The following diagram shows a 4-step process that begins with Operation 1 and en
ID: 461934 • Letter: T
Question
The following diagram shows a 4-step process that begins with Operation 1 and ends with Operation 4. The rates shown in each box represent the effective capacity of that operation.
If you could increase the capacity of only two operations through process improvement efforts, which two operations would you select, how much additional capacity would you strive for in each of those operations? (Enter your answers as whole numbers. Enter the lower operation number in the TOP answer box and the higher operation number in the BOTTOM answer box.)
Operations
Additional capacity
What would the resulting capacity of the entire process be?
Resulting capacity
The following diagram shows a 4-step process that begins with Operation 1 and ends with Operation 4. The rates shown in each box represent the effective capacity of that operation.
a. What is the current capacity of the entire process?If you could increase the capacity of only two operations through process improvement efforts, which two operations would you select, how much additional capacity would you strive for in each of those operations? (Enter your answers as whole numbers. Enter the lower operation number in the TOP answer box and the higher operation number in the BOTTOM answer box.)
Operations
Additional capacity
(Click to select)24573816 (Click to select)74523816What would the resulting capacity of the entire process be?
Resulting capacity
units per hourExplanation / Answer
The limit of any procedure is dictated by those with slightest stream rates i.e ones with the most minimal yield.
For Processes 1-2-3: 2 is bottleneck as it has the most minimal yield of 10 units/hr
For Processes 4-5-6: 4 is bottleneck as it has the most reduced yield of 5 units/hr
As these two units work in paraller the present limit = 10 + 5 = 15 units/hr
These join and encourage into procedure 7 which has the ability to handle both branches as does the last stride
a). Limit of the whole process= 15 units/hr
b). in the event that you can just build two operations, the most extreme yield would be accomplished by enhancing 1 and 2, convey them both up to 20/hr.
We won't increment for the lower arm (4-5-6) as the increases would be littler.
Operartion 2 (lower operation number) , Additional Capacity = 20-10=10 units/hr
Operartion 1 (higher rnumber) , Additional Capacity = 20-15= 5 units/hr
c) Resulting limit of whole process :
For Processes 1-2-3: All have square with yield of 20 units/hr
For Processes 4-5-6: 4 is bottleneck as it has the most reduced yield of 5 units/hr
Therefore coming about capacity=20+5= 25 units/hr
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