Operations Plan

• Main purpose: Specify the optimal combination of– production rate (units completed per unit of

time)– workforce level (number of workers)– inventory on hand (inventory carried from

previous period)• Product group or broad category (Aggregation)• This planning is done over an intermediate-range

planning period of 3 to18 months

4

Required Inputs to the Production Planning System

Planning for

production

External capacity

Competitors’behavior

Raw material availability

Market demand

Economic conditions

Currentphysical capacity

Current workforce

Inventory levels

Activities required for production

External to firm

Internal to firm

5

Key Strategies for Meeting Demand

Chase

Level

Some combination of the two

6

Mixing Options to Develop a Plan

Chase strategyChase strategy• Match output rates to demand Match output rates to demand

forecast for each periodforecast for each period

• Vary workforce levels or vary Vary workforce levels or vary production rateproduction rate

• Favored by many service Favored by many service organizationsorganizations

7

Level strategyLevel strategy• Daily production is uniformDaily production is uniform

• Use inventory or idle time as bufferUse inventory or idle time as buffer

• Stable production leads to better Stable production leads to better quality and productivityquality and productivity

Some combination of capacity Some combination of capacity options, a mixed strategy, might be options, a mixed strategy, might be the best solutionthe best solution

Mixing Options to Develop a Plan

8

Aggregate Planning Examples: Unit Demand and Cost Data

Materials Rs5/unitHolding costs Rs1/unit per mo.Marginal cost of stockout Rs1.25/unit per mo.Hiring and training cost Rs200/workerLayoff costs Rs250/workerLabor hours required .15 hrs/unitStraight time labor cost Rs8/hourBeginning inventory 250 unitsProductive hours/worker/day 7.25Paid straight hrs/day 8

Suppose we have the following unit demand and cost information:

Suppose we have the following unit demand and cost information:

Demand/mo Jan Feb Mar Apr May Jun

4500 5500 7000 10000 8000 6000

9

Jan Feb Mar Apr May JunDays/mo 22 19 21 21 22 20Hrs/worker/mo 159.5 137.75 152.25 152.25 159.5 145Units/worker 1063.33 918.33 1015 1015 1063.33 966.67Rs/worker 1,408 1,216 1,344 1,344 1,408 1,280

Productive hours/worker/day 7.25Paid straight hrs/day 8

Demand/mo J an Feb Mar Apr May Jun

4500 5500 7000 10000 8000 6000

Given the demand and cost information below, whatare the aggregate hours/worker/month, units/worker, and rupees/worker?

Given the demand and cost information below, whatare the aggregate hours/worker/month, units/worker, and rupees/worker?

7.25x22

7.25/0.15=48.33 & 48.33x22=1063.3322x8hrsxRs8=Rs1

408

Cut-and-Try Example: Determining Straight Labor Costs and Output

10

Chase Strategy(Hiring & Firing to meet demand)

JanDays/mo 22Hrs/worker/mo 159.5Units/worker 1,063.33Rs/worker 1,408

JanDemand 4,500Beg. inv. 250Net req. 4,250Req. workers 3.997HiredFired 3Workforce 4Ending inventory 0

Lets assume our current workforce is 7 workers.

Lets assume our current workforce is 7 workers.

First, calculate net requirements for production, or 4500-250=4250 units

Then, calculate number of workers needed to produce the net requirements, or 4250/1063.33=3.997 or 4 workers

Finally, determine the number of workers to hire/fire. In this case we only need 4 workers, we have 7, so 3 can be fired.

11

Jan Feb Mar Apr May JunDays/mo 22 19 21 21 22 20Hrs/worker/mo 159.5 137.75 152.25 152.25 159.5 145Units/worker 1,063 918 1,015 1,015 1,063 967Rs/worker 1,408 1,216 1,344 1,344 1,408 1,280

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250Net req. 4,250 5,500 7,000 10,000 8,000 6,000Req. workers 3.997 5.989 6.897 9.852 7.524 6.207Hired 2 1 3Fired 3 2 1Workforce 4 6 7 10 8 7Ending inventory 0 0 0 0 0 0

Below are the complete calculations for the remaining months in the six month planning horizon


