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1
Lecture6
Inventory ManagementChapter 11
2
Economic production quantity (EPQ) model: variant of basic EOQ model
Production done in batches or lots Replenishment order not received in one lump sum
unlike basic EOQ model Inventory is replenished gradually as the order is
produced hence requires the production rate to be greater than the
demand rate This model's variable costs are
annual holding cost, and annual set-up cost (equivalent to ordering cost).
For the optimal lot size, annual holding and set-up costs are equal.
Economic Production Quantity (EPQ)Economic Production Quantity (EPQ)
3
EPQ = EOQ with Incremental Inventory EPQ = EOQ with Incremental Inventory ReplenishmentReplenishment
4
EPQ Model AssumptionsEPQ Model Assumptions Demand occurs at a constant rate of D items per
year. Production capacity is p items per year.
p > D Set-up cost: $Co per run.
Holding cost: $Ch per item in inventory per year. Purchase cost per unit is constant (no quantity
discount). Set-up time (lead time) is constant. Planned shortages are not permitted.
5
EPQ Model FormulaeEPQ Model Formulae Optimal production lot-size (formula 11-16 of
book)
Run time: Q */p
Time between set-ups (cycle time): Q */D years
Total cost (formula 11.15 of book)
Dp
p
C
DCQ
h
o
20
oh CQ
DQC
P
DTC )1(
2
1
6
Example: Non-Slip Tile Co.Example: Non-Slip Tile Co. Non-Slip Tile Company (NST) has been using production runs of
100,000 tiles, 10 times per year to meet the demand of 1,000,000 tiles annually.
The set-up cost is $5,000 per run Holding cost is estimated at 10% of the manufacturing cost of $1
per tile. The production capacity of the machine is 500,000 tiles per month.
The factory is open 365 days per year. Determine
Optimal production lot size Annual holding and setup costs Number of setups per year Loss/profit that NST is incurring annually by using their present
production schedule
7
Management Scientist SolutionsManagement Scientist Solutions
Optimal TC = $28,868 Current TC = .04167(100,000) + 5,000,000,000/100,000
= $54,167 LOSS = 54,167 - 28,868 = $25,299
8
Only one item is involved Annual demand is known Usage rate is constant Usage occurs continually Production occurs periodically Production rate is constant Lead time does not vary No quantity discounts
Economic Production Quantity AssumptionsEconomic Production Quantity Assumptions
9
Too much inventory Tends to hide problems Easier to live with problems than to eliminate
them Costly to maintain
Wise strategy Reduce lot sizes Reduce safety stock
Operations StrategyOperations Strategy
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28-Jan-00 29-Jan-99 Change Percent
Current assets:Cash $3,809 $1,726 $2,083 121%Short-term investments 323 923 (600) -65%Account receivables, net 2,608 2,094 514 25%Inventories 391 273 118 43%Other 550 791 (241) -30% Total current assets 7,681 5,807 1,874 32%
Property, plant, and equipment, net 765 523 242 46%Long-term investments 1,048 532 516 97%Equity securities and other investments 1,673 --- 1,673Goodwill and others 304 15 289 1927%
Total assets $11,471 $6,877 $4,594 67%
LIABILITIES AND STOCKHOLDERS' EQUITY
Current liabilities:Accounts payable $3,538 $2,397 $1,141 48%Accrued and other 1,654 1,298 356 27% Total current liabilities 5,192 3,695 1,497 41%
Long-term debt 508 512 (4) -1%Other 463 349 114 33%
Total liabilities 6,163 4,556 1,607 35%Stockholders' equity:
Preferred stock --- ---Common stock and capitalin excess of $0.01 per value 3,583 1,781 1,802 101%Retained earnings 1,260 606 654 108%Other 465 (66) 531 Total stockholders' equity 5,308 2,321 2,987 129% Total liabilities and stockholders' equity $11,471 $6,877 $4,594 67%
The Balance Sheet – Dell Computer Co.
