Transportation/Logistic Strategy Introduction to Inventory Total Cost Analysis Computation of Safety Stocks Setting Order Points Joint Replenishment Number of Facilities and Inventory

Transportation/Logistic Strategy Total Cost Analysis Start with EOQ Identify alternatives Vendor discounts Transportation options Other issues Examine only total relevant costs Select least cost alternative

Transportation/Logistic Strategy Economic Order Quantity Order Quantity Cost of Ordering Cost of Carrying Inventory Total Cost $ *

Transportation/Logistic Strategy Interest and Opportunity Cost % Storage and Handling8-14 % Depreciation and Obsolescence6-12 % Insurance and Taxes 2 -3 % Total % Elements of Inventory Carrying Costs

Transportation/Logistic Strategy Order Quantity Cost of Ordering Cost of Carrying Inventory Total Cost $ (With Constant Ordering Costs) Zero Inventory JIT Economic Order Quantity

Transportation/Logistic Strategy Total Cost = OC + CC OC = Order Placement Cost = A(R/Q) CC = Inventory Carrying Cost = 1/2(QVW) Where: Q = Optimal Order Quantity (EOQ) A = Cost of placing an order R = Annual Rate of use V = Value per unit W = Carrying cost as a percentage of average value of inventory Determining EOQ

Transportation/Logistic Strategy Average Inventory Quantity Time Average Inventory Inventory Carrying Cost = 1/2 (QVW) Where: Q=Order Quantity V=Value per Unit W= Carrying Cost as a Percentage of the Average Value of Inventory

Transportation/Logistic Strategy Total Cost = OC + CC OC = A(R/Q) CC = 1/2(QVW) Total Cost = A(R/Q) + 1/2(QVW) and solve for QTo minimize cost, set 1st derivative = zero TC = ARQ QVW ARQVW Q 2 + TC = or TC ’ = - ARQ VW or -ARVW Q2Q2 2 + TC ’ = = 0 ARVW Q2Q2 2 = or 2AR VW Q2Q2 =  Q =Q = 2AR VW Determining EOQ

Transportation/Logistic Strategy Total Cost = OC + CC + Tr + PC + It + SS + Other Where: OC = Order Placement Cost CC = Inventory Carrying Cost Tr = Transportation Cost PC = Product Cost It = Inventory in Transit Cost SS = Safety Stock Cost Total Cost Analysis

Transportation/Logistic Strategy Total Cost = OC + CC + Tr + PC + It + SS + Other OC = A(R/Q) CC = 1/2(QVW) Tr = rRwt/100 PC = VR It = iVRt/365 SS = BVW Where: Q, R, A, V, W = As previously defined r = Transportation rate per 100 pounds (CWT) wt = Weight per unit i = Interest rate or cost of capital t = Lead time in days B = Buffer of inventory to prevent stockouts Total Cost Analysis

Transportation/Logistic Strategy Total Cost without Coordination of Purchasing and Transportation TC CC PC TR OC Quantity ~ ~ $ r1r1 r2r2 p2p2 p1p1

Transportation/Logistic Strategy Total Cost without Coordination of Purchasing and Transportation PC TR OC CC TC $ Quantity ~ ~ r 1/ p 1 r 2/ p 2

Transportation/Logistic Strategy If sales or lead times vary, safety stock is necessary to maintain desired service levels. Computing Safety Stock Begin with Order Point (BVW) If sales are constant and there is no variation in delivery times, no buffer is necessary, i.e., safety stock is not a relevant cost.

Transportation/Logistic Strategy Setting Order Points Order Point = Daily Use X Lead Time in Days Place orders when stock level falls to the level of the order point. Lead Time Quantity Order Point Time No Uncertainty

Transportation/Logistic Strategy Order Point = Mean Daily Use X Lead Time in Days To Protect Against Stockouts, Place Orders Sooner But How Much Sooner? Lead Time Quantity Order Point Time Stockout Setting Order Points With Uncertainty

Transportation/Logistic Strategy Place orders when stock levels fall to the mean, and half the time you have enough to cover sales and half the time you don’t; i.e., you have a 50% Fill Rate. Raise the order point, and you raise the service level. One Standard Deviation of Buffer Stock Provides an 84% Fill Rate Two Standard deviations of Buffer Stock Provides a 97.5% Fill Rate How Do We Know? Example Setting Order Points With Uncertainty

Transportation/Logistic Strategy Average Daily Sales Sales Day (X i ) (X i ‑ X) (X i – X) ‑ ‑ ‑ ‑ SUM Mean = 100; Std Dev = 10

Transportation/Logistic Strategy Mean and Standard Deviation of Average Daily Sales Mean = 1200/12 = 100 ___________ ________ Std Dev = √  (X i – X) 2 = √ 1100/11 = 10 n-1

