Task Task Time (minutes) Predecessor(s) 3 6 7 A 120 A, B B 80 C - 70 5 F D. E H BP Smart Monitor E Your group has been recruited by CMD's operations manager to determine the most optimal (balanced and efficient) assembly line for assembling their BP smart monitor by reviewing and answering the questions provided below. a) Draw an appropriate precedence diagram for the process details provided in the table (4 marks). b) Refer to the group data table provided on page 2 of this assignment. Find your group # listed in the table. Considering your available time for production and your scheduled (required) number of units to assemble per day, determine the TAKT time for your assembly process (4 marks) c) Determine the theoretical minimum number of single-person workstations required for the assembly process (2 marks) d) Balance your assembly line with an appropriate number of workstations accounting for all tasks. Visually show how/where each task is accounted for within its chosen/assigned workstation (8 marks) e) Calculate your assembly line efficiency (2 marks) Group Data Table: Locate your group number in the table below and take note of the assigned production variables for which you will use to determine your answers for b) thru e) above. Failure to use the appropriate group data may result in grade deduction and/or investigation of academic misconduct. Group 1 |Group 2 |Group 3 |Group 4| Group 5|Group 6 |Group 7 |Group 8 Available Production Time (hours) 10 12 14 10 16 12 14 Demand (units per day) 60 80 120 60 50 60 103 60Designing a Product Layout; example 2 Step 3: Group work elements into workstations by combining work elements based on precedence eligibility and until the sum of each elements grouped cycle time reaches (or is just less than) the desired cycle time (TAKT). Create a table to assess the new situation. Recall: TAKT (Cd) = 0.24 min, N=5 workstations Eligible Work Assigned Work Work Element Cumulative Workstation Elements Elements Time (min) Time (min) Idle Time (min) 1 A A 0.1 0.1 (0.24 - 0.1) = 0.14 B, C None. Ca is exceeded by B or C cycle times! Assign a new workstation! 2 B, C B 0.2 0.2 (0.24 - 0.2) = 0.04 C 'Parallel processing net W output: 0.2 (0.24 - 0.2) = 0.04 C (0.4 min)/(2units) = 0.2 min D 'Parallel processing net output: 0.15 (0.24 - 0.15) = 0.09 (0.3 min)/(2units) = 0.15 minDesigning a Product Layout; example 2 Step 4: Envision the process by visually grouping and physically representing the assembly line based on the workstation assignment. 0.2 Workstation Eligible Work Assigned Work Work Element Cumulative Elements Elements Time (min) Time (min) Idle Time (min) A A 0.1 0.1 (0.24 - 0.1) = 0.14 B B, C None. C. is exceeded with B or C cycle times! Assign a new workstation. D N B, C B 0.2 0.2 (0.24 - 0.2) = 0.04 A 0.3 W C C 'Parallel processing net D output: 0.2 (0.24 - 0.2) = 0.04 0.1 C (0.4 min)/[2units) = 0.2 min C D D 'Parallel processing net output 0.15 D (0.24 - 0.15) = 0.09 D (0.3 min)/(2units) =0.15 min C Work-station Work-station 0.4 Work- Work- 3 & 4 5 & 6 station 1 station 2 Flow time (process lead time) must still consider each C D station's processing time. Thus flow time is still = 1.0 min. A B 0.2 min C D Process cycle time (actual cycle time) is governed by the 0.1 min bottleneck. i.e. the longest effective rate of output. 0.4 min 0.3 min Thus process actual cycle time = max CT = 0.2 min. This Effective CT=0.2min Effective CT=0.15min meets TAKT time! But is this truly the best arrangement???Designing a Product Layout; example 2 Step 5: Calculate the Balance Delay and Assembly Line Efficiency, E. Balance Delay = W B l D l (0'31) 0215 21 Sty a ance B a = = . = . 3' (6)(0.24) Line Efficiency,E : (1 Balance Delay) : 1 .215 = 0.785 = 78. 5% Recall that we calculated the minimum number of workstations to be 5 ...? Is there perhaps a better layout design that we could feasibly achieve? Let's look... Revisit Step 3: When grouping elements into workstations, keep in mind the amount of idle time within other workstations. It might be feasible to leverage and utilize the resource (worker or machine) with excess idle time, somewhere else... Recall: TAKT (Cd) = 0.24 min, N=5 workstations Work Element Cumulative Workstation Eligible Work Assigned Work Elements Elements Time (min) Time (min) Idle Time (min) A A 0.1 0.1 (0.24 - 0.1) = 0.14 B, C None. Cd is exceeded with B or C cycle times! Assign a new workstation. B, C B 0.2 0.2 (0.24 - 0.2) = 0.04 C C 'Parallel processing net output: 0.2 (0.24 - 0.2) = 0.04 UI A W C C (0.4 min)/(2units) = 0.2 min O D 0.3 0.3 -0.06 Although we numerically exceed the TAKT time in Station 5 with element 'D' ... we observe that station 1 has ample idle time. Can we use this worker to support station 5 tasks to achieve TAKT without an extra station? The gap of time needing supported is .06 minutes (3.6 sec). Perhaps if we create a layout that permits the motion of workers to support each other, then we can do this and achieve N = 5 ..Revisit Step 4: Envision the process by visually grouping and physically representing the assembly line based on the workstation assi-nment. Work Ele me nt Time {min} Add the surplus of cycle time from element D [station 5) to station 1 element A = U.1+D.i]6 = 0.16 min NEW total cycle time for station 1, which remains below desired cycle time. OKi Cumulative Time [mini Idle Time {min} 1 A :'A ,: 0.240.16 = 0.03 B, C None. FATS exceeded with B or C cycle times! Assign a new workstation. 2 B, c (3):; 0.2 0.2 (0.24 - 0.2) = 0.04 3 C [PE-z: """\"':J:':\"\""" 0.2 (0.24 0.2) = 0.04 . , . . . 