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Supply chain and logistic optimization Road Map Definition and concept of supply chain. Primary tool box at strategic level (software). Models at strategic level. Models at tactical level. Potential of SCM • A box of cereal spends, on average, more than 100 days from factory to sale. - A car spend around 2 weeks from factory to dealer. • National Semiconductor (USA) used air transportation and closed 6 warehouses, 34% increase in sales and 47% decrease in delivery lead time. • Compaq estimates it lost $0.5 billion to $1 billion in sales in 1995 because laptops were not available when and where needed. • When the 1 gig processor was introduced by AMD (Advanced Micro Devices), the price of the 800 meg processor dropped by 30%. • P&G estimates it saved retail customers $65 million (in 18 months) by collaboration resulting in a better match of supply and demand. • IBM claims that it lost a major market share for desktops in 93, for not been able to purchase enough of a display chip. • US companies spent $898 B for SC activities in 98. Out of the above 58% of SC costs were incurred for transportation and 38% for inventory. Advantage of low inventories • Less time in storage– less deteriorateHigh quality. • Effective distribution process (Fast delivery to customers) • Switching from old technology to new technology without scraping lot of products. • Of course less storage cost. Gartner Group: “By 2004 90% of enterprises that fail to apply supply-chain management technology and processes to increase their agility will lose their status as preferred suppliers”. AMR Research: “The biggest issue enterprises face today is intelligent visibility of their supply chains – both upstream and down” Why supply chain: A tutorial Supplier Retailer Retailer Traditional scenario Demand (Q)= A - B*Z Assume A = 120 B= 2 Supplier buys a goods at price X Sells to retailer at a price Y Retailer sells to customer at a price Z Retailers profit (R) = (Z-Y) * (A-B*Z) Retailers profit (R) = (Z-Y) * (A-B*Z) Y=40 Profit 250 200 150 Profit 100 50 Z 60 58 56 54 52 50 48 46 44 42 0 40 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Demand Profit 40 0 38 38 36 72 34 102 32 128 30 150 28 168 26 182 24 192 22 198 20 200 18 198 16 192 14 182 12 168 10 150 8 128 6 102 4 72 2 38 0 0 Profit Price Suppliers profit = selling cost Q * (Y-X) supplier profit Suppliers profit 250 200 150 Suppliers profit 100 50 Selling price 58 54 50 46 42 38 34 0 30 0 29 56 81 104 125 144 161 176 189 200 209 216 221 224 225 224 221 216 209 200 189 176 161 profit 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 Retailers profit w.r.t. Y Retailer profit 500 450 400 350 300 250 Retailer profit 200 150 100 50 Y 60 57 54 51 48 45 42 39 36 33 0 30 Profit selling price (Y) Retailer profit 36 288 37 264.5 38 242 39 220.5 40 200 41 180.5 42 162 43 144.5 44 128 45 112.5 46 98 47 84.5 48 72 49 60.5 50 50 51 40.5 52 32 53 24.5 55 12.5 56 8 57 4.5 58 2 59 0.5 60 0 Combined profit CP=0.25 (A2 / B -2 . A .X + 2 . B. X. Y – B. Y2) Supply chain 500 450 400 350 300 250 Supply chain 200 150 100 50 Y 60 57 54 51 48 45 42 39 36 0 33 SC profit 418 409.5 400 389.5 378 365.5 352 337.5 322 305.5 288 269.5 250 229.5 208 185.5 162 137.5 112 85.5 58 29.5 0 30 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Combined profit Y Retailers profit w.r.t. Y Profit Profit 500 400 300 200 100 Z (Retail price) 48 46 44 42 40 38 36 0 34 0 58 112 162 208 250 288 322 352 378 400 418 432 442 448 450 448 442 432 418 32 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 30 Selling priceZ Summery Traditional scenario Demand 20 units Retail price $ 50.0 Retailer’s profit $ 200.0 Supplier’s profit $ 200.0 Collaborative scenario Demand 30 units Retail price $ 45.0 Retailer’s profit $ 225.0 Supplier’s profit $ 225.0 Example 10000 • A Retailer and a manufacturer. Demand Curve Demand – Retailer faces customer demand. – Retailer orders from manufacturer. P=2000-0.22Q 2000 Price Variable Production Cost=$200 