ISE 453: Design of PLS Systems Michael G. Kay Geometric Mean How - PowerPoint PPT Presentation

Monetary vs. Physical Weight m m ∑ ∑ = = min TC X ( ) w d X P ( , ) f r d X P ( , ) i i i i i  = = i 1 i 1 w i = where TC total transport cost ($/yr) = w monetary weight ($/mi-yr) i = f physical weight rate (ton/yr) i = r transport rate ($/ton-mi) i = d X P ( , ) distance between NF at X an d EF at (mi) P i i i NF = new facility to be located EF = existing facility m = number of EFs Σ < Σ (Montetary) Weight Gaining: w w in out Σ > Σ Physically Weight Losing: f f in out 23

Minisum Location: TC vs. TD • Assuming local input costs are – same at every location or – insignificant as compared to transport costs, the minisum transport-oriented single-facility location problem is to locate NF to minimize TC • Can minimize total distance (TD) if transport rate is same: m m ∑ ∑ = = min TD X ( ) w d X P ( , ) f r d X P ( , ) i i  i i i = = i 1 i 1 w i = where TD total transport distance (mi/yr) = w monetary weight (trip/yr) i = f trips per year (trip/ yr) i = r transport rate = 1 i = d X P ( , ) per-trip distance between NF an E d F (mi/trip) i i 24

Ex 4: Single Supplier and Customer Location 36.5 raw finished Winston-Salem material goods Greensboro Asheville Product A Durham Durham 36 190 220 2 ton 1 ton 270 Raleigh Statesville 295 Asheville 150 1 ton 35.5 50 scrap 40 35 ubiquitous Wilmington inputs 34.5 2 ton raw finished 420 material goods Winston- 34 Wilmington Product B Salem 1 ton 3 ton -83 -82 -81 -80 -79 -78 • The cost per ton-mile (i.e., the cost to ship one ton, one mile) for both raw materials and finished goods is the same (e.g., $0.10). 1. Where should the plant for each product be located? 2. How would location decision change if customers paid for distribution costs (FOB Origin) instead of the producer (FOB Destination)? • In particular, what would be the impact if there were competitors located along I-40 producing the same product? 3. Which product is weight gaining and which is weight losing? 4. If both products were produced in a single m = ∑ TC X ( ) f r d X P ( , ) shared plant, why is it now necessary to  i i i = w i 1 know each product’s annual demand ( f i )? i 25

Ex 5: 1-D Location with Procurement and Distribution Costs unit of Asheville finished good Production System 1 ton Durham Assume: all scrap is disposed of locally A product is to be produced in a plant that will be located along I-40. Two tons of raw materials from a supplier in Ashville and a half ton of a raw material from a supplier in Durham are used to produce each ton of finished product that is shipped to customers in Statesville, Winston-Salem, and Wilmington. The annual demand of these customers is 10, 20, and 30 tons, respectively, and it costs $0.33 per ton-mile to ship raw materials to the plant and $1.00 per ton-mile to ship finished goods from the plant. Determine where the plant should be located so that procurement and distribution costs (i.e., transportation costs to and from the plant) are minimized, and whether the plant is weight gaining or weight losing. 26

Ex 5: 1-D Location with Procurement and Distribution Costs ∑ = × TC w d i i ($/yr) unit of Asheville ($/mi-yr) (mi) finished good Production monetary physical weight weight System   1 ton = × w f r Durham i i i ($/mi-yr) (ton/yr) ($/ton-mi) Assume: all scrap is disposed of locally = r $1.00/ton-mi r = $0.33/ton-mi out in = = = f 10, w f r 10 3 ∑ 3 3 3 out ( ) = = = = = f BOM f 2 60 120, w f r 40 1 1 1 out 1 1 in = = = f 20, w f r 20 4 NF 4 4 4 out ∑ ( ) = = = = = f BOM f 0.5 60 30, w f r 10 2 2 2 out 2 2 in = = = f 30, w f r 30 5 5 5 5 out Asheville Statesville Winston-Salem Durham Wilmington w : 10 20 10 30 40 i j W 40<55 50<55 70>55 ∑ ≤ w : 60>55 40<55 30<55 i 2 = i 1 * Σ = < Σ = (Montetary) Weight Gaining: w 50 w 60 in out Σ = > Σ = Physically Weight Losing: f 150 f 60 in out 27

Metric Distances 28

Great Circle Distances North Pole (90ºN lat) North Pole C P r i m e (Meridian) b M a e International Dateline (180º lon) r i d Latitude ( y ) Longitude ( x ) i a n ( lon , lat ) = ( x , y ) (Parallel) A ( x 1 , y 1 ) = (140ºW, 24ºN) N B c = (–140º, 24º) ( x 2 , y 2 ) W E lat 1 = y 1 0 º lat 2 = y 2 Equator (0º lat) S lon 1 = x 1 Equator Greenwich (Prime) Meridian (0º lon) lon 2 = x 2 South Pole (90ºS lat) 13.35 R mi 29

Circuity Factor d ∑ road = ≤ ≤ Circuity Factor : g , where usually 1.15 g 1.5 i d GC i ≈ ⋅ d g d ( P P , ), estimated road distance from to P P road GC 1 2 1 2 From High Point to Goldsboro: Road = 143 mi, Great Circle = 121 mi, Circuity = 1.19 36.6 36.4 36.2 36 High Point 35.8 35.6 35.4 Goldsboro 35.2 35 -80 -79.5 -79 -78.5 -78 -77.5 30

