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Assignment 9.1 – Kanban Answer the following questions Hinge Manufacturing Compa

ID: 456993 • Letter: A

Question

Assignment 9.1 – Kanban

Answer the following questions

Hinge Manufacturing Company employs a Kanban system for a component part. The daily demand is 900 hinges. Each container has a wait time of 0.05/day, and a processing time of 0.37 days. The Container size is 60 hinges and the safety factor (a) is 10%.

How many Kanban card sets should be authorized?

What is the maximum inventory of this hinge in the system?

A company is using a Kanban system with 2 containers, each holding 50 parts, between a downstream (using) and an upstream (producing) work center. The using work center can handle 250 jobs per day. The average elapsed time for the entire cycle is currently .25 days. The company is concerned with the safety factor for the operation and the size of the containers.

What is the efficiency rating of the current operation?

If the company decides to use only one container, how many parts must it hold?

If the safety factor is changed to .12 and there are still 2 containers, how many parts must each container now hold?

Explanation / Answer

Hinge Manufacturing Company employs a Kanban system for a component part. The daily demand is 900 hinges. Each container has a wait time of 0.05/day, and a processing time of 0.37 days. The Container size is 60 hinges and the safety factor (a) is 10%.

Answer :

How many Kanban card sets should be authorized?

Answer

K = Avg daily demand during lead time plus safety stock / No of units per container

K = d(P+W)(1+ ) /C

Where d = 900 , w =0.05 , p=0.37 ,c= 60 , Safety factor = 0.10   

= 900 (0.37+0.05)(1+0.10)/60

= 7

What is the maximum inventory of this hinge in the system?

Answer :

Maximum Inventory = K * C

   = 7 * 60

   =420

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A company is using a Kanban system with 2 containers, each holding 50 parts, between a downstream (using) and an upstream (producing) work center. The using work center can handle 250 jobs per day. The average elapsed time for the entire cycle is currently .25 days. The company is concerned with the safety factor for the operation and the size of the containers.

What is the efficiency rating of the current operation?

If the company decides to use only one container, how many parts must it hold?

If the safety factor is changed to .12 and there are still 2 containers, how many parts must each container now hold?

Answer :

What is the efficiency rating of the current operation?

Answer :

[KC - d(p+w)]/ [d(p+w) ] = [2*50 - 250*0.25]/[250 * 0.25] = 0.60

If the company decides to use only one container, how many parts must it hold?

Answer :

C = d(P+W)(1+ ) /K = [ 250 *0.25][1+0.6] / 1 = 100

If the safety factor is changed to .12 and there are still 2 containers, how many parts must each container now hold?

Answer :

C = d(P+W)(1+ ) /K = [ 250 *0.25][1+0.12] / 1 = 35 parts