Review for Exam II—Nov

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							Review for Exam II

  This exam will be administered
  Tuesday, October 16, 2012, usual
  time and place
Hws11-34
                                              answer key



                               U.S. Distributors
    Plants         7. Texas       8. Virginia    9. Ohio   Supply    Shipped
 1. Germany            0               0            0       5.2         0
 2. Belgium            0               0            0       6.3         0
    3. Italy          2.1              0            0       4.5        2.1
   Shipped            2.1              0            0

                                  Factories                                                        U.S. Distributors
     Plant          4. PR         5. Mexico    6. Panama   Shipped                 Plant    7. Texas    8. VA       9. Ohio   Shipped
 1. Germany           0              5.2            0        5.2                  4. PR         0         0            0         0
 2. Belgium           0               0            6.3       6.3                5. Mexico       0         0           5.2       5.2
    3. Italy          0               0             0        2.1               6. Panama        0        3.7          2.6       6.3
   Demand            2.1             3.7           7.8                          Demand         2.1       3.7          7.8
   Shipped            0              5.2           6.3                           Shipped       2.1       3.7          7.8

Transshipments:
                      4.PR            0
                   5. Mexico          0
                  6. Panama           0

                   Cost =           27120
        HW #10-2
Homework Problem #10-2                                      answer key

                                         Average              Total
                             Raw                   Unit Cost
                          Materials      Inventory    ($)    Value
                             oak           8000        6     48000
                             pine          4500        4     18000
                            brass
                          fixtures        1200        8      9600
                            stains        3000        2      6000
                           joiners        900         1       900
                                                    Total    82500
                           Work in
                           Process
                           frames          200        30      6000
                           drawers         400        10      4000
                            panels         600        50     30000
                            chests         120       110     13200
                            tables          90        90      8100
                                                    Total    61300
                           Finished
                            Goods
                            chests         300      500     150000
                         coffee tables     200      350     70000
                                                    Total   220000

                                         Cost of goods sold 370000
                                                  =           0
HW 10-6

                                                             answer
      Homework Problem #10-6                                 key

                                   Supplier 1   Supplier 2
              Cost of goods sold   8360000      14800000
                Raw materials       275000       870000
               Work-in-process       62000       550000
               Finished goods        33000       180000

              Inventory turns =      22.6          9.3
              Weeks of supply =      2.3           5.6
                                     Best
HW 14-22
  Homework Problem #S14-22                            answer key
                                                                        Nutrient Contributions
                                                         Calories Fat Cholesterol Iron Calcium Protein Fiber
  Breakfast Food        Serving        Unit     Cost ($)           g     mg        mg      mg     g      g
      Bran cereal =        0       cup            0.18      90     0       0        6      20     3      5
       Dry cereal =        0       cup            0.22     110     2       0        4      48     4      2
       Oatmeal =         1.025 cup                0.10     100     2       0        2      12     5      3
       Oat bran =          0       cup            0.12      90     2       0        3       8     6      4
       Egg =               0       egg            0.10      75     5     270        1      30     7      0
       Bacon =             0       slice          0.09      35     3       8        0       0     2      0
       Orange =            0       orange         0.40      65     0       0        1      52     1      1
       Milk - 2% =       1.241 cup                0.16     100     4      12        0     250     9      0
       Orange juice =      0       cup            0.50     120     0       0        0       3     1      0
       Wheat toast =     2.975 slice              0.07      65     1       0        1      26     3      3
                            Nutritional Requirements       420    20      30        5     400    20     12
                                     Nutritional Levels    420    10    14.90     5.02    400  25.22 12

    Cost Per Meal = $   0.509
Exam Format
 45-50 multiple choice
 3 problems
 Closed-book
 Closed-notes
 Closed-neighbor
 BRING---pencil, calculator, orange
  scantron sheet
Exam details
 Tuesday,    October 16, 2012
 Will start the exam at 2:00 p.m.
Exam Coverage
 Chapter10, Chapter 11, Supplement to
 Ch 11, Chapter 13, Supplement to
 Ch13, and Chapter 15-second half
   LPproblems in the supplement to Chapter
   14, but not the content of Chapter 14—will
   cover that later.
Typical problems—see
Practice Exam II
 Inventory with Independent Demand
 Problems involving calculation of
  inventory turns and days of supply
 Production Scheduling Problem
 Transportation problem
 LP formulation problem
 Interpretation of LP SENSITIVITY
  output
More typical discussion
problems
 Transshipment    problem
 Linear programming formulation
 Be able to draw schematics of
  mainframe/glass architecture,
  client/server architecture and N-tier
  architecture
Chapter 15 – ERP
 Inventoryfor Dependent Demand will
  NOT be covered….
 Exam coverage of this chapter starts on
  page 700
What were five motivations for
transitioning from mainframes
 Absence  of data integration
 36 month backlogs at centralized MIS
  shops
 Idle CPU cycles on desktops
 Mainframes were expensive bottlenecks
 Support for Internet and thin clients
 Quicker, cheaper development times
  through REUSE
What is the information
architecture modern ERP
systems are currently based
on?

