Outline. Page 1. Big Picture LAYOUT EXAMPLES LAYOUT OF MACHINES AND FACILITIES

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1 Andrew Kusiak, Professor Industrial Engineering 9 Seamans Center Iowa City, Iowa - 7 LAYOUT OF MACHINES AND FACILITIES Tel: 9-9 Fax: andrew-kusiak@uiowa.edu Outline LAYOUT EXAMPLES SINGLE-ROW MACHINE LAYOUT DOUBLE-ROW MACHINE LAYOUT MULTIROW FACILITY LAYOUT ALGORITHMS Equipment Selection Cell Formation Layout Cell Layout Latest layout software Proplanner Big Picture Facility planning Facility location Facility design Facility utilization Jobs: Facility Designer Facility Planner Structural design layout design Material handling system design AutoCAD Software LAYOUT EXAMPLES MANUFACTURING FACILITY LAYOUT ASSEMBLY AREA LAYOUT Page

2 Office Layout LAYOUT OF MACHINES Methods: Visual, e.g., using templates Computer based (often an environment for evaluating various layout alternatives) Model based LAYOUT OF MACHINES Visualization types: D D Virtual reality LAYOUT OF MACHINES Model Based Approach Basic Layout Patterns Single-Row Layout Double-Row Layout Multi-row Layout M M M M M M6 M7 M8 Page

3 BASIC DATA UNIT DISTANCE TRAVEL COST (c) - OR FREQUENCY OF TRIPS (f) DISTANCE (d) - OR TRAVEL TIME (t) OBJECTIVE Arrange machines on the shop floor in such a way that the total product of the travel cost per unit distance traveled [$/m] and the distance traveled [m] is minimum. Travel cost is often replaced with the number (frequency - f) of trips per time unit [/sec] between any two machines. Distance is replaced with travel time - t [sec]. n n n n Min fik tjl x xkl i=j=k=l= Example: Intuitive Layout 0 sq.m. sq.m. 0 sq.m. sq.m. 6 sq.m. High Frequency of Traffic SINGLE-ROW MACHINE LAYOUT Frequency of trips (traffic flow) w here: k v n v k f = k = u k = volume of part type k to be moved from machine i to machine j in a given time horizon (for example, year) ni j = number of different parts to be moved from machine i to machine j in a given time horizon u k = number of parts to be moved in a single trip of the material handling equipment = smallest integer greater than or equal to. The travel time involves: loading time acceleration time travel time deceleration time unloading time DISTANCE CALCULATION Consider machines 7 A B x C Site Unidirectional distance (along x axis) r = 0, r = 7, r = r = 7 Rectilinear (city) distance d =, d = 7, d = 7 + = 0 Euclidean distance e = 7, e = 7.8 Page

4 Adjacent unidirectional distance r 0 0 M 6 clearance 0 M 0 Time is the best measure to be used in the optimization of machine layout, however, if the travel speed is constant, an appropriate distance measure can be considered. r =d = = 6 SINGLE ROW MACHINE LAYOUT Algorithm Overview Two machines with the maximum flow between them (e.g., and ) are placed in the adjacent locations. The minimization of the total product of flow x travel time is accomplished by local maxima. Global Minimum Local Maximum Other machines are placed to the left and right of the two machines based on the flow. Min n n n n fik tjl x xkl i=j=k=l= Algorithm Step 0. Set iteration number k =. From the flow matrix [f] compute fi*j* = max {f: i, j =,,..., m}. If there is a tie, fi*j* = max {f t: i, j =,,..., m}. Connect i*, j* and include them in the solution set. Set the solution set, U = {i*, j*}. Step. Compute f p*q* =max{f i*k, f j*l:k,l {{,,..., m} - U}. Set s* = q*. Consider two alternatives: (a) Place machine s* left of machine i*; (b) Place machine s* right of machine j*. Compute: q s*i* = { f s*r t s*r:r U and machine s* is placed left of machine i*}; q j*s* = { f s*r t s*r:r U and machine s* is placed right of machine j*} Page

5 If q s*i* q j* s*, () S elect the alternative (a) above; () S et i* = s*. Step.Set iteration number k = k +. Repeat Step until the final solution is obtained (i.e.., until all the machines are included in the solution set U). If q s*i* > q j* s*, () S elect the alternative (b) above; () S et j* = s*. U = U + s*. Given: [f ] = Single-row machine layout Calculate adjacent distances [d ] = number dimensions x x 6x6 x f = frequency of trips t = travel time r = unidirectional distance d = adjacent distance number dimensions x x 6x6 x Clearance = x x d = M M + + = For simplicity assume time t = distance d 0 0 M Distances 6 0 M 0 Adjacent distance d = = 6 Unidirectional distance r = d = 6 Clearance Iteration Step 0. From the flow matrix [f], max {f: i =,,,, j =,,, } = f = 0 is obtained. Thus i* =, j* = is determined. s and are connected and included in the solution, symbolically denoted as The solution set is updated U = {, } and the columns and of the matrix are marked with asterisks. * * [f ] = Page

