John H. Vande Vate Spring, 2005 Frequency John H. Vande Vate Spring, 2005 1

Review Our initial case study Direct deliveries Consolidation Transportation Costs: $ 460,000 Pipeline Inventory: $ 360,000 Inventory Costs at Plants: $ 232,500 Inventory Costs at DCs: $ 23,250,000 Total: $ 24,302,500! Consolidation Transportation Costs: $ 751,800 Pipeline Inventory: $ 435,000 Inventory Costs at Plants: $ 232,500 Inventory Costs at XDock $ 281,640 Inventory Costs at DCs: $ 4,914,000 Total: $ 6,614,940 The big opportunity 2

The Trade Off Consolidation Increased Inventory at the Warehouse Increased Transportation Decreased Inventory at the DCs! 3

Frequency Ship in less-than-full truck load quantities Why? Increase Transportation … 4

Selecting Frequency Trade off Recognize this? Transportation cost Truck costs the same regardless of load Assuming we don’t change to LTL Inventory cost The more the truck carries the greater the inventory at both ends Recognize this? 5

A Model Start with the direct model Consider CPU’s from Green Bay Q = quantity to send in each truck Constraint: Q  6,000 Annual transportation cost to 1 destination $/mile * Miles/trip * Trips/year $1* 1000 * ? Trips/year = Annual Demand/Q = 2,500/Q 6

More Model Annual transport cost to 1 destination Inventory cost $1*1,000*2,500/Q = 2,500,000/Q $/year Inventory cost Start with 1 destination Expand to many destinations Inventory with 1 destination Holding % * $/item * Average Inventory level Average Inventory level = ? 7

EOQ Model Total Cost With 1 destination $1*1,000*2,500/Q + 0.15*$300*Q (or Q/2) General Form with 1 destination A = fixed cost per trip = $1,000 D = Annual Demand = 2,500 h = Holding percentage = 0.15 C = Cost per item = $300 Total Cost = AD/Q + hC*Q 8

Optimal Frequency Do the math Discrete Thinking dTotalCost/dQ = 0 Discrete Thinking One more item on the truck Inventory cost of that item is hC Transport impact is to save AD/Q – AD/(Q+1) = AD/[Q(Q+1)] ~ AD/Q2 Stop adding when costs = savings hC = AD/Q2 Intuition: Balance Inventory and Transport Cost hCQ = AD/Q Best Answer: Q = AD/hC 9

Optimal Frequency With One destination What if it had been 23,500? A = fixed cost per trip = $1,000 D = Annual Demand = 2,500 h = Holding percentage = 0.15 C = Cost per item = $300 Total Cost = AD/Q + hC*Q Q* = AD/hC = 2,500,000/45 ~ 235 What if it had been 23,500? Remember, Q  6,000 10

Total Cost 11

With Several Destinations item-days inventory at the plant accumulated for each shipment to DC #1, say, if the shipment size is Q? Q2/(2*Production Rate) Q 12 Q/Production Rate

Total Item-Days How many such shipments will there be? Annual Demand at DC #1/Q So, the total item-days per year from shipments to DC #1 will be… Q2/(2*Production Rate)*Demand at DC/Q Q*Demand at DC/(2*Production Rate) So, making shipments of size Q to DC #1 adds what to the average inventory at the plant? 13

Effect on Average Inventory Q*Demand at DC/(2*Production Rate) Example: Q*2500/(2*100*2500) = Q/200 Correct EOQ for Direct Shipments: Total Cost: hC*Q*D/(2*Production Rate) + hC*Q/2 +A*D/Q Q* = 2*A*D/hC P/(D+P) 14

In Our Case Since Demands at the n DCs are equal P/(D+P) = nD/(nD+D) = n/(n+1) Q* = 2*A*D/hC P/(D+P) Q* = 2*A*D/hC n/(n+1) Q* = 2*1000*2500/(0.15*30) 100/101 Q* = 332 The main point is the 2 – that’s 40% larger! Why? 15

Total Cost of Direct Strategy CPU’s Q* = 332 Consoles Q* = 574 Monitors/TV’s Q* = 406 Transport Costs CPU’s = 100*2500*1000/332 = $753,000 Monitors = 100*5000*1000/406 = $1,232,000 Consoles = 100*2500*1000/574 = $436,000 Total $2,421,000 16

Inventory Costs Inventory Costs CPU’s Consoles Monitors At Plant Q/2 Why? At DC Q/2 Total 101*Q/2 CPU’s 15%*$300*101*332/2 = $754,000 Consoles 15%*$100*101*574/2 = $435,000 Monitors 15%*$400*101*406/2 = $1,230,000 Total: $2,419,000 17

Strategies Direct Full Trucks: $ 24,302,500 Consolidate Full Trucks: $ 6,614,940 Direct EOQ: $ 5,300,000 Other strategies? 18

Consolidate & EOQ EOQ from Plant to Indianapolis This is like serving one destination Q* = AD/hC CPU’s from Green Bay Q* = 400*250,000/0.15*300 = 1490 Consoles from Denver Q* = 1100*250,000/0.15*100 = 4281! Monitors/TV’s from Indianapolis Q* = 0*500,000/0.15*100 = 0 19

From the Plants Green Bay to Indianapolis Denver to Indianapolis Transport: 400*250,000/1490 = $67,114 Inventory: The same = $67,114 Total = $134,228 Denver to Indianapolis Transport: 1100*250,000/1,000 = $275,000 Inventory: 0.15*100*1000 = $ 15,000 Total = $290,000 20

From the Warehouse This is a case of serving many D = 2,500 Q* = 2*A*D/hC P/(D+P) What’s P? Q* = 2*A*D/hC n/(n+1) D = 2,500 C = $1,200 Why? A = $1,000 Q* = 2*1,000*2,500/180 100/101 = 165 21

Costs from Warehouse Transportation: Inventory: 1000*2,500/165 = $15,152 per DC Total $1,515,200 Inventory: 0.15*1,200*101*Q/2 = $1,500,000 Pipeline Inventory $435,000 Total Cost of Strategy ~$3,874,000 22

Review of Options Direct Full Trucks: $ 24,302,500 Consolidate Full Trucks: $ 6,614,940 Direct EOQ: $ 5,300,000 Consolidate EOQ: $ 3,439,000 Other strategies? 23

Deterministic Supply Chain Design Consolidation Frequency Higher level look at the financial values Ford Finished Vehicle Case Study Trading off pipeline inventory and cross dock inventory 24