2-D Euclidean Distance   1 1 [ ]   = =  x 2 3 , P 7 1  4 5   3 5   ( ) ( ) 2 2 − + − x p x p d 3   1 1,1 2 1,2   d 1   ( ) ( )   2 2 = = − + − d d x p x p   2 1 2,1 2 2,2     x d   ( ) ( ) 3 3 2 2 − + − x p x p   y 1 3,1 2 3,2   d 2 1 d 1 2 1 1 2 4 7 x 31

Minisum Distance Location   1 1   =  P 7 1  4 5 3   5 ( ) ( ) 2 2 = − + − d ( ) x x p x p i 1 i ,1 2 i ,2 d 3 3 ∑ = TD ( ) x d ( ) x x * y i 2.73 = i 1 d * 2 = x arg min TD ( ) x d 1 x 120° 1 2 1 * = * TD TD ( x ) 1 4 7 x Fermat’s Problem (1629): Given three points, find fourth (Steiner point) such that sum to others is minimized (Solution: Optimal location corresponds to all angles = 120°) 32

Minisum Weighted-Distance Location • Solution for 2-D+ and non-rectangular distances: = m number of EFs – Majority Theorem: Locate NF at m ∑ = TC ( ) x w d ( ) x m W i i = ∑ ≥ EF j if w , where W w = i 1 j i 2 = i 1 * = x arg min TC ( ) x – Mechanical (Varigon frame) x – 2-D rectangular approximation * * = TC TC ( x ) – Numerical: nonlinear unconstrained optimization • Analytical/estimated derivative (quasi-Newton, fminunc ) • Direct, derivative-free (Nelder- Mead, fminsearch ) Varignon Frame 33

Convex vs Nonconvex Optimization 30 25 20 15 10 5 10 10 5 5 0 0 34

Multiple Single-Facility Location 9 Distribution 3 6 Manufacturing 1 10 Customers Suppliers 7 4 2 11 8 5 12 EFs NFs EFs 35

Best Retail Warehouse Locations 36

Optimal Number of NFs TC Facility Fixed + Transport Cost t s o C d e x i F y t i l i c a F Transport Cost 1 2 3 4 5 6 Number of NFs 37

Fixed Cost and Economies of Scale • How to estimate facility fixed cost?  β      f = TPC max TPC , TPC     act min 0 f – Cost data from existing facilities can be <   f f   max  0  used to fit linear estimate  0.62, Hand tool mfg. • y -intercept is fixed cost, k   0.48, Construction β =  – Economies of scale in production 0.41, Chemical processing   0.23, Medical centers ⇒ k > 0 and β < 1  = + TPC k c f est p TPC TPC β − act 0 1 = = APC f act β f f 0 TPC k ( = 0.5) 0 TPC = + Average Production Cost ($/ton) APC c act Total Production Cost ($/yr) est p TPC f est Actual EF cost = k fixed cost APC act APC est = c constant unit production cost p k = f / f min/max feasible scale min max TPC c min = p f Minimum Efficient Scale MES = TPC / f base cost/rate 0 0 f min f MES f 0 f max 38 Production Rate (ton/yr)

MILP [ ] + = max 6 x 8 x c 6 8 LP: max c x ' 1 2 + ≤ s.t. 2 x 3 x 11     2 3 11 ≤ s.t. Ax b 1 2 = = A , b     2 0 7 ≤ 2 x 7     1 ≥ x 0 ≥ x x , 0 1 2 MILP: some x integer i 4 ILP: x integer { } ∈ BLP: x 0,1 3   1 x 3 2   2 2 , * ′ * = = x c x 31   2 1 3   13   1 0 1 2 3 4 5 6 x 1 39

Branch and Bound [ ] + = c 6 8 2 max 6 x 8 x = = 1 2 LP 0 UB 31 , LB 0 3 + ≤ s.t. 2 x 3 x 11     2 3 11 1 2 x ≤ x ≥ 3 4 = = A , b 1 1     ≤ 2 0 7 2 x 7     1 1 Fathomed, = = UB 31 , LB 0 1 8 ≥ 3 infeasible x x , 0 1 2 x ≤ x ≥ 1 2 x x , integer 2 2 4 1 2 Incumbent = = 2 3 UB 31, LB 26 30 1 x ≤ 2 = = x ≥ UB 31 , LB 26 3 3 1 1 6 3 x 2 2 30 Fathomed, 2 3 4 7 = = UB 30 , LB 26 4 infeasible 31 3 2 1 5 3 313 28 x ≥ 3 x ≤ 2 2 2 1 2 31 3 0 Incumbent Incumbent 5 6 26 1 2 2 2 = = = = UB 30 , LB 28 UB 30 , LB 30 3 3 2 = − < ⇒ gap 30 30 1 STOP 0 1 2 3 4 5 6 3 x 1 40

MILP Formulation of UFL ∑ ∑ ∑ + min k y c x i i ij ij ∈ ∈ ∈ i N i N j M ∑ = ∈ s.t. x 1, j M ij ∈ i N ≥ ∈ ∈ y x , i N j , M i ij ≤ ≤ ∈ ∈ 0 x 1, i N j , M ij { } ∈ ∈ y 0,1 , i N i where { } = ∈ = k fixed cost of NF at site i N 1,..., n i { } = ∈ = c variable cost from to serve EF i j M 1,..., m ij  1, if NF established at site i =  y i 0, otherwise  = x fraction of EF demand served from NF at site . j i ij 41

MILP Formulation of p -Median ∑ ∑ min c x ij ij ∈ ∈ i N j M ∑ = s.t. y p i ∈ i N ∑ = ∈ x 1, j M ij ∈ i N ≥ ∈ ∈ y x , i N j , M i ij ≤ ≤ ∈ ∈ 0 x 1, i N j , M ij { } ∈ ∈ y 0,1 , i N i where = p number of NF to establish { } = ∈ = c variable cost from to serve EF i j M 1,..., m ij  1, if NF established at site i =  y i 0, otherwise  = x fraction of EF demand served from NF at site . j i ij 42

