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# note6 - University 0f Southern California Depuﬂmm‘ of...

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Unformatted text preview: University 0f Southern California Depuﬂmm‘ of Industrial and Systems Engineering 33!? jfﬂf. f’rotfm‘fiﬂn I: Facifiﬂ'es and Logistics Spring 2600 April Handouts {PM 2) Computerized Relative Allocation of rhcﬁ'iﬂﬂ Technique (CRA FT) Dr. Ardm'tm Ase} Vazfri We explained that facilities layout algoridtins are divided into construction algorithm and improvement algorithms. Construction algorithms start from scratch and end up 1with a layout. In handout I5 we explained a construction algorithm. Now we explain an improvement algorithm called CRAFT {Computerized Relatith: Allocation of Facilities Technique}. CW starts from an initiaJ layout and end up with a better layout. We also explain MicroCRAJ-T and MULTIPLE which are slight modiﬁcation or'CRAFl'. (lit-en an initial layout. CRAl-T interchanges pairs or triples of departments. We onlyr discuss pairwise interchange. Therefore, it picks Up a pair of departments and changes their locations. The objective is to reduce total flow it distance. Distances are rectilinear and centroid to centtoid. Given the Following layout and FTchart. ED 30 4U 40 First we idemin the centroid of the department and measure the rectiiirtear distance. 5|] 70 . gain}. Total ﬂow it distance in this layout is 1020. Can we by pairwise interchange of department reduce this total ﬂow x distance? CRAFT only changes the locations of a pair of departments which are either adjacent or of the same size. Why? CRAFT interchanges the location of all potential pairs of departments, and ﬁnd the pair of departments which their interchange results in maximum saving in ﬂow x distance. CRAFT assumes when two departments are interchanges their centroids are interchanged. This is true for two departments of the same size but not for two adjacent departments. Therefore, CRAFT indeed estimates the total saving resulted from interchange of two departments. Suppose CRAFT interchanges the locations of A and B. Since it assumes that when A and B are interchanged, their centroids are interchanged. Therefore, the new distance from department-A to any other department is equal to the old distance of department B to that department. Distances A-B unchanged A—C (old B-C) = 65 A-D (old B-D) = 25 (D) (F) (FXD) Estimated (F)(D) is 1060, which means interchange of A and B does not result in a saving. Now let’s check interchange of B and D. Let’s also do not include unnecessary calculations. The distance matrix is as follows. Net result is -75. Therefore the estimated saving is 75. After examining all other potential interchanges. A—C, A-D, C—D we conclude that interchange of B and D results in maximum saving. Note that this is estimated saving, not actual saving, since we have assumed that when B and D are interchanged, their centroras are interchanged. Then we interchange B and D and calculate the actual change in ﬂow x distance. A . (2530) C B . (20,10) . (55,10) D1 . (55,30) D2 0 ) Area ofDI X Area ofD2 XD ‘ Area of (Di+D2) D1 + Area of (D1+D2) 2 @ a _ XD= 800 65 + 80075 — 67.5 \$529 w YD: 800 30* 800 w 25 New actual distance matrix is Total ﬂow x distance in the new layout is 985. Flow x distance in the initial layout was 1020. Therefore interchange of B and D results in 1020 — 985 = 35 units saving. Note that estimated saving was 75. Now we look at this layout as the initial layout, and the interchange process re—starts from the beginning. The candidates for interchange are A-B, A-C, A-D, B-C, C-D. Estimated savings are calculated and A—B shows the highest estimated saving. Again the above layout is assumed as the initial layout. Potential interchanges are A-C, A-D, B- C, C—D. C-D results in maximum estimated saving. The next layout is Again the potential candidates for interchange are A-B, A-C, A-D, B-D. None of them will result in a ﬂow x distance saving. The process stops. Is this layout optimal? No, it is Z—optimal. Can we ﬁx some departments in CRAFT? Yes. ...
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note6 - University 0f Southern California Depuﬂmm‘ of...

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