Problem 1 Figure 1 shows a two dimensional part with five elementary machining features and the corresponding incidence matrix The incidence matrix defines machining features V1 through V8 The costs of machining features V1 V8 are provided by the vector cj 56 166 Production Systems e1 e2 e3 e4 Material Review 3 Determine a set of machining features with the lowest cost using Chavatal s heuristic e5 V 1 V2 V3 V4 V5 V6 V7 V8 e1 e2 a ij e 3 e4 e5 1 1 1 1 1 c j 1 The University of Iowa Intelligent Systems Laboratory 1 1 Figure 1 Part and the corresponding elementary feature feature matrix 1 4 2 2 Intelligent Systems Laboratory Repeat Step 1 Step 0 J Max Step 1 Vj cj e1 e a ij Partial solution J 8 e e e1 1 2 e 1 1 1 1 1 3 4 a ij 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 3 1 1 1 4 c j 1 4 1 2 1 1 2 1 4 2 2 Intelligent Systems Laboratory The University of Iowa Max Max 0 0 1 0 0 1 0 25 0 1 V3 selected Cross out row 3 Recompute Vj J 8 4 1 2 e5 1 2 e e 1 Repeat Step 1 Cj Max 0 0 1 1 0 5 1 0 25 0 1 V 1 V2 V 3 V 4 V 5 V 6 V 7 V 8 c j 1 Vj cj V 1 V2 V 3 V 4 V 5 V 6 V 7 V 8 e5 The University of Iowa Vj V4 selected Cross out row 4 Recompute Vj Max 1 0 5 1 1 1 1 0 75 1 1 V8 selected Cross out rows 1and 2 Recompute Vj Max 2 The University of Iowa Solution Max 1 1 1 1 1 1 1 Vj Cj Intelligent Systems Laboratory Max 0 0 0 0 0 1 0 0 1 V6 selected Cross out row 6 Recompute Vj J 8 4 3 J 8 4 3 6 Vj therefore the solution has been found V 1 V 2 V3 V 4 V 5 V 6 V 7 V 8 e1 e a ij e e 1 2 1 1 1 1 1 1 1 4 The University of Iowa V8 V4 V3 V6 1 e5 c j 1 The machining features selected are 1 1 1 3 2 1 1 2 1 4 2 and the corresponding cost is 1 1 1 2 5 Intelligent Systems Laboratory The University of Iowa Page 1 Intelligent Systems Laboratory Solution Problem 2 Sequence the operations with the Topological Ordering Algorithm to minimize the number of setups The graph in Figure 2 illustrates an electronic assembly with six assembly operations assembly features O1 O6 As all assembly operations are performed on one assembly station setup becomes an issue Three different fixtures F1F3 are needed to produce the assembly Select an F3 F2 appropriate algorithm and O1 O2 F1 F3 determine an assembly plan that O5 O6 F2 F3 minimizes the number of setups O3 O4 F2 O2 The total number of setups not including the initial setup is 3 Intelligent Systems Laboratory The University of Iowa Intelligent Systems Laboratory Solution What is the major difference between schedules a and b in Figure 3 Schedule a is a flow shop schedule Schedule b is a job shop schedule a 3 a 1 2 3 M2 M1 4 1 2 4 Time 4 1 2 3 4 1 2 4 Time b 1 3 M2 3 M2 b M1 F3 O6 The corresponding fixtures are F2 F2 F3 F3 F1 F3 Problem 3 M1 F1 O5 F2 F3 O2 O3 O1 O4 O5 O6 O3 O4 O3 O2 O1 O4 O5 O6 O2 O3 O4 O1 O5 O6 O3 O2 O4 O1 O5 O6 are the best alternative sequences Figure 2 Assembly operations and fixtures The University of Iowa F3 O1 3 M1 1 4 2 3 Time 4 1 3 3 M2 1 4 2 3 Time Figure 3 Two machine schedules The University of Iowa Intelligent Systems Laboratory The University of Iowa Intelligent Systems Laboratory Problem 4 You have been asked to develop a schedule for three products to be assembled at two stations The product structure precedences and operations and the processing times are shown in Figure 4 Develop a good quality schedule by modifying and using a known optimal scheduling algorithm The reason for getting a schedule of good quality is its multiple use a periodic schedule Product 1 1 2 Part 5 Product 2 3 4 6 7 Product 3 9 8 1 2 3 4 5 6 7 8 9 Processing Station 1 4 2 1 5 2 2 3 3 1 time Station 2 5 2 3 4 1 2 2 5 2 Figure 4 Three products and processing data The University of Iowa Intelligent Systems Laboratory The University of Iowa Page 2 Intelligent Systems Laboratory Solution Step 2 Rearrange the sequence considering the precedence constraints in Figure 5 Considering the precedence constraints we modify Johnson s Algorithm to the following steps Step 1 Apply Johnson s algorithm find out the sorted triplets Part Number Min ti1 ti2 Corresponding Machine Number 3 1 1 9 1 1 5 1 2 2 2 1 6 2 2 7 2 1 8 3 1 1 4 1 4 4 2 The University of Iowa Product 1 Station 2 7 8 Product 2 7 The University of Iowa 1 8 2 9 3 4 1 6 2 6 Product 3 4 6 9 8 3 3 1 1 9 1 1 5 1 2 2 2 1 6 2 2 7 2 1 8 3 1 1 4 1 4 4 2 7 2 1 8 3 1 9 1 1 1 4 1 2 2 1 3 1 1 4 4 2 5 1 2 6 2 2 Product 3 Product 1 Product 2 The final schedule is 7 8 9 1 2 3 4 6 5 The University of Iowa 5 3 4 7 Intelligent Systems Laboratory 9 2 5 The final schedule is Station 1 1 5 Intelligent Systems Laboratory Page 3 Intelligent Systems Laboratory
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