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INTRODUCTION
Operations: The production of goods and services; the transformation process that converts inputs to outputs
Operations Management: The direction and control of the processes that transform inputs into finished goods and services
- The purpose of the operations function is to add value during the transformation process.
- Operations can add value by decreasing the cost of inputs or increasing the value of outputs or both.
Manufacturing vs. Service
Operations Management covers issues in both manufacturing and service.
The differences of these two involve the following:
1. degree of customer contact2. uniformity of input3. labor content of jobs4. uniformity of output5. measurement of productivity6. production and delivery7. quality assurance8. amount of inventory9. evaluation of work10. ability to patent design
Productivity Measure
€
Productivity =Output
Input
Partial Measures of productivity:
1
2
OPERATIONS STRATEGY
- The successful implementation of an operations strategy creates value for customers.
Order-Qualifiers and Order-Winners
Order-qualifiers are characteristics that customers perceive as minimum standards of acceptability to be considered as a potential for purchase. Companies need only be as good as competitors.
Order-winners are characteristics of an organization’s goods or services that cause it to be perceived as better than the competition. Companies need to be better than their competitors.
Components of the Operations Strategy
Structural decision categories: CapacityFacilitiesVertical integrationTechnology
Infrastructural decision categories:
WorkforceOrganizationInformation/control systems
Capabilities: Unique to each firm
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Competitive priorities: CostQuality High-performance design Consistent qualityTime Fast delivery time On-time delivery Development speedFlexibility Customization Volume flexibility
Criteria for Evaluating an Operations Strategy
Consistency (internal and external) Between the operations strategy and the overall business
strategy Between the operations strategy and the other functional
strategies within the business Among the decision categories that make up the operations
strategy Between the operations strategy and the business
environment (resources available, competitive behavior, governmental restraints, etc.)
Contribution (to competitive advantage) Making trade-offs explicit, enabling operations to set priorities
that enhance the competitive advantage Directing attention to opportunities that complement the
business strategy Promoting clarity regarding the operations strategy
throughout the business unit so its potential can be fully realized Providing the operations capabilities that will be required by
the business in the future
4
FORECASTING
Judgment Methods- sales force estimates- executive opinion- market research- Delphi Method
Linear Regression
where:
Y = dependent variable
X = independent variable
a = Y-intercept of the line
b = slope of the line
Coefficient of correlation (Multiple R)- strength of dependent and independent variable
Coefficient of determination
Standard error of the estimate
R2- % of the variation of Y that can be explained by X
F-test: tests the overall fit of the model to the dataP-Value: significance of coefficients (a and b)
***P-Value<.05: REJECT hypothesis→ model is GOOD fit***P-Value>.05: ACCEPT hypothesis→ model is NOT GOOD fit
Time Series Methods
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Naïve forecasts- forecast of period is similar to last period’s trend- stable average, trend, seasonal
Ft=At-1
Moving Averages
F5 = A4+ A3 +A2
3
Weighted Moving Average
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F5 = w1A4 + w2A3 + w3A2
Exponential Smoothing
€
Ft = Ft−1 +α (At−1 − Ft−1)
Trend-Adjusted Exponential Smoothing
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St = SmoothingFactor
Tt = TrendFactor
Errors
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Systematic errors --- Bias Random errors --- Variability
Bias:
Average error
Variability:
Mean squared error MSE
MSE= Total Error Sq n-1
Standard deviation s
Mean absolute error MAD
MAD= Total Abs Error n
Mean percent absolute error MAPE
MAPE= Total [(Abs E)/A] x 100 n
Control Chart for Forecast Errors
Upper Control Limit:
Lower Control Limit:
Z = the number of standard deviations from the mean
7
*Where to find “z” given the percentage of the control chart, ? Where to find “z” given the probability for type I error, ?Normal Distribution Table (page 883, Table B.2)Look for “z” corresponds to the probability:
P{Z<= z} = = 0.5+ ,
=1-e.g. A 95% control chart has = 1-95% = 5%, which means its probability for type I error is 5%. Thus probability in the table should be 0.975 (P = 1-0.025 or P = 0.5+ 0.475), which corresponds to z = 1.96.
8
CAPACITY PLANNING
Utilization =
Efficiency =
Effective Capacity = Design Capacity (maximum output rate) – Allowances (e.g. personal time, maintenance, and scrap)
Cushion: the amount of the reserved capacity that a firm maintains to handle sudden increase in demand or temporary losses of production capacity
Capacity cushion =1 - Utilization
Bottleneck operation: An operation in a sequence of operations whose capacity is lower than that of the other operations.
The capacity of a process is the capacity of the bottleneck operation.
Estimate Capacity Requirements
One type of product
where D = number of units (customers) forecast per yearp = processing time (in hours per unit or customer)N = total number of hours per year during which the process operatesC = desired capacity cushion rate (%)
More than one type of product: n types of products
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Q = number of units in each lots = setup time (in hours) per lot
Note: Always round up the fractional part for the number of machines required.
