Question

Summary:

A car part manufacturing company currently produce a suspension assembly for Toyota Motor Corporation. Figure 1 shows the breakdown structure of the suspension assembly and Figure 2 shows the final assembly and how to fit into a car. Table 1 shows the name of components, part numbers and number of components in each suspension assembly.

Figure 1. Breakdown structure of Suspension Assembly

Figure 2. Suspension Assembly Number in Figure 1	Part number	Part name	Amount per assembly
1	PA-T15-001	Shock absorber unit	1
2	PA-T15-002	Lower insulator	1
3	PA-T15-003	Spring bumper	1
4	PA-T15-004	Coil spring	1
5	PA-T15-005	Upper insulator	1
6	PA-T15-006	Spring upper seat	1
7	PA-T15-007	Bushing	1
8	PA-T15-008	Suspension Support	1
9	PA-T15-009	Washer	1
10	PA-T15-010	Stop nut	1
11	GE-001	Bolt	2
12	GE-002	Nut	5

Coil springs are produced in-house. The process to produce the coil spring in the company is as follows.

(a) Heating the wire: to an appropriate temperature for the winding process.

(b) Hot winding process: with a spring winding machine to align the wire into a proper pitch.

(c) Cutting the wire: cut the wire in an appropriate length after the winding process.

(d) Cooling process: cool down the spring immediately by plunging into oil.

(e) Heat treatment: remove internal stress within the material to provide higher resilience. Usually heated in an oven and held at an appropriate temperature for a predetermined time.

(f) Grinding: the spring needs to have flat ends. The coiled spring will be placed into a jig by operators to feed into the grinding machine.

(g) Shot peening: this process requires strengthening the steel to resist fatigue and cracking during its lifetime.

(h) Coating: to prevent corrosion, the entire spring surface will be coated by dipping into a liquid rubber.

(i) Inspection and Transporting: the final coil spring will be inspected and transported to the final assembly work station.

Using the information above, answer the followings

1. The number of defects in the coiled springs has been counted and shown in table 10.

Table 10. Number of defects for coil springs Sample	Number of defects	Sample	Number of defects	Sample	Number of defects	Sample	Number of defects
1	3	11	8	21	1	31	12
2	2	12	4	22	2	32	9
3	4	13	6	23	8	33	7
4	5	14	7	24	3	34	5
5	1	15	13	25	4	35	3
6	2	16	3	26	6	36	5
7	4	17	5	27	9	37	6
8	1	18	6	28	5	38	7
9	2	19	7	29	6	39	4
10	7	20	2	30	1	40	2

c) As you can see the answer in (a), the process is not under control. The operation manager has decided to conduct the total population sampling, i.e., the entire coil spring produced has been inspected for the defects. From the inspection it has found that the defect rate is 25% and only 85% of defected parts can be reworked to make them good. Assume the manufacturing cost for each spring is $450.00 and the rework cost per spring is $45.00. What is the daily good quality product yield? (Ans. 616 parts) The number of coil springs produced per hour is 80 coil springs and the production is made normally 8 hours per day. What is the resultant cost per coil spring? (Ans. $477.46)

d) After some quality improvement measure, the daily percentage of good quality springs has increased to 95% but the rework rate for the defects parts has decreased to 70%. What is the daily good quality product yield? (Ans. 630.4 parts) What is the resultant cost per coli spring? (Ans. $458.45)

Answer 1

(c)

Total production per day = 80 coil springs per hour x 8 hours per day = 640

Good quality with no rework = 640 x (100% - 25%) = 480

Defective coil springs at first instance = 640 - 480 = 160

Successful rework = 160 x 85% = 136

So, good quality yield per day = 480+136 = 616

Cost per coil spring = ($450 x 640 + $45 x 136) / 616 = $477.4

(d)

Total production per day = 80 coil springs per hour x 8 hours per day = 640

Good quality with no rework = 640 x 95% = 608

Defective coil springs at first instance = 640 - 608 = 32

Successful rework = 32 x 70% = 22.4

So, good quality yield per day = 608+22.4 = 630.4

Cost per coil spring = ($450 x 640 + $45 x 22.4) / 630.4 = $458.45