   Mainframe/glass    house
   Client/server
         distributed
   N-tier
   None of these
Every application software
package consists of
1.    Presentation management
     component
2.    Business logic management
     component
3.    Data management component
4.    All of the above
5.    1 and 2 only
ERP
 Issoftware that organizes and
  manages a company’s business
  processes by sharing info across
  functional areas
 Large caps have been there and
  done that—transitioned to ERP
 Mid and small caps are getting there
 The road to implementation has been
  rough
More ERP
 Based  on an N-tier distributed
  architecture
 Not on mainframe glasshouse
•Advantages of N-tier
architecture
 Provides  for data integration
 Better usage of MIPS on both PCs
  and servers
 Solves the 36-month backlog of the
  centralized MIS shop
 Enables a better career path for the
  MIS professional
N-Tier distributed
architecture
  Is decentralized or centralized, or
   some combination of these
   (which?)
  Utilizes thick clients or thin
   clients (which?)
ERP Modules
 Sales &           Project
  distribution       management
 Production &      Accounting and
  Materials          controlling/finance
  Management
                    Supply chain
 Quality
  management         management
 Human resource    Customer
  management         relationship
                     management
ERP Terms
 Best-of-breed
 Collaborativeproduct commerce
 Customer relationship management
 Supply chain management
 XML
Re-engineered Computer
Architectures
 Startedwith mainframe/glasshouse
 Migrated to client/server
 Evolved to N-tier distributed
Why did such re-engineering
occur?
 There was no data integration
 MIPs on mainframes were hugely
  expensive and very much in demand
 MIPs on PCs were idle 95% of the time
  and extremely cheap
 Backlogs for MIS shops were at 36
  months
 Developing new applications were slow
  and expensive
Distributed architectures
solved these problems
 Dataresides behind a single database
 engine
Components of any Software
Application
Components in brief
Mainframe Architecture (circa
1993)
         Mainframe Computer
Problems with Mainframe
Architecture
 Absence of data integration, resulting in little
  enterprise visibility
 The applications are maintainable only by the
  centralized MIS shop, which is overloaded,
  resulting in 36 month lead times to get
  revisions effected
 Every application had to be built from scratch,
  line-by-line, resulting in large cost and long
  lead times to create new applications
More problems with Mainframe
Architecture
 No  reuse was possible
 These mainframe apps were accessed
  on networked PC’s via IBM 3278
  terminal emulation software that was
  completely incompatible with the
  windows GUI applications—meaning no
  cut and paste
 Mainframes were computational
  bottlenecks
 Desktop PCs sat idle 99% of the time
First solution: Client/server
architecture
Clients (PM, BL)




                      Server (DM)


                                    Database
These were known as thick
clients
 Because   they contained both the
  presentation management (PM) and the
  business logic (BL) components of the
  application
 Notice how the application is distributed
  across the network, residing in two
  computing boxes—the client or desktop
  and the server
First solution: Client/server
architecture
Thick Clients (PM, BL)




                         Server (DM)


                                       Database
Advantages of Client/server
architecture
 AllData are all accessible behind the
  Server which runs the data
  management portion of the
  application—usually an Oracle
  Database engine
 Now the marketing guy can see where
  his customer’s job is, and whether the
  customer is current with his payments,
  among other ‘things’
Advantages of client/sever
architecture
 The IT professional could sit shoulder-to-
  shoulder with the end-user and develop
  applications as well as make changes to
  existing software rapidly, without a 36 month
  backlog
 For new applications, there were huge reuse
  opportunities—in particular, the IT
  professional does not have to create a DM
  component—the Oracle engine can be
  reused
Problems with Client/server
 It wasn’t Internet compatible
 It required an IT professional to install
  software on the end-user’s personal
  computer (the client)
 It required an IT professional to work
  closely with the non-IT professional
 There were no career paths for IT
  professional hired in marketing, finance,
  accounting, manufacturing, etc.
Modern solution of today: N-
TIER DISTRIBUTED
ARCHITECTURE

