 उ unemployed 할 नत्सिकСп�, शबके अगे अगे में आइ। शलक राशए भी देशाती उपनी फाजएके अगर नत्सार हो वो जा हैताингा, जो चाहती स्नउंनेukt, इस शार शी के अगर है गोोगी, इरींं के आगरét कके अगरा ही शा�pta richi, cook it.. more than that, the total idle いました रहें मुदन गता, सब यहना सकतerem तो ज़ आदर ताएम कित्ना बन रहें. तेख at the time, with a 10 minute cycle time. तो तोडल आईटल ताएम किता में बन रहें तें किस साए चे वर्खत सेशिन की सा� contouring total idle time is being tied with 6 work stations. Workstation content is basically your total task time which is 50 Minutes divided by number of workstations which is 6 and the cycle time which is 10 So, अफ्यानई धृए यान तो इस बेसेढ के अपर यान यान प्कोवाद, जी आ कर रहाता। अपनी बेसी ज़॥ कि नहीं थे जहाँ मुझा लग भी छईट बाव शेंगों नहीं पियाद घॊन्गार। next is what is the total idle time per cycle total idle time per cycle that basically we have to see that we have actual number of workstation multiplied by cycle time total how much is being used minus total task time total task time is 50 minutes whereas the actual number of workstations is 6 and largest assigned time is 10 so 60 minus 50 will give you 10 minutes as the idle time on this whole assembly line next we want to assign tasks to different workstations and for that we have to use largest number of followers and then we have to calculate the efficiency of this assignment this is the precedence which we have developed according to this we have to see which assignment time of workstation cannot go up from 10 minutes less than 10 minutes now if we look at this we have because we are looking at the largest number of followers so one of the followers are B, C, H, F, I, J, N, K which is 7 BK is 6 so first we will assign A in workstation number 1 and after that then we will see the available C C can be assigned and B also can be but again the largest number of followers is only 3 of C and B is 6 so we will assign B and first in workstation A and B will be assigned and total time of this workstation is 9 minutes next because we have assigned these 2 we have E, D and C these 3 are available D's followers are 3 C's followers are also 3 whereas E's followers are only 2 so in this the tie is breaking if we assign C but when we assign C we have time of 7 minutes of workstation so it has 3 minutes more available next we have D and E we cannot assign D because it will be 12 minutes so it will go up 10 minutes so we will assign D to workstation number 2 C and E these 2 will be assigned and again 9 minutes time of this workstation now we have D whose followers are 3 F's followers are 2 and J is also available because E has been assigned so we will assign D first so we assign D if we assign D then H will also be available because E and D and E both will be assigned so E and H will also be available but if we assign H then it will be 12 minutes whereas if we assign F then it will be 9 minutes so in this case we will assign D we will assign F and we will get 9 minutes time of workstation available is H which is 7 minutes J is 6 minutes G which is 5 minutes so in this first we will assign H because it has largest number of followers then we have these 3 in this we cannot assign G because it will be 12 minutes J will be 13 minutes so this is left and we will assign which should be 8 minutes workstation number 4 now J and G both have same number of followers so we will break time with the longest processing time so we will assign J along with this we cannot assign anything else although K makes it 10 but K cannot be assigned due to precedence relationship till G is not complete we cannot operate so what we will do is we will workstation 5 J and 6 minutes whereas last 2 will be assigned number 6 so this is the assignment of different tasks to different workstations and if we calculate its efficiency total task is 50 minutes workstations are 6 but here the cycle time has changed from 10 minutes to 9 minutes because maximum time if you see here then 9 minutes are coming workstation 1 workstation 2 workstation 3 and workstation 6 so cycle time maximum time is 9 minutes if we calculate its basis then its efficiency comes to 92.59%