Review for Exam II

This exam will be administered Thursday, October 29, 2015, usual time and place

HW 10-4Homework Problem #10-4

  Average   TotalRaw Materials Inventory Unit Cost ($) Value

1 7200 8.5 612002 4500 7.2 324003 3200 15.4 492804 4800 13.7 657605 6900 10.5 72450

    Total 281090Work in Process      

A 100 16200 1620000B 70 13500 945000C 60 6100 366000D 35 14400 504000

  Total 3435000Finished Goods    

X 20 78700 1574000Y 10 65300 653000Z 10 86000 860000

    Total 3087000

Cost of goods sold = 18500000

Average aggregate value of inventory = 6803090

Inventory turns = 2.72

Days of supply = 134.22

HW 10-6

Homework Problem #10-6

  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 S11-12Homework #S11-12

Closing Open Plants Available Employees

Plant A B C Employees Transferred

1 0 60 0 60 60

2 55 10 40 105 105

3 0 10 0 70 10

Demand 55 80 40    

Transferred 55 80 40    

   

Output = 1660

Product Output (units/day):

Closing Open Plant

Plant A B C

1 5 8 6

2 10 9 12

3 7 6 8

HW S11-36Shipping cost ($/container)

  U.S. Port Available Containers   U.S. PortEuropean Port 4. Boston 5. Savannah 6. Mobile 7. Houston Containers Shipped European Port 4. Boston 5. Savannah 6. Mobile 7. Houston

1. Antwerp 85 40 0 0 125 125 1. Antwerp 1,725 1,800 2,345 2,7002. Cherbourg 0 70 15 125 210 210 2. Cherbourg 1,825 1,750 1,945 2,3203. Barcelona 0 0 85 0 160 85 3. Barcelona 2,060 2,175 2,050 2,475

Containers Shipped 85 110 100 125    

Shipping cost($/container)  Inland Port Intermodal Containers   Inland Port

U.S. Port 8. Ohio 9. Texas10. North Carolina Capacity Shipped U.S. Port 8. Ohio 9. Texas

10. North Carolina

4. Boston 0 0 85 85 85 4. Boston 825 545 3205. Savannah 0 55 55 110 110 5. Savannah 750 675 450

6. Mobile 100 0 0 100 100 6. Mobile 325 605 6907. Houston 70 55 0 130 125 7. Houston 270 510 1,050

Intermodal Capacity 170 240 140   550Containers shipped 170 110 140    

Shipping cost ($/container)  Distribution Centers Containers   Distribution Centers

Inland Port11.

Phoenix 12. Columbus 13. Kansas City 14. Louisville15.

Memphis Shipped Inland Port 11. Phoenix 12. Columbus 13. Kansas City14.

Louisville 15. Memphis8. Ohio 85 0 35 50 0 170 8. Ohio 450 830 565 420 960

9. Texas 0 40 70 0 0 110 9. Texas 880 520 450 380 660

10. North Carolina 0 20 0 0 120 14010. North Carolina 1,350 390 1,200 450 310

Demand 85 60 105 50 120  Containers Shipped 85 60 105 50 120  

Transshipment Total cost =

1,179,400.00 Flows

0

HW S14-8

Homework #S14-8

Process:   Molding Smoothing PaintingConstraints:         Available Usage Left over

Budget ($) 8.0 5 6.5 3,000 246 2754Available time (hrs) 1 1 1 120 36 84

Fiberglass (lbs) 630     10,000

10,000 0Process flow 7 -12   - -  Process flow   12 -10 - -  Process flow 7   -10 - -  

Process:Molding= 15.87 hours

Smoothing= 9.26 hoursPainting= 11.11 hours

Profit=

19,444.44

HW S14-24Homework #S14-24

    Plant Mine    Mine 1 2 3 4 Capacity ProductionCincinnati 1 38 0 18 72 220 128  2 30 160 0 0 190 190Pittsburgh 3 42 0 72 108 280 222  Demand (tons) 110 160 90 180      Production 110 160 90 180      Ash content 0 0 0 0      Sulfur content -1 -2 -3 -3.24      Cost= 41,594

Shipping and processing cost ($/ton):