Capacity Planning Problem
You have been asked to put together a capacity plan for a critical bottleneck operation at the Surefoot Sandal Company. Your capacity measure is number of machines. Three products (men’s women’s, and kid’s sandals) are manufactured. The time standards (processing and setup), lot sizes, and demand forecasts are given in the following table. The firm operates two 8-hour shifts, 5 days per week, 50 weeks per year. Experience shows that a capacity cushion of 5 percent is sufficient.
Time Standards
ProductProcessing(hr/pair)
Setup(hr/lot)
Lot Size(pairs/lot)
Demand Forecast
(pairs/yr)Men’s sandals 0.05 0.5 240 80,000Women’s sandals 0.10 2.2 180 60,000Kid’s sandals 0.02 3.8 360 120,000
a. How many machines are needed at the bottleneck?b. If the operation currently has two machines, what is the capacity gap?c. If the operation can not buy any more machines, which products can be made?d. If the operation currently has five machines, what is the utilization?
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Capacity Planning ProblemSolution
Total time available per machine per year: (2 shifts/day)(8 hours/shift)(5 days/week)(50 weeks/year)= 4000 hours/machine/year
With a 5% capacity cushion, the hours/machine/year that are available are:4000(1-0.05) = 3800 hours/machine/year
Total time to produce the yearly demand of each product: (This is equal to the processing time plus the setup time.)Men’s =(0.05)(80,000)+(80,000/240)(0.5)= 4167 hrs
Women’s =(0.10)(60,000)+(60,000/180)(2.2)= 6733 hrs
Kid’s =(0.02)(120,000)+(120,000/360)(3.8)= 3667 hrs
Total time for all products =4167+6733+3667= 14567 hrs
a. Machines needed = (14,567/3800) = 3.83 = 4 machines
b. Capacity gap is 4 - 2 = 2 machines
c. With two machines, we have (3800)(2) = 7600 hours of machine capacity. We can make all of the women’s sandals (6733 hours) and some of the men’s sandals, for example.
d. With five machines, (5)(4000) = 20,000 machine-hours/yr are available. The total number of machine-hours/yr needed for production are 14,567.
Utilization = (14,567/20,000)(100%) = 73%. Thus, the capacity cushion is (100% - 73%) = 27%.
Vertical Integration Problem: Make or Buy
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Hahn Manufacturing has been purchasing a key component of one of its products from a local supplier. The current purchase price is $1,500 per unit. Efforts to standardize parts have succeeded to the point that this same component can now be used in five different products. Annual component usage should increase from 150 to 750 units. Management wonders whether it is time to make the component in-house, rather than to continue buying it from the supplier. Fixed costs would increase by about $40,000 per year for the new equipment and tooling needed. The cost of raw materials and variable overhead would be about $1,100 per unit, and labor costs would go up by another $300 per unit produced.
a. Should Hahn make rather than buy?
b. What is the break-even quantity?
c. What other considerations might be important?
12
PROCESS, LOCATION, LAYOUT (SYSTEM DESIGN)
Location Decisions
Dominant Factors in Manufacturing
• Favorable labor climate (unions, ex. Paris)
• Proximity to markets (transportation cost)
• Quality of life (talent is hard to find in a far away area)
• Proximity to suppliers and resources
• Proximity to other parent companies
• Utilities, taxes and real estate costs
Dominant Factors in Services
• Proximity to customers
• Transportation costs and proximity to markets
• Location of competitors (stay away from competitors, clustering-malls, gas stations)
• Site-specific factors
Breakeven Location Problem
By chance, the Atlantic City Community Chest has to close temporarily for general repairs. They are considering four temporary office locations:
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Property Address Move-in Costs Monthly RentBoardwalk $400 $50Marvin Gardens $280 $24St. Charles Place $350 $10Baltic Avenue $60 $60
a. Can any of these addresses be immediately eliminated from consideration if the goal is to minimize total costs?
b. Use the graph on the next page to help determine for what length of lease each location would be favored. Calculate the breakeven lease lengths between addresses.
Breakeven Location Problem
14
Breakeven calculations:
Layout Types
- Process- organize around process (can handle varied requirements)hospitals, universities, auto repair shops, airlines, public libraries
- Product- organize resources around product/customer (smooth, rapid, high-volume flow)
o assembly lines, cafeteria
- Hybrid- (all four combined)- ex. Hospital
- Fixed-Position- resources come to product, ex. Boeing, build house (product or project remains stationary, and workers, materials, and equipment are moved as needed)
o farming, fire-fighting, road-building, remolding and repair, home building, drilling for oil
Production Layout
- L: space limitation, want straight line out of facility- S: employees stay in circle, access to several workstations,
because inputs come in and outputs leave it can reward product- O: get more out of space, utilize it as much as you can- U: try to organize inbound and outbound transportation
15
LEARNING CURVES
where:
= time for the nth unit
= time for the first unit
b = ln(learning percent)/ln(2)
16
TOTAL QUALITY MANAGEMENT
Customer-Driven Definitions of Quality
Conformance to specifications Value Fitness for use Support Psychological impressions
- Different dimensions of product quality vs. service quality.
