Production & Operations Management Comprehensive Exam Review February 8, 2015 San Beda College Graduate School of Business

1. POM Review for
Comprehensive
Examination
PRODUCTION &
OPERATIONS
MANAGEMENT

2. 1. Introduction
2. Forecasting
3. Design of Goods & Services, Processes & Systems
4. Facilities & Capacity Planning
5. Inventory Planning and Supply Chain Management
6. Materials, Manufacturing & Enterprise Resource
Planning (MRP, ERP)
8. Quality Management, TQM, ISO
9. Quality Improvement, JIT, Lean, Six Sigma
10. Project Management
11. Service Operations Management
12. Operations Strategy, Sustainability, Social & Ethics
13. Plant Tour/Group Project
* 7. Midterm Exam and 14 Final Exam
REVIEW OUTLINE
POM
Syllabus

3. INTRODUCTION

4. ORGANIZATION

5. ORGANIZATION

6. Figure 1.1
ORGANIZATION

7. 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

8. Design of Goods & Services,
Processes & Systems

9. IPO Model

10. SIPOC Model

12. 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

14. ▶ 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

15. PRODUCTIVITY CALCULATIONS
Productivity =
Units produced
Labor-hours used
= = 4 units/labor-hour
1,000
250
Labor Productivity
One resource input  single-factor productivity

16. MULTI-FACTOR PRODUCTIVITY
Output
Labor + Material + Factory OH
Productivity =
► Also known as total factor productivity
► Output and inputs are often expressed in
dollars
Multiple resource inputs  multi-factor productivity

17. COLLINS TITLE PRODUCTIVITY
Staff of 4 workers 8 hrs/day 8 titles/day
Payroll 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

18. 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.

19. Forecasting

21. FORECASTING METHODS
Copyright ©2013 Pearson Education
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

22. 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 Arrivals
1 400
2 380
3 411
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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%
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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
2 380
3 411 415
FNew = FPrevious + α (Actual – FPrevious)

25. LINEAR REGRESSION METHOD
Compute forecast for week 7 using the linear regression
method.
Week Patient Arrivals
1 400
2 380
3 411
4 415
5 421
6 427
Y = A(X) + B
Where A = slope, B = Y-intercept

26. LINEAR REGRESSION METHOD
Formula Result
ΣX 21
ΣY 2,454
n 6
ΣX2 91
ΣY2 1,005,116
ΣXY 8,720

27. LINEAR REGRESSION METHOD
nΣXY – ΣXΣY 6(8720)-(21)(2,454) 786
A = ------------- = ---------------------- = ----- = 7.485714
nΣX2 – (ΣX)2 6(91) – (21)2 105
ΣY – AΣX 2,454 – (7.485714)(21) 2296.8
B = ---------- = --------------------------- = -------- = 382.8
n 6 6
Y7 = A(X7) + B = 7.485714 (7) + 382.8 = 435.2

28. LINEAR REGRESSION METHOD
nΣXY – ΣXΣY
R = -------------------------------------- = 0.828103
√(𝒏𝚺𝑿 𝟐 – (𝚺𝑿) 𝟐)(𝒏𝚺𝒀 𝟐 – (𝚺𝒀) 𝟐)
REGRESSION COEFFICIENT
0 < /r/ < 0.3 = Weak Correlation
.3 < /r/ < 0.7 = Moderate Correlation
/r/ > 0.7 = Strong Correlation

29. Facilities Planning
Copyright ©2013 Pearson Education
11- 030

30.  Proximity to the Market
 Location of the Plant, Warehouse and Office
 Size of the Plant, Warehouse and Office
 Construction and Renovation Costs
 Equipment, Furniture and Fixtures Required
 Organization and Manpower Requirements
 Purchasing of Equipment, Furniture & Fixtures
 Supply Chain for Raw Materials and Consumables
 Environmental Conditions
 Community and Social Responsibility
KEY DECISIONS IN FACILITIES PLANNING

31. Site Selection

32. Site Selection in Physical Distribution

35. Inventory Planning

36. 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?

37. EOQ
a. Economic Order Quantity
b. Optimal No. of Orders Per Year
N = D = 9,000
EOQ 290
N = 31 Orders/Year
Given:
K = Ordering Cost 3,000.00 Pesos
H = Carrying Cost 640.00 Pesos/Yr
D = Demand 9,000 Units/Yr
Working Days 288 Days
Daily Demand = Consumption 31.25 Units/Day
EOQ = 2 D K = 2(9,000)(3,000)
H 640
= 54,000,000 = 84,375
640
EOQ = 290 Units
√ √
√ √

38. EOQ
Given:
K = Ordering Cost 3,000.00 Pesos
H = Carrying Cost 640.00 Pesos/Yr
D = Demand 9,000 Units/Yr
Working Days 288 Days
Daily Demand = Consumption 31.25 Units/Day
c. Minimum Total Annual Inventory Cost
Total Cost = Ordering + Holding
Tc = To + Th
+ H Q/2
+ (640)(290)
2
+ 92,952
Pesos Per Year
92,952
D K/Q
(3000)(9000)
290
185,903.20

39. Quality Management

40. 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 1.12
2 1.15
3 1.16
4 1.12
5 1.15
6 1.18
7 1.14
8 1.14
9 1.19
Given allowance 20%
Assembly Line

41. Quality Management
a. Using an allowance of 20% on
job time, determine the
appropriate standard time for
the operation.
Observation Time (secs.)
1 1.12
2 1.15
3 1.16
4 1.12
5 1.15
6 1.18
7 1.14
8 1.14
9 1.19
Mean 1.15
Allowance (20%) 0.23
Standard Time
Lower limit = 0.92
Upper limit = 1.38

42. Six Sigma Limits
6 Sigma Limits
4 Sigma Limits
2 Sigma Limits
99.73%
95.45%
68.27%

43. Six Sigma Control Chart
Observation Time (secs.)
1 1.12
2 1.15
3 1.16
4 1.12
5 1.15
6 1.18
7 1.14
8 1.14
9 1.19
Mean 1.15
Standard Deviation 0.0240
Six-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.

44. Project Management
Garage
Bed
Room1
Master’s
Bedroom
Living
Room
Kitchen
Bed
Room2
3 – Bedroom Bungalow (One Storey) House
BUILDING
A HOUSE
Watch Video

45. BUILDING A HOUSE RECORD
4 Hours, 18 Minutes
700 People
Cement Mixer and Cranes
Skilled Workers w/ Power Tools
Pre-Fabricated Wood Framing
San Diego California Team A vs Team B will
attempt to break the world record.

46. FinishStart
A
B
C
D
E
F
G
H
I
J
K
A —
B —
C A
D B
E B
F A
G C
H D
I A
J E,G,H
K F,I,J
Immediate
Predecessor
02 -
47
CRITICAL PATH METHOD
Duration
(Days)

47. GANTT CHART

48. FinishStart
A
B
C
D
E
F
G
H
I
J
K
Path Time (days)
A-I-K 33
A-F-K 28
A-C-G-J-K 67
B-D-H-J-K 69
B-E-J-K 43
Paths are the sequence of activities
between a project’s start and finish.
02 -
49
CRITICAL PATH METHOD

49. 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

50. 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.