BU385 Final Exam-AID

Lecture 9Operations Consulting and

Reengineering

Forces for Rethinking Operational Capabilities

• Customer Sophistication• Competitive Pressure• Move to Services

Operations Consulting

• Assists clients in developing operations strategies and improving production processes

• Strategy Development:– Analyze the capabilities of the operations in light

of the firm’s competitive advantage• Process Improvement:

– Employing analytical tools to help managers enhance performance of their departments

5 Main Consulting Areas

• Plant, People, Parts, Processes, Planning and Control Systems

• Consulting Process– Sales and Proposal Development– Analyze Problem– Design, Develop and Test Alternatives– Develop Systematic Performance Measure– Present Final Report– Implement Changes– Assure Client Satisfaction– Assemble Learnings from Study

Consulting “Tool Kit”

• Problem Definition Tools: Customer Surveys, Gap Analysis, Employee Surveys, Five Forces Model

• Data Gathering Tools: Plant Tours/Audits, Work Sampling, Flowcharts, Organization Charts

• Data Analysis Tools: Problem Analysis (SPC Tools), Bottleneck Analysis, Computer Simulation, Statistical Tools

• Cost Impact and Payoff Analysis: Stakeholder Analysis, Balanced Scorecard, Process Dashboards

• Implementation: Responsibility Charts, Project Management Techniques

Business Process Reengineering (BPR)

• Michael Hammer: “The fundamental rethinking and radical redesign of business processes to achieve dramatic improvements in critical, contemporary measure of performance”

• Guidelines for Implementation:– Codification of Reengineering– Clear Goals and Consistent Feedback– High Executive Involvement in Clinical Changes

BPR Principles

1. Organize around Outcomes, Not Tasks2. Have Those Who Use the Output of the Process Perform the

Process3. Merge Information-Processing Work into the Real Work that

Produces the Information4. Treat Geographically Dispersed Resources as Though They

Were Centralized 5. Link Parallel Activities Instead of Integrating Their Results6. Put the Decision Point Where the Work is Performed and

Build Control into the Process7. Capture Information Once – At the Source

Lecture 12

Process Selection: Manufacturing

Types of Processes

• Conversion Processes (ore steel sheets)• Fabrication Processes (sheet metal car door)• Assembly Processes (assembling parts to a

car)• Testing Processes

Process Flow Structures

• Job Shop (Small batches of many different products)

• Batch Shop (Stable line of products produced in periodic batches)

• Assembly Line• Continuous Flow (Processing of

undifferentiated materials like petroleum)

Break Even Analysis

• Cost MinimizationE.g. 2 possible production methodsA) FC= 200, 000; VC = 15$B) FC= 80, 000; VC = 75$To solve for indifference level of production:200, 000 + 15*Demand=80, 000 + 75*DemandDemand = 2, 000 UnitsTherefore: Demand<2000:B Demand>2000: A

Manufacturing Process Flow Design

• Evaluate processes raw materials, parts and subassemblies follow through the plant

• Tools Used:– Assembly Drawings– Assembly Charts– Route Sheets– Flow Process Charts

Lecture 13Facility Layout and Assembly

Line Balancing

Facility Layout

• Goal is to find layout that maximized production efficiency

• Typical Objectives of Plant Layout:– Min. investment required in new equipment– Min. time of production– Utilize existing space more efficiently– Provide safety, comfort and convenience to

employees

Patterns of Flow• Most common goal for quantitative analysis:

– Minimize Materials Handling Cost• With this goal Flow Analysis is necessary• 2 Classes

– Horizontal Flows– Vertical Flow

Patterns of Flow cont’d• Horizontal Flow:

– When all operations are on the same floor• Vertical Flow

– Operations are on different floors of a building• 6 Horizontal Flow and 6 Vertical Flow patterns

exits

Flow Patterns• Straight (straight line operation)• Serpentine (used when space is too small to fit

all operations in straight line)• U flow (allows for shipping and receiving to be

in same location)• Circular Flow• L-Flow (same as straight but building forces

the line to move)• S-Flow (same as serpentine)