12

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250Net req. 4,250 5,500 7,000 10,000 8,000 6,000Req. workers 3.997 5.989 6.897 9.852 7.524 6.207Hired 2 1 3Fired 3 2 1Workforce 4 6 7 10 8 7Ending inventory 0 0 0 0 0 0

Jan Feb Mar Apr May Jun CostsMaterial 21,250.00 27,500.00 35,000.00 50,000.00 40,000.00 30,000.00 203,750.00Labor 5,627.59 7,282.76 9,268.97 13,241.38 10,593.10 7,944.83 53,958.62Hiring cost 400.00 200.00 600.00 1,200.00Firing cost 750.00 500.00 250.00 1,500.00

260,408.62

Below are the complete calculations for the remaining months in the six month planning horizon with the other costs included

13

Level Workforce Strategy (Surplus and Shortage Allowed)

JanDemand 4,500Beg. inv. 250Net req. 4,250Workers 6Production 6,380Ending inventory 2,130Surplus 2,130Shortage

Lets take the same problem as before but this time use the Level Workforce strategy

Lets take the same problem as before but this time use the Level Workforce strategy

This time we will seek to use a workforce level of 6 workers

This time we will seek to use a workforce level of 6 workers

14

Jan Feb Mar Apr May JunDemand 4,500 5,500 7,000 10,000 8,000 6,000Beg. inv. 250 2,130 2,140 1,230 -2,680 -1,300Net req. 4,250 3,370 4,860 8,770 10,680 7,300Workers 6 6 6 6 6 6Production 6,380 5,510 6,090 6,090 6,380 5,800Ending inventory 2,130 2,140 1,230 -2,680 -1,300 -1,500Surplus 2,130 2,140 1,230Shortage 2,680 1,300 1,500

Note, if we recalculate this sheet with 7 workers we would have a surplus

Note, if we recalculate this sheet with 7 workers we would have a surplus



15

Jan Feb Mar Apr May Jun4,500 5,500 7,000 10,000 8,000 6,000

250 2,130 10 -910 -3,910 -1,6204,250 3,370 4,860 8,770 10,680 7,300

6 6 6 6 6 66,380 5,510 6,090 6,090 6,380 5,8002,130 2,140 1,230 -2,680 -1,300 -1,5002,130 2,140 1,230

2,680 1,300 1,500

Jan Feb Mar Apr May Jun8,448.00 7,296.00 8,064.00 8,064.00 8,448.00 7,680.00 48,000.00

31,900.00 27,550.00 30,450.00 30,450.00 31,900.00 29,000.00 181,250.002,130.00 2,140.00 1,230.00 5,500.00

3,350.00 1,625.00 1,875.00 6,850.00

241,600.00



Note, total costs under this strategy are less than Chase at Rs260.408.62

Note, total costs under this strategy are less than Chase at Rs260.408.62

LaborMaterialStorageStockout

16

Chapter 15

Materials Requirements Planning

17

Planning for MaterialsTwo types of inventories

Two types of inventories exist in any operations system– Operating Inventory:

Denotes all the resources (broadly of material and capacity) that are available for the operating system to consume in the production process

Dependant demand attributes

– Distribution Inventory: Meant for market consumption Independent demand attributes

They differ in their demand attributes & therefore require alternative planning methodologies

18

Material Requirements PlanningDefined• Materials requirements planning (MRP) is a

means for determining the number of parts, components, and materials needed to produce a product

• MRP provides time scheduling information specifying when each of the materials, parts, and components should be ordered or produced

• Dependent demand drives MRP• MRP is a software system

19

Independent Demand

20

Dependent Demand

21

21

Demand Characteristics

11 22 33 44 55WeekWeek

400 400 –

300 300 –

200 200 –

100 100 –No

. o

f ta

ble

sN

o.

of

tab

les

Continuous demandContinuous demand

M T W Th F M T W Th FM T W Th F M T W Th F

400 400 –

300 300 –

200 200 –

100 100 –No

. o

f ta

ble

sN

o.

of

tab

les

Discrete demandDiscrete demand

Independent demandIndependent demand

100 tables100 tables

Dependent demandDependent demand

100 x 1 = 100 x 1 = 100 tabletops100 tabletops

100 x 4 = 400 table legs100 x 4 = 400 table legs

Demand Characteristics for Finished Products and Their Components

22

6

E(1)

