4 network design cfvg 2012
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This document discusses a network planning problem involving optimizing supply chain logistics. There are two plants and two warehouses that supply goods to three market areas. The problem is formulated as a linear program to minimize total distribution costs given supply and demand constraints. The optimal solution is found to ship 140,000 units from plant 1 to warehouse 1 and 60,000 units from plant 2 to warehouse 2 to minimize total costs of $740,000.
![77SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
TP: a Linear Programming Model
•The structure of the model is:
Minimize Total Shipping Cost
ST
[Amount shipped from a source] <= [Supply at that
source]
[Amount received at a destination]=[Demand at
that Destination]
• Decision variables
Xij = the number of cases shipped from plant i to warehouse j.
where: i=1 (FARIDABAD), 2 (GURGAON), 3 (GHAZIABAD)
j=1 (N-DELHI), 2 (S-DELHI), 3 (E-DELHI), 4(W-DELHI)](https://image.slidesharecdn.com/4networkdesigncfvg2012-140716030733-phpapp02/85/4-network-design-cfvg-2012-7-320.jpg)
![2525SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
Heuristic #2:
Choose the warehouse where the total delivery costs
to and from the warehouse are the lowest
[Consider inbound and outbound distribution costs]
D = 50,000
D = 100,000
D = 50,000
Cap = 60,000
$4
$5
$2
$3
$4
$5
$2
$1
$2
$0
P1 to WH1 $3
P1 to WH2 $7
P2 to WH1 $7
P2 to WH 2 $4
P1 to WH1 $4
P1 to WH2 $6
P2 to WH1 $8
P2 to WH 2 $3
P1 to WH1 $5
P1 to WH2 $7
P2 to WH1 $9
P2 to WH 2 $4
Market #1 is served by WH1, Markets 2 and 3
are served by WH2](https://image.slidesharecdn.com/4networkdesigncfvg2012-140716030733-phpapp02/85/4-network-design-cfvg-2012-25-320.jpg)
![2626SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
D = 50,000
D = 100,000
D = 50,000
Cap = 60,000
Cap = 200,000
$5 x 90,000
$2 x 60,000
$3 x 50,000
$1 x 100,000
$2 x 50,000
$0 x 50,000
P1 to WH1 $3
P1 to WH2 $7
P2 to WH1 $7
P2 to WH 2 $4
P1 to WH1 $4
P1 to WH2 $6
P2 to WH1 $8
P2 to WH 2 $3
P1 to WH1 $5
P1 to WH2 $7
P2 to WH1 $9
P2 to WH 2 $4
Total Cost = $920,000
Heuristic #2:
Choose the warehouse where the total delivery
costs to and from the warehouse are the lowest
[Consider inbound and outbound distribution
costs]](https://image.slidesharecdn.com/4networkdesigncfvg2012-140716030733-phpapp02/85/4-network-design-cfvg-2012-26-320.jpg)
4 network design cfvg 2012
- 1. 11SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Modeling Supply Chain & Network Planning Prof. Ravi Shankar Department of Management Studies, Indian Institute of Technology Delhi, New Delhi Prof. Ravi Shankar Department of Management Studies, Indian Institute of Technology Delhi, New Delhi “Supply Chain Management”
- 2. 22SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Optimization
- 3. 33SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Transportation Planning
- 4. 44SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Example#1: Transportation, Problems Can be formulated as linear programs and solved by general purpose linear programming codes. However, there are many computer packages, which contain separate computer codes for these models which take advantage of their network structure.
- 5. 55SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Transportation Problem The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si ) to n destinations (each with a demand dj ), when the unit shipping cost from an origin, i, to a destination, j, is cij.