Activity Relationship Chart

• Describes how desirable it is to place 2 activities in close proximity (subjective)– A: absolutely important– E: especially important– I: Important– O: Ordinary Importance– U: Unimportant– X: Undersirable

Activity Relationship Chart Closeness Rating between Departments Department 1 2 3 4 5 61. Burr and grind — E U I U A

2. NC equipment — O U E I

3. Shipping and receiving — O UA

4. Lathes and drills — E X

5. Tool crib — U

6. Inspection —

From-to-Charts• Used to analyze the flow of materials between

departments• Shows distance between departments and the

number of materials handling trips during the day

• It is based on a specific layout

From-to-Chart - example

To Saws Milling Punch PressFromSaws - 18 40

Milling 18 - 38

Punch Press 40 38 -

From-to-Charts can then be adjusted by adding a cost to each unit of distance travelled.

The charts are not meant to determine the layout of the facility, but to just give an understanding of where costs come from.

Types of Layouts

• Fixed Position Layouts• Product Layouts• Process Layouts• Layouts Based on Group Technology

Fixed Position

• Product is too big to be moved• Stays in one spot and work is done around it• E.g. Airplanes, Ships, Construction Sites

Product Layout

• Machines are organized based on the sequence of operations required

• Used for high-volume standardized production• Fastest cycle times in the environment• Main drawback: if one part of the line stops

the rest of the line remains idleProcess 1 Process 2 Process 3 Process 4

Process Layouts• Small-to-Medium volume manufacturers

• For wide variety in product mix

FoundryMilling machines

LathesGrinding

Painting Drills

Office

Welding

Forging

Layouts Based on Group Technology

• Parts are indentified and grouped based on similarities in manufacturing function or design

• Cells are then used to process part families• Advantages: Reduce WIP inventory, Lower Set-

Up Times, Reduce Materials Handling Costs, Better Scheduling

Assembly Line Balancing

• Characterized as set of tasks per item produced• Time required per task: ti

• Goal is to make groups of tasks for individual workstations

• Amount of time per work station is set out beforehand

• Time is based on desired production rate– Cycle Time (C)

Assembly Line Balancing• Factors that affect planning

– Precedence Constraints (task sequence)– Zoning Restrictions (two tasks can’t be performed

at the same station)

• If ti is the amount of time per task, ∑ ti = T is the amount of time for the production of an item

• Minimum # of Workstations = T/C, rounded to the next largest integer

Ranked Positional Weight Technique

Task Immediate Predecessor

Time ( ti)

1 N/A 122 1 63 2 64 2 25 2 26 2 127 3, 4 78 7 59 5 110 9, 6 411 8, 10 612 11 7

• Given– C = 15min

• T = ∑ ti = 70Theoretical Minimum Number of Workstations = [70/15]Min. # of Workstations = [4.67]Min. # of Workstations = 5

• Ranked Positional Method requires solving for the Positional Weight of each task

• Positional Weight of task i = ti + time of each task that follows i

Task Positional Weight

1 70

2 58

3 31

4 27

5 20

6 29

7 25

8 18

9 18

10 17

11 13

12 7

• Tasks are ranked according to weights and assigned in that order

• We know the cycle time for each station is 15 minutes

• This means that the tasks at each station can only take up a maximum of 15 minutes

• Using this method we get

Station 1 2 3 4 5 6Tasks 1 2, 3, 4 5, 6, 9 7, 8 10, 11 12

Idle Time 3 1 0 3 5 8

Using the ranked positional method, we see that the minimum number of work stations is 6, and not the calculated number 5.

This is because the ranked positional method is a heuristic. There may very well be a solution with only 5 stations, but using this method we cannot solve for it.

For any new cycle time, just rebalance the stations using the positional weights already calculated.

