Transcript decision analysis - University of Redlands

Lecture
6
Inventory Management
Chapter 11
1
Economic Production Quantity (EPQ)

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

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
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This model's variable costs are
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hence requires the production rate to be greater than the
demand rate
annual holding cost, and
annual set-up cost (equivalent to ordering cost).
For the optimal lot size,
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annual holding and set-up costs are equal.
2
EPQ = EOQ with Incremental Inventory
Replenishment
3
EPQ 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.
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EPQ Model Formulae

Optimal production lot-size (formula 11-16 of book)
Q0 
2 DC o
Ch
p
pD
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Run time: Q */p
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Time between set-ups (cycle time): Q */D years

Total cost (formula 11.15 of book)
1
D
D
TC  (1  )QCh  Co
2
P
Q
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Example: Non-Slip Tile Co.
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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
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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
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Management Scientist Solutions
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
Optimal TC = $28,868
Current TC = .04167(100,000) + 5,000,000,000/100,000
= $54,167
LOSS
= 54,167 - 28,868 = $25,299
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Economic Production Quantity Assumptions
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 

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Operations Strategy
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Too much inventory
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Tends to hide problems
 Easier to live with problems than to eliminate
them
 Costly to maintain
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Wise strategy

Reduce lot sizes
 Reduce safety stock
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The Balance Sheet – Dell Computer Co.
28-Jan-00 29-Jan-99
Current assets:
Cash
Short-term investments
Account receivables, net
Inventories
Other
Total current assets
Property, plant, and equipment, net
Long-term investments
Equity securities and other
investments
Goodwill and others
Total assets
Change
Percent
$3,809
323
2,608
391
550
7,681
765
1,048
$1,726
923
2,094
273
791
5,807
523
532
$2,083
(600)
514
118
(241)
1,874
242
516
121%
-65%
25%
43%
-30%
32%
46%
97%
1,673
304
$11,471
--15
$6,877
1,673
289
$4,594
1927%
67%
$3,538
1,654
5,192
508
463
6,163
$2,397
1,298
3,695
512
349
4,556
$1,141
356
1,497
(4)
114
1,607
48%
27%
41%
-1%
33%
35%
---
---
3,583
1,260
465
5,308
1,781
606
(66)
2,321
1,802
654
531
2,987
101%
108%
$11,471
$6,877
$4,594
67%
LIABILITIES AND STOCKHOLDERS' EQUITY
Current liabilities:
Accounts payable
Accrued and other
Total current liabilities
Long-term debt
Other
Total liabilities
Stockholders' equity:
Preferred stock
Common stock and capital
in excess of $0.01 per value
Retained earnings
Other
Total stockholders' equity
Total liabilities and
stockholders' equity
129%
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Income Statement – Dell Computer Co.
(in millions, except per share amount)
Net revenue
Cost of revenue
Gross margin
Operating expenses:
Selling, general and administrative
Research, development, and engineering
Total operating expenses
Operating income
Other income
Income before income taxes
Provision for income taxes
Net income
Earnings per common share:
Basic
Diluted
Weighted average shares outstanding:
Basic
Diluted
Retained Earnings:
Balances at beginning of period
Net income
Repurchase of common stocks
Balances at end of period
Fiscal Year Ended
28-Jan-00 29-Jan-99
$25,265 $18,243
20,047
14,137
5,218
4,106
2,387
568
2,955
2,263
188
2,451
785
$1,666
1,788
272
2,060
2,046
38
2,084
624
$1,460
$0.66
$0.61
$0.58
$0.53
2,536
2,728
2,531
2,772
606
1,666
(1,012)
$1,260
607
1,460
(1,461)
$606
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Debt Ratio
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What It Measures: The extent to which a firm uses debt financing
How You Compute: The ratio of total debt to total assets
Total debt
Debt ratio =
Total assets
$6,163

$11,471
 53.73%
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Inventory Turnover Ratio
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What It Measures: How effectively a firm is
managing its inventories.
How You Compute: This ratio is computed by
dividing sales by inventories
Sales
Inventory turnover ratio =
Average inventory balance
$25,265

 76.10 tim es
($273 $391) / 2
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Lecture
6
MGMT 650
Simulation – Chapter 13
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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
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Models can play “what if” experiments
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Extensive software packages available
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ARENA, ProModel
Very popular and powerful method
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Applications

