POM Review forComprehensive

Examination

PRODUCTION & OPERATIONS MANAGEMENT

1. Introduction2. Forecasting3. Design of Goods & Services, Processes &

Systems4. Facilities & Capacity Planning5. Inventory Planning and Supply Chain

Management6. Materials, Manufacturing & Enterprise

Resource Planning (MRP, ERP)8. Quality Management, TQM, ISO9. Quality Improvement, JIT, Lean, Six Sigma10. Project Management11. Service Operations Management12. Operations Strategy, Sustainability, Social

& Ethics13. Plant Tour/Group Project * 7. Midterm Exam and 14 Final Exam

REVIEW OUTLINE

POMSyllabus

INTRODUCTION

ORGANIZATION

ORGANIZATION

Figure 1.1

ORGANIZATION

POM DEFINITION

WHAT IS OPERATIONS MANAGEMENT?

Production is the creation of goods and services

Operations management (OM) is the set of activities that create value in the form of goods and services by transforming inputs into outputs

Design of Goods & Services, Processes &

Systems

IPO Model

SIPOC Model

PRODUCTIVITY CHALLENGE

Productivity is the ratio of outputs (goods and services) divided by the inputs (resources such

as labor and capital)

The objective is to improve productivity!

Important Note!Production is a measure of output only

and not a measure of efficiency

▶ Measure of process improvement

▶ Represents output relative to input

▶ Only through productivity increases can our standard of living improve

PRODUCTIVITY

Productivity =Units produced

Input used

PRODUCTIVITY CALCULATIONS

Productivity =Units produced

Labor-hours used

= = 4 units/labor-hour1,000

250

Labor Productivity

One resource input single-factor productivity

MULTI-FACTOR PRODUCTIVITY

OutputLabor + Material + Factory OHProductivity =

► Also known as total factor productivity► Output and inputs are often expressed in

dollars

Multiple resource inputs multi-factor productivity

COLLINS TITLE PRODUCTIVITY

Staff of 4 workers 8 hrs/day 8 titles/dayPayroll cost = $640/day Overhead = $400/day

Old System:

14 titles/day Overhead = $800/day

New System:

8 titles/day

$640 + 400

14 titles/day

$640 + 800

=Old multifactor

productivity

=New multifactor

productivity

= .0077 titles/dollar

= .0097 titles/dollar

IMPROVING PRODUCTIVITY AT STARBUCKS

A team of 10 analysts continually look for ways to shave time. Some improvements:

Stop requiring signatures on credit card purchases under $25

Saved 8 seconds per transaction

Change the size of the ice scoop

Saved 14 seconds per drink

New espresso machines Saved 12 seconds per shot

Operations improvements have helped Starbucks increase yearly revenue per outlet by $250,000 to $1,000,000 in seven years.

Productivity has improved by 27%, or about 4.5% per year.

Forecasting

Copyright ©2013 Pearson Education

FORECASTING METHODS

14 - 22

Naïve Method

The forecast for the next period is the demand for the current period

Moving Average Method

Weighted Moving Average Method

Exponential Smoothing Method

Linear Regression Method

MOVING AVERAGE METHOD

Compute a three-week moving average forecast for the arrival of medical clinic patients in week 4. The numbers of arrivals for the past 3 weeks were:

Week Patient Arrivals1 4002 3803 411

14 - 23

WEIGHTED AVERAGE METHOD

Compute the forecast for the arrival of medical patients in week 4 using the weighted average method. The numbers of arrivals were as follows:

Week

Patient Arrivals Weight

1 400 20%2 380 30%3 411 50%

14 - 24

EXPONENTIAL SMOOTHING METHOD

Compute the forecast for the arrival of patients in week 4 using the exponential smoothing method. The smoothing constant is α = 0.10:

Week

Patient Arrivals

Previous Forecast

1 400

2 3803 411 415

FNew = FPrevious + α (Actual – FPrevious)

LINEAR REGRESSION METHOD

Compute forecast for week 7 using the linear regression method.

Week Patient Arrivals1 4002 3803 4114 4155 4216 427

Y = A(X) + BWhere A = slope, B = Y-intercept

LINEAR REGRESSION METHOD

Formula

Result

ΣX 21

ΣY 2,454

n 6

ΣX2 91

ΣY2 1,005,116

ΣXY 8,720

LINEAR REGRESSION METHOD

nΣXY – ΣXΣY 6(8720)-(21)(2,454) 786A = ------------- = ---------------------- = ----- = 7.485714 nΣX2 – (ΣX)2 6(91) – (21)2 105

