Question

Suppose a workstation receives parts automatically from a conveyor. An accumulation line has been provided at the workstation and has a storage capacity for 5 parts (N=6). Parts arrive randomly at the switching junction for the workstation; if the accumulation line is full, parts are diverted to another workstation. Parts arrive at a Poisson rate of 1 per minute; service time at the workstation is exponentially distributed with a mean of 45 seconds.

a. What is the rate at which parts enter the system?

b. Find the average number of parts in the accumulation line.

c. Suppose the production manager desires that no more than 2% of the arriving packages be diverted, how long should the accumulation line be>

Answer 1

When the 1st machine fails in the 10 minute of the hour for 10 minutes

in first 10 minutes, the first machine would have produced 10 parts and 2nd machine have produced 9 parts as the second machine has to wait for 1 minute two fill the buffer

In the 11th minute, the first machine fails but there is a part n the buffer which is produced in the 10th minute so it can go to the second machine and processed so that 10 parts are produced till 11th minute(9partstill 10 minutes + 1 part in the 11th minute)

Now till 20 minutes, there is no production from machine 1 so machine 2 can't also produce and from 21 to 60 the production of machine 1 resumes so total 39 units can be produced in these 40 minutes. Total production = 39+10 =49

b) now kanban count increased to 10 so till 10 minutes there will be no processing from machine 2

the machine 1 will fill the buffer to 10 in the first 10 minutes now the machine 1 fails up to 20 minutes but as there is a buffer available for machine 2 to produce it can process and produce parts

Till 20 minutes the machine 1 fails but machine 2 produces from the buffer 10 parts now when the machine 1 resumes its operation in the 21st minute it will produce 10 parts up to the 30th minute to fill the buffer and machine 2 waits

machine 2 resumes its operation from 31st minute to 60th minute which will produce 30 parts

so total production is 30 +10 =40 parts

Part(3)

Machine 2 fails 40th minute of the hour for 10 minute

Machine 1 failed in 10th minute of the hour for 10 minutes

kanban size =2

first 2 minutes the machine 2 can't produce

so till 10th-minute production = 8

but in the 11th-minute machine 1 fails but machine 2 can produce as there is still one part

M/C 1 - (Production) (0-1)-1,(1-2)-1,(2-3)-1,(3-4)-1,(4-5)-1,(5-6)-1,(6-7)-1,(7-8)-1,(8-9)-1,(9-10)-1, fails

M/C2-(Production) (0-1)-W,(1-2)-W,(2-3)-1,(3-4)-1,(4-5)-1,(5-6)-1,(6-7)-1,(7-8)-1,(8-9)-1,(9-10)-1,(10-11)-1,(11-12)-1

M/C1-(Production resumes)(20th minute

Mc/2 fails in the 40th minute for 10 minutes

till 40th minute the production considering 2 units in buffer = 18 units

41st to 50 th minute there is no processing from M/C 2 but M/C 1 can fill the buffer i.e. 2 units so that M/C 2 no need to wait when it resumes the operation at 50th minute. So, 50th minute to 60 th minute there are 10 units produced

Total production = 10+18+10 =38

b) Kanban size increases to 10

Now in the first-hour production = 1st to 10th minute M/C1 10 parts and M/C 2 waits till the buffer gets filled

The 11th minute the M/C 1 fails till 20 the but MC2 can produce as the buffer is available

10 parts

20 to 40-minute production = 10 units as first 10-minute m/c 2 has to wait till the buffer gets filled

Agin in 40th to 50th the machine 2 fails but machine 2 continues to fill the buffer

so 40th 60th production = 10 but buffer is full 10 as m/c 1 is continuing its work from 40th to 50th

From the second hour as the buffer of 1o units are available both machine can simultaneously start from the 1st minute itself

till 10th-minute production = 10

11th -20 M/c1 fails but machine 2 can produce = 10 units

21st to 30 M/C 1 has to wait since the buffer has to be filled production =0

3oth-40 minute both machine work to produce = 10 units

40th minute M/C-2 fails but M/C2 working

40th -50 M/C 1 working and M/C 2 fails production 0 (buffer is full)

50th to 60th both machine starts and produce 10 units more

Total production = 10+10+0+10+0+10 = 40 uits