Independent vs. Dependent Demand

Independent Demand(Demand not related to other items)

Dependent Demand(Derived)

Drives MRP

23

23

Major Inputs to MRP Process:1. Bill of Material

• Product structure file• Determines which component items need to be scheduled

Product Structure RecordProduct Structure Record

ClipboardLevel 0Level 0

Level 1Level 1

Level 2Level 2Spring (1)

Bottom Clip (1)

Top Clip (1)

Pivot (1)

Rivets (2)

Clip Assembly

(1)

Pressboard (1)

Top clip (1) Bottom clip (1)

Pivot (1) Spring (1)

Rivets (2)Finished clipboard Pressboard (1)

Clipboard

24

Example of MRP Logic and Product Structure Tree

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

Product Structure Tree for Assembly A Lead TimesA 1 dayB 2 daysC 1 dayD 3 daysE 4 daysF 1 day

Total Unit DemandDay 10 50 ADay 8 20 B (Spares)Day 6 15 D (Spares)

Given the product structure tree for “A” and the lead time and demand information below, provide a materials requirements plan that defines the number of units of each component and when they will be needed


25

LT = 1 day

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50

First, the number of units of “A” are scheduled backwards to allow for their lead time. So, in the materials requirement plan below, we have to place an order for 50 units of “A” on the 9th day to receive them on day 10.

First, the number of units of “A” are scheduled backwards to allow for their lead time. So, in the materials requirement plan below, we have to place an order for 50 units of “A” on the 9th day to receive them on day 10.

26

Next, we need to start scheduling the components that make up “A”. In the case of component “B” we need 4 B’s for each A. Since we need 50 A’s, that means 200 B’s. And again, we back the schedule up for the necessary 2 days of lead time.

Next, we need to start scheduling the components that make up “A”. In the case of component “B” we need 4 B’s for each A. Since we need 50 A’s, that means 200 B’s. And again, we back the schedule up for the necessary 2 days of lead time.

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50B Required 20 200

Order Placement 20 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

SparesLT = 2

4x50=200

27

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

LT=1 Order Placement 50B Required 20 200

LT=2 Order Placement 20 200C Required 100

LT=1 Order Placement 100D Required 55 400 300

LT=3 Order Placement 55 400 300E Required 20 200

LT=4 Order Placement 20 200F Required 200

LT=1 Order Placement 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

40 + 15 spares

Part D: Day 6

Finally, repeating the process for all components, we have the final materials requirements plan:

Finally, repeating the process for all components, we have the final materials requirements plan:

©The McGraw-Hill Companies, Inc., 2001

27

28

Example of MRP Logic and Product Structure Tree

B(3)

E(2)D(1)

C(4)

F(3)D(2)

A

Product Structure Tree for Assembly A Lead TimesA 2 dayB 1 daysC 2 dayD 3 daysE 2 daysF 3 day

Total Unit DemandDay 10 60 ADay 8 15 B (Spares)Day 6 20 D (Spares)



29

Material Requirements Planning System

• Based on a master production schedule, a material requirements planning system:– Creates schedules identifying the specific parts

and materials required to produce end items

– Determines exact unit numbers needed

– Determines the dates when orders for those materials should be released, based on lead times

3030

©The McGraw-Hill Companies, Inc., 2004

Firm orders from knowncustomers

Forecastsof demand

from randomcustomers

Aggregateproduct

plan

Bill ofmaterial

file

Engineeringdesign

changes

Inventoryrecord file

Inventorytransactions

Master productionSchedule (MPS)

Primary reportsSecondary reports

Planned order schedule for inventory and production control

Exception reportsPlanning reportsReports for performance control

Materialplanning(MRP

computer program)

31

Bill of Materials (BOM) FileA Complete Product Description

• Materials• Parts• Components• Production sequence • Modular BOM

– Subassemblies

• Super BOM– Fractional options

32

Inventory Records File

• Each inventory item carried as a separate file– Status according to “time buckets”

• Pegging– Identify each parent item that created demand

33

Primary MRP Reports• Planned orders to be released at a future time• Order release notices to execute the planned

orders• Changes in due dates of open orders due to

rescheduling • Cancellations or suspensions of open orders due

to cancellation or suspension of orders on the master production schedule

• Inventory status data

34

Secondary MRP Reports

• Planning reports, for example, forecasting inventory requirements over a period of time