ΣY – AΣX 2,454 – (7.485714)(21) 2296.8B = ---------- = --------------------------- = -------- = 382.8 n 6 6

Y7 = A(X7) + B = 7.485714 (7) + 382.8 = 435.2

LINEAR REGRESSION METHOD

nΣXY – ΣXΣY R = -------------------------------------- = 0.828103

REGRESSION COEFFICIENT

0 < /r/ < 0.3 = Weak Correlation 

.3 < /r/ < 0.7 = Moderate Correlation 

/r/ > 0.7 = Strong Correlation

Facilities PlanningCopyright ©2013 Pearson Education

11- 030

Proximity to the MarketLocation of the Plant, Warehouse and Offi ceSize of the Plant, Warehouse and Offi ceConstruction and Renovation CostsEquipment, Furniture and Fixtures RequiredOrganization and Manpower RequirementsPurchasing of Equipment, Furniture & FixturesSupply Chain for Raw Materials and

ConsumablesEnvironmental ConditionsCommunity and Social Responsibility

KEY DECISIONS IN FACILITIES PLANNING

Site Selection

Site Selection in Physical Distribution

Inventory Planning

Economic Order Quantity

A local company expects to sell 9000 tires of a certain tire next year, Annual carrying cost is P640 per tire and ordering cost is P3,000. The distributor operates 288 days a year.

a. What is the EOQ?b. How many times does the store have

to re-order per year?c. What is the total annual cost, if the

EOQ is ordered?

EOQ

a. Economic Order Quantity

b. Optimal No. of Orders Per YearN = D = 9,000

EOQ 290N = 31 Orders/Year

Given:K = Ordering Cost 3,000.00 PesosH = Carrying Cost 640.00 Pesos/YrD = Demand 9,000 Units/YrWorking Days 288 DaysDaily Demand = Consumption 31.25 Units/Day

EOQ = 2 D K = 2(9,000)(3,000)H 640

= 54,000,000 = 84,375640

EOQ = 290 Units

ξ ξξ ξ

EOQGiven:K = Ordering Cost 3,000.00 PesosH = Carrying Cost 640.00 Pesos/YrD = Demand 9,000 Units/YrWorking Days 288 DaysDaily Demand = Consumption 31.25 Units/Day

c. Minimum Total Annual Inventory CostTotal Cost = Ordering + Holding

Tc = To + Th

+ H Q/2+ (640)(290)

2+ 92,952

Pesos Per Year92,952

D K/Q (3000)(9000)

290

185,903.20

Quality Management

Quality Management

A time study in an assembly line yielded the following observation times and the performance rating is 1.13 seconds.

a. Using an allowance of 20% on job time, determine the appropriate standard time for the operation.

b. Develop a quality control chart for the process, using six sigma limits.

Observation Time (secs.)

1 1.122 1.153 1.164 1.125 1.156 1.187 1.148 1.149 1.19

Given allowance 20%

Assembly Line

Quality Management

a. Using an allowance of 20% on job time, determine the appropriate standard time for the operation.

Observation Time (secs.)1 1.122 1.153 1.164 1.125 1.156 1.187 1.148 1.149 1.19

Mean 1.15Allowance (20%) 0.23Standard Time Lower limit = 0.92 Upper limit = 1.38

Six Sigma Limits

6 Sigma Limits

4 Sigma Limits

2 Sigma Limits

99.73%

95.45%

68.27%

Six Sigma Control Chart

Observation Time (secs.)1 1.122 1.153 1.164 1.125 1.156 1.187 1.148 1.149 1.19

Mean 1.15Standard Deviation 0.0240Six-Sigma Limits Lower limit = 1.08 Upper limit = 1.22

1.221.08 1.15

b. Develop a quality control chart for the process, using six sigma limits.

Project Management

Garage

BedRoom1

Master’s Bedroom

Living Room

Kitchen

BedRoom2

3 – Bedroom Bungalow (One Storey) House

BUILDING A HOUSE

Watch Video

BUILDING A HOUSE RECORD

4 Hours, 18 Minutes700 PeopleCement Mixer and CranesSkilled Workers w/ Power ToolsPre-Fabricated Wood FramingSan Diego California Team A vs Team B

will attempt to break the world record.

FinishStart

A

B

C

D

E

F

G

H

I

J

KA

—B

—C

AD

BE

BF

AG

CH

DI

AJ

E,G,HK

F,I,J

Immediate Predecessor

02 - 47

CRITICAL PATH METHOD

Duration (Days)

GANTT CHART

FinishStart

A

B

C

D

E

F

G

H

I

J

KPath Time (days)

A-I-K 33A-F-K 28A-C-G-J-K 67B-D-H-J-K 69B-E-J-K 43

Paths are the sequence of activities between a

project’s start and finish.

02 - 49

CRITICAL PATH METHOD

02 - 50

Latest finish time

Latest start time

Activity

Duration

Earliest start time

Earliest finish time

0

2

12

14

A

12

CRITICAL PATH METHOD

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 63 690 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

PERT/CPM

S = 0

S = 2

S = 26

S = 0

S = 36

S = 2

S = 2

S = 41 S = 0

S = 0 S = 0

The critical path is B–D –H

–J - K with a project

duration of 69 days.