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Income Statement – Dell Computer Co.(in millions, except per share amount)
28-Jan-00 29-Jan-99Net revenue $25,265 $18,243Cost of revenue 20,047 14,137Gross margin 5,218 4,106Operating expenses: Selling, general and administrative 2,387 1,788 Research, development, and engineering 568 272 Total operating expenses 2,955 2,060Operating income 2,263 2,046Other income 188 38Income before income taxes 2,451 2,084Provision for income taxes 785 624Net income $1,666 $1,460Earnings per common share: Basic $0.66 $0.58 Diluted $0.61 $0.53Weighted average shares outstanding: Basic 2,536 2,531 Diluted 2,728 2,772Retained Earnings: Balances at beginning of period 606 607 Net income 1,666 1,460 Repurchase of common stocks (1,012) (1,461) Balances at end of period $1,260 $606
Fiscal Year Ended
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Debt RatioDebt Ratio What It Measures: The extent to which a firm uses debt financing How You Compute: The ratio of total debt to total assets
Debt ratio =Total debt
Total assets
$6,
$11,
.
163
471
53 73%
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Inventory Turnover RatioInventory Turnover Ratio
What It Measures: How effectively a firm is managing its inventories.
How You Compute: This ratio is computed by dividing sales by inventories
Inventory turnover ratio =
times
balanceinventoryAverage
Sales
10.762/)391$273($
265,25$
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Lecture6
MGMT 650Simulation – Chapter 13
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Simulation Is …Simulation Is … Simulation – very broad term
methods and applications to imitate or mimic real systems, usually via computer
Applies in many fields and industries
Simulation models complex situations
Models are simple to use and understand
Models can play “what if” experiments
Extensive software packages available
ARENA, ProModel Very popular and powerful method
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ApplicationsApplications Manufacturing facility Bank operation Airport operations (passengers, security, planes,
crews, baggage, overbooking) Hospital facilities (emergency room, operating
room, admissions) Traffic flow in a freeway system Waiting lines - fast-food restaurant, supermarkets Emergency-response system Military
17
Example – Simulating Machine BreakdownsExample – Simulating Machine Breakdowns
The manager of a machine shop is concerned about machine breakdowns.
Historical data of breakdowns over the last 100 days is as follows
Simulate breakdowns for the manager for a 10-day period
Number of Breakdowns Frequency
0 10
1 30
2 25
3 20
4 10
5 5
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Simulation ProcedureSimulation ProcedureNumber of Breakdowns Frequency Probability Cum Prob Corresponding Random Numbers
0 10 0.10 0.10 01 to 101 30 0.30 0.40 11 to 402 25 0.25 0.65 41 to 653 20 0.20 0.85 61 to 854 10 0.10 0.95 86 to 955 5 0.05 1.00 96 to 00
100
Day Random Number Simulated # of Breakdowns1 90 42 73 33 82 34 16 15 94 46 92 47 68 38 5 09 84 310 91 4
19
Expected number of breakdowns = 1.9 per day
19
Statistical AnalysisStatistical AnalysisDay # Replication 1 Replication 2 Replication 3 Replication 4 Replication 5 Replication 6 Replication 7 Replication 8Replication 9 Replication 10
1 1 4 4 1 0 2 5 1 1 02 3 5 0 2 0 4 3 2 0 33 3 1 1 1 2 1 0 1 1 54 1 2 2 3 2 3 1 1 1 15 2 0 1 1 1 5 2 2 0 26 0 2 3 1 3 1 4 3 2 27 1 1 2 2 1 2 1 1 1 08 3 3 2 2 0 4 2 1 3 29 1 1 1 2 2 1 4 0 1 4
10 5 1 3 2 3 2 1 0 1 1
2.00 2.00 1.90 1.70 1.40 2.50 2.30 1.20 1.10 2.00
95 % confidence interval for mean breakdowns for the 10-day period is given by:
]955.1,665.1[)10
458.0(262.281.1
21;1
n
stx
n
20
Monte Carlo SimulationMonte Carlo Simulation
Monte Carlo method: Probabilistic simulation technique used when a process has a random component
Identify a probability distribution Setup intervals of random numbers to match
probability distribution Obtain the random numbers Interpret the results
21
Example 2 – Simulating a Reorder Example 2 – Simulating a Reorder Policy Policy
The manager of a truck dealership wants to acquire some insight into how a proposed policy for reordering trucks might affect order frequency
Under the new policy, 2 trucks will be ordered every time the inventory of trucks is 5 or lower
Due to proximity between the dealership and the local office, orders can be filled overnight
The “historical” probability for daily demand is as follows
Simulate a reorder policy for the dealer for the next 10 days Assume a beginning inventory of 7 trucks
Demand (x) P(x)
0 0.50
1 0.40
2 0.10
22
Example 2 SolutionsExample 2 Solutions
x P(x) Cum P(x) RN Day RN Demand Begin Inv End Inv Reorder0 0.5 0.5 01 to 50 1 81 1 7 6 01 0.4 0.9 51 to 90 2 20 0 6 6 02 0.1 1.0 91 to 00 3 82 1 6 5 2
4 34 0 7 7 05 85 1 7 6 06 35 0 6 6 07 10 0 6 6 08 14 0 6 6 09 84 1 6 5 210 92 2 7 5 2
23
In-class Example 3 using MS-ExcelIn-class Example 3 using MS-Excel The time between mechanics’ requests for tools in a
AAMCO facility is normally distributed with a mean of 10 minutes and a standard deviation of 1 minute.
