BM2: OPERATIONS MANAGEMENT & TQM
FIRST SEMESTER: MIDTERM
PRODUCT AND SERVICE DESIGN
Major Factors in Design Strategy
1. Cost
2. Quality
3. Time-to-Market
4. Customer Satisfaction
5. Competitive Advantage
Idea Generation (Sources of Ideas for New or Redesigned Products and Services)
1. Internal Sources
a. Research and Development (R&D)
 Organized efforts to increase scientific knowledge or product
innovation.
2. External Sources
a. Customers
b. Suppliers
c. Competitors
 Reverse engineering
o Dismantling and inspecting a competitor’s product to
discover product improvements.
Design Considerations
1. Product/Service Life Cycle
 Incubation, growth, maturity, saturation, and decline.
2. Standardization
 Extent to which there is an absence of variety in a product, service or
process.
 Standardized products are immediately available to customers.
 Mass Customization
o A strategy of producing standardized goods or services but
incorporating some degree of customization.
o Tactics in Mass Customization
a. Delayed Differentiation
 A postponement tactic, producing but not quite
completing a product or service until customer
preferences or specifications are known.
b. Modular Design
 A form of standardization in which component parts are
subdivided into modules that are easily replaced or
interchanged.
BM2: OPERATIONS MANAGEMENT & TQM
FIRST SEMESTER: MIDTERM
PRODUCT AND SERVICE DESIGN
It allows: (1) easier diagnosis and remedy of failures;
(2) easier repair and replacement; (3) simplification of
manufacturing and assembly.
c. Reliability
 The ability of a product, part, system to perform its
intended function under a prescribed set of condition.
 Failure is a situation in which a product, part,
or system does not perform as intended.
 Normal Operating Condition is a set of
conditions under which an items reliability is
specified.
d. Robust Design
 Design results in products or services that can function
over a broad range of condition.

Designing for Production
1. Concurrent Engineering
 Bringing together of engineering design and manufacturing personnel early
in the design phase.
2. Computer Aided Design (CAD)
 Designing products using computer graphics.
 Advantages
o Increases productivity of designers, 3 to 10 times.
o Creates a database for manufacturing information on product
specifications.
o Provides possibility of engineering and cost analysis on proposed
designs.
Designing for Manufacturing
3. Recycling
 Recovering materials for future use.
 Reasons for recycling
o Cost savings
o Environment concerns
o Environment regulations
4. Remanufacturing
 Refurbishing used products by replacing worn-out or defective components.
BM2: OPERATIONS MANAGEMENT & TQM
FIRST SEMESTER: MIDTERM
PRODUCT AND SERVICE DESIGN
Reliability
1. Independent Events
 Events whose occurrence or non-occurrence do not influence each other.
2. Redundancy
 The use of backup components to increase reliability.
Rule #1
If two or more events are independent and success is defined as the probability
that all these events occur, then the probability of success is equal to the product of the
probabilities of the events.
Rule #2
If two events are independent and success is defined as the probability that at least
one of the events will occur, the probability of success is equal to the probability of either
one plus 1.00 minus the probability multiplied by the other probability.
Rule #3
If three events are involved and success is defined as the probability that at least
one of them occurs, the probability of success is equal to the probability of the first one
(any of the events), plus the product of 1.00 minus that probability and the probability of
the second event (any of the remaining events), plus the product of 1.00minus each of
the first two probabilities and the probability of the third event, and so on. This rule can
be expanded to cover more than three events.
BM2: OPERATIONS MANAGEMENT & TQM
FIRST SEMESTER: MIDTERM
PRODUCT AND SERVICE DESIGN
PRACTICE SET
Rule #1
.92 x .89 = 0.8188
Rule #2
.92 + (1 - .92) x .80  .92 + 0.08 x .80  .92 + 0.064 = 0.984
Rule #3
1 – [(1 - .90)(1 - .85)(1 - .70)]  1 – [(0.1)(0.15)(0.3)]  1 – 0.0045 = 0.9955
1 – [(1 - .93)(1 - .80)(1 - .79)]  1 – [(0.07)(0.2)(0.21)]  1 – 0.00294 = 0.9970
Total System Reliability
0.8188 x 0.984 x 0.9955 x 0.9970 = 0.7996
Rule #1
.92 x .92 x .89 = 0.7532
Rule #2
.93 + (1 - .93) x .80  .93 + 0.07 x .80  .93 + 0.056 = 0.986
Rule #3
1 – [(1 - .90)(1 - .85)(1 - .70)(1 - .70)]  1 – [(0.1)(0.15)(0.3)(0.3)]  1 – 0.00135 = 0.9986
Total System Reliability
0.7532 x 0.986 x 0.9986 = 0.7416