Computational Tools Data Simple Calculator Spreadsheet Processing Complex Hybrid Scripting (toolbox, Language addon) Structured Unstructured 43

Logistics Software Stack MIP Solver Standard Library Data Report (Gurobi, etc.) (C,Java) (csv,Excel,etc.) (GUI,web,etc.) User Library Scripting (in script language) (Python,Matlab,etc.) Commercial Software MIP Solver Standard Library (Lamasoft,etc.) (Gurobi,Cplex,etc.) (in compiled C,Java) • New Julia (1.0) scripting language – (almost?) as fast as C and Java (but not FORTRAN) – does not require compiled standard library for speed – uses multiple dispatch to make type-specific versions of functions 44

PharmaCo Case Study 45

Logistics Engineering Design Constants Circuity Factor: 1.2 ( g ) 1. 1.2 × GC distance ≈ actual road distance – 2. Local vs. Intercity Transport: Local: < 50 mi ⇒ use actual road distances – Intercity: > 50 mi ⇒ can estimate road distances – 50-250 mi ⇒ return possible (11 HOS) • > 250 mi ⇒ always one-way transport • > 500-750 mi ⇒ intermodal rail possible • Inventory Carrying Cost ( h ) = funds + storage + obsolescence 3. – 16% average (no product information, per U.S. Total Logistics Costs) (16% ≈ 5% funds + 6% storage + 5% obsolescence) • – 5-10% low-value product (construction) – 25-30% general durable manufactured goods – 50% computer equipment – >> 100% perishable goods (produce) 46

Logistics Engineering Design Constants Value $2,620 Shanghai-LA/LB shipping cost 4. ≈ 3  1: $1 ft 3 Transport Cost 2,400 ft 40’ ISO container capa city TL Weight Capacity: 25 tons ( K wt ) 5. – (40 ton max per regulation) – (15 ton tare for tractor-trailer) = 25 ton max payload – Weight capacity = 100% of physical capacity TL Cube Capacity: 2,750 ft 3 ( K cu ) 6. – Trailer physical capacity = 3,332 ft 3 – Effective capacity = (8'6" - 9'2" = 102" - 110") Interior Height: Max Height: 3,332 × 0.80 ≈ 2,750 ft 3 13'6" = 162" Truck Trailer Cube = 3,332 - 3,968 CFT – Cube capacity = 80% of Max Gross Vehicle Wt = 80,000 lbs = 40 tons Max Payload Wt = 50,000 lbs = 25 tons Width: 8'6" = 102" physical capacity (8'2" = 98") Length: 48' - 53' single trailer, 28' double trailer 47

Logistics Engineering Design Constants TL Revenue per Loaded Truck-Mile: $2/mi in 2004 ( r ) 7. – TL revenue for the carrier is your TL cost as a shipper Raleigh Gainesville 532 mi L L L U U Greensboro Jacksonville 15%, average deadhead travel $1.60, cost per mile in 2004 $1.60 = $1.88, cost per loaded-mile − 1 0.15 6.35%, average operating margin for trucking $1.88 ≈ $2.00, revenue per loaded-mile − 1 0.0635 48

One-Time vs Periodic Shipments • One-Time Shipments ( operational decision): know shipment size q – Know when and how much to ship, need to determine if TL and/or LTL to be used – Must contact carrier or have agreement to know charge • Can/should estimate charge before contacting carrier • Periodic Shipments ( tactical decision): know demand rate f , must determine size q – Need to determine how often and how much to ship – Analytical transport charge formula allow “optimal” size (and shipment frequency) to be estimated • U.S. Bureau of Labor Statistic's Producer Price Index (PPI) for TL and LTL used to estimate transport charges 49

Truck Shipment Example • Product shipped in cartons from Raleigh, NC (27606) to Gainesville, FL (32606) • Each identical unit weighs 40 lb and occupies 9 ft 3 (its cube ) – Don’t know linear dimensions of each unit for TL and LTL • Units can be stacked on top of each other in a trailer • Additional info/data is presented only when it is needed to determine answer 50

Truck Shipment Example: One-Time 1. Assuming that the product is to be shipped P2P TL, what is the maximum payload for each trailer used for the shipment? wt = = q K 25 ton max wt 3 = K 2750 ft cu 40 lb/unit = = 3 s 4.4444 lb/ft 3 9 ft /unit cu q sK max cu cu = ⇒ = K q cu max   s 2000    2000  { }   sK wt cu cu = = q min q , q min K , 2000   max max max wt     4.4444(2750) = = min 25, 6.1111 ton    2000  51

Truck Shipment Example: One-Time 2. On Jan 10, 2018, 320 units of the product were shipped. How many truckloads were required for this shipment?     40 q 6.4 = = = = q 320 6.4 ton, 2 truckloads     2000 q 6.1111     max 3. Before contacting the carrier (and using Jan 2018 PPI ), what is the estimated TL transport charge for this shipment? = d 532 mi Jan 2018 PPI PPI TL TL = × = × r r $2.00 / mi TL 2004 2004 102.7 PPI TL 131.0 = × = $2.00 / mi $2.5511/ mi 102.7   q  6.4  = = = c r d (2.5511)(532) $2,714.39     TL TL q  6.1111    max 52