Lecture 14

Process Selection: Services

Service Process Selection

• Generally classified according to – who the customer is– the creation of the service – what processes are

involved – the presence of customer contact– The extent of the contact

Competitive Strategic Focus

• Treatment of the customer• Speed and convenience• Price • Variety• Quality of the tangible goods that accompany

the service• Unique skills that constitute the service

Service-System Design Matrix

• Three degrees of customer contact1) Buffered core: no contact2) Permeable System: penetrable by the customer via phone or face-to-face

contact3) Reactive System: penetrable and reactive to the customer’s requirements

Buffered core

Permeable System

Reactive System

Sales Opportunity

Production Efficiency

Service-System Design Matrix

• Obviously, there is a trade-off between sales opportunities and production efficiencies

• Operational uses:– Identification of worker requirements– Focus of operations– Innovations

• Strategic uses:– Enables a systematic integration of operations and marketing strategy– Clarifies exactly which combination of service delivery the firm is providing– Permits comparison between firms & identification of competitive

advantage– Indicates evolutionary or life cycle changes that might be in order as the

firm grows

Fail-Safing – Poka-yokes

• Poka-yokes are procedures that block the inevitable mistake from becoming a service defect

• Help reduce errors caused by both the customer and the server

• 3 types– Warning methods– Physical/visual contact methods– Three T’s: Task, Treatment(of the customer), and

Tangible (features of the service facility)

Well-Designed Service Systems7 Characteristics

1. Each element of the service system is consistent with the operating focus of the firm

2. It is user-friendly 3. It is robust 4. It is structured so that consistent performance by its people

and systems is easily maintained5. It provides effective links between the back office and the

front office so that nothing falls between the cracks 6. It manages the evidence of service quality in such a way that

customers see the value of the service provided 7. It is cost-effective

Lecture 15

Waiting Lines/ Queuing Analysis

Queuing Theory• N(t) = number of customers in the system at time t• Pn = probability of n customers in the system

• λn = Arrival rate when there are n customers in the system; usually constant λn = λ– Number of arrivals/unit of time i.e. 5 people/minute

• µ n = service rate; usually a constant µ n = µ– Number of people or customers/unit of time

• c = # of servers in the system; c≥1 • p = utilization rate; proportion of time each server is

busy

More Notation

• A(t) = the number of arrivals up until time t• L = Expected number of customers in the system • Lq = Expected number of customers in the queue• W = Expected total waiting time in the system of a

customer– From arrival to being fully served

• Wq = Expected waiting time in the queue • K = system’s capacity – maximum number of people;

i.e. n<K

Useful Relations

• W = Wq + 1/µ• L = λW = (customers/time) * (waiting time in

system)• Lq = λWq = (customers/time) * (waiting time in

queue)

M/M/#

• Arrival process / Service Process / Number ofis random is random servers

• Arrival Process has a Poisson Distribution:P{A(t) = n} = e(-λt) (λt)n/n!

M/M/1 – 1 serverP0 = P{Wq = 0}

P1 =P0*(λ/µ) = (1-p)*(λ/µ)

Pn = pi(1-p)

P{L>k} = probability of there being more than k people in the system

= = pk+1

P{Wq > t} = probability of waiting more than t minutes/seconds/etc in the queue

= pe-(µ-λ)t

M/M/c

• c servers in parallel

Pn = (λ/µ)nP0/n! if n≤cPn = (λ/µ)nP0/(c!cn-c) if n>c

Lecture 16

Strategic Capacity Planning

Capacity Planning

• Capacity: The amount of output that a system is capable of achieving over a specific period of time

• Strategic capacity planning: An approach for determining the overall capacity level of capital intensive resources, including– Equipment – Facilities – Labour Force

Capacity Utilization

• Best operating level: capacity for which the process was designed

• Capacity cushion: % of capacity held in reserve for unexpected occurrences

level operating BestusedCapacity rate nutilizatioCapacity

Underutilization

Best OperatingLevel

Volume

Overutilization

Best Operating Level

During one week of production, a plant produced 83 units of a product. Its best utilization recorded was 120 units per week. What is this plant’s capacity utilization rate?