Quality as a Competitive Weapon
The costs of poor quality are estimated to be 20% - 30% of product or service costs.
Companies can improve their bottom line through better quality in several ways:
Lower costs (through consistent quality) Higher prices (through high-performance design) Greater market share (through both consistent quality and
high-performance design)
Consistent quality has become (or is becoming) an order qualifier in many markets.
3 Elements in TQM
- Source - Errors or defects should be caught and corrected at the source, not passed along to an internal or external customer. In other words, “Do It Right the First Time”!
o Why??1. It costs less.
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2. Inspection and sorting are often not effective.
- Employee Involvement - an important component of TQM because perceived and actual quality is assessed throughout the process, involving all employees
o This suggests that: Quality perceptions can be negatively affected at one
point in the process, even if the rest of the process is fine.
All employees can participate in improving quality.
- Work Teams - small groups of people who have a common purpose, performance goals, and accountability
o Types of teams: Problem-solving teams Special-purpose teams Self-managing teams
o How can work teams help improve quality? Products and services are becoming more complex
and interrelated Quality can not be ensured by individual efforts. Workers producing products or services often have
the best ideas how the processes can be improved.
-Costs of Quality
Cost categories:
Prevention Quality assurance costs (Increasing in level of quality)
Appraisal
Internal failure Nonconformance costs (decrease in level of quality)
External failure
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There is a tradeoff between quality assurance and nonconformance costs.
As the product or service moves farther along in the process, the cost to address a quality problem rises steeply.
19
STATISTICAL PROCESS CONTROL
Quality MeasurementsControl charts can be used to monitor processes where output is measured as either variables or attributes.
Variables measures - Characteristics of a product or service that can be measured on a continuous scale. Examples include length, width, and time.
Attributes measures - Characteristics of a product or service that can be quickly counted for acceptable quality. Examples include the number of defects in a product or service.
Sources of Process Variation
Sources of process variation can be categorized as:
Common causes - Random, natural, or unavoidable, sources of variation within a process. A process with only common causes of variation is stable (i.e. the mean and spread do not change over time). Such a process is said to be “in a state of statistical control” or “in-control”.
Assignable causes - Any cause of variation that can be identified and eliminated, originating from outside the normal process
Variable Control Charts
Standard Deviation of the Process, , Known
The control limits are:
- = center line of the chart and the average of several past sample means
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- z is the standard normal deviate (number of standard deviations from the average)- and is the standard deviation of the distribution of sample means- n is the sample size
[z = 2.17 for 97%]
Standard Deviation of the Process, , Unknown
Control charts for variables monitor the mean ( X chart) and variability (R chart) of the process distribution. (n # top rows in table)
R chart: To calculate the range of the data, subtract the smallest from the largest measurement in
the sample.
The control limits are:
UCL D R LCL D RR R 4 3 and
- R = average of several past R values and is the central line of the control chart, and- D D3 4, = constants that provide three standard deviation (three-sigma) limits for a given sample size
X chart: The control limits are:
UCL X A R LCL X A RX X 2 2 and
- X = central line of the chart and the average of past sample means- A2 = constant to provide three sigma limits for the process mean.
**ALWAYS look at R-chart first
21
Control Chart for Variables ExampleX , R Charts
Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces.
Tube numberSample 1 2 3 4 5 6 7 8 Avg. Range
1 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.762 8.23 8.12 7.98 8.41 8.31 8.18 7.99 8.06 8.160 0.433 7.89 7.77 7.91 8.04 8.00 7.89 7.93 8.09 7.940 0.324 8.24 8.18 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.415 7.87 8.13 7.92 7.99 8.10 7.81 8.14 7.88 7.980 0.336 8.13 8.14 8.11 8.13 8.14 8.12 8.13 8.14 8.130 0.03
8.050 0.38
X = 8.050, R = 0.38, n = 8
From Table 10.3:
b) We delete the sixth observation and recalculate the control limits. The ranges, including the range for the first sample are now all within the revised control limits, and the process average for the second sample now falls just inside of the revised control limits.
Control Charts for Attributesp-Chart
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A p-chart is a commonly used control chart for attributes, whereby the quality characteristic is counted, rather than measured, and the entire item or service can be declared good or defective.
The standard deviation of the proportion defective, p, is: p p p n ( ) /1 ,
where n = sample size, and
€
p =p∑
n= average of several past p values
and central line on the chart.
Using the normal approximation to the binomial distribution, which is the actual distribution of p,
UCL p z LCL p zp p p p and
where z is the normal deviate (number of standard deviations from the average).
Calculating P
= proportion of defective products= number of the defected products
n = sample sizem = number of samples
p-Chart Example
A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following number of leaking tubes are found:
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Sample Tubes Sample Tubes Sample Tubes1 3 8 6 15 52 5 9 4 16 03 3 10 9 17 24 4 11 2 18 65 2 12 6 19 26 4 13 5 20 17 2 14 1
Total 72Calculate p-chart three-sigma control limits to assess whether the capping process is in statistical control.