Truck Shipment Example: One-Time 53

Truck Shipment Example: One-Time 4. Using the Jan 2018 PPI LTL rate estimate, what was the transport charge to ship the fractional portion of the shipment LTL (i.e., the last partially full truckload portion)? = − = − = q q q 6.4 6.1111 0.2889 ton frac max   2 s + 14   8   = r PPI LTL LTL   1 15   7 ( ) 2 − + +   q 7 d s 2 s 14 29   frac    2    2 4.44 + 14   8   = = 177.4 $3.8014 / ton-mi   1 15   ( ) 7 2 + + − 4.44 2(4.44) 14   0.2889 532 7 29      2  = = = c r q d 3.8014(0.28 89)(532) $584.23 LTL LTL frac 54

Truck Shipment Example: One-Time 5. What is the change in total charge associated with the combining TL and LTL as compared to just using TL? ( ) ∆ = − + c c c c − TL TL 1 LTL       q q = − + r d r d r q d       TL TL LTL frac q q       max max = $772.96 55

Truck Shipment Example: One-Time 6. What would the fractional portion have to be so that the TL and LTL charges are equal? Indifference Point between TL and LTL 1400   q =  c ( ) q r d 1200  TL TL q   max 1000   Transport Charge ($) 2 s 800 + 14   8 600   = r ( ) q PPI LTL LTL   1 15   7 400 ( ) 2  −  + + q d 7 29 s 2 s 14      2  200 0 = c ( ) q r ( ) q qd 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 LTL LTL Shipment Size (ton) ( ) = − q arg min c ( ) q c ( ) q I TL LTL q = 0.7960 ton 56

Truck Shipment Example: One-Time 7. What are the TL and LTL minimum charges? Indifference Point between MC and LTL   r TL = = MC 45 $57.40 160   TL  2  140 120   28     PPI d 19 Transport Charge ($) 100 LTL = + MC 45     LTL 80   104.2  1625 60   28 40     177.4 532 19 c = + = LTL 45 $87.51 20     c  0  104.2  1625 0 0 0.01 0.02 0.03 0.04 0.05 0.06 Shipment Size (ton) • Why do these charges not depend on the size of the shipment? • Why does only the LTL minimum charge depend of the distance of the shipment? 57

Truck Shipment Example: One-Time • Independent Transport Charge ($): { } { } { } = c q 0 ( ) min max c ( ), q MC ,max c ( ), q MC TL TL LTL LTL Independent shipment charge: Class 200 from 27606 to 32606 2500 2000 Transport Charge ($) 1500 1000 500 0 0 1 2 3 4 5 6 7 Shipment Size (ton) 58

Truck Shipment Example: One-Time 8. Using the same LTL shipment, find online one-time (spot) LTL rate quotes using the FedEx LTL website = q 0.2889 ton frac Class-Density Relationship = = 0.2889(2000) 578 lb   = 0.2889(2000) no. = 15 cartons   units  40  • Most likely freight class: 40 lb/unit 3 = = s 4. 4444 lb/ft 3 9 ft /unit ⇒ Class 0 2 0 • What is the rate quote for the reverse trip from Gainesville (32606) to Raleigh (27606)? 59

Truck Shipment Example: One-Time • The National Motor Freight Classification (NMFC) can be used to determine the product class • Based on: 1. Load density 2. Special handling 3. Stowability 4. Liability 60

Truck Shipment Example: One-Time • CzarLite tariff table for O-D pair 27606-32606 100 1 cwt = = = = hundredweight 100 lb ton 2000 20 Tariff (in $/cwt) from Raleigh, NC (27606) to Gainesville, FL (32606) (532 mi, CzarLite DEMOCZ02 04-01-2000, minimum charge = $95.23) 4719 3224 1638 999 $ 638 TC tariff w/o Break TC tariff 0.25 0.5 1 2.5 5 ton 61

Truck Shipment Example: One-Time 9. Using the same LTL shipment, what is the transport cost found using the undiscounted CzarLite tariff? = = q 0.2889, class 200 = = disc 0, MC 95.23 { } B B = ≤ < B i arg q q q q i − i i 1 { } B B = ≤ < arg q B q q q 2 2 1 { } B = ≤ < = arg q 0.25 0.2889 0.5 2 2 { } { } ( ) B = − + c 1 disc max MC ,min OD class i ( , )20 , q OD class i ( , 1)20 q tariff i { } ( ) { } = − 1 0 max 95.23,min OD (200,2)20(0.2889), OD (200,3)20(0.5) { } { } = max 95.23,min (127.69)20(0.2889), (99.92)20(0.5) { } { } = = max 95.23,min 737.76, 999.20 $737.76 62

Truck Shipment Example: One-Time 10. What is the implied discount of the estimated charge from the CzarLite tariff cost? − c c tariff LTL = disc c tariff 4719 − 737.76 584.23 3224 = 737,76 1638 = 20.81% 999 $ • What is the weight 638 break between the rate breaks? TC tariff w/o Break + OD class i ( , 1) W B = TC q q tariff i i OD class i ( , ) 0.25 0.5 1 2.5 5 ton 99.92 (0.5) = = 0.3913 ton 127.69 63

Truck Shipment Example: One-Time ( ) = c R wt , zone • PX: Package Express PX chrg – (Undiscounted) charge c PX based { } =    wt max wt , wt (lb)  chrg act dim rate tables, R , for each service (2- = wt actual weight (1 to 150 lb) day ground, overnight, etc.) act – Rate determined by on chargeable 3 × × l w d (in ) (lb) = wt weight , wt chrg , and zone dim 3 sf (in / lb) – All PX carriers (FedEX, UPS, USPS, = DHL) use dimensional weight , wt dim l w d , , length, width, depth (in) – wt dim > 150 lb is prorated per-lb rate ≥ × × ≥ l w , l w d actual cube – Actual weight 1 – 70 lb (UPS, FedEx home), 1 – 150 lb (FedEx commercial) = 3 sf shipping factor (in / lb) – Carrier sets a shipping factor , which 3 = 12 s , invers e of density is min cubic volume per pound = – Zone usually determined by O-D 139 FedEx (2019) distance of shipment 3 ⇒ = s 12.43 lb/ft (Class 85) – Supplemental charges for home 3 delivery, excess declared value, etc. = ⇒ = 194 USPS s 8.9 lb/ft 64

