Stracener_EMIS 7305/5305_Spr08_04.07.08 1 Supply Chain Analysis Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7305/5305 Systems Reliability,

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Stracener_EMIS 7305/5305_Spr08_04.07.08

Supply Chain Analysis

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7305/5305Systems Reliability, Supportability and Availability Analysis

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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Objectives Supply Chain Analysis

• Analysis provides understanding of critical tradeoffs and alternatives in practical decision-making for a range of inter-related supply chain management issues: – Structure of the Supply Chain:

• “Optimal" numbers• Location

based on considerations such as customer service requirements, leadtimes, operational costs, and capacities.

– Supply Uncertainty: • Relationship with suppliers• Selection of suppliers

based on cost, flexibility in supply contracts, expected learning curves of suppliers, and agreements on cost and information sharing.

– Operational Policies: • Inventory control policies• Information-sharing strategies.

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What is a Supply Chain?

• Network of Support Resources including:– Material– People– Infrastructure

• To provide essential services to:– Store serviceable parts– Repair unserviceable parts– Transport parts between various sites

• That sufficiently and efficiently satisfies customer need for:– Asset Management– Asset Demands.

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Basing Site

Depot/Warehouse Site

Manufacturing Site

Operational Unit

What is a Supply Chain?

Parts and Information

LocalPart

Repair

DepotPart

Repair

Remanufacture

Product Use

Product Maintenance

LocalStock

Warehouse

Manufacture

InformationSystem

Good Parts DataBad

Dispose

Dispose

Dispose

Dispose

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SC Metrics

• Availability– Fill rate = # issued divided by number requested

• Delay– Backorders = number ordered from next level– Backorder time = average time to issue from next level– Supply Downtime = average time user waits on spare– Transport time = average time to move item from warehouse to user– Repair time = average time to remanufacture item– Retrograde time = average time to return item for remanufacturer– Production lead time = average time to manufacture new item

• Efficiency– Turn rate = # issues/ # stocked for period of time (usually annual)– Shelf cost = average value of items on shelf times cost of money

factor

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Analysis Methodologies

• Static– One commodity at one operational site with one supply

warehouse– Multi-indentured system at one operational site one

supply warehouse– Multi-indentured system at multiple operational sites

with a multi-echelon supply chain• Dynamic

– Fleet changes– Repairs– Builds– Reallocations

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Simple Supply Chain Model

• One Item

• Driver is MTBD (mean time between demand) for item

– AKA MTBR (mean time between removal)

• Fill rate(FR) is assessed using POISSON distribution

• Given:

– Operate time: T

– Stockage (# of spares): N

• Mean demand: λ = T/MTBD

• FR =probability that demand ≤ N = POISSON(N,λ,1)

Number of Failures Model:

• DefinitionIf T ~ E() and if X is the number of failures occurringin an interval of time, t, then X ~ P(t/ ), the Poisson Distribution with Probability Mass Function given by:

for x = 0, 1, ... , n

Where = 1/ is the Failure Rate

• The expected number of failures in time t is

= t = t/

!x

et)xX(P)x(P

tx


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Poisson Distribution

• Mean or Expected Value

• Variance and Standard Deviation of X

tXE

tXVar 2

t

The Poisson Model:


The Poisson Model: Probability Distribution Function

0 1 2 3 4 5 6 7 8

1.0

0.5

0.0

x

0y

ypF(x)

.135

.405

.675

.855

XxPF(x)


The Poisson Model:


The Poisson Model - Example Application:

An item has a failure rate of = 0.002 failures per hour if the item is being put into service for a period of 1000 hours. What is the probability that 4 spares in stock will be sufficient?


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Expected number of failures (spares required) = t = 2

P(enough spares) = P(x 4) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.945or about a 5% chance of not having enough spares!

The Poisson Model - Example Application - Solution

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Poisson Distribution - Example

When a company tests new tires by driving themover difficult terrain, they find that flat tiresexternally caused occur on the average of once every 2000 miles. It is found also that the Poissonprocess yields a useful model. What is the probabilitythat in a given 500 mile test no more than one flatwill occur?

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Poisson Distribution - Example Solution

Here the variable t is distance, and the random variable of interest is

X = number of flats in 500 miles

Since E(X) is proportional to the time interval involvedin the definition of X, and since the average is givenas one flat is 2000 miles, we have

miles 500in flat 4

1)( XEt

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The values assigned to and t depend on the unitof distance adopted. If we take one mile as the unit,then t = 500, = 0.0005, and t = 1/4. If we take1000 miles as the unit, then t = 1/2, = 1/2, and again t = 1/4, and so on. The important thing is that t = 1/4, no matter what unit is chosen.

)1()miles 500in flat 1( XPP

)1()0( pp

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97.0

4

54/1

e

!1

4/1

!0

4/1 104/1e

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Poisson Model - Example

An item has a MTBD of 500 hours. If the item is being put into service for a period of 1000 hours, what is the fill rate probability if 4 spares are in stock?

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• Expected demands (spares required) = =1000/500= 2

• P(enough spares) = P(x 4) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.945

• About 95% fill rate.

Poisson Model – Example Solution

Documents

Stracener_EMIS 7305/5305_Spr08_04.07.08 1 Supply Chain Analysis Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7305/5305 Systems Reliability,