Selling Price=? Retailer Manufacturer Wholesale Price=$900 Example • Retailer profit=(PR-PM)(1/0.22)(2,000 - PR) • Manufacturer profit=(PM-CM) (1/0.22)(2,000 PR) Retailer takes PM=$900 Sets PR=$1450 to maximize (PR -900) (1/0.22)(2,000 - PR) Q = (1/0.22)(2,000 – 1,450) = 2,500 units Retailer Profit = (1,450-900)∙2,500 = $1,375,000 • Manufacturer takes CM=variable cost Manufacturer profit=(900-200)∙2,500 = $1,750,000 Example: Discount • Case with $100 discount New demand Q = (1/0.22) [2,000 – (PR-Discount)] = (1/0.22) [2,000 – (1450-100)] = 2954 Retailer Profit = (1,450-900)∙2,955 = $1,625,250 Manufacturer profit=(900-200-100)∙2,955 = $1,773,000 Wholesale discount • $100 wholesale discount to retailer • Retailer takes PM=$800 Sets PR=$1400 to maximize (PR -800) (1/0.22)(2,000 - PR) Q = (1/0.22)(2,000 – 1,400) = 2,727 units Retailer Profit = (1,400-800)∙2,727 = $1,499,850 • Manufacturer takes PR=$800 and CM=variable cost • Manufacturer profit=(800-200)∙2,727 = $1,499,850 Global Optimization • What happens if both collaborate (SCM)? Manufacturer sets PR=$1,100 to maximize (PM 200) (1/0.22)(2,000 - PM) Q = (1/0.22)(2,000 – 1,100) = 4,091 units Net profit =(1100-200)∙ 4,091 = $3,681,900 Strategy Comparison Strategy No discount Wth ($100) discount Discount to retailer only SCM scenario Retailer Manufacturer 1,370,096 1,750,000 1,625,250 1,773,000 1,499,850 1,499,850 - Total 3,120,096 3,398,250 2,999,700 3,681,900 Market Requirement • • • • • Low price. High quality. Product customization. Fast delivery. Fast technology induction. Strategy: Right product + Right quantity + Right customer + Right time Cost reduction and value addition at each stage Sharing information will lead to reducing uncertainty for all the partners and hence: - reducing safety stocks, - reducing lead times, - improving brand name. Result - Value added product/services - large market segment (Leading brand), - long lasting supply chain, - consistent and growing profits, - Satisfied customer. It is better to collaborate and co-ordinate to achieve a win-win situation. Definition Supply Chain refers to the distribution channel of a product, from its sourcing, to its delivery to the end consumer. (some time referred as the value chain) A supply chain is a global network of organization that cooperate to improve the flows of material and information between suppliers and customers at the lowest cost and the highest speed. Objective is customer satisfaction and to get competency over others Supply chain Information flow Material Flow Cooperation Value addition Value added product Customer satisfaction Coordination Information sharing Assets Management (Inventory) Difficulties • • • • Increasing product variety. Shrinking product life cycle. Fragmentation of supply chain. Soaring randomness because of new comers. Responsiveness (High cost) Low cost Deterministic Uncertainty FMCG (Tooth paste, soaps etc ) Price sensitive, Low uncertainty Fashionable products (Garments), customize products etc High responsiveness, high uncertainty Ineffective marketing Wrong material High Inventories Low order fill rates Supply shortages Inefficient logistics High stock outs Local objectives vs Global objective Marketing Production - High inventories levels. - Low production cost. - Low prices. - High utilization. - Ability to accept every customer order. - High quality raw material. - Short product delivery lead times. - Stable production. -Various order sizes and product mixes. - Big order size. -Short delivery lead times (Raw materials) SCM is a tool which integrate necessary activities and decomposing irrelevant activities. Management by projects Management by departments Orders Finished products Primary decisions Buy Secondary decisions Store Sell Recruitment of employees Training, relocation, etc Salary, TA, DA Strategic models Selection of providers - Provider provides the quantity between two limits. - Cost is a concave function of the shipped quantity. - Objective is to select one or more provider to satisfy the demand. •Case