1. 1 1 Frequency: Inventory vs Transportation John H. Vande Vate Spring, 2007

2. 2 2 Review Our initial case study –Direct deliveries Transportation Costs: $ 460,000 Pipeline Inventory: $ 360,000 Inventory Costs at Plants: $ 232,500 Inventory Costs at Stores: $ 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 Stores: $ 4,914,000 Total: $ 6,614,940 The big opportunity

3. 3 3 The Trade Off Consolidation –Increased Inventory at the Warehouse –Increased Transportation –Decreased Inventory at the Stores!

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

5. 5 5 Selecting Frequency Trade off –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?

6. 6 6 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

7. 7 7 More Model Annual transport cost to 1 destination –$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 = ?

8. 8 8 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

9. 9 9 Optimal Frequency Do the math –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/Q 2 –Stop adding when costs = savings hC = AD/ Q 2 Intuition: Balance Inventory and Transport Cost –hCQ = AD/Q Best Answer: Q =  AD/hC You often see this as  2AD/hC We don’t have the 2 because we considered inventory at both ends.

10. 10 Optimal Frequency With One 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 –Q* =  AD/hC =  2,500,000/45 ~ 235 What if it had been 23,500? Remember, Q  6,000

11. 11 Total Cost

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

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

14. 14 Effect on Average Inventory Q*Demand at Store/(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)

15. 15 In Our Case Since Demands at the n Stores 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? What if n = 1? What if n is large, ignore inventory at the plant.

16. 16 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

17. 17 Inventory Costs –At Plant Q/2 Why? –At Store 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

18. 18 Strategies Direct Full Trucks: $ 24,302,500 Consolidate Full Trucks: $ 6,614,940 Direct EOQ: $ 5,200,000 Other strategies?

19. 19 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

20. 20 From the Plants Green Bay 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

21. 21 From the Warehouse This is a case of serving many 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

22. 22 Costs from Warehouse Transportation: –1000*2,500/165 = $15,152 per Store –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

23. 23 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?

24. 24 Big Question What roles do facilities play in the transportation network –Change of Mode (Ports, Airports, Rail terminals, cross docks,…) –Consolidation/Deconsolidation Sorting Sequencing Repackaging –Product completion Postponement –Storage Reduce lead time by keeping product near market –…

25. 25 Summary Frequency –Trade-off between transportation and inventory –EOQ model Do we consider inventory at both ends or not 1-to-Many vs 1-to-1 model (special case)

26. 26 Next Time Network Models