• Performance reports used to determine agreement between actual and programmed usage and costs

• Exception reports used to point out serious discrepancies, such as late or overdue orders

35

General Format of an MRP Report

• Item Identification• Lead Time• Available Inventory• Lot size• Safety Stock• Allocated• Low-level-code:• Report date:

36

Period Week 1 2 3 4 5 6 7 8

• Gross Requirements

• Scheduled receipts

• Projected available balance

• Net requirements

• Planned order receipt

• Planned order release

37

MRP Example

A(2) B(1)

D(5)C(2)

X

C(3)

Item On-Hand Lead Time (Weeks)X 50 2A 75 3B 25 1C 10 2D 20 2

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

38

A(2)

X

Day: 1 2 3 4 5 6 7 8 9 10X Gross requirements 95

LT=2 Scheduled receipts Proj. avail. balance 50 50 50 50 50 50 50 50 50 50

On- Net requirements 45hand Planned order receipt 4550 Planner order release 45A Gross requirements 90

LT=3 Scheduled receipts Proj. avail. balance 75 75 75 75 75 75 75 75

On- Net requirements 15 hand Planned order receipt 15 75 Planner order release 15 B Gross requirements 45


On- Net requirements 20 hand Planned order receipt 20 25 Planner order release 20 C Gross requirements 45 40

LT=2 Scheduled receipts Proj. avail. balance 10 10 10 10 10

On- Net requirements 35 40 hand Planned order receipt 35 40 10 Planner order release 35 40 D Gross requirements 100

LT=2 Scheduled receipts Proj. avail. balance 20 20 20 20 20 20 20

On- Net requirements 80 hand Planned order receipt 80 20 Planner order release 80

It takes 2 A’s for each X

It takes 2 A’s for each X

39












B(1)A(2)

X

It takes 1 B for each X

It takes 1 B for each X

40

A(2) B(1)

X

C(3)












It takes 3 C’s for each A

It takes 3 C’s for each A

41

A(2) B(1)

C(2)

X

C(3)












It takes 2 C’s for each B

It takes 2 C’s for each B

42

A(2) B(1)

D(5)C(2)

X

C(3)












It takes 5 D’s for each B

It takes 5 D’s for each B

43

Calculation of Order Size in MRP

• Lot-for-lot Method

• EOQ Method

• Least Total Cost Method

• Least Unit Cost Method

44

Operation Scheduling

45

What is Scheduling?

• Last stage of planning before production occurs

• Specifies when labor, equipment, and facilities are needed to produce a product or provide a service

17-45

46

Scheduled Operations

• Process Industry– Linear programming– EOQ with non-instantaneous

replenishment• Mass Production

– Assembly line balancing• Project

– Project -scheduling techniques (PERT, CPM)

• Batch Production– Aggregate planning– Master scheduling– Material requirements

planning (MRP)– Capacity requirements

planning (CRP)

17-46

47

Objectives in Scheduling

• Meet customer due dates• Minimize job lateness• Minimize response time• Minimize completion time• Minimize time in the system

• Minimize overtime• Maximize machine or labor

utilization• Minimize idle time• Minimize work-in-process

inventory

17-47

48

Shop Floor Control (SFC)• scheduling and monitoring of day-to-day production in a

job shop• also called production control and production activity

control (PAC)• usually performed by production control department

– Loading• Check availability of material, machines, and labor

– Sequencing• Release work orders to shop and issue dispatch lists for individual

machines– Monitoring

• Maintain progress reports on each job until it is complete

17-48

49

Loading

• Process of assigning work to limited resources• Perform work with most efficient resources• Use assignment method of linear

programming to determine allocation

17-49

50

Assignment Method

1. Perform row reductions– subtract minimum value in each

row from all other row values2. Perform column reductions

– subtract minimum value in each column from all other column values

3. Cross out all zeros in matrix use minimum number of

horizontal and vertical lines

4. If number of lines equals number of rows in matrix, then optimum solution has been found. Make assignments where zeros appear– Else modify matrix