of single manufacturing unit. •Case of several manufacturing unit. Cost function Problem to be solved Algorithm Select the cheapest provider if the quantity to be supplied is maximum. Adjust the remaining quantity among rest of the providers so that the solution remain feasible. Exact algorithm for the integer demand 1. Compute the cost for the first provider for q=1,2,…A. 2. For provider=2,…N - Compute the Q=1,2,…,A F(Pro,Q)=Min0<=x<=Q (F(Pro-1, Q-x), fpro(x) Case of several manufacturing units: Approches • Heuristic approach. • Piece wise linearization for integer programming. • Bender decomposition. Capacity Planning Providers Manufacturers Retailers 1. Idle processing capacity with providers in different periods. 2. Idle transportation capacity with providers in different periods. 3. Idle manufacturing capacity with manufacturers. 4. Idle transportation capacity with manufacturers. 5. Demand at various retailers and the demand of one period may differ from another period. Objective is to decide how much and where to invest. Supply <= Available Capacity + Added Capacity Added capacity <= BigNumber* Binary Variable {0,1} Investment Cost BinaryVariable*FixedCost + Slope*Added capacity Formulation Results 1. There exist at least one optimal solution in which all the binary constraints are saturated. 2. Replacing big number by corresponding capacity, if greater than zero, and denote new problem by P2m, then m is definite. Approach - Construct the several instances P20 , P21, ... of the problem from the relaxed solution of the problem P1 . These instance converge towards the solution of the problem P. Unfortunately, we do not know the conversion time. - For this reason, we derive sub-optimal solution from the instances and select the best solution. Algorithm Optimal solution Error 1. 276518 275781 0.267 2. 257359 257359 0.0 3. 324934 322973 0.607 4. 292007 292007 0.0 5. 354365 354116 0.070 6. 322126 325881 1.910 7. 334013 333499 0.154 8. 321299 319871 0.446 9. 291253 290190 0.366 10. 310869 310869 0.0 Other approaches - Langrangean heuristic approach to find good lower bound. - Branch and price approach. Presented approach was similar to the langrangean approach. Short term supply chain formation • • • • Multi-echelon system. Selection of a partner from each echelon. Expected demand is known for a given horizon. Objective is not to invest at any location. - Utilization of idle capacity (Production, transportation, storage etc). • Solution should be feasible for entire horizon. • Decision: Whether the new chain exists and profitable? Demand Costs • Storage cost at entry and at exit (running). • Production cost (running). • Connection cost (Fixed) Note: It is possible that solution may not exist Minimisation Path relaxation approach 1. Resolve two echelon problem for each pair of nodes using final demand. 2. Consider the cost corresponding to these arcs as surrogate length. 3. Solve the k-shortest path problem and compute the k-shortest path. 4. For each shortest path, compute the real cost. 5. If the relax path length is bigger than best real cost, stop. D1 D2 D3 D1 D2 D3 Demand Insertion of new project 1 1 2 3 2 3 Problem data 1 (2, 1) 2 (5,1) 3 (2,0) Solution 1 5 1 3 Formulation Min xm1 s.t. An optimal algorithm is known for the above case. Simple assembly Algorithm 1. 2. Computing two times: Early start time – Latest start time Select the common interval. Complex schedule 3 4 1 7 5 6 14 2 8 9 12 10 11 13 Two approaches • Simple- easy to program but time consuming. • Little tidy – difficult to program but on average performance is better. • Both gives the optimal schedule. • Worst case complexity is also same. Simple approach S1 1 3 4 7 14 S2 2 3 4 7 14 S3 8 9 12 13 14 5 6 7 14 11 12 13 14 S4 S5 10 For each assembly operation • If the idle windows for two different lines are not the same then select the window which has higher lower limit (beginning time) and restart the calculation. • If the windows are the same but starting time are different, then set the lower limit of this window as the greatest starting time of the two and restart the calculation. If neither of the above case is present, then the solution is optimal. The algorithm converge towards an optimal solution. Second approach 1. Decompose the assembly into subassemblies. 