Solution: n = 144, p 72
20 1440 025
( ).
p
p p
p p
p p
n
UCL p z
LCL p z
( ) . ( . ).
. ( . ) .
. ( . ) .
1 0 025 1 0 025
1440 013
0 025 3 0 013 0 064
0 025 3 0 013 0 014 (adjusted to zero)
The highest proportion of defectives occurs in sample #10, but is still within control limits: p10 9 144 0 0625 / . . Therefore, the process is in statistical control.
c-ChartA c-chart is another type of control chart for attributes, whereby the quality characteristic is counted as the number of defects/ unit.
Using the normal approximation to the Poisson distribution, which is the actual distribution of c,
UCL c z c LCL c z cc c and
where cis the average number of defects/unit and the center line of the c-chart.
Control Chart for Attributes Examplec-Chart
At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are the results of the study:
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Tube # Lumps Tube # Lumps Tube # Lumps1 6 5 6 9 52 5 6 4 10 03 0 7 1 11 94 4 8 6 12 2
Determine the c-chart two-sigma upper and lower control limits for this process.
Solution:
c
c
UCL c z
LCL c z
c
c c
c c
( )
( )( )
( )( )
6 5 0 4 6 4 1 6 5 0 9 2
124
4 2
4 2 2 8
4 2 2 0
The eleventh tube has too many lumps (9), so the process is probably out of control.
Process Capability Exercise
Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces 0.60 ounces. The target process capability ratio is 1.33. The current distribution of the filling process is centered on 8.054 ounces with a standard deviation of 0.192 ounces. Compute the process capability index to assess whether the filling process is capable and set properly.
Solution:
Process capability ratio:
Cp
Upper specification - Lower specification
6
8 6 7 4
6 019210417
. .
( . ).
Process capability index:
25
Cpk 0 948.
The process is not capable of consistently meeting specifications according to the minimum capability level set by Webster. ** IF IT is LESS than 1.33, it is NOT CAPABLE
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INVENTORY MANAGEMENT
Definition of Independent Demand Inventory
Independent demand inventory consists of items for which demand is influenced by market conditions and is not related to production decisions for any other item held in stock.
Contrast this with dependent demand inventory, consisting of items required as components or inputs to a product or service. We will talk about managing dependent demand inventory in manufacturing using a material requirements planning (MRP) system.
Total Annual Relevant Cost
Annual holding cost =
Annual ordering cost =
Total annual relevant cost:
Q= Order Quantity (units)H= Hold, carrying cost per unitD= YEARLY demandS= ordering cost
Optimal # Orders= 1/TBO
Overland Motors Example
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Overland Motors uses 25,000 gear assemblies each year (i.e. 52 weeks) and purchases them at $3.40 per unit. It costs $50 to process and receive each order, and it costs $1.10 to hold one unit in inventory for a whole year. Assume demand is constant.
Ralph U. Reddie has been ordering 1,000 gear assemblies at a time, but can adjust his order quantity if it will lower costs.
a. What is the annual cost of the current policy of using a 1,000-unit lot size?
b. What is the order quantity that minimizes cost?
c. What is the time between orders for the quantity in part b?
d. If the lead time is two weeks, what is the reorder point, R?
Economic Production Quantity
Similar to the EOQ but used for batch production. A complete order no longer received at once and inventory is replenished gradually (i.e., non-instantaneous replenishment).
1. Maximum Cycle Inventory
2. Total cost = Annual holding cost + Annual ordering or setup cost
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3. Economic Production Quantity (EPQ) aka ECONOMIC LOT SIZE (ELS)
Economic Production Quantity ExampleA domestic automobile manufacturer schedules 12 two-person teams to assemble 4.6 liter DOHC V-8 engines per work day. Each team can assemble five engines per day. The automobile final assembly line creates an annual demand for the DOHC engine at 10,080 units per year. The engine and automobile assembly plants operate six days per week, 48 weeks per year. The engine assembly line also produces SOHC V-8 engines. The cost to switch the production line from one type of engine to the other is $100,000. It costs $2,000 to store one DOHC V-8 for one year.
a. What is the economic lot size?
b. How long is the production run?
c. What is the average quantity in inventory?
d. What are the total annual relevant costs?
Quantity DiscountsIn the case of quantity discounts (price incentives to purchase large quantities), the price, P, is relevant to the calculation of total annual cost (since the price is no longer fixed).
Total cost = Annual holding cost + Annual ordering cost + Annual cost of materials
29
GRAPH
Quantity DiscountsTwo-Step Procedure
Step 1: Beginning with lowest price, calculate the EOQ for each price level until a feasible EOQ is found. It is feasible if it lies in the range corresponding to its price.
Step 2: If the first feasible EOQ found is for the lowest price level, this quantity is best. Otherwise, calculate the total cost for the first feasible EOQ and for the larger price break quantity at each lower price level. The quantity with the lowest total cost is optimal.