The time to fill requests is also normal with a mean of 9 minutes and a standard deviation of 1 minute.
Mechanics’ waiting time represents a cost of $2 per minute.
Servers represent a cost of $1 per minute. Simulate arrivals for the first 9 mechanic requests and
determine Service time for each request Waiting time for each request Total cost in handling all requests
Assume 1 server only
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AAMCO SolutionsAAMCO SolutionsInter-request time Cum Int-req time Service Time Service Begins Service Ends Wait Time
8.76 8.76 9.57 8.76 18.33 0.005.49 14.25 9.25 18.33 27.58 4.085.21 19.46 8.63 19.46 28.09 0.003.98 23.44 8.19 23.44 31.63 0.004.66 28.10 6.09 28.10 34.19 0.006.71 34.81 9.15 34.81 43.96 0.006.31 41.12 9.93 41.12 51.05 0.005.57 46.70 8.66 46.70 55.36 0.006.38 53.08 9.67 53.08 62.75 0.00
Sum of wait 4.08
Server cost/min 1Waiting cos/min 2
Total cost 70.91
25
Simulation Models Are BeneficialSimulation Models Are Beneficial
Systematic approach to problem solving Increase understanding of the problem Enable “what if” questions Specific objectives Power of mathematics and statistics Standardized format Require users to organize
26
Different Kinds of SimulationDifferent Kinds of Simulation
Static vs. Dynamic Does time have a role in the model?
Continuous-change vs. Discrete-change Can the “state” change continuously or only at
discrete points in time? Deterministic vs. Stochastic
Is everything for sure or is there uncertainty? Most operational models:
Dynamic, Discrete-change, Stochastic
27
Discrete Event SimulationDiscrete Event SimulationExample 1 - A Simple Processing Example 1 - A Simple Processing
SystemSystem
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Advantages of SimulationAdvantages of Simulation
Solves problems that are difficult or impossible to solve mathematically
Flexibility to model things as they are (even if messy and complicated)
Allows experimentation without risk to actual system
Ability to model long-term effects
Serves as training tool for decision makers
29
Limitations of SimulationLimitations of Simulation
Does not produce optimum solution
Model development may be difficult
Computer run time may be substantial
Monte Carlo simulation only applicable to random systems
30
Fitting Probability Distributions to Existing Fitting Probability Distributions to Existing DataData
Data Summary
Number of Data Points = 187Min Data Value = 3.2
Max Data Value = 12.6Sample Mean = 6.33Sample Std Dev = 1.51
Histogram Summary
Histogram Range = 3 to 13
Number of Intervals = 13
31
ARENA – Input AnalyzerARENA – Input AnalyzerDistribution Summary
Distribution: Gamma Expression: 3 + GAMM(0.775, 4.29)
Square Error: 0.003873
Chi Square Test Number of intervals = 7 Degrees of freedom = 4
Test Statistic = 4.68 Corresponding p-value = 0.337
Kolmogorov-Smirnov Test Test Statistic = 0.0727
Corresponding p-value > 0.15
Data Summary
Number of Data Points = 187Min Data Value = 3.2
Max Data Value = 12.6Sample Mean = 6.33Sample Std Dev = 1.51
Histogram Summary
Histogram Range = 3 to 13Number of Intervals = 13
32
Simulation in IndustrySimulation in Industry
33
Course ConclusionsCourse Conclusions Recognize that not every tool is the best fit for every problem Pay attention to variability
Forecasting Inventory management - Deliveries from suppliers
Build flexibility into models Pay careful attention to technology
Opportunities Improvement in service and response times
Risks Costs involved Difficult to integrate Need for periodic updates Requires training
Garbage in, garbage out Results and recommendations you present are only as reliable as the model and its
inputs Most decisions involve tradeoffs Not a good idea to make decisions to the exclusion of known information