6 [f ] = * * Step. Compute Max { fk, fl: k =, ; l =, } = f = f =. Set s* =. Consider two alternatives: (a) Place machine on the left hand side of machine Min n n n n fik tjl x xkl i=j=k=l= qs*i* = q = f t + f t = f r + f r = f d + f (d + d) = + ( + ) = 9. (b) Place machine on the right hand side of machine qj*s* = q = f t + f t = f r + f r = f d +f (d + d) = + ( + ) = 6. Compute (repeated from the previous slide): qs*i* = q = f t + f t = f r + f r = f d + f (d + d) = + ( + ) = 9. qj*s* = q = f t + f t = f r + f r = f d +f (d + d) = + ( + ) = 6. As q > q, () Alternative (b) (min) is selected where machine is placed to the right of machine () Set j* = ; () The set U is updated to U + s* = {,, } and column is marked with an aster. Step. Since machine is not included in the solution set U, go to Step. [f ] = Iteration Step.Compute Max { fk, fl : k =, l = } = = f = 0. Set s* =. Consider two alternatives: (a) Place machine left of machine * * * (b) Place machine right of machine Compute: qs*i* = q = f t + f t + f t = f r + f r + f r = f d + f (d + d) + f (d + d + d) = (6 + ) + 0 (6 + + ) = 9. qj*s* = q = f t + f t + f t = f r + f r +f r = f d + f (d + d) + f (d + d + d) = 0 + ( + ) + 0 ( + + ) = 7. Since q > q, () Alternative (b) [min]is selected, i.e., () Row and column are deleted from [f]; () Set j* = ; () The solution set U is updated to U + s* = {,,, }. Step.Since all machines have been included in the solution set U, stop. Page 6 6

7 Solution Algorithm Summary Pair (, ) selected Look around M M M M selected placed Algorithm Summary Algorithm Summary Alternative selected Look around Min n n n n fik tjl x xkl i=j=k=l= selected placed Lowest cost alternative selected Objective Function Revisited Double-Row Layout Example Layout Min n n n n fik tjl x xkl i=j=k=l= M M M M7 M6 M M f No trips/sec t sec fxt = No of trips A rrang em ent num ber pair (M, M6) (M, M) (M, M7) (M) Page 7 7

8 Computation of Unidirectional Distances Between Pairs of s d M M M M M M M M d -----: Referenceline r Unidirectional distances r =0 r =d r =d r =d r =d r = r -r = d -d r = r -r = d -d Algorithm : Double-row machine layout A pair of machines with the largest value of the flow is selected The machines selected are placed on the opposite sites of the isle (or an track) Other pairs of machines are placed to the left and right of the pairs selected. Algorithm Step 0. Set iteration number k =. From the flow matrix [f], compute fi*j* = Max {f: i, j =,,..., m }. If there is a tie, fi*j* = Max {f t: i, j =,,..., m}. Place i*, j* on the opposite sites of the path and include them in the solution. Set the solution set, U = {i*, j*}. Remove columns i* and j* of matrix [f] from further consideration. Step. Compute fp*q* = Max {fi*k, fj*l: k, l {,,..., m} - U}. Set s* = q* and remove q* from further consideration. Compute fx*y* = Max { fi*k, fj*l, fs*v: k, l, v {{,,..., m} - U - q* }. Set t* = y* and remove y* from further consideration. Consider two alternatives: (a) Place s* right of i* and t* right of j*. (b) Place t* right of i* and s* right of j*. Compute: qi*s*j*t* = { fs*r ts*r + ft*r tt*r: r U } + fs*t* ts*t*, where s* is placed right of i* and t* is placed right of j*; qi*t*j*s* = { fs*r ts*r + ft*r tt*r: r U } + fs*t* ts*t*, where t* is placed right of i* and s* is placed right of j*. If qi*s*j*t* < qi*t*j*s*, select the alternative (a); Otherwise, select the alternative (b). Set U = U + {s*, t*}. Step. If only one machine has not been assigned, go to Step ; Otherwise go to Step. Step. Compute fp*q* = Max { fi*k, fj*l, fs*v, ft*w: k, l, v, w {,,..., m} - U}. Set c* = q* and remove q* from further consideration. Compute fx*y* = max { fi*k, fj*l, fs*v, ft*w, fc*z: k, l, v, w, z {{,,..., m} -U -q*}. Set d* = y* and remove y* from further consideration. Consider four alternatives: (a) Place c* left of i* and d* left of j*; (b) Place d* left of i* and c* left of j*; (c) Place c* right of s* and d* right of t*; (d) Place d* right of s* and c* right of t*. Page 8 8

