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

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

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

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

5. 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

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 6

7. 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

8. EOQ Model Total Cost With 1 destination
$1*1,000*2,500/Q *$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

9. 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

10. 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

11. Total Cost 11

12. 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

13. 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

14. 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

15. 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

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 16

17. 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

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

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 19

20. 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* = $ 15,000
Total = $290,000 20

21. 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

22. 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

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

24. 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