Quantity Discounts ExampleOrder Quantity Price Per Unit
0-99 $50100 or more $45
If the ordering cost is $16 per order, annual holding cost is 20 percent of the per unit purchase price, and annual demand is 1,800 items, what is the best order quantity?
Step 1. =
=
Step 2. =
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=
Reorder Point (Continuous Review System)
A continuous review system tracks the remaining inventory of an item each time a withdrawal is made, to determine if it is time to reorder.
Decision rule: Whenever a withdrawal brings the inventory down to the reorder point (ROP), place an order for Q (fixed) units.
Reorder Point
Demand pattern
Lead time for ordering
ROP
Known and constant
None ROP = 0
Known and constant
Known and constant
ROP =
Variable, normally distributed,
known
Known and constant
ROP =
Known and constant
Variable, normally distributed,
known
ROP =
Variable, normally distributed,
known
Variable, normally distributed,
known
ROP =
Uncertain, discrete probability distribution
unknown Determine ROP for a given service level based on the cumulative probabilities of demand during lead time.
31
Shortage and Service Levels
Expected Shortage per order cycle: E (n) = E (z)
E(z) = standardization parameter obtained from Table 12.3.
= standard deviation of lead time demand
Expected shortage per year:
E (N) = E (n)
Annual Service Level:
Continuous Review System Example
You are reviewing the company’s current inventory policies for its continuous review system (Q system), and began by checking out the current policies for a sample of items. The characteristics of one item are:
Average demand = 10 units/wk (assume 52 weeks per year) Ordering and setup cost (S) = $45/order Holding cost (H) = $12/unit/year Average lead time (L) = 3 weeks Standard deviation of demand during lead time = 17 units Service-level = 70%
a) What is the EOQ for this item?
b) What is the desired safety stock?
c) What is the desired reorder point R?
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d) What is the decision rule for replenishing inventory?
e) What is the expected shortage per year?
Single Period (Newsvendor) Model
Used to handle ordering of perishables and items that have a limited useful life.
shortage costs = unrealized profit per unitexcess costs = the unit cost less the salvage value
1. Calculate the shortage and excess costs:
2. Calculate the service level (SL), which is the probability that demand will not exceed the stocking level:
SL =
3. Determine the optimal stocking level, , using the service level and demand distribution information.
= Mean demand + zSL*demand
Example of the Newsvendor Model
The concession manager for the college football stadium must decide how many hot dogs to order for the next game. Each hot dog is sold for $2.25 and makes a profit of $0.75. Hot dogs left over after the game are sold to the student cafeteria for $0.50 each. Based on previous games, the demand is normally distributed with an average of 2000 hot dogs
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sold per game and a standard deviation of 400. Find the optimal stocking level for hot dogs.
34
SUPPLY CHAIN MANAGEMENT
The bullwhip effect is characterized by fluctuations in inventory and order levels that tend to increase as one moves back up the channel from the final customer.
Demand variations are amplified when moving up the supply chain!
- Causes of bullwhip effect:
1. Demand forecast updating
2. Long lead time/Order batching
3. Price fluctuations
4. Rationing and shortage gaming
5. Lack of visibility/communication throughout the supply chain
- The bullwhip effect can be alleviated by:
Reducing the number of stages in the supply chain
Communicating consumer demand directly up the supply chain talk to retailer like Wal-Mart and ask to share info. Get EDI inventory system and send to P&G
CustomerRetai
lerWholesalerDistri
butorManufacturer
Demand
35
Reducing ordering and shipping delays Cuts LT so can reduce SS
Reducing demand destabilizing practices sales and promotions (OP ppl don’t like b/c distort
consumer’s behavior) to stop: EDLP (everyday low price)- encourages ppl to
buy at regular price, ppl stop buying in bulk
Counter “gaming” during shortages must have enough capacity but, companies worry about cost of capacity and
uncertain demand so, use wait&see- but always have a shortage with this
strategy
Make vs. BuyPros Cons
Make
Increased control over price, quality, etc.
Economies of combined operations
Proprietary products protected
Capital costs Capability limits Time limits Opportunity costs Reduced flexibility to change
partners Reduced volume flexibility
Buy
Low capital costs Specialization Competition Increased flexibility
Unfavorable allocation of product
Lack of control over price, quality, etc.
Lock-in from specialized contracts and assets
Transaction (coordination) costs
Supplier Relations
Competitive Orientation
36
The view that negotiations between buyer and seller is a zero-sum game. Often used when a firm represents a significant share of the supplier’s sales or many substitutes are available. Example: WalMart
Cooperative OrientationThe view that the buyer and seller are partners. Includes sole sourcing. Often used with strategically important and/or high value-added components. Example: McDonald’s or Toyota
Mixed strategySeeks to combine the advantages of the competitive orientation (e.g. low prices) with the cooperative orientation (e.g. few suppliers). Example: Dell Computer
Strategic Management of the Supply Chain
Efficient Supply Chains:The purpose of efficient supply chains is to coordinate the flow of materials and services so as to minimize inventories and maximize the efficiency of the manufacturers and service providers in the chain. Efficient supply chains work best when demand is predictable and products/services are stable. Examples of competitive priorities: low cost.