Truck Shipment Example: One-Time • (Undisc.) charge to ship a FedEx Standard List Rates (eff. Jan. 7, 2019) single carton via FedEx? = = 3 wt 40 lb, cu 9 ft act = ⇒ = d 532 mi zone 4 ⇒ × × = ⇒ carton l w d actual cube 3 3 × × = × = = × × Note: No Zone 1 l w d 9 12 15,552 in 32 27 18 (usually < 50 mi local) × × l w d 15,552 = = = wt 111.9 lb dim sf 139 { }   =  wt max wt , wt  chrg act dim { } =   = max 40,111.9 112 lb   ( ) = c R wt , zone PX chrg ( ) = = R 112,4 $64.2 7 65

Truck Shipment Example: Periodic 11. Continuing with the example: assuming a constant annual demand for the product of 20 tons, what is the number of full truckloads per year? = f 20 ton/yr = = ⇒ ≡ q q 6.1111 ton/ TL (full truckload q q ) max max f 20 = = = n 3.2727 TL/yr, average shipment frequency q 6.1111 • Why should this number not be rounded to an integer value? 66

Truck Shipment Example: Periodic 12. What is the shipment interval? 1 q 6.1111 = = = = t 0.3056 yr/TL, average shipment interval n f 20 • How many days are there between shipments? 365.25 day/yr 365.25 × = = t 365.25 111.6042 day/TL n 67

Truck Shipment Example: Periodic 13. What is the annual full-truckload transport cost? = = d 532 mi, r $2.5511/ mi TL r 2.5511 TL = = = r $0.4175 / ton-mi FTL q 6.1111 max ( ) = = = = TC f r d nr d wd w , monetary weight in $/mi FTL FTL TL = = 3.2727(2.5511)532 $4,441.73/yr • What would be the cost if the shipments were to be made at least every three months? 3 1 f = ⇒ = = ⇒ = t yr/TL n 4 TL/yr q max min { } 12 t max n n , max min { } ′ = TC max n n , r d FTL min TL { } = = max 3.2727, 4 2.5511(532) $5,428.78/yr 68

Truck Shipment Example: Periodic • Independent and allocated full-truckload charges: [ ] [ ] ≤ ⇒ = q q UB LB , c ( ), q qr d max 0 FTL Transport Charge for a Shipment 3 TL 4072 2 TL 2714 1 TL 1357 Transport Charge ($) MC 87.51 150/2000 0.7960 6.11 12.22 69 Shipment Size (tons)

Truck Shipment Example: Periodic • Total Logistics Cost (TLC) includes all costs that could change as a result of a logistics-related decision = + + TLC TC IC PC = TC transport cost = IC inventory cost = + + IC IC IC cycle pipeline safety = PC purchase cost • Cycle inventory : held to allow cheaper large shipments • Pipeline inventory : goods in transit or awaiting transshipment • Safety stock : held due to transport uncertainty • Purchase cost : can be different for different suppliers 70

Truck Shipment Example: Periodic • Same units of inventory can serve multiple roles at each position in a production process • Working stock : held as part of production process • (in-process, pipeline, in-transit, presentation) • Economic stock: held to allow cheaper production • (cycle, anticipation) • Safety stock: held to buffer effects of uncertainty • (decoupling, MRO (maintenance, repair, and operations) ) 71

Truck Shipment Example: Periodic 14. Since demand is constant throughout the year, one half of a shipment is stored at the destination, on average. Assuming that the production rate is also constant, one half of a shipment will also be stored at the origin, on average. Assuming each ton of the product is valued at $25,000, what is a “reasonable estimate” for the total annual cost for this cycle inventory? = IC (annualcost of holding one ton)(average annual inventory level) cycle = α ( vh )( q ) = v unit value of shipment ($/ton) = h inventory carrying rate, the cost per dollar of inventory per year (1/yr) α = average int er-shipment inventory fraction at Origin and Destination q = shipment size (ton) 72

Truck Shipment Example: Periodic • Inv. Carrying Rate ( h ) = interest + warehousing + obsolescence • Interest: 5% per Total U.S. Logistics Costs • Warehousing: 6% per Total U.S. Logistics Costs • Obsolescence: default rate ( yr ) h = 0.3 ⇒ h obs ≈ 0.2 (mfg product) – Low FGI cost ( yr ): h = h int + h wh + h obs – High FGI cost ( hr ): h ≈ h obs , can ignore interest & warehousing • ( h int + h wh )/ H = (0.05+0.06)/2000 = 0.000055 ( H = oper. hr/yr) – Estimate h obs using “percent-reduction interval” method: given time t h when product loses x h -percent of its original value v , find h obs x x h h = ⇒ = ⇒ = = h t v x v h t x h , and t obs h h obs h h obs h t h h obs – Example: If a product loses 80% of its value after 2 hours 40 minutes: 40 x 0.8 h = + = ⇒ = = = t 2 2.67 hr h 0.3 h 60 t 2.67 h – Important: t h should be in same time units as production time, t CT 73

Truck Shipment Example: Periodic q q  1 1  • Avg. annual cycle inventory level = + = + = ⇒ α = q (1) q 1   2 2  2 2  q q In- q q Transit 2 2 0 0 Destination Origin • Note: Cycle inventory is FGI at Origin and RMI at Destination 74