- 6. 66SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) N- DELHI S-DELHI E-DELHI W-DELHI Destinations Sources FARIDABAD GURGAON GAHZIABAD S1=250 S2=200 S3= 250 D1=100 D2=150 D3=250 D4=100 17 19 16 17 13 19 16 15 15 17 17 16 CASE STUDY#1
- 7. 77SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) TP: a Linear Programming Model •The structure of the model is: Minimize Total Shipping Cost ST [Amount shipped from a source] <= [Supply at that source] [Amount received at a destination]=[Demand at that Destination] • Decision variables Xij = the number of cases shipped from plant i to warehouse j. where: i=1 (FARIDABAD), 2 (GURGAON), 3 (GHAZIABAD) j=1 (N-DELHI), 2 (S-DELHI), 3 (E-DELHI), 4(W-DELHI)
- 8. 88SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) N-Delhi S-Delhi E-Delhi W-Delhi D1=100 D2=150 D3=250 D4=100 The supply constraints Faridabad S1=250 X11 X12 X13 X14 Supply from FARIDABAD X11+X12+X13+X14 = 250 Gurgaon S2=200 X21 X22 X23 X24 Supply from GURGAON X21+X22+X23+X24 = 200 Ghaziabad S3= 250 X31 X32 X33 X34 Supply from Ghaziabad X31+X32+X33+X34 = 250 CASE STUDY
- 9. 99SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) The complete mathematical model Minimize 13X11+16X12+19X13+ 17X14 +17X21+19X22+16X23+15X24+ 15X31+17X32+17X33+16X34 ST Supply constrraints: X11+ X12+ X13+ X14 250 X21+ X22+ X23+ X24 200 X31+ X32+ X33+ X34 250 Demand constraints: X11+ X21+ X31 100 X12+ X22+ X32 150 X13+ X23+ X33 250 X14+ X24+ X34 100 All Xij are nonnegative ≤ ≤ ≤ = = = = Total shipment out of a supply node cannot exceed the supply at the node. Total shipment received at a destination node, must equal the demand at that node. CASE STUDY
- 10. 1010SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) The complete mathematical model Minimize 13X11+16X12+19X13+ 17X14 +17X21+19X22+16X23+15X24+ 15X31+17X32+17X33+16X34 ST Supply constrraints: X11+ X12+ X13+ X14 250 X21+ X22+ X23+ X24 200 X31+ X32+ X33+ X34 250 Demand constraints: X11+ X21+ X31 100 X12+ X22+ X32 150 X13+ X23+ X33 250 X14+ X24+ X34 100 All Xij are nonnegative ≤ ≤ ≤ = = = = Total shipment out of a supply node cannot exceed the supply at the node. Total shipment received at a destination node, must equal the demand at that node. CASE STUDY#6
- 11. 1111SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
- 12. 1212SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
- 13. 1313SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) – Reduced costs • The unit shipment cost between Gurgaon and N-Delhi must be reduced by at least Rs.5, before it would become economically feasible to utilize it • If this route is used, the total cost will increase by Rs. 5 for each case shipped between Gurgaon and N-Delhi .
- 14. 1414SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
- 15. 1515SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) – Allowable Increase/Decrease • This is the range of optimality. • The unit shipment cost between Gurgaon and N-Delhi may increase up to any level or decrease up to Rs. 5 with no change in the current optimal transportation plan.
- 16. 1616SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Another solution: Yet optimal
- 17. 1717SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
- 18. 1818SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)
- 19. 1919SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Network Planning
- 20. 2020SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Solution Techniques Mathematical optimization techniques: 1. Exact algorithms: find optimal solutions 2. Heuristics: find “good” solutions, not necessarily optimal Simulation models: provide a mechanism to evaluate specified design alternatives created by the designer.
- 21. 2121SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Example Single product Two plants p1 and p2 • Plant p2 has an annual capacity of 60,000 units. The two plants have the same production costs. There are two warehouses w1 and w2 with identical warehouse handling costs. There are three markets areas c1,c2 and c3 with demands of 50,000, 100,000 and 50,000, respectively.
- 22. 2222SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Unit Distribution Costs Facility warehouse p1 p2 c1 c2 c3 w1 0 4 3 4 5 w2 5 2 2 1 2
- 23. 2323SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) CASE : NETWORK PROBLEM Network Representation ARNOLD WASH BURN ZROX HEWES 200,000200,000 60,000060,0000 50,00050,000 100,000100,000 50,00050,000 00 55 44 22 33 44 55 22 11 22 P1P1 P2P2 C2C2 C1C1 W1W1 W2W2 C3C3
- 24. 2424SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Heuristic #1: Choose the Cheapest Warehouse to Source Demand D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 $5 x 140,000 $2 x 60,000 $2 x 50,000 $1 x 100,000 $2 x 50,000 Total Costs = $1,120,000
- 25. 2525SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Heuristic #2: Choose the warehouse where the total delivery costs to and from the warehouse are the lowest [Consider inbound and outbound distribution costs] D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 $4 $5 $2 $3 $4 $5 $2 $1 $2 $0 P1 to WH1 $3 P1 to WH2 $7 P2 to WH1 $7 P2 to WH 2 $4 P1 to WH1 $4 P1 to WH2 $6 P2 to WH1 $8 P2 to WH 2 $3 P1 to WH1 $5 P1 to WH2 $7 P2 to WH1 $9 P2 to WH 2 $4 Market #1 is served by WH1, Markets 2 and 3 are served by WH2
