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Ch 17 - Maintenance and Reliability
- Reliability
(p670)
- Def - Reliability - probability that a product (or component) will function
properly for a specified time under stated conditions.
- If a system consists of a series of three independently operating components,
then the system reliability, R s , is:
R s = R1 * R2 * R3,
where R i is the reliability of the i-th component.
- Example 1 - National Bank loan processing - p658 - 3 clerks
R s = R1 * R2 * R3 = 0.90 * 0.80 * 0.99 = 0.71
- *Redundancy - a system can improve its reliability by adding redundant components
in parallel.
- If component 1 has an identical backup, then the reliability, R b, 1 ,
of the two components is:
R b, 1 = Prob(first component works) + Prob(first component fails) *
Prob(backup works)
= R1 + (1.0 - R1)
* R1
- Example 3 - National Bank provides identical redundancy for clerks 1 and 2. What
is overall reliability?
R b, 1 = 0.90 + 0.10 * 0.90 = 0.99
R b, 2 = 0.80 + 0.20 * 0.80 = 0.96
R s = R b, 1 * R b,
2 * R3 = 0.99 * 0.96 * 0.99 = 0.94
- Maintenance
(p674)
- Def - Maintenance - activities involved in keeping equipment (or other assets) in
acceptable operating condition.
- Types of Maintenance
Preventive - before failure
Breakdown - after failure
- *Objective of Maintenance - minimize total cost of:
Preventive maintenance
Breakdown maintenance
Equipment downtime
Idle labor
Customer dissatisfaction
- Analysis of Maintenance
- Preventive maintenance - Example 4 - Should CPA firm sign
contract for preventive maintenance of computer? - p677.
- *Breakdown maintenance - use Waiting-Line Models
- Single-Channel Queuing Model (Module D - page 754)
- Arrival Characteristics
- Size of source population infinite
- Pattern of arrivals - Poisson distribution.
- Behavior of arrivals patient
- Waiting Line Characteristics
- Length of line infinite
- Discipline - first-in, first-out (FIFO)
- Service Characteristics
- Service time - random with exponential distribution.
- Configurations - single-channel, single-phase (Fig D.3, p757)
- Relationships
(p761)
- Ls - average number of units in system (waiting and in service)
Ls = l /( µ - l )
[ = lambda/(mu - lambda) ]
- Ws - average time in system (waiting and in service)
Ws = 1
/( µ - l )
[ =
1/(mu - lambda) ]
- Golden Muffler Shop, p761. What is cost with one mechanic?
l = average
arrival rate of customers = 2/hr
[ lambda ]
µ = average service rate = 3/hr
[ mu ]
- Ws - average time in system (waiting and in service)
Ws = 1
/( µ - l ) = 1 /(3-2)
= 1 hr
- Cost of single customer time is $10/hr (waiting and in service)
In 8 hr day, there are 2*8=16 customers.
Customer time cost = 16 ($10/hr) Ws
= $160
Note: Use Ws
because it is more meaningful than Wq.
- Labor cost = $56/day
- *Total cost = Customer time cost + Labor cost
= 160 + 56 = $216
Do assigned HW
Multiple-Channel Queuing Model
(p763)
- Arrival Characteristics - same as single-channel
- Waiting Line Characteristics - same as single-channel
- Service Characteristics
- Service Time - same as single-channel
- Configurations - multichannel, single-phase
- Relationships - see Table D.4, p763
- Golden Muffler Shop. What is cost with two mechanics?
- Ws - average time in system (waiting and in service)
Ws = 0.375 hr
(see p764)
- Customer time cost = 16 ($10/hr) Ws = $60
- Labor cost = 2 ($56/day) = $112
- *Total cost = Customer time cost + Labor cost
= 60 + 112 = $172
It is better to have two mechanics than one.
(This
page was last edited on
April 19, 2010
.)