Truck Shipment Example: Periodic • Inter-shipment inventory fraction alternatives: α = α + α O D q q q 2 2 2 ≈ 0 Constant Constant Constant Immediate Consumption Consumption Production Production 1 1 1 1 α = + = α = + = 0 1 2 2 2 2 q 2 ≈ 0 ≈ ≈ 0 0 Batch Constant Immediate Batch Consumption Production Consumption Production 1 1 α = + = α = + = 0 0 0 0 2 2 75

Truck Shipment Example: Periodic • “Reasonable estimate” for the total annual cost for the cycle inventory: = α IC vhq cycle = (1)(25,000)(0.3)6.1111 = $45,833.33/ yr where 1 1 α = = at Origin + at Destination 1 2 2 = = v $25,000 unit value of shipment ($/ton) = = h 0.3 estimated carrying rate for manufactured products (1/yr) = = q q = 6. 111 FTL shipment size (ton) max 76

Truck Shipment Example: Periodic 15. What is the annual total logistics cost (TLC) for these (necessarily P2P) full-truckload TL shipments? = + TLC TC IC FTL FTL cycle = + α nr d vhq TL = + 3.2727(2.5511)532 (1)(25,000)(0.3)6.1111 = + 4,441.73 45,833.33 = $50,275.06 / yr • Problem: FTL may not minimize TLC ⇒ Can assume, for any periodic shipment, q ≤ q max ⇒ Assuming P2P TL, what to find q , q * , that minimizes TLC   q ⇒ = = c ( ) q r d r d   TL TL TL q   max 77

Truck Shipment Example: Periodic 16. What is minimum possible annual total logistics cost for P2P TL shipments, where the shipment size can now be less than a full truckload? f f = + = + α = + α TLC ( ) q TC ( ) q IC q ( ) c ( ) q vhq rd vhq TL TL TL q q dTLC ( ) q f r d 20(2.5511)532 TL TL * = ⇒ = = = 0 q 1.9024 ton TL α dq vh (1)25000(0.3) f * * = + α TLC ( q ) r d vhq TL TL TL TL * q TL 20 = + (2.5511)532 (1)25000(0.3)1.8553 1.8553 = + 14,268.12 14,268.12 = $28,536.25 / yr 78

Truck Shipment Example: Periodic • Including the minimum charge and maximum payload restrictions:  { }  f max r d MC ,   f r d TL TL * TL = ≈ q min , q   TL max α α vh vh     • What is the TLC if this size shipment could be made as a (not-necessarily P2P) allocated full-truckload? ( ) f r * * * TL * = + α = + α TLC ( q ) q r d vhq f d vhq AllocFTL TL TL FTL TL TL * q q max TL 2.5511 = + 20 532 (1)25000(0.3)1.9024 6.1111 = + 4,441.73 14,268.12 ( ) = $18,709.85 / yr vs. $28,536.25 as independent P2P TL 79

Truck Shipment Example: Periodic 17. What is the optimal LTL shipment size? f = + = + α TLC ( ) q TC ( ) q IC q ( ) c ( ) q vhq LTL LTL LTL q * = = q arg min TLC ( ) q 0.7622 ton LTL LTL q • Must be careful in picking starting point for optimization since LTL formula only valid for limited range of values: ≤ ≤  37 d 3354 (dist)   2 s  + 14 150 10,000 (wt)   ≤ ≤  q 8   = r PPI ,  2,000 2,000 LTL LTL   1 15   7 ( )  q 2  −  + + q d 7 29 s 2 s 14   ≤ 3  2000 650 ft (cube)     2  s   150 10,000 650 s ≤ ≤ ⇒ ≤ ≤ q min , 0.075 q 1.44   2000 2,000 2000   80

Truck Shipment Example: Periodic 18. Should the product be shipped TL or LTL? * * * = + = + = TLC ( q ) TC ( q ) IC q ( ) 34,349.19 5,716.40 $40,065.59 / yr LTL LTL LTL LTL LTL TLC TL TLC LTL 40066 TC LTL 28536 $ per year TC TL IC 0.76 1.90 Shipment weight (tons) 81

Truck Shipment Example: Periodic 19. If the value of the product increased to $85,000 per ton, should the product be shipped TL or LTL? TLC TL TLC LTL 40066 TLC TL TC LTL $ per year 28536 $ per year TLC LTL TC 52618 TL 47801 TC TL TC LTL IC IC 0.76 1.90 0.27 1.03 Shipment weight (tons) Shipment weight (tons) (a) $25000 value per ton (b) $85000 value per ton 82

Truck Shipment Example: Periodic • Better to pick from separate optimal TL and LTL because independent charge has two local minima:   ! f { } * * = + α = q arg min c q ( ) vhq q arg min TLC ( ), q TLC ( ) q   0 0 0 TL LTL q   q q 10.9 11.05 11 10.8 10.95 10.7 10.9 10.6 10.85 10.5 10.8 10.4 10.75 10.3 10.7 10.2 10.65 1 2 3 4 5 6 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 83

Truck Shipment Example: Periodic 20. What is optimal independent shipment size to ship 80 tons per year of a Class 60 product valued at $5000 per ton, with the same inventory fraction and carrying rate, between Raleigh and Gainesville? = 3 s 32.16 lb/ft { } * = = q arg min TLC ( ), q TLC ( ) q 8.5079 ton 0 TL LTL q * * = < TLC ( q ) $25,523.60 / yr TLC ( q ) TL 0 LTL 0 84