Utilization = 83/120 = 69%

Example of Capacity Utilization

The LearningCurve

As plants produce more products, they gain experience in the best production methods and reduce their costs per unit

Total accumulated production of units

Cost per unit

Yesterday

TodayTomorrow

• Flexible Plants • Flexible Processes• Flexible People

Ways to achieve flexibility

Capacity Flexibility

Maintaining System Balance: Output of one stage is the exact input requirements for the next stage

Stage 1 Stage 2 Stage 3Unitsper

month

6,000 7,000 5,000

Unbalanced stages of production

Stage 1 Stage 2 Stage 3Unitsper

month

6,000 6,000 6,000

Balanced stages of production

Capacity Planning: Balance

1. Forecast sales within each individual product line

2. Calculate equipment and labor requirements to meet the forecasts

3. Project equipment and labor availability over the planning horizon

Determining Capacity Requirement

A manufacturer produces mustard, which is sold in small and family-size plastic bottles.

Year: 1 2 3 4Small (000s) 150 170 200 240Family (000s) 115 140 170 200

• Three 100,000 units-per-year machines are available for small-bottle production. Two operators required per machine.

• Two 120,000 units-per-year machines are available for family-sized-bottle production. Three operators required per machine.

Example of Capacity Requirement

Year: 1 2 3 4Small (000s) 150 170 200 240Family (000s) 115 140 170 200

Small Mach. Cap. 300,000 Labor 6Family-size Mach. Cap. 240,000 Labor 6

SmallPercent capacity used 50.00%Machine requirement 1.50Labor requirement 3.00Family-sizePercent capacity used 47.92%Machine requirement 0.96Labor requirement 2.88

60

56.67% 67% 80%

1.70 2 2.40

3.40 4 4.80

58.33% 70.83% 83.33%

1.16 1.42 13.67

3.50 4.25 5

Capacity Planning: Decision Tree

• Time: Capacity must be available to provide a service when it is needed

• Location: Capacity must be located near the customer

• Volatility of Demand: Much greater than in manufacturing

Capacity Planning: Service

Lecture 17

Facility Location

Distance Based Methods of Location Determination

• Euclidean Distance– Given A(xA, yA) and B(xB, yB)

• Rectilinear Distance– Given A(xA, yA) and B(xB, yB)

dAB = ( xA - xB )2 + ( yA - yB )2

A

B

dAB = | xA - xB | + | yA - yB |

A

B

Single-Facility Rectilinear Distance Location Problem

• Goal: Minimize weighted sum of rectilinear distances from the new facility to existing facilities.

)(),(1

i

n

iii byaxwyxf

Single-Facility Rectilinear Distance Location Problem

• Solution Method1) Rank all possible locations based on x-values. (Lowest to

Highest)2) Choose the smallest x for which the cumulative weight is

greater than half the total weight.3) Rank all possible locations based on y-values. (Lowest to

Highest)4) Choose the smallest y for which the cumulative weight is

greater than half the total weight.• Contour Lines are used to determine cost levels of

individual locations

ExampleLocation Weight

A (1,2) 5

B (4,1) 2

C (2,5) 4

D (5,9) 9

E (3, 10) 8

Location X-Value Weight

A 1 5

C 2 9

E 3 17

B 4 19

D 5 28

1)

2)

Given Info:

Location Y-Value Weight

B 1 2

A 5 7

C 5 11

D 9 20

E 10 28

Final Location: (3,9)

Other Methods for Single-Facility Rectilinear Problems

• Center of Gravity Method– Process

1. Divide the market into Market Areas2. Estimate the demand (load) for each Area

– Determine the coordinates of each market area– Calculate the Centre of Gravity – Locate facility at the Centre of Gravity

x* = ( Lixi)/( Li)y* = ( Liyi)/( Li)

Minimax Method

Process:Solve for: c1 = min (xi + yi); c2 = max (xi + yi); c3 = min (-xi +yi); c4 = max (-xi

+ yi);c5 = max(c2 – c1, c4 – c3)X1 = (c1 – c3)/2; Y1 = (c1 + c3 + c5)/2X2 = (c2 – c4)/2; Y2 = (c2 + c4 – c5)/2

Solution: All the points along the line connection (x1, y1) and (x2, y2) will be optimal locations.