• subtract minimum uncrossed value from all uncrossed values

• add it to all cells where two lines intersect

• other values in matrix remain unchanged

5. Repeat steps 3 and 4 until optimum solution is reached

17-50

51

Assignment Method: Example

17-51

Row reductionRow reduction Column reduction Column reduction Cover all zeros Cover all zeros

55 00 11 55 33 00 11 44 33 0 1 4444 00 22 44 22 00 22 33 22 0 2 3322 11 00 11 00 11 00 00 0 1 0 055 11 00 66 33 11 00 55 33 1 0 55

Number lines Number lines number of rows so modify matrix number of rows so modify matrix

Initial PROJECTInitial PROJECT

MatrixMatrix 11 22 33 44

BryanBryan 1010 55 66 1010KariKari 66 22 44 66NoahNoah 77 66 55 66ChrisChris 99 55 44 1010

52

Assignment Method: Example (cont.)

17-52

Modify matrixModify matrix Cover all zerosCover all zeros11 00 11 22 1 0 1 200 00 22 11 0 0 2 100 33 22 00 0 3 2 011 11 00 33 1 1 0 3

Number of lines = number of rows so at optimal solutionNumber of lines = number of rows so at optimal solution

11 22 33 44


PROJECTPROJECT

11 22 33 44


PROJECTPROJECT

Project Cost = (5 + 6 + 4 + 6) X $100 = $2,100Project Cost = (5 + 6 + 4 + 6) X $100 = $2,100

53

Sequencing

17-53

Prioritize jobs assigned to a resourceIf no order specified use first-come first-served (FCFS)Other Sequencing Rules

FCFS - first-come, first-servedLCFS - last come, first servedDDATE - earliest due dateCUSTPR - highest customer prioritySETUP - similar required setupsSLACK - smallest slackCR - smallest critical ratioSPT - shortest processing timeLPT - longest processing time

54

Minimum Slack andSmallest Critical Ratio

17-54

CR = =CR = =

If CR > 1, job ahead of scheduleIf CR > 1, job ahead of scheduleIf CR < 1, job behind scheduleIf CR < 1, job behind scheduleIf CR = 1, job on scheduleIf CR = 1, job on schedule

time remainingtime remaining due date - today’s datedue date - today’s date

work remainingwork remaining remaining processing timeremaining processing time

SLACK considers both work and time remainingSLACK considers both work and time remaining

CR recalculates sequence as processing continues and CR recalculates sequence as processing continues and arranges information in ratio formarranges information in ratio form

SLACK = (due date – today’s date) – (processing time)SLACK = (due date – today’s date) – (processing time)

55

Sequencing Jobs through One Process• Flow time (completion time)

– Time for a job to flow through system

• Makespan– Time for a group of jobs to be completed

• Tardiness– Difference between a late job’s due date

and its completion time

17-55

56

Simple Sequencing Rules

17-56

PROCESSINGPROCESSING DUEDUEJOBJOB TIMETIME DATEDATE

AA 55 1010BB 1010 1515CC 22 55DD 88 1212EE 66 88

57

Simple Sequencing Rules: FCFS

17-57

FCFSFCFS STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

58

Simple Sequencing Rules: FCFS

17-58

AA 00 55 55 1010 00BB 55 1010 1515 1515 00CC 1515 22 1717 55 1212DD 1717 88 2525 1212 1313EE 2525 66 3131 88 2323

TotalTotal 9393 4848AverageAverage 93/5 = 18.60 93/5 = 18.60 48/5 = 9.6 48/5 = 9.6

FCFSFCFS STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

59

Simple Sequencing Rules: DDATE

17-59

DDATEDDATE STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

60

Simple Sequencing Rules: DDATE

17-60

CC 00 22 22 55 00EE 22 66 88 88 00AA 88 55 1313 1010 33DD 1313 88 2121 1212 99BB 2121 1010 3131 1515 1616

TotalTotal 7575 2828AverageAverage 75/5 = 15.00 75/5 = 15.00 28/5 = 5.628/5 = 5.6

DDATEDDATE STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

61

Simple Sequencing Rules: SLACK

17-61

A(10-0) – 5 = 5B(15-0) – 10 = 5C(5-0) – 2 = 3D(12-0) – 8 = 4E(8-0) – 6 = 2

SLACK considers both work and time remainingSLACK considers both work and time remainingSLACK = (due date – today’s date) – (processing time)SLACK = (due date – today’s date) – (processing time)

