上海交通大学：《生产计划与控制 Production Planning and Control》课程教学资源（课件讲稿）chap07_Class5

Production Planning and Control Chapter 7 Production Planning Professor JIANG Zhibin Dr GEAN Na Department of Industrial Engineering Management Shanghai Jiao Tong University 上浒充通大粤

Production Planning and Control Professor JIANG Zhibin Dr GEAN Na Department of Industrial Engineering & Management Shanghai Jiao Tong University Chapter 7 Production Planning

图 Chapter 7 Production Planning Contents ·Introduction Master Production Scheduling (MPS) 。 Material Requirement Planning (MRP) ·Capacity Planning 。Improvement in MRP 上游充道大睾

Chapter 7 Production Planning Contents • Introduction • Master Production Scheduling (MPS) • Material Requirement Planning (MRP) • Capacity Planning • Improvement in MRP

Optimal Lot Sizing for Time-Varying Demand Optimal in this context means the policy that minimizes the total holding and setup cost over the planning horizon. .The optimal policies can be determined by casting the problem as a shortest-path problem. √Assume that Forecasted demands over the next n periods are known and given by the vector r(r1,r2,...,rn) Costs are charged against holding at Sh per unit per period and $K per setup.We will assume that the holding cost is charged against ending inventory each period. 上游充道大睾

Optimal Lot Sizing for Time-Varying Demand •Optimal in this context means the policy that minimizes the total holding and setup cost over the planning horizon. •The optimal policies can be determined by casting the problem as a shortest-path problem.  Assume that  Forecasted demands over the next n periods are known and given by the vector r=( r1, r2, …, rn )  Costs are charged against holding at $ h per unit per period and $ K per setup. We will assume that the holding cost is charged against ending inventory each period

Optimal Lot Sizing for Time-Varying Demand .Example:The forecast demand for an electronic assembly produced at Hi-Tech,a local semiconductor fabrication shop,over the next four weeks is 52,87,23,56.There is only one setup each week for production of these assemblies,and there is no back- ordering of excess demand.Assume that the shop has the capacity to produce any number of the assemblies in a week. Let y,...be the order (or Production)quantity in each of the four weeks ·52≤y1≤52+87+23+56=218 10200 ·139-y1y2218-y1ify1139 pairs ·0sy2≤218-y1ify1>139 上游充道大睾

Optimal Lot Sizing for Time-Varying Demand •Example: The forecast demand for an electronic assembly produced at Hi-Tech, a local semiconductor fabrication shop, over the next four weeks is 52, 87, 23, 56. There is only one setup each week for production of these assemblies, and there is no back￾ordering of excess demand. Assume that the shop has the capacity to produce any number of the assemblies in a week. • Let y1,…,y 4 be the order (or Production) quantity in each of the four weeks. • 52 ≤ y1≤ 52+87+23+56=218 • 139- y1 ≤y 2 ≤218- y1 if y1≤139 • 0 ≤y 2 ≤218- y1 if y1>139 10200 pairs

Optimal Lot Sizing for Time-Varying Demand .Search all the feasible policies is unreasonable. The Wagner-Whitin algorithm observation: An optimal policy has the property that each value ofy is exactly the sum of a set of future demands. 、y1=1,0ry1=1+r22…,0ry1=r1+r2+.+rn ·y2=0,y2=72,0r2-r2+r3,,0ry2=f2+3+.+rn ·yn=0,yn=rn The number of exact requirements policies is much smaller than the total number of feasible policies. 上泽充鱼大皇

Optimal Lot Sizing for Time-Varying Demand •Search all the feasible policies is unreasonable. •The Wagner-Whitin algorithm observation: • An optimal policy has the property that each value of y is exactly the sum of a set of future demands. • y1= r1, or y1= r1+ r2, …, or y1= r1+ r2 + … + rn • y 2=0, y 2 = r2, or y 2 = r2 + r 3, …, or y 2 = r2 + r 3 + … + rn • . • . • y n=0, y n = rn. • The number of exact requirements policies is much smaller than the total number of feasible policies

Optimal Lot Sizing for Time-Varying Demand .Example (continued):We continue with the four-period scheduling problem. ·y1=52,0ry1=139,y1=162,ory1=218 ·y2=0,y2=87,ory2=110,ory2=166 ·y3=0,y3=23,0ry3=78 ·y4=0,0ry4=56 .It is easy to see that every exact requirements policy is completely determined by specifying in what periods ordering should take place. 上泽充鱼大皇