Euclidean Distance Problems• Gravity Problem

– Solve x* = ∑wixi/ ∑wi

– Solve y* = ∑wiyi/ ∑wi

• Straight Line Distance Problem

– Use gravity problem solutions to solve for first g value. Gravity Solutions will give you (x0, y0)

– For every subsequent x and y solution, keep plugging the new answers into the g equation until the values of x and y begin to converge

Locating Multiple Facilities• Linear Program• Minimize

Existing Locations (an ,bn)New Facility Locations (xm, ym)Where, , such that , ,

And: ,

Locating Multiple Facilities

• Constraints:

• Don’t forget non-negativity constraint

Further Extensions

• Facilities Having Positive Areas– Locating machines within a shop while leaving

enough space between them for operations• Location-Allocation Problems

– Solving for which existing facility will be served by each new facility

• Discrete Location Problems– Limitation on where new facilities can be located

• Network Location Models

Lecture 18

Transportation and Transshipment

Transportation Problems

Purpose: to schedule the flow of goods at the minimum cost

Typically involves a set of sending locations (sources/origins) and a set of receiving locations (sinks/destinations)

Transhipment Problems

A transhipment problem is a transportation problem in which some locations are used as intermediate shipping points (i.e. goods are shipped to and from these locations)

Solving Transportation Problems

• Use Linear Programming in Excel• Goal is to minimize shipping costs of goods

from m sources to n destinations

Set Up

Objective Function• Min ij cij * xij

• Cij- cost to ship from source i to sink j• Xij- amount shipped from source i to sink j

• Two sources and three destinations...1

2

3

1

2

c11c12

c13

c21 c22

c23

d1

d2

d3

s1

s2

SOURCES DESTINATIONS

s2

X12

X21

X12

X22

X12

X23

Constraints • Supply Constraints

– For every source i – j xij ≤ max supply at source i

• Demand Constraints– For every sink j– i xij = demand at sink j

• Non Negativity – xij ≥ 0

Solving Transhipment Problems

Same set up as Transportation problems with the following additional constraint

– Amount shipped to and from intermediate nodes net out to zero

i xik - j xjk = 0

Transshipment Problem

2

3

4

5

6

7

1c13

c14

c23c24

c25

c15

s1

c36

c37

c46

c47

c56

c57

d1

d2

INTERMEDIATE NODES

SOURCES DESTINATIONS

s2

ExampleSource: BU275 lecture notes, by Trent Tucker

• Thomas Industries and Washburn Corporation supply three firms (Zrox, Hewes, Rockwright) with customized shelving for its offices.

• They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc.

• Current weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockwright.

• Both Arnold and Supershelf can supply at most 75 units to its customers.

Unit costs from the manufacturers to the suppliersTomas Washburn

Arnold 5 8 Supershelf 7 4

Zrox Hewes RockwrightTomas 1 5 8

Washburn 3 4 4

Cost to install the shelving at the various locations

Decision Variables

• xij = amount shipped from manufacturer i to supplier j

• xjk = amount shipped from supplier j to customer k

• where i = 1 (Arnold), 2 (Supershelf) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7

(Rockwright)

Objective Function• Minimize Overall Shipping Costs:• Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 +

5x36 + 8x37+ 3x45 + 4x46 + 4x47

Constraints

• Amount out of Arnold: x13 + x14 ≤ 75• Amount out of Supershelf: x23 + x24 ≤ 75• Amount through Thomas: x13 + x23 - x35 - x36 - x37 = 0• Amount through Washburn: x14 + x24 - x45 - x46 - x47 = 0• Amount into Zrox: x35 + x45 = 50• Amount into Hewes: x36 + x46 = 60• Amount into Rockwright: x37 + x47 = 40• Non-negativity of variables: xij ≥ 0, for all i and j.

