Question $1(3+4+4+3=14$ points$)$

Consider the following process. The processing time $($sec$/$unit$)$ is given for each worker.

Designed CT for Stage $140$ sec $/$ unit

Designed CT for Stage $2120$ sec $/$ unit

Bottleneck Stage: $\_\_2\_\_\_$

Process Designed CT $300$ sec $/$ unit

Process Capacity $3600$ units $/$ hour

a$)$ Calculate the following design parameters for this process:

$$

b$)$ Customers arrive at this process in the following way: Every $240$ secs two customers arrives at M$1$ starting from time $0$ and every $240$ secs one customer arrives starting from time $100.$ Once a customer arrives at stage $2,$ if both servers are empty, the customer is served at the server with lower processing time. Fill in the below Gantt chart.

Each grid in the following chart represents $20$ sec$.$ You can extend the below chart as needed.

c$)$ Under the arrival schedule given in $($b$),$ calculate the following run$-$time parameters:

Demand Rate $45$ units$/$hour

Flow Rate $90$ units$/$hour

Actual Cycle Time $240$ sec$/$unit

WIP for Buffer $12$ units

WIP for M$11$ units

WIP for Buffer $21$ units

WIP for M$21$ units

WIP for M$31$ units

Total WIP $6$ units

Flow time $240$ secs

d$)$ Use the total WIP and the flow time calculated in $($c$),$ then apply the Little$$s formula to get the flow rate and the actual CT$.$ Specifically, show that

a$.$ The flow rate calculated using the Little$$s law is the same as the flow rate calculated in $($c$)$ using the Gantt chart.

$90$ units $/$hr

b$.$ The actual CT calculated using the Little$$s law is the same as the actual CT calculated in $($c$)$ using the Gantt chart.

$240$ seconds