Optimal Lot Sizing for Time-Varying Demand •Example (continued): We continue with the four-period scheduling problem. • y1=52, or y1=139, y1=162, or y1=218 • y 2=0, y 2 =87, or y 2 =110, or y 2 =166 • y 3=0, y 3=23, or y 3=78 • y 4=0, or y 4=56 •It is easy to see that every exact requirements policy is completely determined by specifying in what periods ordering should take place

Optimal Lot Sizing for Time-Varying Demand Example (continued):We continue with the four-period scheduling problem. if the production takes place in periodj else .This problem can be regarded as a one-way network with the number of nodes equal to exactly one more than the number of periods. .Every path through the network corresponds to a specific exact requirements policy 上游充道大睾

Optimal Lot Sizing for Time-Varying Demand •Example (continued): We continue with the four-period scheduling problem. •Define •This problem can be regarded as a one-way network with the number of nodes equal to exactly one more than the number of periods. •Every path through the network corresponds to a specific exact requirements policy. 1 if the production takes place in period 0 else j j i     1 2 3 4 5

Optimal Lot Sizing for Time-Varying Demand .Example(continued):We continue with the four-period scheduling problem. .The next is to assign a value to each arc in the network. .The value or"length"of the arc (,)called c is defined as the setup and holding cost of ordering in period i to meet the requirements through j-1. "Finally,dynamic programming is used to determine the minimum- cost production schedule,or shortest path through the network. 上游充道大睾

Optimal Lot Sizing for Time-Varying Demand •Example (continued): We continue with the four-period scheduling problem. •The next is to assign a value to each arc in the network. • The value or “length” of the arc ( i, j), called cij, is defined as the setup and holding cost of ordering in period i to meet the requirements through j-1. •Finally, dynamic programming is used to determine the minimum￾cost production schedule, or shortest path through the network

Optimal Lot Sizing for Time-Varying Demand Example(continued):We continue with the four-period scheduling problem. .Path enumeration is used to solve the Hi-Tech problem.Recall that r=(52,87,23,56).In addition,h=$1,andK=$75. c12=75 (setup cost only) ℃13=75+87=167 j 2 3 4 5 ℃1475+87+23*2=208 1 75 162 208 376 2 75 98 ℃15=75+87+23*2+56*3=376 210 3 75 131 ℃23=75 4 75 c24=75+23=98 C25=75+23+56*2=210 ℃34-75 C35=75+56=131 C45-75 1-2-3-4-5:75+75+75+75=300:1-2-4-5:75+98+75=248 上游充鱼大粤

Optimal Lot Sizing for Time-Varying Demand •Example (continued): We continue with the four-period scheduling problem. •Path enumeration is used to solve the Hi-Tech problem. Recall that r=(52, 87, 23, 56). In addition, h=$1, and K=$75. • c12=75 (setup cost only) • c13=75+87=167 • c14=75+87+23*2=208 • c15=75+87+23*2+56*3=376 • c23=75 • c24=75+23=98 • c25=75+23+56*2=210 • c34=75 • c35=75+56=131 • c45=75 1-2-3-4-5: 75 ＋75 ＋75 ＋75 ＝300；1-2-4-5: 75+98+75=248 i\j 1 2 3 4 5 1 75 162 208 376 2 75 98 210 3 75 131 4 75

Optimal Lot Sizing for Time-Varying Demand Example(continued):We continue with the four-period scheduling problem. .Path enumeration is used to solve the Hi-Tech problem.Recall that r=(52,87,23,56).In addition,h=$1,andK=$75 Path Cost($) 1 2 3 4 5 1 1-2-3-4-5 75 162 208 376 300 2 75 98 210 1-2-4-5 248 3 75 131 1-2-5 285 4 75 1-2-3-5 281 1-3-4-5 312 The optimal path is 1-2-4-5 at cost 248. 1-3-5 293 1-4-5 283 .The optimal ordering is 1-5 376 ·y1=52,y2=110,y3=0,y4=56 上游充道大睾

Optimal Lot Sizing for Time-Varying Demand •Example (continued): We continue with the four-period scheduling problem. •Path enumeration is used to solve the Hi-Tech problem. Recall that r=(52, 87, 23, 56). In addition, h=$1, and K=$75. •The optimal path is 1-2-4-5 at cost 248. •The optimal ordering is • y1=52, y 2=110, y 3=0, y 4=56 i\j 1 2 3 4 5 1 75 162 208 376 2 75 98 210 3 75 131 4 75 Path Cost($) 1-2-3-4-5 300 1-2-4-5 248 1-2-5 285 1-2-3-5 281 1-3-4-5 312 1-3-5 293 1-4-5 283 1-5 376