Question

1. The process is performed by several resources.

2. Resources are shared across assembly lines (i.e., workers must multitask and perform multiple activities for every flow unit.)

3. The resource that limits hourly production is not the same resource that limits the process lead-time.

From 2002 to 2009, Honda’s 1800cc cruiser-style motorcycle, the VTX1800, was assembled in Marysville, Ohio from four major subassemblies. The following is a simplified explanation of the assembly process. All resources are indicated in bold.  

• The first subassembly produces the engine from three activities: a left and right part of the engine block come out of the automatic mold every 2 minutes. These two parts are welded together, requiring 1 minute of the continuous welding machine. Finally, it takes the engine assembler 3 minutes to insert the two pistons and valves.

• The second subassembly produces the frame in two steps: first heavy metal bars are bent in the 10,000 lb press in 1 minute. Then it takes 3 minutes to setup and weld the multiple bars together with the same continuous welding machine to produce the frame.

• The third subassembly consists of the front and rear fenders, both of which are formed using the same 10,000 lb press for 1 minute per fender.

• The fourth subassembly makes the seat. It takes a seat assembler 7 minutes to cut the padding, put it on a preformed piece of plastic, and wrap the two with synthetic leather.

• The first four subassemblies are put together by final assemblers. Adding the handlebar, front and tail lights, radiator, wheels, drive shaft, brakes, cables, fluids and electrical wiring produces a new bike that is ready for test drive. A final assembler requires 30 minutes per motorcycle and uses a work cell layout (10 final assemblers simultaneously complete 10 motorcycles).

The staffing is one automatic molding machine operator, one continuous welding machine operator, one 10,000-lb. press operator, one engine assembler, two seat assemblers, and 10 final assemblers.

Analysis:

1. Draw a process flow diagram identifying resources, activity times, and any potential work-in-process (WIP) storage buffers.  

2. Compute the theoretical process lead time2 (aka, flow time or throughput time) of the VTX1800 production process.

3. What is the constraint (aka, bottleneck) in this process?

4. What is the “process capacity” (aka, flow rate)?

5. How much time should there be, on average, between completed motorcycles rolling off the assembly line?

6. Suppose all employees work at the rate of the bottleneck and that there is at least as much market demand as the rate of the bottleneck. On average, how many motorcycles (at various stages of completion) will be in the facility?

7. On average, what percentage of an hour is the welding machine operator busy? What about the average final assembler?

8. Honda would like to reduce the theoretical process lead time of the VTX1800 assembly. What specific action(s) do you recommend?

9. Honda would like to increase the hourly production capacity to 20 motorcycles per hour. What specific action(s) do you recommend?

10. Suppose market demand for VTX1800’s is actually 12 motorcycles/hr. What is the bottleneck in the process?

Answer 1

4 sub-assemblies +one final assembly producing 10 motor cycles in 30 mts= 3mts/m/cycle.=20 /hr.

Sub assly 1

Operation no.	Op name	time in mts	No. of op.	machine /equipment	Potential inventory buffer reqd.
10	Molding	2.00	1	Molding m/c	X
20	Welding	1.00	1	Cont welding m/c. @	X
30	Assly	3.00	1	Assly table	X

Cycle time===3.0 mts. Throughput time =6.0

Inventory expected before opn 30

Before opn 20 .....

SA-2 Frame

Operation no.	Op name	time in mts	No. of op.	machine /equipment	Potential inventory buffer reqd.
40	Bending	1.00	1	10,000 lb bending press.*	X
50	Welding	1.00	--	Cont welding m/c.@	yes, may be

Cycle time = 1.00 mt.throughput time=2.0 mins

Cont welding m/c -opn 20 @1.00 min. + opn 50 @1.00 min=2.00 min /unit

Before opn 50 . Yes

SA 3

Operation no.	Op name	time in mts	No. of op.	machine /equipment	Potential inventory buffer reqd.
60	Pressing	2.00	--	10,000 bending press.*	yes -may be

Cycle time = 2.00

10,00 lb bending press opn 40 @ 1.00 min + opn 60 @2.00= 3.00

Inventory expected before opn 60......
SA 4

Operation no.	Op name	time in mts	No. of op.	machine /equipment	Potential inventory buffer reqd.
70	Seat assembly	7.00	2	Assly table	X

cycle time =3.5 mts

Final assembly

Operation no.	Op name	time in mts	No. of op.	machine /equipment	Potential inventory buffer reqd.
80	Final assembly	30.00	10	Assly table	yes

cycle time =1.00 m

Work station	SA1	SA2	SA3	SA4	Final
Cycle time	3.0	1.0	2.0	3.5	1.00	total cycle time=3.5 mins.
Lead time for 1st piece	(6)	(2)	(2)	7	30	7+30=37 minutes

Throughput time (there are no movements and inspections) = 6+2+2+7+30=47 mins.

Constraint in the process ==sub assembly 4 is the constraint (at 3.5 mts per unit)

With a cycle time of 3.5 mins per unit, the capacity=60/3.5=17 units per hour .

A motor cycle will roll off the assembly line every 3.5 minutes.

Q. No. 6......

Assumption=No inventory is assumed on the output side of any operation.

Motorcycles at final assly=10 waiting for assly + 10 under process of assly=20

At other stations there may/will be components and SA's onl(cannot call these motor cycles )

7. welding machine operator busy in % =(2/3.5)8100=57.1 %

Final assembler busy for what % =(3/3.5)*100=85.7 %
8. :Process lead time can be reduced by: Improving the final assembly process which takes 30 minutes.

9. Increase the no. of assemblers of SA4 from 2 to 3. This will bring down the bottleneck to 3 mins (SA1) The capacity shall increase to 20 units per hour.

10. For a demand of 12 units per hour, there is no bottleneck in the entire process.

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