9 Compute: qc*i*d*j* = { fc*r tc*r + fd*r td*r: r U } + fc*d* tc*d*, where c* is placed left of i* and d* is placed left of j*; qd*i*c*j* = { fc*r tc*r + fd*r td*r: r U } + fc*d* tc*d*, where d* is placed left of i* and c* is placed left of j*; qs*c*t*d* = { fc*r tc*r + fd*r td*r: r U } + fc*d* tc*d*, where c* is placed right of s* and d* is placed right of t*; qs*d*t*c* = { fc*r tc*r + fd*r td*r: r U } + fc*d* tc*d*, where d* is placed right of s* and c* is placed right of t*. If qc*i*d*j* = min { qc*i*d*j*, qd*i*c*j*, qs*c*t*d*, qs*d*t*c* }, () Select alternative (a); () Set i* = c* and j* = d*. If qd*i*c*j* = min { qc*i*d*j*, qd*i*c*j*, qs*c*t*d*, qs*d*t*c* }, () Select alternative (b); () Set i* = d* and j* = c*. If qs*c*t*d* = min { qc*i*d*j*, qd*i*c*j*, qs*c*t*d*, qs*d*t*c* }, () Select alternative (c); () Set s* = c* and t* = d*. If qs*d*t*c* = min { qc*i*d*j*, qd*i*c*j*, qs*c*t*d*, qs*d*t*c* }, () Select alternative (d); () Set s* = d* and t* = c*. Set U = U + {c*, d*}. Go to Step. Step. Set c* to the last machine which has not been assigned. Consider four alternatives: (a) Place c* left of i*; (b) Place c* left of j*; (c) Place c* right of s*; (d) Place c* right of t*. Compute: qc*i*= { fc*r tc*r: r U }, where c* is placed left of i*; qc*j* = { fc*r tc*r: r U }, where c* is placed left of j*; qs*c* = { fc*r tc*r: r U }, where c* is placed right ofs*; qt*c* = { fc*r tc*r: r U }, where c* is placed right of t*. If qc*i* = min { qc*i*, qc*j*, qs*c*, qt*c* }, select alternative (a); If qc*j* = min { qc*i*, qc*j*, qs*c*, qt*c* }, select alternative (b); If qs*c* = min { qc*i*, qc*j*, qs*c*, qt*c* }, select alternative (c); If qt*c* = min { qc*i*, qc*j*, qs*c*, qt*c* }, select alternative (d). Set U = U + { c* }. Go to Step. Step.Set iteration number k = k +. Repeat Steps through until the final solution is obtained (i.e., all the machines are included in the solution set U). [f ] = Frequency of trips [c ] = Dimensions Number Dimension 0x0 0x0 x0 60x 0x Clearances [d ] = Distances Distance d is calculated from machine dimensions and clearance Dimensions Number Dimension 0x0 0x0 x0 60x 0x Based on d, r is computed in the algorithm Assume t = r Page 9 9

10 Recall? 0 Adjacent distance d = = 6 Unidirectional distance r = d = 6 0 M 6 0 M Clearance 0 [f ] = Iteration Step 0. For the flow matrix above, compute max {f: i, j =,,..., } = f = f = is determined. The flow value f is selected because f t = f d = 8 = 0 > f t = f * d = 6 = 80. Thus i* = and j* =. s and are assigned to the opposite sites of the path and are included in the solution. The solution set is updated, U = {, }. Exclude columns and from further consideration. [f ] = * * Step. Compute max { fk, fl: k, l =,, } = f. Set s* = and exclude column from further consideration. * * * [f ] = Compute: max {fl, fv: l, v =, } = f. Set t* = and exclude column from further consideration. * * * * Consider two alternatives: 0 0 (a) Place machine right of machine and machine right of machine (b) Place machine right of machine and machine right of machine (a) [f ] = (b) Distance Pattern (a) Compute: (a) [d ] = t = r [f ] = qs*i*t*j* = q = f t + f t + f t + f t + f t = f r + f r + f r + f r + f r = f d + f d + f d + f d + f d - d = = Page 0 0