4 C '. E : :n.aminm2umu1=u.2 min Station 5 s new cycle time becomes 0.24, meeting desired TAKT 5 D (\"6\": 024 0'3 (100 perfectly, rendering the idle time now to zero {0). Workstation 1: Workstation 2, 0.2min 0.16min "' This layout has been arranged to facilitate the movement of the worker from workstation 1 to support workstation 5. l This layout formation is known as a "cell". Which allows for a better utilization of resources, when properly trained and _ Workstation 3 8r 4, setup. *7 0.4min each The previous layout was linear, and is not conducive to Work station 5' Eedive CEO-2min facilitating the reverse/forward motion of product or 0.24min l Cclnv+ynr Line people. Designing a Product Layout; example 2 Re-Visit Step 5: Calculate the Balance Delay and Assembly Line Efficiency, E. Balance Delay = E(idle time) nc a (0.16) Balance Delay = = 0.215 = 13. 3% (5) (0.24) Line Efficiency, E = (1 - Balance Delay) = 1 -.133 = 0.867 = 86. 7% Leveraging the idle time from available workstations should always be considered. Sometimes it is not feasible, but often ... it is ... or can be arranged.Task Task Time (Seconds) A 15 Take time = Available working Time/ Customer Demand B 23 C 17 D 42 E 15 In this example let's say that our target is to produce 500 units per day 37 within an 8 hour shift. Therefore, the Takt Time would be as follows: 5 Takt Time = 480 minutes / 500 units = 0.96 minutes = 57.6 seconds H 12 34 27 Each station should at least have The steps have been labeled [A-L] K 18 a 57.6 second design cycle time and each have a unique cycle F to meet market demand of 500 time associated to that specific Total Time 252 units. In order to know how many task. These cycle times could have stations are required we need been captured using MODAPTS to know some detailed insight or actual stop watch calculations. into the underlying product, bill We now have the two key pieces of material, and bill of process. of information to calculate the This is how we can establish the required number of stations. required tasks to assemble the The number of stations is simply product. Let's assume that this calculated by the below equation. new line has 12 required steps to complete the assembly.Now we have 12 tasks that need to be accomplished within 5 stations it becomes a question of which tasks to include within a specific station. This is where the "Bill of Process" comes in; we need to know some information of the precedence or the order of the tasks. Certain tasks must be completed prior to taking action on other tasks. The "Bill of Process" is where each step or task is described to assemble the unit. It should clearly demonstrate the order of steps, including synchronous and simultaneous tasks. This is often captured in a "Precedence Diagram" A Precedence Diagram is a lot like a process flow diagram; with shapes and arrows describing significant and critical steps within assembly of the product or service. In our example we will assume that we have been supplied with the following Precedence Diagram for our 12 tasks (A-L): Precedence Diagram A K L G IE This clearly shows that task A must be completed Task Task Time (Seconds] Precedence before task B can be started. It also shows that that A 15 tasks C, D, and E can be started simultaneously after 23 task B has been completed. Moreover, both tasks F and C 17 G must be completed before task H can start. We can D 42 B now add a Precedence column to our initial Task Table: E 15 B F 37 C Given the precedence we can now start assigning tasks G 5 D, E to stations. One common approach is to use a "Task H 12 F, G Assignment Table" This table will look at all eligible 34 H tasks to be included within a station, and keeping 27 H track of the accumulated cycle time within the station. 18 1, J A common scheme is to start in the order of the 7 precedence diagram and seek the longest cycle time. Total Time 252We start out with Station 1; the only Eligible Assigned Task Cumulative Remaining eligible task is Task A. We then assign Str Task Task Time Time Time Task A using its 15 second cycle time. The A 15 15 12.6 remaining time left within the station 23 19.6 is the Takt Time minus the assigned GE 17 55 task; [57.6 seconds = 15 seconds] = 2 D 42 42 15.6 42.6 seconds. Therefore, we have 42.6 m 15 57 D.6 seconds remaining in Station 1 to assign F, G F 37 37 20.6 additional tasks. The only qualified task is G G 5 42 15.6 Task B, so we then assign Task B with its H H 12 54 3.6 23 second cycle time to Station 1. We now 1, J 34 34 23.6 have a remaining time of 19.6 seconds. J 27 27 30.6 K K 18 45 12.6 7 52 5.6 The next tasks are C, D, and E; but D does not qualify because its cycle time is greater than the remaining cycle time. Therefore, C and E become the next eligible tasks. Since we are using the longest cycle time rule; we will then select Task C using its 17 second cycle time. This now completes station 1 with a remaining time of 2.6 seconds; as there are no other identified tasks less than 2.6 seconds. We are now ready to assign tasks to the 2nd station; the eligible tasks are D and E. Since D has a higher cycle time of 42 seconds it will be the first task assigned to Station 2. Task E can now close out the station with a remaining idle time of 0.6 seconds. We then complete the remaining 3 stations using the same eligible task scheme to fill out the completed table. The newly designed line will then appear on the layout as depicted below. We can see that Station 4 is under-cycle with 23.6 seconds of idle time; but it could be the best available design according to the process precedence rules. Stn 1 Stn 2 Stn 3 Stn 4 Stn 5 55 sec or 57 sec or 54 sec or 34 sec or 52 sec or 65.5 JPH 63.2 JPH 66.7 JPH 105.9 JPH 69.3 JPH