Responsive Supply Chains:The purpose of responsive supply chains is to react quickly to market demands by positioning inventories and capacities in order to hedge against uncertainties in demand. Responsive supply chains work best when demand is unpredictable, new product introduction is frequent, and product variety is high. Examples of competitive priorities: development speed, fast delivery, customization, volume flexibility.
In addition: Innovations in information technology and other practices are facilitating the integration of the supply chain for greater efficiency and responsiveness and enabling “orchestrated” networks.
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Choice of supply chain should serve a firm’s competitive priority
Case study: One Industry, Two Supply Chains
1. Fashion predictor: Gap, J.Crew, etc
Long lead time: 3 to 6 months Sourcing and manufacturing in Asian and South American
countries If prediction is wrong, huge inventory markdown or salvage
Choice: efficient supply chain
2. Fashion follower: Zara
Short lead time: 2 to 3 weeks Sourcing and manufacturing in Europe, mainly in Spain Intentionally create shortage, very few inventory
markdown or salvage
Choice: responsive supply chain
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WAITING LINE MODELS
€
units
time=
arrival
service=
λ
μ
Single-Server Model (M/M/1)Assumptions:
Number of servers = 1
Number of phases = 1
Input source: infinite, no balking or reneging
Arrival distribution: Poisson; mean arrival rate =
Service distribution: Exponential; mean service rate = ; mean service time =
Waiting line: single line; unlimited length
Priority discipline: FCFS
Single-Server Operating Characteristics
Average utilization:
Probability that n customers are in the system: - has to wait: prob n>1= 1-Pr[n=0]
Average number of customers in the system(line and being served):
Average number of customers in line:
Average time spent in the system:
Average time spent in line:
Single-Server Application
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Customers arrive at a checkout counter at an average of 20 per hour, according to a Poisson distribution. They are served at an average rate of 25 per hours, with exponential service times. Use the single-server model to estimate the operating characteristics of this system.
What service rate is required to have customers average only 10 minutes in the system?
Multiple-Server Model (M/M/s)Assumptions:
Number of servers = M
Number of phases = 1
Input source: infinite, no balking or reneging
Arrival distribution: Poisson; mean arrival rate =
Service distribution: Exponential; mean service rate = ; mean service time =
Waiting line: single line; unlimited length
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Priority discipline: FCFS
Multiple-Server Operating Characteristics (TABLE)
Average utilization:
Idle Time= 1-
€
ρ
Probability that zero customers are in the system:
Probability that n customers are in the system:
Average number of customers in line:
Average time spent in line:
Average time spent in the system:
Average number of customers in the system:
Multiple-Server ApplicationSuppose the manager of the checkout system decides to add another counter. The arrival rate is still 20 customers per hour, but now each checkout counter will be designed to service customers at the rate of 12.5 per hour. Use the multiple-server model to estimate the operating characteristics of this system.
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How do the single-server and multiple-server models compare?
Cost AnalysisQuestion: How many servers do we need?
Minimize: Total cost = Customer waiting cost + Capacity cost
Example 6 on Page 846:
Trucks arrive at a warehouse at a rate of 15 per hour during business hours. Crews can unload the trucks at a rate of 5 per hour. Recent changes in wage rates have caused the warehouse manager to reexamine the question of how many crews to use. The new rates are: Crew and dock cost is $100 per hour; truck and driver cost is $120 per hour. How many crews does the warehouse need to minimize the total cost per hour?
First question: At least how many crews does this warehouse need? If we have 5 crews:
Crew/Dock cost per hour =
Average number of drivers in system
Driver/Truck cost per hour =
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Total cost per hour =
# of crews
Crew/Dock cost per hour
Avg. # of drivers in system
Driver/Truck cost per hour
Total cost per hour
4 $400 4.528 $543.36 $943.3656 $600 3.099 $371.88 $971.887 $700 3.028 $363.36 $1063.36
Finite-Source, Single-Server Model
Assumptions:
Number of servers = 1
Number of phases = 1
Input source: finite, equals N customers
Arrival distribution: Exponential inter-arrival times; mean =
Service distribution: Exponential; mean service rate = ; mean service time =
Waiting line: single line; no more than N - 1
Priority discipline: FCFSFinite-Source, Single-Server Operating Characteristics
Probability that zero customers are in the system:
Average utilization:
Average number of customers in line:
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Average number of customers in the system:
Average time spent in line:
Average time spent in the system:
Finite-Source, Single-Server ApplicationDBT Bank has 8 copy machines located in various offices throughout the building. Each machine is used continuously and has an average time between failures of 50 hours. Once failed, it takes an average of 4 hours for the engineer from the service company to have it fixed. What is the average number of copy machines in repair or waiting to be repaired?
Hilltop Produce ProblemThe Hilltop Produce store is staffed by one checkout clerk. The average checkout time is exponential distributed around an average of two minutes per customer. An average of 20 customers arrive per hour.
a. What is the average utilization rate?
b. What is the probability that three or more customers will be in the checkout area?
