1 1 In-Bound Logistics John H. Vande Vate Fall, 2002.

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In-Bound Logistics

John H. Vande Vate

Fall, 2002

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Exam 2

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Scores

• Average 81

• Std. Dev: 11

• Remember: All questions about grading submitted IN WRITING

• Probably made mistakes

• Happy to review grading

• Unwilling to discuss grading

44

Causes of Bullwhip Today

• Product Proliferation/Mass Customization– More varieties of products

• Build-to-Order – Prohibits pooling orders to smooth

requirements

• Lean– Prevents pooling releases to smooth

demand on the supply chain

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Why Lean (Just-In-Time)?

• Reduces inventory– Capital requirements– Etc

• Reduces handling – Direct-to-Line

• Improves Quality – See problems quickly

• Increases launch speed

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Why Not Lean?

Capacity

• Changes in requirements create upstream inventory

• Changes in requirements raise transport costs

Reliability

• Distant supplies subject to disruption

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Lean Works

When

• Total volume is relatively constant

• Product variety is limited

• Changeovers are fast and cheap

• Suppliers are nearby

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US Auto Industry

• Total volume is relatively constant

• Product variety is limited

• Changeovers are fast and cheap

• Suppliers are nearby

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How Lean Works

• Manufacturer has standing PO with supplier– Releases permission to supply against that PO

– Daily quantities or even more frequent to match planned production

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How Lean Works

Daily Reciept

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01-Apr-02 06-Apr-02 11-Apr-02 16-Apr-02 21-Apr-02 26-Apr-02

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A Financial Model

Cash Acct

From Revenues

Cash Expenses

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Invest

Sell Assets

A Financial Model

Cash Acct

From Revenues

Cash Expenses

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Controls

When Cash balance reaches here

Invest enough to bring it to here

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Controls

When Cash balance falls to here

Sell assets to bring it to here

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Controls

T

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Trade-offs

• Opportunity cost of Cash Balance• Transaction costs of investing and

selling assets• Set the controls, T, t and b to balance

these costs

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Inventory Analogy

• Cash Expenses Daily Production reqs.• From Revenue Constant supplies• Sell Assets Expedited order• Invest Excess Curtailed order

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Trade-offs

• Opportunity cost of Cash Balance

• Transaction costs of investing and selling assets

• Cost of holding Inventory

• Supply chain costs of expediting and curtailing orders

• Set the controls, T, t and b to balance these costs

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Cost Components

Cost of Inventory

• H: an interest rate on the value of the goods ($/item/year)

Cost of Expediting• E: extra transport costs (above std) ($/event)

Cost of Curtailing• C: disruption costs - savings over std transport

($/event)

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Total Cost

• Minimize H*Average Inventory Level E*Expected number of times we expedite C*Expected number of times we curtail

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Simple Model

• Simple model of production requirements

Probability

• Avg demand: a units/day 1 – 2p

• Above avg demand: a + units p

• Below avg demand: a - units p

• Standard Supply: a units/day

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A Markov Process

0 2 b Tt… … …

“Cash Balance”

p p p p p

p p p p p

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1-2p

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Transition Matrix

0 … b … t … T0 0 0 0 0 0 1 0 0 0 0 p 1-2p p 0 0 0 0 0 0 0 0 p 1-2p p 0 p 1-2p p… 0 p 1-2p p

b 0 p 1-2p p… 0 p 1-2p pt 0 p 1-2p p

… 0 p 1-2p pT 0 0 0 0 0 0 0 1 0 0

Drop . So k refers to k

Instead of b, t, T we find b, t, T

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Steady State Probabilities (i): Steady state probability “cash balance is

i* (0) = p(1) • 2p(1) = p(2) • 2p(i) = p(i-1)+p(i+1), i=2, 3, …, b-1• 2p(b) = (0) + p(b-1)+p(b+1) • 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t• 2p(t) = (T) + p(t-1)+p(t+1)• 2p(i) = p(i-1)+p(i+1), i=t+1, t+2,…, T-2• 2p(T-1) = p(T-2) (T) = p(T-1)

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Steady State Probabilities (i): Steady state probability “cash balance is i* (0) = p(1) => (1) = (0)/p• 2p(1) = p(2) => (2) = 2(1) = 2(0)/p• 2p(i) = p(i-1)+p(i+1), i=2, 3, …, b-1• 2p(b) = (0) + p(b-1)+p(b+1) • 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t• 2p(t) = (T) + p(t-1)+p(t+1)• 2p(i) = p(i-1)+p(i+1), i=t+1, t+2,…, T-2• 2p(T-1) = p(T-2) == (T-2) = 2(T-1) = 2(T)/p (T) = p(T-1) == (T-1) = (T)/p

