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- Politecnico di Milano
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- Management Engineering
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- Industrial Technologies
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University study material for Industrial Technologies in the Management Engineering degree programme at Politecnico di Milano. The document covers: 01. Manufacturing Cells 1 EXERCISE 1 The following matrix identifies the bill of process of the j-th products on the i-th machines. The cycles have been defined with the objective of designing a manufacturing system organized in cells. There are 5 types of products to be made,
University study material for Industrial Technologies in the Management Engineering degree programme at Politecnico di Milano. The document covers: 01. Manufacturing Cells 1 EXERCISE 1 The following matrix identifies the bill of process of the j-th products on the i-th machines. The cycles have been defined with the objective of designing a manufacturing system organized in cells. There are 5 types of products to be made,
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01. Manufacturing Cells 1 EXERCISE 1 The following matrix identifies the bill of process of the j-th products on the i-th machines. The cycles have been defined with the objective of designing a manufacturing system organized in cells. There are 5 types of products to be made, and for the realization of all bills of process of different products are required 7 different types of machine. The objectives are: - the number of cells for the system; - the types of machine to be installed in each cell. SOLUTION: We have to apply the ROC (Rank Ordering Clustering technique) algorithm j-th product 1 2 3 4 5 16 8 4 2 1 i-th machine 1 64 0 1 1 0 1 13 2 32 1 0 0 1 0 18 3 16 0 1 1 0 0 12 4 8 1 0 0 1 0 18 5 4 1 0 0 0 0 16 6 2 1 0 0 1 0 18 7 1 0 0 1 0 1 5 We have to calculate the decimal number from the binary number referred to the single row First row = 2(5−1) ∗ 0 + 2(5−2) ∗ 1 + 2(5−3) ∗ 1 + 2(5−4) ∗ 0 + 2(5−5) ∗ 1 = = 24 ∗ 0 + 23 ∗ 1 + 22 ∗ 1 + 21 ∗ 0 + 20 ∗ 1 = = 0 + 23 + 22 + 0 + 20 = = 8 + 4 + 1 = 𝟏𝟑 … Then we have to order the rows from the highest to the lowest result. NB: the pink column and row remain always in the same order! j-th product 1 2 3 4 5 16 8 4 2 1 i-th machine 2 64 1 0 0 1 0 18 4 32 1 0 0 1 0 18 6 16 1 0 0 1 0 18 5 8 1 0 0 0 0 16 1 4 0 1 1 0 1 13 3 2 0 1 1 0 0 12 7 1 0 0 1 0 1 5 120 6 7 112 5 Now we have to repeat the steps also for columns. First column = 2(7−1) ∗ 1 + 2(7−2) ∗ 1 + 2(7−3) ∗ 1 + 2(7−4) ∗ 1 + 2(7−5) ∗ 0 + 2(7−6) ∗ 0 + 2(7−7) ∗ 0 = = 26 ∗ 1 + 25 ∗ 1 + 24 ∗ 1 + 23 ∗ 1 + 22 ∗ 0 + 21 ∗ 0 + 20 ∗ 0 = = 26 + 25 + 24 + 23 + 0 + 0 + 0 = = 64 + 32 + 16 + 8 = 𝟏𝟐𝟎 … In this case both rows and columns are ordered after only one step, fortunately. j-th product 1 4 3 2 5 16 2 4 8 1 i-th machine 2 64 1 1 0 0 0 18 4 32 1 1 0 0 0 18 6 16 1 1 0 0 0 18 5 8 1 0 0 0…
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