  PlantMine 1 2 3 4

1 69 71 72 742 76 74 75 793 86 89 80 82

HW 13-12

Homework Problem #13-12

Demand = 21600Carrying cost = 2.40

Order cost = 80Lead time = 5

Q = 1200.0

Total cost = 2880.00

Reorder point = 300.00

HW 13-20

Homework Problem #13-20

 Carrying

cost = $ 1.90

 Ordering

cost = $ 800

  Demand = 40000

Quantity Price Q Discount Q Total Cost

1 3.40 5803.81 5803.81 $ 147,027.24

10000 3.20 5803.81 10000.00 $ 140,700.00

20000 3.00 5803.81 20000.00 $ 140,600.00 optimal

30000 2.8 5803.81 30000.00 $ 141,566.67

40000 2.6 5803.81 40000.00 $ 142,800.00

50000 2.4 5803.81 50000.00 $ 144,140.00

HW S13-10Homework Problem #S13-10

Probability of Sales Volume:

  Volume   Price  Variable

Cost   "Z"

P(x) Cumulative Volume Months RN1 Volume RN2 Price RN3 Cost Profit

0.12 0 300 1 0.8981 700 0.589391 25 0.8809 11 800

0.18 0.12 400 2 0.8980 700 0.211427 23 0.4370 9 800

0.20 0.30 500 3 0.7251 600 0.043591 22 0.5938 10 -1800

0.23 0.50 600 4 0.5663 600 0.939051 27 0.6266 10 1200

0.17 0.73 700 5 0.4494 500 0.031211 22 0.9135 11 -3500

0.10 0.90 800 6 0.5676 600 0.73029 26 0.0620 8 1800

7 0.9881 800 0.357548 24 0.0667 8 3800

8 0.6308 600 0.210563 23 0.1806 9 -600

Probability of Price: 9 0.7925 700 0.095235 23 0.7663 10 100

P(x) Cumulative Price 10 0.1375 400 0.129264 23 0.2409 9 -3400

0.07 0 22 11 0.5043 600 0.863569 26 0.8842 11 0

0.16 0.07 23 12 0.4145 500 0.615934 25 0.4715 9 -1000

0.24 0.23 24 13 0.1253 400 0.597582 25 0.6948 10 -3000

0.25 0.47 25 14 0.9106 800 0.867387 26 0.5985 10 3800

0.18 0.72 26 15 0.0177 300 0.788519 26 0.8847 11 -4500

0.10 0.90 27 16 0.8517 700 0.779089 26 0.5655 10 2200

      17 0.9296 800 0.029418 22 0.7785 10 600

18 0.1208 400 0.765968 26 0.9694 12 -3400

19 0.6711 600 0.287881 24 0.9563 12 -1800

Probability of Variable Cost: 20 0.0718 300 0.772055 26 0.4707 9 -3900

P(x) Cumulative Cost

0.17 0 8 Average Z = -590

0.32 0.17 9

0.29 0.49 10 P(BE) = 0.50

0.14 0.78 11

0.08 0.92 12

1.00    

Exam Format50 multiple choice3 problemsClosed-bookClosed-notesClosed-neighborBRING---pencil, calculator, orange

scantron sheet

Exam detailsThursday, October 29, 2015Will start the exam at 8:00 a.m.

Exam CoverageChapter 10, Chapter 11, Supplement to

Ch 11, Chapter 12, Chapter 13, Supplement to Ch13, and Chapter 15-second half LP problems in the supplement to Chapter 14,

but not the content of Chapter 14—will cover that later.

Chs. 10, 11, 12, 13, second half of 15Supplements to chs. 11,13,14

Typical problems—see Practice Exam II Inventory with Independent DemandProblems involving calculation of

inventory turns and days of supplyProduction Scheduling ProblemTransportation problemLP formulation problem Interpretation of LP SENSITIVITY output

Typical Discussion ProblemsTransshipment problemLinear programming formulationBe able to draw schematics of

mainframe/glass architecture, client/server architecture and N-tier architecture

Chapter 15 – ERP Inventory for Dependent Demand will

NOT be covered….Exam coverage of this chapter starts on

page 700

What were five motivations for transitioning from mainframesAbsence of data integration36 month backlogs at centralized MIS

shops Idle CPU cycles on desktopsMainframes were expensive bottlenecksSupport for Internet and thin clientsQuicker, cheaper development times

through REUSE

What is the information architecture modern ERP systems are currently based on?