62


17-62

SLACKSLACK STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

63


17-63

EE 00 66 66 88 00CC 66 22 88 55 33DD 88 88 1616 1212 44AA 1616 55 2121 1010 1111BB 2121 1010 3131 1515 1616

TotalTotal 8282 3434AverageAverage 82/5 = 16.40 34/5 = 6.882/5 = 16.40 34/5 = 6.8

SLACKSLACK STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

64

Simple Sequencing Rules: SPT

17-64

SPTSPT STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

65

Simple Sequencing Rules: SPT

17-65

CC 00 22 22 55 00AA 22 55 77 1010 00EE 77 66 1313 88 55DD 1313 88 2121 1212 99BB 2121 1010 3131 1515 1616

TotalTotal 7474 3030AverageAverage 74/5 = 14.80 74/5 = 14.80 30/5 = 6 30/5 = 6

SPTSPT STARTSTART PROCESSINGPROCESSING COMPLETIONCOMPLETION DUEDUESEQUENCESEQUENCE TIMETIME TIMETIME TIMETIME DATEDATE TARDINESSTARDINESS

66

Example of Job Sequencing: Critical Ratio MethodJobs (in order Processing Due Date

of arrival) Time (days) (days hence)A 4 5B 7 10C 3 6D 1 4

Suppose you have the four jobs to the right arrive for processing on one machine


What is the CR schedule?What is the CR schedule?

No, but since there is three-way tie, only the first job or two will be on time

No, but since there is three-way tie, only the first job or two will be on time

In order to do this schedule the CR’s have be calculated for each job. If we let today be Day 1 and allow a total of 15 days to do the work. The resulting CR’s and order schedule are:CR(A)=(5-4)/15=0.06 (Do this job last)CR(B)=(10-7)/15=0.20 (Do this job first, tied with C and D)CR(C)=(6-3)/15=0.20 (Do this job first, tied with B and D)CR(D)=(4-1)/15=0.20 (Do this job first, tied with B and C)

Do all the jobs get done on time?Do all the jobs get done on time?

67

Example of Job Sequencing:Last-Come First-Served

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5B 7 10C 3 6D 1 4

Answer: Last-Come First-Served ScheduleJobs (in order Processing Due Date Flow Time

of arrival) Time (days) (days hence) (days)D 1 4 1C 3 6 4B 7 10 11A 4 5 15

No, Jobs B and A are going to be late

No, Jobs B and A are going to be late



What is the LCFS schedule?What is the LCFS schedule?Do all the jobs get done on time?Do all the jobs get done on time?

68

Simple Sequencing Rules: Summary

17-68

AVERAGEAVERAGE AVERAGEAVERAGE NO. OFNO. OF MAXIMUMMAXIMUMRULERULE COMPLETION TIMECOMPLETION TIME TARDINESSTARDINESS JOBS TARDYJOBS TARDY TARDINESSTARDINESS

69

Simple Sequencing Rules: Summary

17-69

FCFSFCFS 18.6018.60 9.69.6 3 2323DDATEDDATE 15.0015.00 5.6 3 16SLACKSLACK 16.4016.40 6.86.8 44 1616SPTSPT 14.80 6.06.0 3 16

AVERAGEAVERAGE AVERAGEAVERAGE NO. OFNO. OF MAXIMUMMAXIMUMRULERULE COMPLETION TIMECOMPLETION TIME TARDINESSTARDINESS JOBS TARDYJOBS TARDY TARDINESSTARDINESS

70

Guidelines for Selecting a Sequencing Rule

1. SPT most useful when shop is highly congested2. Use SLACK for periods of normal activity3. Use DDATE when only small tardiness values can be

tolerated4. Use LPT if subcontracting is anticipated5. Use FCFS when operating at low-capacity levels6. Do not use SPT to sequence jobs that have to be

assembled with other jobs at a later date

17-70

71

n-jobs m-machines

17-71

JOBJOB PROCESS 1PROCESS 1 PROCESS 2PROCESS 2

AA 66 88BB 1111 66CC 77 33

72

Sequencing Jobs Through Two Serial Process

17-72

Johnson’s RuleJohnson’s Rule1.1. List time required to process each job at each machine. Set List time required to process each job at each machine. Set

up a one-dimensional matrix to represent desired sequence up a one-dimensional matrix to represent desired sequence with # of slots equal to # of jobs.with # of slots equal to # of jobs.