Truck Shipment Example: Periodic 21. What is the optimal shipment size if both shipments will always be shipped together on the same truck (with same shipment interval)? = = = α = α = α d d , h h h , 1 2 agg 1 2 agg 1 2 = + = + = f f f 20 80 100 ton agg 1 2 ( ) aggregate weight, in lb f 100 agg 3 = = = = s 14.31lb/ft ( ) agg f f 20 80 3 aggregate cube, in ft 1 + 2 + s s 4.44 32.16 1 2 f f 20 80 1 2 = + = + = v v v 85,000 5000 $21,000 / ton agg 1 2 f f 100 100 agg agg f r d 100(2.5511)532 agg TL * = = = q 4.6414 ton TL α v h (1)21000(0.3) agg agg agg 85

Truck Shipment Example: Periodic • Summary of results: 86

Ex 6: FTL vs Interval Constraint • On average, 200 tons of components are shipped 750 miles from your fabrication plant to your assembly plant each year. The components are produced and consumed at a constant rate throughout the year. Currently, full truckloads of the material are shipped. What would be the impact on total annual logistics costs if TL shipments were made every two weeks? The revenue per loaded truck-mile is $2.00; a truck’s cubic and weight capacities are 3,000 ft 3 and 24 tons, respectively; each ton of the material is valued at $5,000 and has a density of 10 lb per ft 3 ; the material loses 30% of its value after 18 months; and in-transit inventory costs can be ignored. 1 1 = = α = + = = = = = = f 200, d 750, 1, r 2, K 3000, K 24, v 5000, s 10 TL cu wt 2 2   x 0.3 sK h cu = = = ⇒ = + + = = = = h 0.2 h 0.05 0.06 0.2 0.31, q q min K , 15   obs FTL max wt t 1.5  2000  h f = = = + α = ⇒ n 13.33, TLC n r d vhq 43,250, 2-wk TL LTL not considered FTL FTL FTL TL FTL q FTL ⋅ 2 7 f = ⇒ = = = = + α = t n 26.09, q 7.67, TLC n r d vhq 51,016 max min 2wk 2wk min TL 2wk 365.25 n min ∆ = − = TLC TLC TLC $7,766 per year increase with two-week interval constraint 2wk FTL 87

Ex 7: FTL Location • Where should a DC be located in order to minimize transportation costs, given: 36.5 Winston-Salem Greensboro Durham 1. FTLs containing mix of products 36 190 220 270 Raleigh Statesville 295 Asheville 150 A and B shipped P2P from DC to 35.5 50 customers in Winston-Salem, 40 35 Wilmington Durham, and Wilmington 34.5 420 2. Each customer receives 20, 30, 34 and 50% of total demand -83 -82 -81 -80 -79 -78 3. 100 tons/yr of A shipped FTL P2P to DC from supplier in Asheville 4. 380 tons/yr of B shipped FTL P2P to DC from Statesville Each carton of A weighs 30 lb, and occupies 10 ft 3 5. Each carton of B weighs 120 lb, and occupies 4 ft 3 6. 7. Revenue per loaded truck-mile is $2 Each truck’s cubic and weight capacity is 2,750 ft 3 and 25 tons, 8. respectively 88

Ex 7: FTL Location ∑ = × TC w d i i Winston- ($/yr) ($/mi-yr) (mi) Salem Asheville = × = × w f r n r i i FTL i , i i DC ($/mi-yr) (TL/yr) ($/TL-mi) (ton/yr) Durham ($/ton-mi) 30% r f = = r , n Statesville FTL i , q q Wilmington ($/ton-mi) max max f   480 10.4348(2750 agg 3 = = + = + = = = = = = r $2 / TL-mi, f f f 100 380 480 ton/yr, s 10.4348 lb/ft , q 25, 14.3478   agg A B agg max f f 100 380  2000  A B + + s s 3 30 30  3(2750)  A B = = 3 = = s 3 lb/ft , q min 25,   4.125 ton 96 1 max = = = = = = f 0.20 f 96, n 6.69, w 6.69(2) 13.38 10  2000  3 agg 3 3 14.3478 3 100 1 = = = = = 144 f 100, n 24.24, w 24.24(2) 48.48 = = = = = = 1 1 1 f 0.30 f 144, n 10.04, w 10.04(2) 20.07 4.125 DC 4 4 agg 4 4 14.3478 120  30(2750)  2 = = 3 = = 240 s 30 lb/ft , q min 25,   25 ton 5 2 max = = = = = = f 0.50 f 240, n 16.73, w 16.73(2) 33.45 4  2000  5 agg 5 5 14.3478 380 = = = = = f 380, n 15.2, w 15.2(2) 30.4 2 2 2 25 w : W 48 30 13 20 33 W = = i 146, 73 2 48 < 73 78 > 73 ( ) Σ = > Σ = Σ = > Σ = (Montetary) Weight Losing: w 79 w 67 n 39 n 33 in out in out Σ = = Σ = Physically Weight Unchanging (DC): f 480 f 480 in out 89

Ex 7: FTL Location • Include monthly outbound frequency constraint: – Outbound shipments must occur at least once each month – Implicit means of including inventory costs in location decision 1 1 = ⇒ = = t yr/TL n 12 TL/yr max min 12 t max { } ′ = TC max n n , rd FTL min { } = = = = n max 6.69,12 12, w 12(2) 24 3 3 { } = = = = n max 10.04,12 12, w 12(2) 24 4 4 { } = = = = n max 16.73,12 16.73, w 16.73(2) 33.45 5 5 W w : W = = 160, 80 48 30 24 24 33 i 2 48 < 80 78 < 80 ( ) Σ = < Σ = Σ = < Σ = (Montetary) Weight Ga ining : w 79 w 81 n 39 n 41 in out in out Σ = = Σ = Physically Weight Unchanging (DC) : f 480 f 48 0 in out 90