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
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Example – Simulating Machine Breakdowns
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The manager of a machine shop is concerned about
machine breakdowns.
Historical data of breakdowns over the last 100 days is as
follows
Number of Breakdowns
Frequency
0
10
1
30
2
25
3
20
4
10
5
5
Simulate breakdowns for the manager for a 10-day period
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Simulation Procedure
Number of Breakdowns
0
1
2
3
4
5
Day
1
2
3
4
5
6
7
8
9
10
Frequency
10
30
25
20
10
5
100
Random Number
90
73
82
16
94
92
68
5
84
91
Probability Cum Prob
0.10
0.10
0.30
0.40
0.25
0.65
0.20
0.85
0.10
0.95
0.05
1.00
Corresponding Random Numbers
01 to 10
11 to 40
41 to 65
61 to 85
86 to 95
96 to 00
Simulated # of Breakdowns
4
3
3
1
4
4
3
0
3
4
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Expected number
of breakdowns
= 1.9 per day
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Statistical Analysis
Day #
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
0
2
3
5
0
2
0
4
3
2
0
3
3
3
1
1
1
2
1
0
1
1
5
4
1
2
2
3
2
3
1
1
1
1
5
2
0
1
1
1
5
2
2
0
2
6
0
2
3
1
3
1
4
3
2
2
7
1
1
2
2
1
2
1
1
1
0
8
3
3
2
2
0
4
2
1
3
2
9
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:
xt
n 1;1

2
s
0.458
 1.81 2.262(
)  [1.665,1.955]
n
10
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Monte Carlo Simulation
Monte Carlo method: Probabilistic simulation
technique used when a process has a random
component
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Identify a probability distribution
Setup intervals of random numbers to match
probability distribution
Obtain the random numbers
Interpret the results
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Example 2 – Simulating a Reorder Policy
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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
Demand (x)
P(x)
0
0.50
1
0.40
2
0.10
Simulate a reorder policy for the dealer for the next 10 days
Assume a beginning inventory of 7 trucks
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Example 2 Solutions
x
0
1
2
P(x)
0.5
0.4
0.1
Cum P(x) RN
0.5 01 to 50
0.9 51 to 90
1.0 91 to 00
Day
1
2
3
4
5
6
7
8
9
10
RN
81
20
82
34
85
35
10
14
84
92
Demand Begin Inv End Inv Reorder
1
7
6
0
0
6
6
0
1
6
5
2
0
7
7
0
1
7
6
0
0
6
6
0
0
6
6
0
0
6
6
0
1
6
5
2
2
7
5
2
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In-class Example 3 using MS-Excel
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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
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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 Solutions
Inter-request time Cum Int-req time Service Time Service Begins Service Ends Wait
8.76
8.76
9.57
8.76
18.33
5.49
14.25
9.25
18.33
27.58
5.21
19.46
8.63
19.46
28.09
3.98
23.44
8.19
23.44
31.63
4.66
28.10
6.09
28.10
34.19
6.71
34.81
9.15
34.81
43.96
6.31
41.12
9.93
41.12
51.05
5.57
46.70
8.66
46.70
55.36
6.38
53.08
9.67
53.08
62.75
Sum of wait
Server cost/min
Waiting cos/min
Total cost
Time
0.00
4.08
0.00
0.00
0.00
0.00
0.00
0.00
0.00
4.08
1
2
70.91
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Simulation 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
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Different Kinds of Simulation

Static vs. Dynamic
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Continuous-change vs. Discrete-change
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Can the “state” change continuously or only at
discrete points in time?
Deterministic vs. Stochastic

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Does time have a role in the model?
Is everything for sure or is there uncertainty?
Most operational models:

Dynamic, Discrete-change, Stochastic
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Discrete Event Simulation
Example 1 - A Simple Processing System
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Advantages 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
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Limitations of Simulation

Does not produce optimum solution

Model development may be difficult
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Computer run time may be substantial

Monte Carlo simulation only applicable to
random systems
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Fitting Probability Distributions to Existing Data
Data Summary
Number of Data Points
Min Data Value
Max Data Value
Sample Mean
Sample Std Dev
= 187
= 3.2
= 12.6
= 6.33
= 1.51
Histogram Summary
Histogram Range
Number of Intervals
= 3 to 13
= 13
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ARENA – Input Analyzer
Distribution 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
Min Data Value
Max Data Value
Sample Mean
Sample Std Dev
= 187
= 3.2
= 12.6
= 6.33
= 1.51
Histogram Summary
Histogram Range
Number of Intervals
= 3 to 13
= 13
31
Simulation in Industry
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Course Conclusions
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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

33