Mgt 3325 - Home   Spring 2010   Email to Dr. Lyons     PatLyons Home
[ Calendar12:20 | 1:25 | Class Participation AI | App of OM | Table of Contents | Search
[ Ch 1 | 2 | 3 | 4 | 5 | 6 | 6S | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | | HW1 | 2 | 3 | 4 | | Career1| 2 | 3 | 4 ]
[
SJU | TCB | CareerCenter | StudentInfo | CareerLinks | Internships ] [NYC Teaching Fellows] [ SJU Closing ] [H1N1SelfAssessment]


Ch 17 - Maintenance and Reliability

  1. Reliability            (p670)
    1. Def - Reliability - probability that a product (or component) will function properly for a specified time under stated conditions.
      1. 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.
      2. Example 1 - National Bank loan processing - p658 - 3 clerks
        R s = R1 * R2 * R3 = 0.90 * 0.80 * 0.99 = 0.71
    2. *Redundancy - a system can improve its reliability by adding redundant components in parallel.
      1. 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
      2. 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
  2. Maintenance              (p674)
    1. Def - Maintenance - activities involved in keeping equipment (or other assets) in acceptable operating condition.
    2. Types of Maintenance
          Preventive - before failure
          Breakdown - after failure
    3. *Objective of Maintenance - minimize total cost of:
          Preventive maintenance
          Breakdown maintenance
          Equipment downtime
          Idle labor
          Customer dissatisfaction
    4. Analysis of Maintenance
      1. Preventive maintenance - Example 4 - Should CPA firm sign contract for preventive maintenance of computer? - p677.
      2. *Breakdown maintenance - use Waiting-Line Models
  3. Single-Channel Queuing Model (Module D - page 754)
    1. Arrival Characteristics
      1. Size of source population – infinite
      2. Pattern of arrivals - Poisson distribution.
      3. Behavior of arrivals – patient
    2. Waiting Line Characteristics
      1. Length of line – infinite
      2. Discipline - first-in, first-out (FIFO)
    3. Service Characteristics
      1. Service time - random with exponential distribution.
      2. Configurations - single-channel, single-phase (Fig D.3, p757)
    4. Relationships   (p761)
      1. Ls - average number of units in system (waiting and in service)
                Ls = l /( µ - l )           [ = lambda/(mu - lambda) ]
      2. Ws - average time in system (waiting and in service)
                Ws = 1 /( µ - l )
                 [ = 1/(mu - lambda) ]
    5. Golden Muffler Shop, p761. What is cost with one mechanic?
      1. l = average arrival rate of customers = 2/hr      [ lambda ]  
        µ = average service rate = 3/hr                            [ mu ]
      2. Ws - average time in system (waiting and in service)
                Ws = 1 /( µ - l ) = 1 /(3-2) = 1 hr
      3. 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.
      4. Labor cost = $56/day
      5. *Total cost = Customer time cost + Labor cost
                = 160 + 56 = $216
                            Do assigned HW
  4. Multiple-Channel Queuing Model         (p763)
    1. Arrival Characteristics - same as single-channel
    2. Waiting Line Characteristics - same as single-channel
    3. Service Characteristics
      1. Service Time - same as single-channel
      2. Configurations - multichannel, single-phase
    4. Relationships - see Table D.4, p763
    5. Golden Muffler Shop. What is cost with two mechanics?
      1. Ws - average time in system (waiting and in service)
                Ws = 0.375 hr                (see p764)
      2. Customer time cost = 16 ($10/hr) Ws = $60
      3. Labor cost = 2 ($56/day) = $112
      4. *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 .)