GEOP 4355 Distribution Problems Outline Pooling effect problem Distrbution problem Sources/references used in the preparation of this presentation are listed in the Introduction presentation

Distribution models Variety of problems in distribution Models have been developed to support strategic, tactical, and operational distribution decision making. Pooling effect problem: tradeoff between inventory and transportation Distribution/ transportation problem: How to move material effectively given customer requirements and source limitations.

Pooling effect problem month dc1 dc2 dc3 dc4 1 141 300 373 385 2 207 280 314 112 3 199 194 354 259 4 273 201 206 360 5 166 135 276 125 6 293 339 7 109 303 133 118 8 115 250 291 9 157 333 223 378 10 102 244 139 11 235 352 114 343 12 278 265 327 288 Total 2,354 3,076 3,048 3,133 Monthly demand for product X Serves by 4 DCs

Pooling effect problem month dc1 dc2 dc3 dc4 total 1 141 300 373 385 1199 2 207 280 314 112 913 3 199 194 354 259 1006 4 273 201 206 360 1040 5 166 135 276 125 702 6 293 339 1121 7 109 303 133 118 663 8 115 250 291 850 9 157 333 223 378 1091 10 102 244 139 824 11 235 352 114 343 1044 12 278 265 327 288 1158 Total 2,354 3,076 3,048 3,133 11,611 mean 196.2 256.3 254.0 261.1 967.6 std dev 64.4 80.4 91.2 108.8 176.3 Column to the left shows the total demand per month Rows at the bottom show average demand per month (d) standard deviation of demand (sd)

Pooling effect problem month dc1 dc2 dc3 dc4 total 1 141 300 373 385 1199 2 207 280 314 112 913 3 199 194 354 259 1006 4 273 201 206 360 1040 5 166 135 276 125 702 6 293 339 1121 7 109 303 133 118 663 8 115 250 291 850 9 157 333 223 378 1091 10 102 244 139 824 11 235 352 114 343 1044 12 278 265 327 288 1158 Total 2,354 3,076 3,048 3,133 11,611 mean 196.2 256.3 254.0 261.1 967.6 std dev 64.4 80.4 91.2 108.8 176.3 Assuming monthly replenishments to DC Replenishment lead time to each DC is L = 0.5 months, sL = 0 sddlt = 𝐿 × 𝑠 𝑑 2 + 𝑑 2 × 𝑠 𝐿 2 sddlt 46 57 65 77 125

Pooling effect problem month dc1 dc2 dc3 dc4 total 1 141 300 373 385 1199 2 207 280 314 112 913 3 199 194 354 259 1006 4 273 201 206 360 1040 5 166 135 276 125 702 6 293 339 1121 7 109 303 133 118 663 8 115 250 291 850 9 157 333 223 378 1091 10 102 244 139 824 11 235 352 114 343 1044 12 278 265 327 288 1158 Total 2,354 3,076 3,048 3,133 11,611 mean 196.2 256.3 254.0 261.1 967.6 std dev 64.4 80.4 91.2 108.8 176.3 Desired TSL = 99% Z = 2.33 SS = SS  sddlt Aggregate Safety stock needed with 4 DCs = (106 + 132 + 150 + 179) SS = 567 If all orders were to be served by a single DC. SS = 290 sddlt 46 57 65 77 125 SS 106 132 150 179 290

Pooling effect problem Assume each unit costs $1,000. h = 20% ASC = SS  h  cu = $113,400 with 4DCs, ASC = $58,000 with one DC dtc/unit units/yr transp c dc1 $ 4.50 2,354 $ 10,593 dc2 $ 5.10 3,076 $ 15,688 dc3 $ 4.20 3,048 $ 12,802 dc4 $ 3.40 3,133 $ 10,652 Delivery Annual Transportation costs = $49,735 Delivery Annual Transportation costs = $80,116 one DC $ 6.90 11,611 $ 80,116 TRC (4DCs) = $163,135 TRC(oneDC) = $138,116

Distribution problem How many units to ship from each warehouse to each of the customers based on availability, needs and transportation costs. $5/u Store 1 Need = 60 How many units would you ship in each of the four routes? DC1 Available = 80 $3/u $11/u Store 2 Need = 40 $6/u DC2 Available = 25

Distribution problem Could you optimally solve a larger problem? Store 1 Need = 190 DC1 Available = 356 Could you optimally solve a larger problem? Store 2 Need = 98 DC2 Available = 408 Store 3 Need = 141 DC3 Available = 230 Store 4 Need = 330 Total Available = 994 Total Needs = 759 Store1 Store 2 Store 3 Store 4 DC1 7.2 8.1 6.4 8.4 DC2 6.0 5.0 7.3 9.4 DC3 5.4 3.6 6.3 3.1

