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number 14

osvevation liling 73 12. A work-sampfe study conducted over lerty houss o a one week period yielded the tofflenine n operator preduced 125 parts and was perlormance rated as 110 percent. The opera was To precenl; the company establihed allowance was 1s percent tor shis job class Detenhine the standard time for this part. tor's idle ie s and part 13. Several laboralory technicians in a hospitlare primariäy responsible for sunning the was work sampled on this 356 blood samples. 1h aulomated "Chemistry 12" blodd prolile tesl. An eaperiented technician v job over a two.week peried (seventy hows). The lab technician produced 35t and idle she analyit studying the job lound that the technician was working 60 petea ent of the time and complete analysis. The rest. Some idleness was dug to waiting for the automated equipment 1o performańce iating was 90 percent, but the analyst wat uncertain about this be ta) Defermine a standard tine for a slandard blood prefile b) How could the analyst be more cestain about the performance rating e rating 14 A manulactur is considering the p two lypes of equipmenl, type A or lype B, to n in opcration. Initial equipment cost is 510,000 lor either A or B. Opèraling care estimated as follows: $500 052 9 750 Maintenance (per moth) Supplies (per uail Operator (per,hour) 334 both arranged for experimental demonstrstions, in which stopwatch The equipment monu actu time sługies of mnu operator/mac followint tmacine performance for five crçles were measured. Demonstrations Cycles for B 2 Gycles for A 3 0.32 0.29 0.28 0.31 0.30 Activity Load machine Machine time 0.30 0.27 0.25 0.22 0.21 2.62 2.57 2.59 2.51 2.54 0.12 0.09 0.10 0.11 0.09 0.92 0.94 0.86 0.79 0.87 (machine paced) Unload machine Inspect product Apply label to 2.73 2.61 2.68 2.7 2.63 0.14 0.10 0.09 0.12 o.11 1.21 1.08 1.29 .15 1.20 producr 0.05 0.04 0.05 0.05 0.04 *tabel application is automatic for A วnd manual for B. The time study expert iated the operator as performing at 115 percent of normal on A and 110 rcent of normal on B during the observation cycles. it is estimated that operators will receive per two 15-minute coffee breaks daily. Unavoidable delays are eslimated to-be 40 minutes for A and 25 minutes for B during an eight-hour day: Evaluate the two alternatives and justily your recommendation.ol.A or.B.--一

Answer 1

Initial Equipment Cost for A or B = 10000$

Operating Costs:

	A	B
Maintenance/month	750$	500$
Supplies / Unit	-	0.52
Operator/hour	9	9

Option 1: Calculations for A

We will be taking the average of all the cycles for A for a particular parameter as below:

	1	2	3	4	5	Average
Load Machine	0.32	0.29	0.28	0.31	0.30	0.3
Machine Time	2.73	2.61	2.68	2.71	2.63	2.672
Unload time	0.14	0.10	0.09	0.12	0.11	0.112
Inspect Product	1.21	1.08	1.29	1.15	1.20	1.186
Apply Label to Product	-	-	-	-	-	-
Total time take per unit						4.27

The above time is in mins

Since the worker was working at 115% of the normal, if he works at 100% then the time per unit will be equal to 1.15*4.27 = 4.91 mins

A day is of 8hrs = 8*60 = 480 mins

Two 15 min coffee breaks = 30 mins

Unavoidable delays = 40 mins for A

Effective working time = 480 - 30 - 40 = 410 mins

No. of units that can be produced through machine A = 410 / 4.91 = ~83 units

Option 2: Calculations for B

We will be taking the average of all the cycles for B for a particular parameter as below:

	1	2	3	4	5	Average
Load Machine	0.30	0.27	0.25	0.22	0.21	0.25
Machine Time	2.62	2.57	2.59	2.51	2.54	2.566
Unload time	0.12	0.09	0.10	0.11	0.09	0.102
Inspect Product	0.92	0.94	0.86	0.79	0.87	0.876
Apply Label to Product	0.05	0.04	0.05	0.05	0.04	0.046
Total time take per unit						3.84

The above time is in mins

Since the worker was working at 110% of the normal, if he works at 100% then the time per unit will be equal to 1.10*3.84 = 4.22 mins

A day is of 8hrs = 8*60 = 480 mins

Two 15 min coffee breaks = 30 mins

Unavoidable delays = 25 mins for B

Effective working time = 480 - 30 - 25 = 425 mins

No. of units that can be produced through machine A = 425 / 4.22 = ~100 units

Ignoring the initial investment and Operator Cost since it is same for both A and B, the other cost calculation is as follows:

Calculation is being done per day

Cost = Maintenance Cost (per day) + Supply Cost (per unit)

For A: (750/30) + 0 = 25 $

For B: (500/30) + (0.52*100) = 68.67 $

I believe there is some discrepancy in the data, i have considered the supply cost as 0.52$ per unit

Since we are producing 17 units more per day through B, we will be able to make more profit which should cover the extra cost of the operations. It will depend completely in the profit per unit.

Machine B is a better option