Location and Transport Costs • Monetary weights w used for location are, in general, a function of the location of a NF – Distance d appears in optimal TL size formula – TC & IC functions of location ⇒ Need to minimize TLC instead of TC – FTL (since size is fixed at max payload) results in only constant weights for location ⇒ Need to only minimize TC since IC is constant in TLC m m f ∑ ∑ i = + α = + α TLC ( ) x w ( ) ( ) x d x vhq ( ) x rd ( ) x vhq ( ) x TL i i i i i q ( ) x i = = i 1 i 1   m m f f rd ( ) x 1 ∑ ∑ i i i = + α = + α   rd ( ) x vh f rd ( ) x vh i i i   α vh α f rd ( ) x vh   = = i 1 i i i 1 α vh m m f ∑ ∑ i = + α = + α = + TLC ( ) x rd ( ) x vhq w d ( ) x vhq TC ( ) x con stant FTL i max i i max FTL q = max = i 1 i 1 91

Transshipment • Direct : P2P shipments from Suppliers to Customers A 1 3 1 3 AA AB Customers Customers A Suppliers Suppliers DC AB B BB 4 4 2 2 B • Transshipment : use DC to consolidate outbound shipments – Uncoordinated : determine separately each optimal inbound and outbound shipment ⇒ hold inventory at DC – (Perfect) Cross-dock : use single shipment interval for all inbound and outbound shipments ⇒ no inventory at DC (usually only cross-dock a selected subset of shipments) 92

Uncoordinated Inventory • Average pipeline inventory level at DC: α = α + α O D 3 1 AA AB Customers  1 , Suppliers  α + inbound DC AB =  O BB 2  + α 2 4 0 , outbound  D 4 Supplier 1 3 ≈ q 2 1.55 2 1 0 Supplier 2 2 ≈ q 1.11 1 2 0 Customer 3 Customer 4 93

TLC with Transshipment • Uncoordinated: = TLC TLC of supplier/customer i i * = q arg min TLC q ( ) i i q = ∑ ( ) * * TLC TLC q i i q = t , shipment interval • Cross-docking: f   c t ( ) f 0 = + α = + α TLC t ( ) vhft cf. TLC q ( ) c q ( ) vhq   i i 0 t q   = c t ( ) independent transport charge as function of t 0 α +  0, inbound O α =  + α 0 , outbound  D ∑ * = t arg min TLC t ( ) i t ( ) ∑ * * = TLC TLC t i 94

Economic Analysis • Two aspects of economic analysis are important in production system design: 1. Costing : determine the unit cost of a production activity (e.g., $2 per mile for TL shipments (actually $1.60/mi) ) • Termed “should-cost” analysis when used to guide procurement negotiations with suppliers 2. Project justification : formal means of evaluating alternate projects that involve significant capital expenditures 95

Costing • Capital recovery cost used to make one-time investment costs and salvage values commensurate with per-period operating costs via discounting ( ) − N eff = − + Effective cost : IV IV SV 1 i     i i ( ) ef f = = − + ⋅ Capital recovery cost : K IV IV SV SV i     − − N N − + − + 1 (1 i ) 1 (1 i )     e ff + + K OC IV PV of O C = ≠ Average Cost : AC q Nq where = IV initial one-time investment cost = SV one-time salvage value at time N = OC operating cost per period = q units per period 96

Project Justification • If cash flows are uniform, can use simple formulas; otherwise, need to use spreadsheet to discount each period’s cash flows • In practice, the payback period is used to evaluate most small projects: IV 0 = > Payback period , for OP 0 OP where = − IV IV SV , net intital investment expenditure at time 0 for project 0 new current = IV initial investment cost at time 0 for (new) project new = SV salvage value of curren t project (if any) at time 0 current −  OR OC , uniform operating profit per period from project =  OP − OC OC , net uniform operating cost savings per period  current new = OR uniform operating revenue per period from proje ct OC = uniform operating cost per period of project 97

Discounting • NPV and NAV equivalent methods for evaluating projects • Project accepted if NPV ≥ 0 or NAV ≥ 0 = + Weighted Average Cost of Capital : i (% debt) i (% equity) i debt equity = + = (0.5)0.06 (0.5)0.30 0.18 98

Project with Uniform Cash Flows 99

Cost Reduction Example Common Cost of Capital ( i ) 8% 8% Economic Life ( N , yr) 15 15 Annual Demand ( q /yr) 500,000 500,000 Sale Price ($/q) Project Current New Net Investment Cost ( IV , $) 2,000,000 5,000,000 3,000,000 Salvage Percentage 25% 25% Salvage Value ( SV , $) 500,000 1,250,000 750,000 ( IV ef f , $ ) Eff. Investment Cost 1,842,379 4,605,948 2,763,569 Cost Cap Recovery ( K , $/yr) 215,244 538,111 322,866 Oper Cost per Unit ($/q) 1.25 0.50 (0.75) Operating Cost ( OC , $/yr) 625,000 250,000 (375,000) Operating Revenue ( OR , $/yr) 0 0 0 Operating Profit ( OR - OC ) ( OP , $/yr) (625,000) (250,000) 375,000 Analysis Payback Period ( IV / OP ) (yr) 8.00 PV of OP ($) (5,349,674) (2,139,870) 3,209,805 NPV (PV of OP - IV ef f ) ($) (7,192,053) (6,745,818) 446,236 NAV ( OP - K ) ($/yr) (840,244) (788,111) 52,134 Average Cost (( K + OC )/ q ) ($/ q ) 1.68 1.58 100