Distribution problem Mathematical modelling: use of equations to represent/model/solve a complex business problem. Three components Decision variables: what must be determined Objective functions: the key metric of the problem Constraints: the system characteristics representing its limitations and operations

Distribution problem Decision variables The number of units to ship from each warehouse to each customer Let U indicate units and Us,d = Units shipped from source s to destination d Thus UW1,C2 means the number of units shipped from warehouse 1 to customer 2. There are 12 decision variables (3 × 4) We use sets to represent all possible warehouses (S) and customers (D) S = {1…3}, D = {1…4}

Distribution problem Objective function Let T represent the transportation cost per unit and Ts,d = transportation cost per unit shipped from warehouse s to customer d. This assumes a fixed rate per unit, rate is independent of the total number of units shipped. Therefore TW1,C1 = $7.1, TW1,C2 = $8.1, … C1 C2 C3 C4 W1 7.2 8.1 6.4 8.4 W2 6.0 5.0 7.3 9.4 W3 5.4 3.6 6.3 3.1

Distribution problem Objective Function Transportation costs for each route equals U × T If UW1,C1 = 100 and TW1,C1 = $7.1 then transportation costs for that route are $710. Total system transportation costs are then UW1,C1 × TW1,C1 + UW1,C2 × TW1,C2 + UW1,C3 × TW1,C3 + …. + UW3,C4 × TW3,C4 General form:  sS, dD Us,d × Ts,d Objective of the model is to minimize total system transportation costs

Distribution problem Constraints Supply side constraints establishes the limits on what can be shipped from each source. It’s a maximum, not all units must be shipped. W1 Available = 356 UW1,C1 + UW1,C2 + UW1,C3 + UW1,C4 ≤ 356 UW2,C1 + UW2,C2 + UW2,C3 + UW2,C4 ≤ 408 W2 Available = 408 W3 Available = 230 UW3,C1 + UW3,C2 + UW3,C3 + UW3,C4 ≤ 230

Distribution problem Constraints Destination side constraints establish the amount that must be received at each destination. UW1,C1 + UW2,C1 + UW3,C1 = 190 C1, Need = 190 UW1,C2 + UW2,C2 + UW3,C2 = 98 C2, Need = 98 UW1,C3 + UW2,C3 + UW3,C3 = 141 C3, Need = 141 UW1,C4 + UW2,C4 + UW3,C4 = 330 C4, Need = 330 Non negativity constraints establishes that the number of units shipped from any source to any destination must not be a negative number. Us,d ≥ 0 for all sources and destinations.

Distribution problem Complete mathematical formulation Decision Variables Us,d = Units shipped from warehouse s to customer d. s = 1…3 and d =1…4. define sets : S = {1…3}, D = {1…4} Objective Function Minimize total transportation costs Ts,d = transportation cost per unit from warehouse s to customer d.  sS, dD Ts,d ×Us,d Constraints  s, dD Us,d ≤ Units available at warehouse s.  s  S.  sS,d Us,d = Units needed at customer d.  d  D. Us,d ≥ 0  s  S,  d  D

Distribution problem OF Constraints = Minimize (7.2Uw1,c1 + 6.0Uw2,c1 + …… Constraints W1: Uw1,c1 + Uw1,c2 + Uw1,c3 + Uw1,c4 ≤ 356 W2: Uw2,c1 + Uw2,c2 + Uw2,c3 + Uw2,c4 ≤ 408 W3: Uw3,c1 + Uw3,c2 + Uw3,c3 + Uw3,c4 ≤ 230 C1: Uw1,c1 + Uw2,c1 + Uw3,c1 = 190 C2: Uw1,c2 + Uw2,c2 + Uw3,c2 = 98 C3: Uw1,c3 + Uw2,c3 + Uw3,c3 = 141 C4: Uw1,c4 + Uw2,c4 + Uw3,c4 = 330 Us,d ≥ 0 for all s in S and d in D

Distribution problem Mathematical models are implemented/solved using special software applications. Excel can be used to solve small/medium sized problems. Finding the optimal solution is the first step. Second step is to perform sensitivity analysis. Analysis of how changes to some parameters change the decisions and the objective function.

Distribution problem Extensions Route constraints: Us,d ≤ Cs,d where Cs,d is the maximum that can be shipped in route s,d during the decision timeframe. Transshipment problem Transshipment nodes between warehouses and customers with a capacity Missing requirements problem Units available from the warehouses cannot satisfy the demand. Let md be the unmet demand at d. Unmet requirements have a cost per unit per customer. UW1,C4 + UW2,C4 + UW3,C4 = 330 – mC4