2. Solve the sub-assemblies. 3. Coordinate the timing of sub-assemblies. • Recursive approach. • Each time early start time (EST) and latest start time (LST) of assembly has to be computed. • Advantage: If the time lies between EST and LST then re-computation is not required. Second approach 3 4 1 7 5 6 14 2 8 9 12 10 11 13 3 4 7 1 5 5 6 2 7 8 9 12 10 14 13 11 13 WIP Control (Extension) First case Number of finished jobs at the exit of each machine is not limited in quantity but limited by time. Approach Introduce one virtual machine, following the real machine with operation time 0 and flexibility [0, T] • Case 2 Number of finished jobs are limited in time and in quantity too. Approach Introduce as many virtual machines as the number of finished jobs permitted. In both the cases, the algorithm presented before are applicable. Delivery date Instant of ordering Delivery instant Inventory holding cost Three cots are to be considered I1, Inventory cost between the arrival of first component and last component. I2, Inventory cost between the arrival of last component and the delivery date. B, Backlogging cost if the last component arrives after the delivery date. Next W(Z)I2 B is continuous and differentiable in [R, +∞] Basic property n Fk (Z rk ) k 1 h n h sk k 1 1 n Fk (Z rk ) Propriété fondamentale: k 1 h n h sk 1 k 1 Algorithme général • Partir d’une solution admissible. R r1,r2 ,...,rn • Chercher une solution admissible R1 du voisinage de R qui vérifié (1) • Conserver ou rejeter R1 (recuit simulé) • Retour à 2. Algorithm 1. Start with one feasible solution. (First feasible solution can easily be generated considering the same ordering time for all components.) 2. Define new solution R1 in the neighborhood of R that satisfies relation (1) (Use gradient method) 3. Conserve or reject R1 (Simulated annealing) 4. Go to 2. The behavior of cost I1 depends upon the density function. The I1 may be convex or concave based on the nature of density functions. Hence, with an exact information of density functions an specialize algorithm could have been devised. With uniform densities, the problem can be solved using gradient method only. For general problem we proposed an approach based on simulated annealing which looks, in each iteration, for a closer solution which satisfies the relation 1. Partnership formation: A model Assumptions Customer is price sensitive. Average customer demand depleted as price increases. Customer demand is stochastic. Backordering cost at supplier is higher than at retailer. Inventory cost at supplier is cheaper than retailer. Inventory can be transferred between supplier and retailer in negligible time. In other words customer is ready to wait during the lead time. Example Model • Each maintain a stock of Ip and Ir. • Retailer pays for the holding and also pays for stock out. • Supplier pays for holding and also pays for stock out. This stock out penalty goes to retailer (Compensation). • Profit is shared according to their relative risk i.e. investment. Objective is to maximize total benefit. Fractional demand Fractional demand is given by f(wr), a concave function of selling price. Lost sell due to high price Price wr Algorithm 1.Take any starting price wr 2. Optimize Ip and Ir => Ip and Ir , keeping Wr fix. Total profit function is convex w.r.t stocks and wr constant. 3. Optimize Wr, Ip and Ir are fixed. Combined profit function is concave w.r.t Wr. 4. Go to 2 until profit increases. Numerical illustration Continue Hot areas in SCM • Dynamic pricing - Online bidding, (Ebay.com) - Price setting and discount for perishable goods in supermarket. - Customize pricing: different prices for different customer segment. • Inventory management in advance demand information sharing.