11 (b) qs*j*t*i* = q = f t + f t + f t + f t + f t = f r + f r + f r + f r + f r = f d + f d + f d + f d + f d - d = = 97. Since q < q, alternative (b) is selected. Set U = U + {, }. Step. Set c* =. [f ] = Consider four alternatives: (a) Place machine left of machine (c) Place machine right of machine (d) Place machine right of machine * * * * (b) Place machine left of machine Step. Since only machine has not been assigned, go to Step. (a) qc*i* = q = f t + f t + f t + f t = f r + f r + f r + f r = f d + f d + f (d + d) + f (d + d) = (. + ) + (. + ) = = 9. (b) Compute qc*j* = q = f t + f t + f t + f t = f r + f r + f r + f r = f d + f d + f (d + d) + f (d + d) = (8. + ) + (8. + ) = 79. (c) Compute (continued) qs*c* = q = f t + f t + f t + f t = f r + f r + f r + f r = f d + f d + f (d + d) + f (d + d) = (. + ) + 0 (. +) = 9. (d) qt*c* = q = f t + f t + f t + f t = f r + f r + f r + f r = f d + f d + f (d + d) + f (d + d) = (8. + ) + 0 (8. +) =. Since q = min {q, q, q, q }, alternative (c) is selected Solution M M M Set U = U + {}. Step. Since all machines have been included in the set U, stop. M M : Reference line Page

12 Algorithm Summary Algorithm Summary Pair (, ) selected Looking for Looking around selected Evaluate cost selected a) b) Alternative (b) selected Algorithm Summary Looking around selected Four alternatives evaluated Algorithm Summary Lowest cost alternative selected n n n n n n Min fik cjl x xkl + a x i=j=k=l= i=j= n = number of facilities and their locations a = cost of locating and operating facility i at location j fik = flow of material between facility i and facility k cjl = cost per unit flow of material between location j and location l. if facility i is at location j x = { 0 otherwise subject to: n x = j= n x = i= x = 0 or i =,..., n j =,..., n i =,..., n, j =,..., n Page

13 MULTIROW FACILITY(MACHINE) LAYOUT Example Determine the layout of four facilities (machines),,, and at sites A, B, C, and D Location of Four Facilities (s) at Four Sites A B C,,, = facilities (machines) A, B, C, D = sites (beacons) = site center D Site Flow Matrix (between facilities (machines)) Assumptions: Symmetric flow matrix Square facilities A B C D [f] = Matrix of Rectilinear Distances (between sites) A B C D A - [d] = B C - - D - Begin with an initial assignment Site Cost 0 A B C D [f] = COST = = fdad + fdab + fdac + fddb + fddc + fdbc = = 0. = A B C D A - B - [d] = C - D - A B C D Pairwise Exchanges A B C A B C Cost 0 A B C D Exchange facilities (machines) and D D Cost = fdad + fdbd + fdcd + fdab + fdac+ fdbc = = 0. Cost = 0 A B C D The cost change is then 0-0 = -00. Page

14 The Result Exchanged Pair (, ) (, ) (, ) (, ) (, ) (, ) Cost Change Cost 0 A B C Exchange Pair (,) (,) (,) (,) (,) (,) D Solution with facilities (machines) and exchanged is selected Cost 70 Cost Change A B C D A B C D Exchange Pair (,) (,) (,) (,) (,) (,) Cost Change Final Solution Relationship Constraints This box shows the relationship between machine (facility) and Cost 70 A B C D E Top half shows importance of relationship Lower half shows reason(s) A B C D of importance Closeness ratings Rating Closeness Example of reasons behind Absolutely the closeness rating A Necessary Code Reason Especially E Important Personal contact Important I Convenience Ordinary O Closeness Noise, Disturbance Unimportant U Light Not desirable X EXAMPLE: Flow data and relationship constraints From To Process Layout: Medical Application E O I Closeness rating E O Rating Closeness 00 O A E Absolutely Necessary Especially Important 0 O 00 A I O Important Ordinary Closeness U X Unimportant Not desirable Page

15 Data Collection Process Possible Groupings Resource a Station b Station c Station M O O Resources Note: Precedences are ignored in this matrix Possible Groupings Station System Layout Based on the process-station grouping Process Process Process Page

General Scheduling Model and

General Scheduling Model and Algorithm Andrew Kusiak 2139 Seamans Center Iowa City, Iowa 52242-1527 Tel: 319-335 5934 Fax: 319-335 5669 andrew-kusiak@uiowa.edu http://www.icaen.uiowa.edu/~ankusiak MANUFACTURING