 This is a distributed architecture like
 client/server, but now the application is
 distributed across three or more
 computing boxes on the network
N-Tier distributed Architecture




                                                          Database
                                                         Data Server (DM)
                                                          Application   Application
                                                          Server 1      Server 2
                                  Thin Clients (1/2PM)
Take a closer look at the




                            Application Server runs the business
                            logic component and half ot the
Application Servers




                            presentation management component—the
                            portion the serves out the web pages
Comments on N-Tier Distributed
Architecture
 Clients  are called ‘thin’ because the only
  thing running on them is the Internet
  Browser
 The IT professional doesn’t have to
  install anything on the client
 More re-use is possible—specifically
  that browser
Advantages of N-Tier Distributed
Architecture
 Like Client/server, it accommodates
  enterprise visibility because the data are
  integrated
 Applications can be built rapidly because
  there is abundant reuse
       The DM module is reused
       Half of the PM component is reused
       There are reuse opportunities within the rest of the
        PM component and the BL component as well
More advantages of N-Tier
 IT professionals don’t have to be
  remotely loaned out to marketing,
  management, accounting and finance
 They can now be centrally located and
  managed where career paths will exist
  for them
Application Servers do Two
things
 They serve out web pages upon request
 They do all of the business logic
  processing.
ERP Modules
 Finance/Accounting
 SalesMarketing
 Production/Materials Management
 Human Resources
 Supply Chain Management
 Customer Relation Management
These modules would be placed
in a
 Thin client
 Data server
 Application server
 Mainframe


WHICH??
ERP Implementation
 Analyze  business processes
 Choose modules to implement
 Align level of sophistication
 Finalize delivery and access
 Link with External Partners
Customer Relationship
Management
 CRM   software plans and executes
  business processes that involve
  customer interaction, such as
  marketing, sales, fulfillment, and service
  (not manufacturing)
 CRM is focused on customers, not
  products
Collaborative Product Commerce
 Software  concerned with new product
 design and development, as well as
 product lifecycle management
Connectivity
A  common data management
  component
 API’s (Application Programming
  Interfaces)
 EAI (Enterprise Application Integration)
 XML (Extensible Markup Language)
   Dr.  Viator (accounting) teaches a course in
     this language
Chapter 10--Supply Chain
Management
 Plants/warehouses/distribution/
  information infrastructure
 Most of America’s product gets moved
  by _____ (air, water, rail, truck,
  pipeline).
 What is COVISINT??
 What benefits accrue from SCM?
What’s new and exciting in
SCM??
 InformationTechnology (specifically
 enterprise visibility)
   Has changed everything
   SCM Software modules within ERP
    systems
        I2 Technologies
   Has    reduced uncertainty
      Which has reduced _____________
      Which is a form of _______________
Inventory turns
 Calculated on an annual basis
 The more, the better
 Inputs:
   Costof goods sold
   Average aggregate value of inventory
Average aggregate value of
inventory
 Calculated by taking the product of the
 unit cost with the number of units and
 then summing these products for all
 inventory categories
Days of supply
 Avgagg value of Inv*365/Ann cost of
 goods sold

 Or   simply…   365/inventory turns
Manufacturing Inventory Types
 Raw  materials inventory
 Work-in-process inventory
 Finished goods inventory
Supply Chain Management
Terms
   Bullwhip effect              Landed cost
   Collaborative planning,      Logistics
    forecasting and              Order fulfillment
    replenishment                RFID
   Continuous                   Sourcing
    replenishment
                                 Vendor-management
   Core competencies             inventory
   Cross-docking                Warehouse
   E-business                    management system
   E-marketplaces
   E-procurement
   EDI
   Inventory turns
Chapter 13 – Inventory
Management
 Inventory for Independent demand
 { Not manufacturing inventory, usually—
  more like retail inventory}
Carrying costs
 Rent
 Lighting/heating
 Security
 Interest (on borrowed capital tied up in
  inventory)
 Taxes
 Shrink/obsolescence/theft

    Can also be expressed as a % of product cost
    A rule of thumb is 30%
Ordering costs—costs related
to
Transportation
Shipping
Receiving
Inspection
Shortage costs
 This  is an opportunity cost
 Is ignored in the simple models you will
  be using, by assuming that there are no
  shortages
Back-order costs
 Willassume impatient customers who
  must have the product they wish to buy
  NOW.
 So back-ordering is not considered in
  the simple models we looked at
Continuous Inventory Systems
 Constant order amount, called the EOQ
 EOQ = Economic Order Quantity
 Fixed annual deterministic demand
 Minimizes
   Holding (carrying) costs
   Ordering costs