Transportation/Logistic Strategy Assume constant delivery time of one day, i.e., no variation Order Point = Mean Daily Use X Lead Time in Days + Buffer Stock Mean Sales = 100 units Standard Deviation of Sales = 10 units Order Point = = 110 Setting Order Points With Uncertainty

Transportation/Logistic Strategy Order Point = Daily Use X Lead Time in Days + Buffer Stock By Ordering Sooner, Stockouts are Avoided Lead Time Quantity 110 Time 100 What is the service level (fill rate) associated with this order point? Setting Order Points With Uncertainty

Transportation/Logistic Strategy One Standard Deviation = 68% of the area. Half of the time sales are below 100; 84% of the time sales are below % 34% 10090Sales110 Computing Safety Stock

Transportation/Logistic Strategy Two Standard Deviations = 95% of the area. Half the time sales are below 100; 97.5% of the time sales are below % 47.5% 10080Sales 120 Computing Safety Stock

Transportation/Logistic Strategy z \\\ \\\\ Areas under the standard normal probability distribution between the mean and successive values of z

Transportation/Logistic Strategy Effects of Delivery Time Variation Where: S Dt = Standard Deviation of Demand over time(Units of Safety Stock required to satisfy 68 percent of sales levels during lead time) t = Average delivery time S t = Standard Deviation of delivery time D =Average Demand S D =Standard Deviation of Demand S Dt = (t)(S D ) 2 + (D) 2 (S t ) 2 

Transportation/Logistic Strategy Example ASSUMPTIONS Average Sales (D) = 100 units Standard Deviation of Sales (S D ) = 10 units CARRIER A CARRIER B Average Transit Time (t) 5 Days 5 Days Standard Deviation of Time (S t ) 1 Day 2 Days

Transportation/Logistic Strategy Combined standard deviations CARRIER A S Dt =  (5) (10) 2 + (100) 2 (1) 2 = 102 units S Dt = √ (5) (10) 2 + (100) 2 (2) 2 = 201 units CARRIER B (Continued) Example

Transportation/Logistic Strategy Assumptions: Cost of Carrying Safety Stock = BVW (Buffer Stock) X (Value) X (Carrying Cost) Value per Unit = $ Carrying Cost = 20 percent Service Level = 97.5 percent a ________________________________________________ a To attain a 97.5% service level, the shipper will have to add 2 standard deviations of safety stock to the order point. (Continued) Example

Transportation/Logistic Strategy CARRIER A: S Dt = 102 units Cost = 2 X 102 X 100 X.2 = $4,080 CARRIER B: S Dt = 201 units Cost = 2 X 201 X 100 X.2 = $8,040 Additional cost of using a less reliable carrier: ($8,040 ‑ $4,080) = $3,960 (Continued) Example

Transportation/Logistic Strategy Basic Trade-off More facilities = more inventory More facilities = shorter shipments Tie in with ABC Analysis Consolidate slow moving items in central locations Trade increased transportation for reduced inventory or vice versa Inventory and the Number of Facilities

Transportation/Logistic Strategy Inventory and the Number of Facilities Number of Facilities (X) $ (Y) Y= b X 

Transportation/Logistic Strategy Joint Replenishment Order multiple items at the same time. When any item hits an order point, place orders for all items. Constraint: Order only full truckloads (e.g., 40,000 lbs).

Transportation/Logistic Strategy Q i = R i Q + I - P R - I i + P i ( ) Where: Qi = Order Quantity of the ith item (to be determined) Ri = Annual Rate of Use of the ith item Ii = Inventory on hand for the ith item Pi = Reorder point for the ith item Q = Total Shipment Size R = Total of Individual Usage Rate I = Total of Individual Inventory Items on Hand P = Total of Individual Reorder Points Joint Replenishment

Transportation/Logistic Strategy Example: Corn, Beans, Peas Units R i P i I i Corn 6, Beans 9, Peas 5, Totals 21, Cases =25 pounds R i P i I i Corn160,000 4,075 3,850 Beans245,000 5,450 6,250 Peas127,000 3,750 4,500 Totals 532,50013,27514,600 R P I Joint Replenishment

Transportation/Logistic Strategy Q i = R i Q + I - P R - I i + P i ( ) Corn: Qi = 160,000 (40,000+14,600-13,275) - 3, ,075 = 12, ,500 Beans: Qi = 245,000(40,000+14,600-13,275) - 6, ,450 = 18, ,500 Peas: Qi = 127,500(40,000+14,600-13,275) - 4, ,750 = 9, ,500 40,000 Example: Corn, Beans, Peas Joint Replenishment