Mainframe/glass house Client/serverN-tier distributedNone 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 Is software 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 thereThe road to implementation has been

rough

More ERPBased on an N-tier distributed

architectureNot on mainframe glasshouse

•Advantages of N-tier architectureProvides for data integrationBetter usage of MIPS on both PCs

and serversSolves the 36-month backlog of the

centralized MIS shopEnables 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 &

distribution Production &

Materials Management

Quality management

Human resource management

Project management

Accounting and controlling/finance

Supply chain management

Customer relationship management

ERP TermsBest-of-breedCollaborative product commerceCustomer relationship managementSupply chain managementXML

Re-engineered Computer ArchitecturesStarted with mainframe/glasshouseMigrated to client/serverEvolved to N-tier distributed

Why did such re-engineering occur?There was no data integrationMIPs on mainframes were hugely

expensive and very much in demandMIPs on PCs were idle 95% of the time

and extremely cheapBacklogs for MIS shops were at 36 monthsDeveloping new applications were slow

and expensive

Distributed architectures solved these problemsData resides 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 ArchitectureNo reuse was possibleThese 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

Server (D

M)

Clients (PM, BL)

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

Server (D

M)

Thick Clients (PM, BL)

Database

Advantages of Client/server architecture

All Data 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

Data S

erver (DM

)

Thin Clients (1/2PM)

Database

Application

Server 1

Application

Server 2

Take a closer look at the Application Servers

Application Server runs the business

logic component and half ot the

presentation managem

ent 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 ModulesFinance/AccountingSales MarketingProduction/Materials ManagementHuman ResourcesSupply Chain ManagementCustomer Relation Management

These modules would be placed in a

Thin clientData serverApplication serverMainframe

WHICH??

ERP Implementation Analyze business processesChoose modules to implementAlign level of sophisticationFinalize delivery and accessLink 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

ConnectivityA common data management

componentAPI’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 ManagementPlants/warehouses/distribution/

information infrastructureMost 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?? Information Technology (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 turnsCalculated on an annual basisThe more, the better Inputs:

Cost of 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 supplyAvg agg value of Inv*365/Ann cost of

goods sold

Or simply… 365/inventory turns

Manufacturing Inventory Types

Raw materials inventoryWork-in-process inventoryFinished goods inventory

Supply Chain Management Terms Bullwhip effect Collaborative planning,

forecasting and replenishment

Continuous replenishment

Core competencies Cross-docking E-business E-marketplaces E-procurement EDI Inventory turns

Landed cost Logistics Order fulfillment RFID Sourcing Vendor-management

inventory Warehouse

management system

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 costA rule of thumb is 30%

Ordering costs—costs related to

Transportation

Shipping

Receiving

Inspection

Shortage costsThis is an opportunity cost Is ignored in the simple models you will

be using, by assuming that there are no shortages

Back-order costsWill assume 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 SystemsConstant order amount, called the EOQEOQ = Economic Order QuantityFixed annual deterministic demandMinimizes

Holding (carrying) costs Ordering costs

Uses re-order point to determine when to order

Time between orders is not fixed

EOQ models also haveNo shortages/back-orderingConstant lead time Instantaneous or finite replenishmentCan 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 systemsNOT

EOQ ModelsThe time between orders is fixed, the

re-order point is fixed, but the order amount is not

Which gives you lowest holding cost? Instantaneous replenishmentFinite (non-instantaneous)

replenishmentQuantity discounts

WHICH OF THE ABOVE GIVES YOU LOWEST TOTAL ORDERING COST?

How do we calculate a re-order point?Lead time in days times the daily

demand plus the safety stockSafety 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 ordersRun length

EOQ or order quantity / daily Production rate

Safety Stocks and Service LevelsSafety stock = Z value * std. dev. of

daily demand * sqrt(lead time)

For 95% service level, use Z value of 1.65

For 99% service level, use Z value of 3

Inventory Terms

ABC system Carrying costs Continuous inventory

system Dependent demand EOQ Fixed-order quantity

system Fixed time period

system Capacity

Independent demand

Inventory In-process inventory Non-instantaneous

receipt Order cycle Quantity discount Stockout Service level Efficiency

SimulationTwo types—

Continuous deterministic VENSIM is an example

Discrete stochastic PROMODEL is an example

Each of 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—The computer-generation of random

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 variatesGamma-distributed random variatesUniformly-distributed random numbersExponentially-distributed random

variates

To get a non-uniform random variate, we often start with

A normal random variateA lognormal random variateA uniform random numberA triangular random variate

To get a non-uniform random variate, we often use…

The central limit theoremThe law of large numbersThe inverse function theoremAll of the above

In discrete/stochastic simulation, we are interested in

Entity idlenessEntity travel time Entity time in the systemResource utilizationAll of the above

In discrete/stochastic simulation, which of the following components has time duration?EventsActivitiesEntitiesResourcesAll of the above

Discrete/stochastic simulation is appropriate for which of the following three decision environmentsDecision Making (DM) under CertaintyDM under risk and uncertaintyDM under change and complexity

Math programming models, like the transportation and transshipment models we looked at, are appropriate for which decision making environment

Decision Making (DM) under CertaintyDM under risk and uncertaintyDM under change and complexity