 Uses  re-order point to determine when
  to order
 Time between orders is not fixed
EOQ models also have
 No  shortages/back-ordering
 Constant lead time
 Instantaneous or finite replenishment
 Can take into consideration price
  discounting
   When    doing so, three costs are minimized
    jointly: Ordering costs, holding costs and
    purchase costs taken over a year’s time
If the quantity ordered is less
than the EOQ, then
 Ordering costs will be greater than
 holding (carrying) costs
ABC Classification—what is
the point??
   To concentrate, focus on the those items in
    inventory that constitute the highest dollar
    value to the firm
       Class A items constitute 5-15% of the items and
        70 to 80% of the total dollar value to the firm
       Class B items constitute 30% of the inventory
        items but only 15% of the dollar value
       Class C items constitute 50 to 60% of the items
        but only 5 to 10% of the dollar value
ABC Classification..
 Class A items are tightly controlled
 Class B items less so
 Class C items even less


   Dollar values are computed by multiplying the
    unit cost by the annual demand for the item

   This technique is used in all auto parts
    inventory control systems and has been for
    15 years
Periodic inventory systems
are….
 Fixed   Time period systems
 NOT
   EOQ   Models
 The time between orders is fixed, the
 re-order point is fixed, but the order
 amount is not
Which gives you lowest
holding cost?
 Instantaneous   replenishment
 Finite (non-instantaneous)
  replenishment
 Quantity discounts


      OF THE ABOVE GIVES YOU
 WHICH
 LOWEST TOTAL ORDERING COST?
How do we calculate a re-
order point?
 Lead    time in days times the daily
  demand plus the safety stock
 Safety stock equals the service level
  (usually 3 for z) times the standard
  deviation of daily demand times the sq.
  rt. of lead time.
 (You will be given the formulas)
How do we calculate…
 Time   between orders?
   Production   days in a year / # of orders
 Run   length
   EOQ   or order quantity / daily Production
   rate
Safety Stocks and Service
Levels
 Safetystock = Z value * std. dev. of
 daily demand * sqrt(lead time)

 For95% service level, use Z value of
 1.65

 For   99% service level, use Z value of 3
Inventory Terms
                        Independent
 ABC system             demand
 Carrying costs        Inventory
 Continuous inventory  In-process inventory
  system                Non-instantaneous
 Dependent demand       receipt
 EOQ                   Order cycle
                        Quantity discount
 Fixed-order quantity
  system                Stockout
                        Service level
 Fixed time period
  system                Efficiency

 Capacity
Simulation
 Two   types—
   Continuous    deterministic
       VENSIM is an example
   Discrete   stochastic
       PROMODEL is an example
 Eachof these two types differ by
 method of time advance
Time advance in continuous
deterministic simulation
 Time  is advanced in small, equidistant
  increments
 The simulation engine is really
  integrating differential equations
Time advance in discrete
stochastic simulation
 Time is advanced from event to event
 The simulation engine maintains a stack of
  discrete events chronologically ordered in
  time, called an events calendar
 The next event to occur is popped off the
  stack and processed.
 The result of processing the event is that
  more events are generated and subsequently
  get saved on the events calendar
MONTE CARLO—
    computer-generation of random
 The
 numbers using an
Which simulation gestalt uses
activities, events, entities and
their attributes?

 Continuous  deterministic?
 Discrete stochastic?
The Excel function RAND()
generates…
 Normally-distributed random variates
 Gamma-distributed random variates
 Uniformly-distributed random numbers
 Exponentially-distributed random
  variates
To get a non-uniform random
variate, we often start with
A  normal random variate
 A lognormal random variate
 A uniform random number
 A triangular random variate
To get a non-uniform random
variate, we often use…
 The  central limit theorem
 The law of large numbers
 The inverse function theorem
 All of the above
In discrete/stochastic simulation,
we are interested in
 Entity  idleness
 Entity travel time
 Entity time in the system
 Resource utilization
 All of the above
In discrete/stochastic
simulation, which of the
following components has
time duration?
 Events
 Activities
 Entities
 Resources
 All   of the above
Discrete/stochastic simulation is
appropriate for which of the
following three decision
environments
         Making (DM) under Certainty
 Decision
 DM under risk and uncertainty
 DM under change and complexity
Math programming models, like
the transportation and
transshipment models we looked
at, are appropriate for which
decision making environment

         Making (DM) under Certainty
 Decision
 DM under risk and uncertainty
 DM under change and complexity

						
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