Question

There is a well-known bakery in the town, named Paris Bakery

A). what is the daily revenue of the bakery based on the following assumptions? (10 points)

- there are 10 customers in/out in 10 minutes

-each costumer buys 3 biscuits on average

-The price of biscuit is $1.50 each

-The bakery open from 8 A.M to 6 P.M

B) Owner of the Bakery wants to adopt Kanban system to reduce the inventory. Howmany Kanban containers will be needed to support the bakery based on the following assumptions?

-The daily demand is 100 batches

-The production lead time is 2 days

-Management has decided to have 1 day of safety stock

-One container fits 20 batches

C) At the Bakery, it takes 120 seconds to assemble one sandwich through 20 activities. If 30 customers arrive at the restaurant every hour on average, what is the Target Manpower the Restaurant needs to meet the demand of the sandwich?(10 points)

D) The weight of one batch has to be 3.6 lbs. to make constant size of biscuits. The Bakery has a low specification limit of 3.5 lbs. and upper specification limit of 3.7 lbs. The standard deviation is 0.1 lbs. Owner wants to reduce its defect probability 1.1%. To what level would the have to reduce the standard deviation in the process to meet this target? (10 points)

Answer 1

A)

Every minute we have 1 customer.

Each customer buys 3 biscuit each costing 1.5. This means the overall revenue per minute is 4.5.

The bakery is open for 10 hours. That is 10*60 = 600 minutes. That is 600*4.5 = 2700

B)

Kanban formula is

C = D * (1+SF) * KCT / K

Where C = number of Kanban, SF = safety factor, KCT = cycle time (or lead time), K = Kanban container capacity

C = 100 * (1 + 1) * 2 / 20 = 20

Bakery needs 20 kanban

C)

30 customers arrive every hour. Each customer requires 120 seconds. This means total labor hour demand is 30*120 = 3600 seconds.

A single worker will be able to work 60*60 = 3600 seconds per hour. This means the target manpower is 1.

D)

For a defect probability of 1.1% the upper side the limit will be at 1.1/2 = 0.55%. This means the upper limit is 100-0.55 = 99.45%. Let us consider the average to be 3.6 lbs. Then for 0.9945 probability, the z value should be 2.5427

Thus the standard deviation σ can be equated as 2.5427 = (3.7-3.6)/ σ

or σ = (3.7-3.6)/2.5427 = 0.03932

The standard deviation should be 0.039