2.2. Select smallest processing time at either machine. If that Select smallest processing time at either machine. If that time is on machine 1, put the job as near to beginning of time is on machine 1, put the job as near to beginning of sequence as possible.sequence as possible.

3.3. If smallest time occurs on machine 2, put the job as near to If smallest time occurs on machine 2, put the job as near to the end of the sequence as possible.the end of the sequence as possible.

4.4. Remove job from list.Remove job from list.5.5. Repeat steps 2-4 until all slots in matrix are filled and all jobs Repeat steps 2-4 until all slots in matrix are filled and all jobs

are sequenced.are sequenced.

73

Johnson’s Rule

17-73

JOBJOB PROCESS 1PROCESS 1 PROCESS 2PROCESS 2

AA 66 88BB 1111 66CC 77 33DD 99 77EE 55 1010

CE A BD

74

Johnson’s Rule (cont.)

17-74

A B CDE

E A D B C Process 1Process 1(sanding)(sanding)

55 1111 2020 3131 3838

E A D B C Process 2Process 2(painting)(painting)

55 1515 2323 3030 3737 4141

Idle timeIdle time

Completion time = 41Idle time = 5+1+1+3=10

75

Monitoring

• Work package– Shop paperwork that travels with a job

• Gantt Chart– Shows both planned and completed activities

against a time scale• Input/Output Control

– Monitors the input and output from each work center

17-75

76

Gantt Chart

17-76

11 22 33 44 55 66 88 99 1010 1111 1212 DaysDays

11

22

33

Today’s DateToday’s Date

Job 32BJob 32B

Job 23CJob 23C

Job 11CJob 11C Job 12AJob 12A

Fac

ilit

yF

acil

ity

Key:Key: Planned activityPlanned activity

Completed activityCompleted activity

Behind scheduleBehind schedule

Ahead of scheduleAhead of schedule

On scheduleOn schedule

77

Input/Output Control

17-77

Input/Output ReportInput/Output Report

Planned inputPlanned input 6565 6565 7070 7070 270270Actual inputActual input 00DeviationDeviation 00Planned outputPlanned output 7575 7575 7575 7575 300300Actual outputActual output 00DeviationDeviation 00BacklogBacklog 3030

PERIODPERIOD 11 22 33 44 TOTALTOTAL

20 10 5 020 10 5 0

78

Input/Output Control (cont.)

17-78

Input/Output ReportInput/Output Report

Planned inputPlanned input 6565 6565 7070 7070 270270Actual inputActual input 6060 6060 6565 6565 250250DeviationDeviation -5-5 -5-5 -5-5 -5-5 -20-20Planned outputPlanned output 7575 7575 7575 7575 300300Actual outputActual output 7575 7575 6565 6565 280280DeviationDeviation -0-0 -0-0 -10-10 -10-10 -20-20BacklogBacklog 3030 1515 00 00 00

PERIODPERIOD 11 22 33 44 TOTALTOTAL

79

Supply Chain Management:

80

Some Definitions

Supply Chain Management encompasses every effort involved in producing and delivering a final product or service, from the supplier’s supplier to the customer’s customer. Supply Chain Management includes managing supply and demand, sourcing raw materials and parts, manufacturing and assembly, warehousing and inventory tracking, order entry and order management, distribution across all channels, and delivery to the customer.

The Supply Chain Council, U.S.A.

81

Supply

Sources:plantsvendorsports

RegionalWarehouses:stocking points

Field Warehouses:stockingpoints

Customers,demandcenterssinks

Production/purchase costs

Inventory &warehousing costs

Transportation costs Inventory &

warehousing costs

Transportation costs

82

Flows in a supply chain

Customer

Information

Product

Funds

83

Philosophy of SCM

• The entire supply chain is a single, integrated entity.

• The cost, quality and delivery requirements of the customer are objectives shared by every company in the chain.

• Inventory is the last resort for resolving supply and demand imbalances.