Variations of Transportation Problem

• Unacceptable routes• Fixed charge problems• ….

Transportation from m warehouses to n customers. There is a unit transportation cost cij between each warehouse i and customer j, and a fixed charge Fi if warehouse i is leased. Each warehouse has a fixed capacity, si. The problem is to decide which set of warehouses to lease to meet all customer demands, dj.

Fixed Charge Problem: LP formulation

Decision Variables: xij (qty shipped from i to j), yi (=1, if warehouse i is leased and 0 otherwise)

Objective Function: Min cost of transportation + leasing

= Min i j cij xij + i Fi yi

Constraints:

Supply: j xij si yi for all i = 1, …, m

Demand: i xij = dj for all j = 1, …, n

Binary (yi), Non-negativity (xij)

Lecture 19 & 20

Total Quality Management

Control Charts

• X-chart – used to monitor the average• R-chart – used to monitor the variance• p-chart – used to monitor the proportion of

defectives• Each chart has an upper (UCL) and lower

control limit (LCL), which easily identify points that are “out of control”, indicating production issues

Notation• R = range = maximum value-minimum value• R = average of ranges• X = average output for sample• n = number of units per sample = sample size• = average defective rate = average fraction defectivep̄• CL = center line = target value• UCL = upper control limit = maximum acceptable value• LCL = lower control limit = minimum acceptable value

– Note: usually LCL ≥ 0

X-chart̄

• CL = X̄• UCL = X + ̄ 3σ/√n = X + ̄ R ̄ /d2

• LCL = X - 3̄ σ/√n = X - ̄ R /̄ d2

Note: d2 is from Table A-5 on p. 770;

σ = population standard deviation

R-Chart̄

• CL = R ̄• LCL = d3*R ̄• UCL = d4*R ̄

The values for d3 & d4 can be found in table A-6 on p. 771

p-chart

• CL=p̄• LCL = - 3p̄

• UCL = +3 p̄

Process Capability Index - Cpk

• Tests the capability of a process to produce products that meet the required design specifications

• Cpk =

• An in-control process has a Cpk≥1• Note: The estimate of σ=σ̂ = R ̄/d2 =

Total Quality Management

• Should consider quality along eight basic dimensions, that can be grouped into two categories:– Performance– Features– Reliability– Conformance– Durability– Serviceability– Aesthetics– Perceived quality

Conformance to Requirements

Customer Satisfaction

TQM - Benchmarking

• Can measure performance against competitors’ performance by using benchmarking:

1) Product benchmarking – comparing the product attributes

2) Functional benchmarking – compare processes3) Best practices benchmarking – compare

management practices4) Strategic benchmarking – compare each firm’s

strategic focus

Lecture 21

Acceptance Sampling

Acceptance SamplingN = Number of items in a lotM = actual number of defectives in the lotp = Proportion of defectives in the lotn = Number of items in a sampleX = Number of defectives in the samplec = Rejection level; reject if X>cp0 = Acceptable Quality Level (AQL); desired or target proportion

of defectives; accept lot if true proportion of defectives is ≤ p0

p1= Lot Tolerance Percent Defective (LTPD); reject lot if true proportion of defectives is > p1

α = Producer’s risk. The probability of rejecting a good lot.

=

Β = Consumer’s risk. The probability of accepting a bad lot.

