b$)$ Silvercrest Manufacturing Ltd$.$ is using Kanban containers. There are two

adjacent work centres, a downstream $($using$)$ and an upstream $($producing$)$

one. The using work centre has a production rate of $200$ parts per day and

each container holds $21$ parts. It takes $0\mathrm{.}5$ days for a container to make the

entire cycle from the time it leaves the upstream centre until it is returned,

filled with production, and leaves again. The manager wants a risk factor of

$20\%.$

i$.$ What is the maximum inventory of parts in the system?

$(10$ marks$)$

ii$.$ If the company wants the number of containers $($holding $21$ parts each$)$

to be $5,$ what must the risk factor become?

$(10$ marks$)$ The following formula is used to calculate the number of bins for a given part

needed for a single Kanban loop between two cells in a production system:

$N=\frac{D(Tw+Tp)(1+r)}{c}$

Where:

$N=$ Total number of bins for a given part.

$D=$ Planned usage rate for the part $($parts per day$)$

$Tw=$ Average waiting time for replenishment of parts $($fraction of a day$)$

$Tp=$ Average production time for a container of parts $($fraction of a day$)$

$c=$ Capacity of the standard container $($number of parts$)$

$r=$ Risk factor set by management $($expressed as a decimal fraction$)$

a$)$ At Schneider Electric Corp. the boxing station at the end of an electric switch

production line has a planned daily usage rate of $150,000$ switches per day

and receives the switches to be boxed from the preceding station in

containers which hold $3\%$ of the day's demand. It takes $90$ minutes for the

preceding station to fill a bin and $9$ minutes for the filled bin to be delivered to

the boxing station.

How many such bins are needed for the kanban loop to work efficiently in this

production cell, given a single $10-$hour shift per day and a risk factor of $6\%?$