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c. What is the average number of customers in the waiting line?
d. If the customers spend an average of 10 minutes shopping for produce, what is the average time customers spend in the store?
Using Waiting Line Models to Analyze Service OperationsBalance costs against benefits of improving service system. Also, consider the costs of not making improvements. There are different measures of service operations:
Line length - Long lines indicate poor customer service, inefficient service, or inadequate capacity.
Number of customers in system - A large number causes congestion and dissatisfaction.
Waiting time in line - Long waits are associated with poor service.
Total time in system - May indicate problems with customers, server efficiency, or capacity.
Service facility utilization - Control costs without unacceptable reduction in service.
Using Waiting Line Models to Design Service Operations Arrival:
o Number of arrivalso Adjust through advertising, promotions, pricing, and
appointmentso Reduce variability
Service: o Number of servers - adjust service system capacity
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o Server efficiency - training, incentives, performance evaluation
o Number of phases - consider splitting service taskso Reduce variability
Priority rule - decide whether to allow preemption
Customer perceptions - shorter waiting time
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AGGREGATE PLANNING
Planning Strategies
Chase strategy – Matching capacity to demand; the planned output for a period is set at the expected demand of the period.
Level strategy – Maintaining a steady rate of regular time output while meeting the variations in demand by a combination of inventories, overtime, part-time workers, subcontracting, and back-orders.
Pros Cons
Chase strategy
Low inventory investment and backlogs
Expense of adjusting output rates and/or work-force
Alienation of work-force Loss of productivity Lower quality
Level strategy
Level output rates Stable work-force
Increased inventory investment
Increased undertime and overtime expense
Increased backlogs
Level and Chase Strategies Example
Bob Carlton’s golf camp estimates the following workforce requirements for its services over the next two years.
Quarter 1 2 3 4Demand 4200 6400 3000 4800
Quarter 5 6 7 8Demand 4400 6240 3600 4800
Each certified instructor puts in 480 hours per quarter regular time and can work an additional 120 hours overtime. Regular-time wages and
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benefits cost Carlton $7200 per employee per quarter for regular time worked up to 480 hours, with an overtime cost of $20 per hour. Unused regular time for certified instructors is paid at $15 per hour. There is no cost for unused overtime capacity. The cost of hiring, training, and certifying a new employee is $10,000. Layoff costs are $4000 per employee. Currently, eight employees work in this capacity.
Level Strategy
Find a level work-force plan that allows for no delay in service and minimizes undertime. What is the total cost of this plan?
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Chase Strategy
Use a chase strategy that varies the work-force level without using overtime or undertime. What is the total cost of this plan?
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RESOURCE PLANNING
Material Requirements Planning (MRP)- Information system that translates master schedule requirements
for finished products into time-phased requirements for subassemblies, parts and raw materials.
Inputs to the MRP Bill of Materials (BOM) The Master Production Schedule (MPS) Inventory recordOutputs from the MRP Planned orders Order releases Secondary reports
- Bill of Materials A record of all the components of an item, the parent-component
relationships, and usage quantities derived from engineering and process designs.
- The Master Production Schedule (MPS) A detailed plan that states how many END items will be produced
within specified periods of time.- Inventory record Records status of each item by time period (time bucket) with
respect to gross requirements, schedule receipts, expected amount on hand, lot-size policy, lead time, supplier, etc.
BOM Example:
Parents BOM of X
Gross requirements
For 10 Xs
On-hand inventory
ofindividual
items
Available Inventory
Net requirements
B 5C 0D 0E 30F 0Lead time (in weeks): X- 1, B-1, C- 2, D-2, E- 1, F-2
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C
D(3) E
E(4)
E(2)
B(2)
X
F(2)
One scheduled receipt of 50 units of Item E to arrive at the beginning of week 1.
Net
Requirements = Gross Requirements –Available Inventory
Available Inventory = Projected on-hand inventory – Safety stock – Inventory allocated to other items
Projected on-hand inventory = On-hand inventory + Scheduled
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Receipt
- Manufacturing Resources Planning (MRP II)- Extension of MRP which includes MRP as well as capacity
requirements planning, financial planning, marketing planning etc…
- Enterprise Resources Planning (ERP)- Information system that integrates all functional areas including
MRP II, finance, accounting, HR, marketing etc…- all info integrated
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LEAN SYSTEM
JIT System- a dependent demand production control system designed to produce goods or services as needed and minimize inventories- stable and predictable environments with high volume
production, demand is certain- philosophy: inventories are wasteful and non-value added
Environments for Effective JIT* Manufacturing: Firms that tend to have highly repetitive
manufacturing processes, well-defined material flows, and reasonably high volumes use JIT systems because the pull method allows closer control of inventory and production at the work stations.
* Services: Firms that tend to have repetitive operations, reasonably high volumes, and deal with tangible items can benefit from JIT systems.
* In general, JIT works well in stable and predictable environments because there is little forward visibility.