=

OC Curve

• OC(p) = P{Accepting the lot | True proportion of defectives=p} = P{X≤c | Proportion of defectives in lot=p}

=

• Shows the relationship between the probability of accepting the lot and its quality level

• Shows how well a sampling plan discriminates between good and bad lots

OC Curve

Probability of Acceptance

Percent defective

| | | | | | | | |0 1 2 3 4 5 6 7 8

100 –95 –

75 –

50 –

25 –

10 –

0 –

LTPDAQLBad lotsIndifference

zoneGood lots

α

β

Average Outgoing Quality (AOQ)

• As acceptance sampling doesn’t involve testing every unit of output, it is possible that – Bad lots will be passed– Good lots will be rejected

• AOQ tests the effectiveness of the sampling plan • AOQ is the average percent defective• Two situations:

– defective items are not replaced– Defective items are replaced

AOQWithout Replacement With Replacement

AOQ

Approximation of AOQ when N>>n Pap

Note: • it is also assumed that rejected lots are subject to 100%

inspection• Pa = OC(p) = P{lot is accepted|p}

AOQL

• Average Outgoing Quality Limit

• The maximum AOQ• Can find the AOQL by

plotting the AOQ curve & determining the maximum of the curve

AOQL

Lecture 22

Supply Chain Management

Supply Chain Management

Goal: maximize profits across whole supply chain

Supply Chain: describes how organizations (suppliers, manufacturers, distributors and consumers) are linked together

Supply Chain Management: a total system approach to managing the entire flow of information, materials and services from raw materials through factories and warehouses to the end consumer

*Competition is no longer between companies but between supply chains

Elements in Supply Chain Management

Deciding how to best move and store materialsLogistics

Determining location of facilitiesLocation

Monitoring supplier quality, delivery, and relationsSuppliers

Evaluating suppliers and supporting operationsPurchasing

Meeting demand while managing inventory costsInventory

Controlling quality, scheduling workProcessing

Incorporating customer wants, mfg., and timeDesign

Predicting quantity and timing of demandForecasting

Determining what customers wantCustomers

Typical IssuesElement

Supply Chain Strategy

Supply Chain Measurements

valueinventory aggregate Averagesold goods ofCost turnoverInventory

In situations where distribution inventory is dominant, “Weeks of Supply” is preferred and measures how many weeks’ worth of inventory is in the system at a particular time

weeks52 sold goods ofCost

valueinventory aggregate Averagesupply of Weeks

Suppose a company’s new annual report claims their costs of goods sold for the year is $160 million and their total average inventory (production materials + work-in-process) is worth $35 million. This company normally has an inventory turn ratio of 10. What is this year’s Inventory Turnover ratio? What does it mean?

IT = 160/35 = 4.6

Meaning: Inventory Turnover is poor, half as good as before, you are now selling inventory at a slower rate

Example

Bullwhip Effect: occurs when slight demand variability is magnified as information moves back upstream

Supply Chain Enablers- EDI- Bar Codes- Radio Frequencies- Internet

Methods of Distribution

1. Railroads2. Trucking – most common3. Air4. Water5. Pipelines

Cost of Shipping Alternatives

• Product in transit is a form of inventory and has carrying costs

• Faster shipping is generally more expensive than slow

• We can evaluate the two costs to better understand the trade offs

Value Density

Defined: value density is the value of an item per unit of weight

• Important measure when deciding how items should be shipped

• Low value density = cheapest method• High value density = fastest and safest method

Lecture 23

Supply Chain Contracts

Fixed Production Cost =$100,000

Variable Production Cost=$35

Manufacturer Manufacturer DCRetail DC

Stores

Selling Price=$125Salvage Value=$20

Wholesale Price =$80

A Basic Supply Chain

Variables To SolveWin-Win Contracts

F: Fixed Production CostP: Variable Production CostQ: Quantity SuppliedQs: Quantity SoldC: Wholesale PriceS: Selling PriceV: Salvage Value

Formulas

Retailer’s Profit: (S-V)Qs – (C-V)QRevenue: S*Qs + V(Q-Qs)Cost: C*Q

Manufacturer's Profit: (C-P)Q – FRevenue: CQCosts: PQ – F

Supply Chain Profit: (S-V)Qs – (P-V)Q – F

Supply Chain Contracts

Key Insights• Effective supply constraints allow supply chain

partners to replace sequential optimization by global optimization

• Buy back and revenue sharing contracts achieve this objective through sharing risks and rewards

• Other types of contracts can also be designed to achieve this (Quality Flexibility, Sales Rebate)