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Steady State Probabilities (i): Steady state probability “cash balance is i* (0) = p(1) => (1) = (0)/p• 2p(1) = p(2) => (2) = 2(1) = 2(0)/p• 2p(i) = p(i-1)+p(i+1), i=2, 3, …, b-1

• 2p(2) = p(1)+p(3) == 4(0) = (0)+p(3) (3) = 3(0)/p

• 2p(i) = p(i-1)+p(i+1) • 2i(0) = (i-1)(0)+p(i+1) • (i+1) = (i+1)(0)/p

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Steady State Probabilities (i): Steady state probability “cash balance is i* (T) = p(T-1) => (T-1) = (T)/p• 2p(T-1) = p(T-2) => (T-2) = 2(T-1) = 2(T)/p• 2p(i) = p(i-1)+p(i+1), i=t+1, t+2,…, T-2• 2p(T-2) = p(T-1)+p(T-3) == 4(T) = (T)+p(T-3)

(T-3) = 3(T)/p

• 2p(T-i) = p(T-i+1)+p(T-i+1) • 2i(T) = p(T-i-1)+(i-1)(T) • (i+1)(T) = p(T-i-1) (T-i-1) = (i+1)(T)/p

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Summary

(i) = i(0)/p, i=1, 2, …, b (T-i) = i(T)/p, i=1, 2, …, T-t

• Still have to solve• 2p(b) = (0) + p(b-1)+p(b+1) • 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t• 2p(t) = (T) + p(t-1)+p(t+1)

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Between b and t

• 2p(b) = (0) + p(b-1)+p(b+1)

• 2b(0) = (0) + (b-1)(0)+p(b+1) • b(0) = p(b+1) (b+1) = b(0)/p

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Summary

(i) = i(0)/p, i=1, 2, …, b (T-i) = i(T)/p, i=1, 2, …, T-t

• We just solved• 2p(b) = (0) + p(b-1)+p(b+1) (b+1) = b(0)/p• Now we have to solve…• 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t• 2p(t) = (T) + p(t-1)+p(t+1)

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Between b and t

• 2p(t) = (T) + p(t-1)+p(t+1)• 2(T-t)(T) = (T) + p(t-1)+(T-t-1)(T)• (T-t)(T) = p(t-1) (t-1) = (T-t)(T)/p

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Summary

(i) = i(0)/p, i=1, 2, …, b (T-i) = i(T)/p, i=1, 2, …, T-t (b+1) = b(0)/p

• We just solved• 2p(t) = (T) + p(t-1)+p(t+1) to get (t-1) = (T-t)(T)/p

• Now we have to solve…• 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t

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Between b and t

• 2p(i) = p(i-1)+p(i+1), i=b+1, b+2,…, t• 2b(0) = b(0)+p(i+1) (i+1) = b(0)/p

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Summary (i) = i(0)/p, i=1, 2, …, b (T-i) = i(T)/p, i=1, 2, …, T-t (i) = b(0)/p, i = b, b+1, …, t

• And (t-1) = (T-t)(T)/p

• So• b(0)/p = (T-t)(T)/p (T) = b(0)/ (T-t) and (T-i) = i(T)/p, i=1, 2, …, T-t becomes (T-i) = [ib/(T-t)](0)/p

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Summary (i) = i(0)/p, i=1, 2, …, b (i) = b(0)/p, i = b, b+1, …, t (i) = [(T-i)/(T-t)]b(0)/p, i=t, t+1, …, T-1 (T) = b(0)/(T-t)

• 1 = (i) = • = (0) + (i(0)/p: i = 1, .., b-1) + (b(0)/p: i = b, .., t)

+ (ib(0)/[p(T-t)]: i = 1, .., T-t-1) + b(0)/[T-t]

= (0) + b(b-1) (0)/2p + 2(t-b+1)b(0)/2p

+ (T-t-1)b (0)/2p + b(0)/[T-t]

= (0) + [T+t-b]b(0)/2p + b(0)/[T-t] = (0)[2p(T-t+b) + b(T-t)(T+t-b)]/[(T-t)2p]

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Calculating (0)

• 1 = (0)[2p(T-t+b) + b(T-t)(T+t-b)]/[(T-t)2p] (0) = 2p(T-t)/[2p(T-t+b) + b(T-t)(T+t-b)]

