The output from an assembly line is 1,600 units in an 8-hour shift. The item is

Question

The output from an assembly line is 1,600 units in an 8-hour shift. The item is moved in containers, each of which holds 50 units. Each container spends an average of 30 minutes in the production part of a cycle, and 20 minutes in the demand part. How many kanbans should be used for the item? What is the stock of work in progress? A scheme has been proposed to reduce the time in the demand part of the cycle to 15 minutes. How many kanbans should be used now? What is the margin of safety in each case? The output from an assembly line is 1,600 units in an 8-hour shift. The item is moved in containers, each of which holds 50 units. Each container spends an average of 30 minutes in the production part of a cycle, and 20 minutes in the demand part. How many kanbans should be used for the item? What is the stock of work in progress? A scheme has been proposed to reduce the time in the demand part of the cycle to 15 minutes. How many kanbans should be used now? What is the margin of safety in each case?

Explanation / Answer

The number of kanbans required = Demand in the cycle/Container size

Demand in the cycle = demand x (Time (TP) spend by container in production part of cycle + Time (TD) spend by container in demand part of cycle) = d x (TP + TD)

Given:

TA = 30 minutes = 0.5 hours

TD = 20 minutes = 20/60 = 0.33 hours

Demand = 1600 units in 8-hour shift = 1600/8 = 200 per hour

Container size = C = 50 units

Number of Kanbans required = K = d x (TP + TD)/C = 200 X (0.5+0.3)/50 = 3.32

Number of containers cannot be in fraction, actual number of Kanbans required = 4

The stock of work in progress = K x C = 4 x 50 = 200 units

By reducing the TD to 15 minutes, the improved TD = 15 minutes = 0.25 hours

Number of Kanbans required = K = d x (TP + TD)/C = 200 X (0.5 + 0.25)/50 = 3

Number of Kanbans required = 3

If safety margin is considered then the number of kanbans formula is modified as follows:

Number of Kanbans required = K = [d x (TP + TD) x (1 + SF)]/C

In First case, K = 4

4 = [200 x (0.5 + 0.33) x (1 + SF)]/50

4 = 3.32 x (1 + SF)

1 + SF = 1.2048

SF = 0.205

Safety margin in first case is 20.5%

In second case, K = 3

3 = [200 x (0.5 + 0.25) x (1 + SF)]/50

3 = 3 x (1 + SF)

1 + SF = 1

SF = 0

Safety margin in second case is 0%

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