- 26. 2626SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 Cap = 200,000 $5 x 90,000 $2 x 60,000 $3 x 50,000 $1 x 100,000 $2 x 50,000 $0 x 50,000 P1 to WH1 $3 P1 to WH2 $7 P2 to WH1 $7 P2 to WH 2 $4 P1 to WH1 $4 P1 to WH2 $6 P2 to WH1 $8 P2 to WH 2 $3 P1 to WH1 $5 P1 to WH2 $7 P2 to WH1 $9 P2 to WH 2 $4 Total Cost = $920,000 Heuristic #2: Choose the warehouse where the total delivery costs to and from the warehouse are the lowest [Consider inbound and outbound distribution costs]
- 27. 2727SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) The Optimization Model The problem described earlier can be framed as the following linear programming problem. Let x(p1,w1), x(p1,w2), x(p2,w1) and x(p2,w2) be the flows from the plants to the warehouses. x(w1,c1), x(w1,c2), x(w1,c3) be the flows from the warehouse w1 to customer zones c1, c2 and c3. x(w2,c1), x(w2,c2), x(w2,c3) be the flows from warehouse w2 to customer zones c1, c2 and c3
- 28. 2828SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) The problem we want to solve is: min 0x(p1,w1) + 5x(p1,w2) + 4x(p2,w1) + 2x(p2,w2) + 3x(w1,c1) + 4x(w1,c2) + 5x(w1,c3) + 2x(w2,c1) + 2x(w2,c3) subject to the following constraints: x(p2,w1) + x(p2,w2) ≤ 60000 x(p1,w1) + x(p2,w1) = x(w1,c1) + x(w1,c2) + x(w1,c3) x(p1,w2) + x(p2,w2) = x(w2,c1) + x(w2,c2) + x(w2,c3) x(w1,c1) + x(w2,c1) = 50000 x(w1,c2) + x(w2,c2) = 100000 x(w1,c3) + x(w2,c3) = 50000 all flows greater than or equal to zero. The Optimization Model
- 29. 2929SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) CASE : NETWORK DESIGN Network Representation ARNOLD WASH BURN ZROX HEWES 200,000200,000 60,00060,000 50,00050,000 100,000100,000 50,00050,000 0 (A)0 (A) 5 (B)5 (B) 4 (C)4 (C) 2 (D)2 (D) 3 (E)3 (E) 4 (F)4 (F) 5 (G)5 (G) 2 (H)2 (H) 1 (I)1 (I) 2 (J)2 (J) P1P1 P2P2 C2C2 C1C1 W1W1 W2W2 C3C3
- 30. 3030SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) The problem we want to solve is: min 0x(p1,w1) + 5x(p1,w2) + 4x(p2,w1) + 2x(p2,w2) + 3x(w1,c1) + 4x(w1,c2) + 5x(w1,c3) + 2x(w2,c1) + 2x(w2,c3) subject to the following constraints: x(p2,w1) + x(p2,w2) ≤ 60000 x(p1,w1) + x(p2,w1) = x(w1,c1) + x(w1,c2) + x(w1,c3) x(p1,w2) + x(p2,w2) = x(w2,c1) + x(w2,c2) + x(w2,c3) x(w1,c1) + x(w2,c1) = 50000 x(w1,c2) + x(w2,c2) = 100000 x(w1,c3) + x(w2,c3) = 50000 all flows greater than or equal to zero. The Optimization Model
- 31. 3131SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Optimal Solution Facility warehouse p1 p2 c1 c2 c3 w1 140,000 0 50,000 40,000 50,000 w2 0 60,000 0 60,000 0 Total cost for the optimal strategy is $740,000
- 32. 3232SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Optimal Solution Facility warehouse p1 p2 c1 c2 c3 w1 140,000 0 50,000 40,000 50,000 w2 0 60,000 0 60,000 0 Total cost for the optimal strategy is $740,000
- 33. 3333SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) CASE : NETWORK PROBLEM Network Representation ARNOLD WASH BURN ZROX HEWES 200,000200,000 60,00060,000 50,00050,000 100,000100,000 50,00050,000 00 55 44 22 33 44 55 22 11 22 P1P1 P2P2 C2C2 C1C1 W1W1 W2W2 C3C3 140,000140,000 60,00060,000 50,000 50,000 40,,000 40,,000 50,000 50,000 60,000 60,000
- 34. 3434SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) New Supply Chain Strategy OBJECTIVES: • Reduce inventory and financial risks • Provide customers with competitive response times. ACHIEVE THE FOLLOWING: • Determining the optimal location of inventory across the various stages • Calculating the optimal quantity of safety stock for each component at each stage Hybrid strategy of Push and Pull • Push Stages produce to stock where the company keeps safety stock • Pull stages keep no stock at all. Challenge: • Identify the location where the strategy switched from Push-based to Pull-based • Identify the Push-Pull boundary Benefits: • For same lead times, safety stock reduced by 40 to 60% • Company could cut lead times to customers by 50% and still reduce safety stocks by 30%
- 35. 3535SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Three Different Product Categories High variability - low volume products Low variability - high volume products, and Low variability - low volume products.
- 36. 3636SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13) Supply Chain Strategy Different for the Different Categories High variability low volume products • Position them mainly at the primary warehouses demand from many retail outlets can be aggregated reducing inventory costs. Low variability high volume products • Position close to the retail outlets at the secondary warehouses • Ship fully loaded tracks as close as possible to the customers reducing transportation costs. Low variability low volume products • Require more analysis since other characteristics are important, such as profit margins, etc.
- 37. 3737SlideSlideDr. R. Shankar, DMS, IIT Delhi (2012-13)