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Dedicated Storage Assignment (DSAP) The assignment of items to slots is termed slotting With randomized storage, all items are assigned to all slots DSAP (dedicated storage assignment problem): Assign N items to slots to minimize


  1. Dedicated Storage Assignment (DSAP) • The assignment of items to slots is termed slotting – With randomized storage, all items are assigned to all slots • DSAP (dedicated storage assignment problem): – Assign N items to slots to minimize total cost of material flow • DSAP solution procedure: 1. Order Slots: Compute the expected cost for each slot and then put into nondecreasing order 2. Order Items: Put the flow density (flow per unit of volume, the reciprocal of which is the “cube per order index” or COI) for each item i into nonincreasing order f f f [1] [2] [ N ] ≥ ≥ ≥  M s M s M s [1] [1] [2] [2] [ N ] [ N ] Assign Items to Slots: For i = 1, … , N , assign item [ i ] to the first 3. slots with a total volume of at least M [ i ] s [ i ] 162

  2. 1-D Slotting Example A B C Max units M 4 5 3 Space/unit s 1 1 1 Flow f 24 7 21 Flow Density f/(M x s) 6.00 1.40 7.00 Flow Expected Total 1-D Slot Assignments Density Distance Flow Distance 21 C C C 3 = 7.00 2(0) + 3 = 3 × 21 = 63 I/O 0 3 24 A A A A 4 = 6.00 2(3) + 4 = 10 × 24 = 240 I/O -3 0 4 7 B B B B B 5 = 1.40 2(7) + 5 = 19 × 7 = 133 I/O -7 0 5 C C C A A A A B B B B B 436 I/O 0 3 7 12 163

  3. 1-D Slotting Example (cont) Dedicated Random Class-Based A B C ABC AB AC BC Max units M 4 5 3 9 7 7 8 Space/unit s 1 1 1 1 1 1 1 Flow f 24 7 21 52 31 45 28 Flow Density f/(M x s) 6.00 1.40 7.00 5.78 4.43 6.43 3.50 Total Total 1-D Slot Assignments Distance Space Dedicated 436 12 C C C A A A A B B B B B (flow density) I/O Dedicated A A A A C C C B B B B B 460 12 (flow only) I/O Class-based C C C AB AB AB AB AB AB AB 466 10 I/O Randomized ABC ABC ABC ABC ABC ABC ABC ABC ABC 468 9 I/O 164

  4. 2-D Slotting Example A B C 8 7 6 5 4 5 6 7 8 Max units M 4 5 3 7 6 5 4 3 4 5 6 7 Space/unit s 1 1 1 Flow f 24 7 21 6 5 4 3 2 3 4 5 6 Flow Density f/(M x s) 6.00 1.40 7.00 5 4 3 2 1 2 3 4 5 4 3 2 1 0 1 2 3 4 Distance from I/O to Slot C C B B B B C A A B B B A C A B A A I/O B B A C I/O C A Original Assignment (TD = 215) Optimal Assignment (TD = 177) 165

  5. DSAP Assumptions 1. All SC S/R moves 2. For item i , probability of move to/from each slot assigned to item is the same 3. The factoring assumption : a. Handling cost and distances (or times) for each slot are identical for all items b. Percent of S/R moves of item stored at slot j to/from I/O port k is identical for all items • Depending of which assumptions not valid, can determine assignment using other procedures     f ( ) i ⋅ ⊂ ⊂ ⊂ d x DSAP LAP LP QAP c x x     j ij ijkl ij kl ∪ M     ( ) i c x TSP ij ij 166

  6. Example 5: 1-D DSAP • What is the change in the minimum expected total distance traveled if dedicated, as compared to randomized, block stacking is used, where a. Slots located on one side of 10-foot-wide down aisle b. All single-command S/R operations c. Each lane is three-deep, four-high d. 40 × 36 in. two-way pallet used for all loads e. Max inventory levels of SKUs A, B, C are 94, 64, and 50 f. Inventory levels are uncorrelated and retrievals occur at a constant rate g. Throughput requirements of A, B, C are 160, 140, 130 h. Single I/O port is located at the end of the aisle 167

  7. Example 5: 1-D DSAP • Randomized: ABC I/O 0 33 + + + +     M M M 1 94 64 50 1 A B C = + = + = M 104      2 2   2 2  − −       D 1 H 1 + + M NH   N     = L  2   2  rand     DH − −      3 1 4 1  + + 104 3(4)   N     = = 2 2     11lanes     3(4)   = = = X xL 3(11) 33 ft rand = = d X 33 ft SC ( ) ( ) = + + = + + = TD f f f X 160 140 130 33 14 ,190 ft rand A B C 168

  8. Example 5: 1-D DSAP • Dedicated: C B A I/O 0 15 33 57 f 160 f 140 f 130 A B C = = = = = = ⇒ > > 1.7, 2.19, 2.6 C B A M 94 M 64 M 50 A B C        M  94  M  64  M  50 A B C = = = = = = = = = L 8, L 6, L 5             A B C  DH  3(4)  DH  3(4)  DH  3(4)       = = = = = = = = = X xL 3(5) 15, X xL 3(6) 18, X xL 3(8) 24 C C B B A A C = = = d X 3(5) 15 ft SC C B = + = + = d 2( X ) X 2(15) 18 48 ft S C C B A = + + = + + = d 2( X X ) X 2(15 18) 24 90 ft SC C B A A B C = + + = + + = TD f d f d f d 160(90) 140(48) 130 ( 1 5) 23,0 7 0 ft ded A SC B SC C SC 169

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