84

Efficiency: Basis of Production Management

• Efficiency leads to lower costs• Lower cost implies

Lower Price => Greater demand => Better market growth => Higher profits => Product/ Process development => Better market share

• 1980s and 1990s: Era of achieving excellence at the firm level (JIT, TQM, TPM, BPR, ERP, etc)

• 2000s: Era of achieving excellence at the value chain level (SCM, CRM, E-Commerce, etc.)

85

Evolution of SCM

Stage 1: Vendor – Purchase – Production - Distribution – Retailer

Stage 2: Materials Management -

Logistics Management

Stage 3: Supply Chain Management

86

Why is SCM Important?

• Strategic Advantage – It Can Drive Strategy* Manufacturing is becoming more efficient* SCM offers opportunity for differentiation (Dell) or cost reduction (Wal-Mart or Big Bazaar)

• Globalization – It Covers The World* Requires greater coordination of production and distribution* Increased risk of supply chain interruption* Increases need for robust and flexible supply chains

87

Why is SCM Important?(continued)

• At the company level, supply chain management impacts* COST – For many products, 20% to 40% of total product costs are controllable logistics costs. * SERVICE – For many products, performance

factors such as inventory availability and speed of delivery are critical to customer satisfaction.

88

Conflicting Objectives in the Supply Chain

1. Purchasing• Stable volume requirements • Flexible delivery time• Little variation in mix• Large quantities

2. Manufacturing• Long run production• High quality• High productivity• Low production cost

89

Conflicting Objectives in the Supply Chain3. Warehousing

• Low inventory • Reduced transportation costs• Quick replenishment capability

4. Customers• Short order lead time• High in stock• Enormous variety of products• Low prices

90

Decision Phases in a Supply Chain

• Supply chain strategy or design• Supply chain planning• Supply chain operation

91

Process view of a supply chain

• Cycle view

• Push/pull view

92

Cycle View of Supply Chains

Customer Order Cycle

Replenishment Cycle

Manufacturing Cycle

Procurement Cycle

Customer

Retailer

Distributor

Manufacturer

Supplier

93

Customer order cycle

• Customer arrival• Customer order entry• Customer order fulfillment• Customer order receiving

94

Replenishment cycle

• Retail order trigger• Retail order entry• Retail order fulfillment• Retail order receiving

95

Manufacturing cycle

• Order arrival from the distributor, retailer, or customer

• Production scheduling• Manufacturing and shipping• Receiving at the distributor, retailer, or

customer

96

Push/Pull View of Supply Chains

• Pull processes: execution is initiated in response to a customer order

• Push processes: execution is initiated in anticipation of customer orders

97

Push/Pull View of Supply Chains

Procurement,Manufacturing andReplenishment cycles

Customer OrderCycle

CustomerOrder Arrives

PUSH PROCESSES PULL PROCESSES

98SUPPLY CHAIN DESIGN:Three Components

1. Insourcing/OutSourcing(The Make/Buy or Vertical Integration Decision) 2. Partner Selection(Choice of suppliers and partners for the chain) 3. The Contractual Relationship(Arm's length, joint venture, long-term contract,strategic alliance, equity participation, etc.)

99

Supply chain objective• Maximize overall value generated• Value strongly correlated to supply chain

profitability – the difference between the revenue generated from the customer and the overall cost across the supply chain

• Example: A customer purchasing a computer from Dell pays $ 700 (the revenue)

• Dell and other stages of the supply chain incur cost to convey information, produce the components, store them, transport them, transfer funds, etc.

100

Examples of Supply Chains

• Dell / Compaq• Toyota / GM / Ford• Milk Distribution System of NDDB• Merry-Go-Round System of NTPC• Dabbawalas of Mumbai• Amazon / Borders / Barnes and Noble

101O

rde

r S

ize

Time

Source: Tom Mc Guffry, Electronic Commerce and Value Chain Management, 1998

CustomerDemand

CustomerDemand

Retailer OrdersRetailer OrdersDistributor OrdersDistributor Orders

Production PlanProduction Plan

The Dynamics of the Supply Chain

102

The Dynamics of the Supply Chain O

rde

r S

ize

Time

Source: Tom Mc Guffry, Electronic Commerce and Value Chain Management, 1998

CustomerDemand

CustomerDemand

Production PlanProduction Plan

1 aggregate planning. 2 master scheduling material requirements planning order scheduling weekly...

Documents