Characteristics of JIT Systems Pull method Consistently high quality Small lot sizes Short setup times Uniform workstation loads Standardization of components and work methods Close supplier ties Flexible work force Product focus Automated production Preventive maintenance
Customization & Standardized goal is to produce large # production with an efficient system
Product or service customization has negative effects on both: Predictability of demand Predictability of operations
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Since uncertainty in operations requires extra resources, customization is inherently less efficient than standardization.
However, it is sometimes possible to increase operational efficiency even with customization using standardization strategies.
Standardization strategies include: Part standardization – Maximize component commonality across
products (ex. Dell) Process standardization – Delay customization as late as possible
(aka postponement, ex. HP) Product standardization – Carry a limited number of products in
inventory (SERVICE industry, car dealerships- i.e. wait for colors, etc.)
Types of Scheduling Operations scheduling - Assigns workers to tasks or jobs to machine
work centers. Operations schedules are short-term plans designed to implement the master production schedule. (look at low volume with high variety)
Work-force scheduling - Determines when human resources are available for work - biggest difference b/w service/mfg. industry is inventory back
logs
Scheduling in Services
Characteristics of services that have an impact on scheduling:
* Services can not buffer demand uncertainties with inventory./ backlog
* Demand for services is difficult to predict. (should have a good idea of demand- always match capacity with demand)
* Scheduling systems can facilitate the capacity management of service providers.
Two approaches:* Schedule customer demand (capacity remains fixed and demand
is leveled)
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* Appointments* Reservations* Backlogs
* Schedule the work force to meet forecasted demand (adjust capacity to demand)
Priority Sequencing Rules
Single dimension rules:
EDD = Select Job With Earliest Due Date
FCFS = Job That Arrives First is Processed First
SPT = Select Job With Shortest Processing Time
Multiple dimension rules:
Performance of Priority Sequencing Rules Earliest due date (EDD) - Performs well with respect to: minimizing
percentages of jobs past due, minimizing the maximum amount of time a job is late. Performs poorly with respect to: job flow time, work-in-process inventory, utilization.
First come, first serve (FCFS) - Perceived as being fair. Performs poorly with respect to all performance measures.
Shortest processing time (SPT) - Performs well with respect to: average job flow time, work-in-process inventory, minimizing percentages of jobs past due, utilization. Performs poorly with respect to: minimizing the maximum amount of time a job is late, minimizing total inventory (it pushes work to finished goods before it is needed), adjusting schedules when due date changes (due date is not used in the calculation of priority).
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Critical ratio (CR) - Performs well when we are concerned with global operation of a system of work centers.
Slack per operations (S/O) - Performs similarly to EDD with added advantages of a global view and accounting for the duration of the jobs.
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PROJECT MANAGEMENT
Total slack- allowable slippage for entire activity without stopping project
Free slack- slack that does not affect the following activity (ex. Taking an extra week to do an activity- will it affect the following nodes?)
Network Time Calculations
Earliest finish time (EF) for an activity
EF = ES + t
Earliest start time (ES) for an activity
ES = Max [EF times of all immediately preceding activities]
Latest start time (LS) for an activity without delaying the project
LS =LF – t
Latest finish time (LF) for an activity without delaying the project
LF = Min[LS times for all immediately following activities]
Crash RulesStep 1: Determine the project’s critical path(s).Step 2: Find the cheapest activity or activities on the critical path(s) to crash.Step 3: Reduce the time for this activity until the first of (a) it cannot be further reduced, (b) another path becomes critical, or (c) the increase in direct costs exceeds the savings that result from shortening the project. If more than one path is critical, the time for an activity on each path may have to be reduced simultaneously.Step 4: Repeat this procedure until the increase in direct costs is less than the savings generated by shortening the project.
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Calculating Time Estimatesa) Optimistic time ( ):
Shortest time during which an activity can be completed.
b) Most likely time ( ):
Best estimate of average time.c) Pessimistic time ( ):
Longest time an activity can take.
d) Activity’s time (te) and variance (2) with beta distribution
,
Analyzing Probabilities
Probabilities can be assessed using the z-transformation formula:
T = specific time, TE = expected time (path mean), = standard deviation of path mean. Assuming the activity times are
independent, the path standard deviation is the square root of the sum of the activity time variances.
To determine the probability of completing a project in a specified amount of time:
Calculate the probability of each of the paths being completed in that amount of time based on the value of z. For any value of z that is greater than 3, the probability that the corresponding path will be completed in that amount of time can be considered to be 100%.
If all paths are independent, then the probability of completing a project in the specified amount of time is the product of the individual path probabilities.
Hospital Project Completion Probabilities
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How likely is it that the hospital project will be completed in 72 weeks?
PathProbability of completion in
72 weeksA-F-K (72 – 28)/2.05 = 21.5 100%A-I-K (72 – 33)/3.04 = 12.8 100%
A-C-G-J-K
(72 – 67)/3.94 = 1.27 90%
B-D-H-J-K
(72 – 69)/3.45 = 0.87 81%
B-E-J-K (72 – 43)/3.80 = 7.6 100%
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** UPDATE paths and show new critical time after each path
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