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Costs

• Expected number of times we expedite (0)*Number of “Days” in Year

• Expected number of times we curtail (T)*Number of “Days” in Year

• Average Inventory Level (i(i): i = 0, .., T)

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Average Inventory

i(i) = • = (i2(0)/p: i = 1, .., b-1) + (ib(0)/p: i = b, .., t-1) +

(i(T-i)b(0)/[p(T-t)]: i = t, .., T-1) +Tb(0)/[T-t]

= b[3T(T-t) – 2(T-t)2 + 3(t2 – b2) + 6b-4] (0)/6p

5 points extra credit for first to find any errors

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Average Inventory

i(i) = b[3T(T-t) – 2(T-t)2 + 3(t2 – b2) + 6b-4] (0)/6p

(0) = 2p(T-t)/[2p(T-t+b) + b(T-t)(T+t-b)]

i(i) = b[3T(T-t) – 2(T-t)2 + 3(t2 – b2) + 6b-4] (T-t)/

[6p(T-t+b) + 3b(T-t)(T+t-b)]

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Costs

• Expected number of times we expedite (0)*Number of “Days” in Year• 2p(T-t)N/[2p(T-t+b) + b(T-t)(T+t-b)]• Expected number of times we curtail (T)*Number of “Days” in Year• 2pbN/[2p(T-t+b) + b(T-t)(T+t-b)]• Average Inventory Level (i(i): i = 0, .., T) b[3T(T-t) – 2(T-t)2 + 3(t2 – b2) + 6b-4] (T-t)/

[6p(T-t+b) + 3b(T-t)(T+t-b)]

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Convert Variables

• Focus on the differences

• b, x, yT

b

t

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y

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Optimize• Expected number of times we expedite• E*(0)*Number of “Days” in Year• E2pyN/[2p(y+b) + by(y+2x+b)]• Expected number of times we curtail• C*(T)*Number of “Days” in Year• C2pbN/[2p(y+b) + by(y+2x+b)]• Average Inventory Level• H(i(i): i = 0, .., T) • Hby[3Ty – 2y2 + 3x(x+2b) + 6b-4]/ [6p(y+b) + 3by(y+2x+b)]

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Optimize• Total Cost• {6pN(Ey+Cb) + Hby[3(b+x+y)y – 2y2 + 3x(x+2b) + 6b-4]}/ [6p(y+b) + 3by(y+2x+b)]

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DellParameter Value Solution Solve forp 0.5 b 4 b 4 69 t 10 x 6 Max YN 250 T 12 y 2 2E 500C 300H 10a 147State Probability Inventory Adjust Total Cost Average Inventory Avg. Days of Supply

0 0.013333333 0 1666.666667 8156.267 448.96 3.05414971 0.026666667 1.84 02 0.053333333 7.36 0 Max Inventory Max Days of Supply3 0.08 16.56 0 828 5.63265314 0.106666667 29.44 05 0.106666667 36.8 0 Total Adjustment Costs Total Inventory Costs6 0.106666667 44.16 0 3,666.67$ 4,489.60$ 7 0.106666667 51.52 08 0.106666667 58.88 09 0.106666667 66.24 0

10 0.106666667 73.6 011 0.053333333 40.48 012 0.026666667 22.08 200013 0 0 014 0 0 015 0 0 016 0 0 017 0 0 0

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Additional Constraint

• Can’t curtail more than one days usage

• On average y a

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Parameter Value Solution Solve forp 0.5 b 4 b 4 69 t 10 x 6 Max YN 250 T 12 y 2 2E 500C 300H 10a 147State Probability Inventory Adjust Total Cost Average Inventory Avg. Days of Supply

0 0.013333333 0 1666.666667 8156.267 448.96 3.05414971 0.026666667 1.84 02 0.053333333 7.36 0 Max Inventory Max Days of Supply3 0.08 16.56 0 828 5.63265314 0.106666667 29.44 05 0.106666667 36.8 0 Total Adjustment Costs Total Inventory Costs6 0.106666667 44.16 0 3,666.67$ 4,489.60$ 7 0.106666667 51.52 08 0.106666667 58.88 09 0.106666667 66.24 0

10 0.106666667 73.6 011 0.053333333 40.48 012 0.026666667 22.08 200013 0 0 014 0 0 015 0 0 016 0 0 017 0 0 0

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Differences

• Constant Stream of Releases punctuated by Expediting and Curtailing

• If supplier can see inventory and knows T, can anticipate and plan for coming expedited and curtailed orders

• Have to set a lower bound > 0 to protect against disruptions – safety stock

• Complicates the calculation of cost of Expediting

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Example: Shipments

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Example: Inventory

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