Page 1The University of Iowa Intelligent Systems Laboratory56:166 Production SystemsMaterial Review 3The University of Iowa Intelligent Systems LaboratoryProblem 1Figure 1 shows a two-dimensional part with fiveelementary machining features and the correspondingincidence matrix. The incidence matrix defines machiningfeatures V1 through V8. The costs of machining featuresV1 - V8 are provided by the vector [cj].eeeeVVVV1234VV12345611111[a ] =ij1eeee1243e5e5V7V8111111[1211214][c ]j2Figure 1. Part and the correspondingelementary feature – feature matrixDetermine a set of machiningfeatures with the lowest cost usingChavatal’s heuristic.The University of Iowa Intelligent Systems LaboratorySolutionMaxjjcV= Max {1, 0.5, 1, 1, 1, 1, 0.75, 1} = 1Step 0. J* = { ∅ }V8 selected. Cross out rows 1and 2Recompute |Vj|eeeeVVVV1234VV12345611111[a ] =ij1e5V7V8111111[1211214][c ]j2Step 1.Partial solution J* = { 8 }The University of Iowa Intelligent Systems LaboratoryMaxjjcV=Max {0, 0, 1, 1, 0.5, 1, 0.25, 0} = 1V4 selected. Cross out row 4Recompute |Vj|J* = { 8, 4 }eeeeVVVV1234VV12345611111[a ] =ij1e5V7V8111111[1211214][c ]j2Repeat Step 1.The University of Iowa Intelligent Systems LaboratoryMaxjjCV=Max { 0, 0, 1, 0, 0, 1, 0.25, 0 } = 1V3 selected. Cross out row 3Recompute |Vj|J* = { 8, 4, 3 }eeeeVVVV1234VV12345611111[a ] =ij1e5V7V8111111[1211214][c ]j2Repeat Step 1.The University of Iowa Intelligent Systems LaboratoryMaxjjCV=Max { 0, 0, 0, 0, 0, 1, 0, 0 } = 1V6 selected. Cross out row 6Recompute |Vj|J* = { 8, 4, 3, 6 }Vj= ∅, therefore the solution has been found.The machining features selected areV8, V4, V3, V6and the corresponding cost is 1 + 1 + 1 + 2 = 5Page 2The University of Iowa Intelligent Systems LaboratoryProblem 2The graph in Figure 2 illustrates an electronic assemblywith six assembly operations (assembly features) O1-O6.As all assembly operations are performed on one assemblystation, setup becomes an issue. Three different fixtures F1-F3 are needed to produce the assembly.O1 O2O3 O4O5 O6F3F3F2F2 F3F1Figure 2. Assembly operations and fixturesSelect anappropriatealgorithm anddetermine anassembly plan thatminimizes thenumber of setups.The University of Iowa Intelligent Systems LaboratorySolutionSequence the operations with the Topological OrderingAlgorithm to minimize the number of setups :O2, O3, O1, O4, O5, O6O3, O2, O1, O4, O5, O6O2, O3, O4, O1, O5, O6O3, O2, O4, O1, O5, O6 are the best alternative sequences.The corresponding fixtures are:F2, F2, F3, F3, F1, F3The total number of setups (not including the initial setup) is 3.O1 O2O3 O4O5 O6F3F3F2F2 F3F1The University of Iowa Intelligent Systems LaboratoryWhat is the major difference between schedules (a) and(b) in Figure 3?Problem 333112 42 4TimeM1M243143 12 3TimeM1M2(a)(b)Figure 3. Two-machine schedulesThe University of Iowa Intelligent Systems LaboratorySolution Schedule (a) is a flow shop schedule. Schedule (b) is a job shop schedule.33112 42 4TimeM1M243143 12 3TimeM1M2(a)(b)The University of Iowa Intelligent Systems LaboratoryYou have been asked to develop a schedule for three productsto be assembled at two stations. The product structure(precedences and operations) and the processing times areshown in Figure 4. Develop a good quality schedule bymodifying and using a known optimal scheduling algorithm.The reason for getting a schedule of good quality is itsmultiple use (a periodic schedule).Problem 4The University of Iowa Intelligent Systems LaboratoryProcessingtimePartStation 1 Station 2145222313454521622732835912123Product 1:456789Product 2:Product 3:Figure 4. Three products and processing dataPage 3The University of Iowa Intelligent Systems LaboratorySolutionConsidering the precedence constraints, we modifyJohnson’s Algorithm to the following steps:Step 1. Apply Johnson’s algorithm; find out the sortedtriplets(Part Number, Min{ti1, ti2}, Corresponding MachineNumber):(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 Intelligent Systems Laboratory(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)The final schedule is {7, 8, 9, 1, 2, 3, 4, 6, 5}Step 2. Rearrange the sequence considering theprecedence constraints in Figure 5.Product 3Product 1Product 2123Product 1:456789Product 2:Product 3:The University of Iowa Intelligent Systems Laboratory7 8 1 49 2 6 534187 9 2 6 53The final schedule is:Station 1Station
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