Question

Problem 3(a)

The average part produced in a certain batch manufacturing plant must be processed through an average of six processes. Twenty (20) new batches of parts are launched each week. Average operation time = 6 min, average setup time = 5hr, average batch size = 25 parts, and average non-operation time per batch = 10hr/ machine. There are 18 machines in the plant. The plant operates an average of 70 production hours per week. The Scrap rate is negligible. Determine:

(i) manufacturing lead time for an average part,

(ii) plant capacity,

(iii) plant utilisation.

(iv) How would you expect the non-operation time to be affected by the plant utilization?

Problem 3(b)

An average of 20 new orders are started through a certain factory each month. On average, an order consists of 50 parts to be processed through 10 machines in the factory. The operation time per machine for each part = 15 min. The non-operation time per order at each machine averages 8 hr, and the required setup time per order = 4 hr. There are 25 machines in the factory, 80% of which are operational at any time (the other 20% are in repair or maintenance). The plant operates 160 hr/month. However, the plant manager complains that a total of 100 overtime machine-hours must be authorized each month to keep up with the production schedule.

(i) What is the manufacturing lead time for the average order?

(ii) What is the plant capacity (on a monthly basis) and why must the overtime be authorized?

(iii) What is the utilization of the plant according to the definition in the text?

(iv) Determine the average level of work-in-process (number of parts-in-process) in the plant.

COURSE: Intelligent Production Systems.

Answer 1

Arrangement:3 (a)

1 . manufacturing lead time = Average activity time(average arrangement time+ normal bunch size*0.1 + normal nonoperation time per group)

In the inquiry:-

Normal activity time =6

normal arrangement time=25

normal nonoperation time per batch=10

along these lines,

fabricating lead time = 6(5 + 25(0.1) + 10) = 105 hr

(2) Production rate = (5 + 25 x 0.1)/25 = 0.30 hr/pc,

Rp = 3.333 pc/hr.

Plant limit = 70(18)(3.333)/6 = 700 pc/week

(3) Parts propelled every week = new groups of parts propelled every week * normal bunch size =20*25 = 500 pc/week.

Use U = Parts propelled every week/Plant limit =500/700 = 0.7143 = 71.43%

(4) The nonoperation time is relied upon to ascend with the usage coming to 100%. When there is an expansion in the outstanding burden in the shop, the shop will have numerous errands and along these lines longer time will be required for a vocation to come out. Also, non activity time is relied upon to fall with diminishing usage.

3 (b)

: (1) MLT = 10(4 + 50x0.25 + 8) = 245 hr/request

(2) Tp = (4 + 50x0.25)/50 = 16.5/50 = 0.33 hr/pc, Rp = 3.0303 pc/hr

PC = (25x0.80)(160)(3.0303)/10 = 969.7 pc/month

Parts propelled every month = 20x25 = 1000 pc/month

Calendar surpasses plant limit by 1000 - 969.7 = 30.3 pc.

This requires additional time in the sum = (30.303 pc x 10 machines)/(3.0303 pc/hr) = 100 hr.

(3) Utilization U = (1000 pc)/(969.7 pc) = 1.03125 = 103.125%

(4) WIP = (245 hr)(969.7 pc/mo)(1.03125)/(160 hr/mo) = 1531.25 parts

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