A company is about to begin production of a new product

Tun 11. A company is about to begin production of a new product. The manager of the department that will produce one of the components for the product wants to know how often the machine used to produce the item will be available for other work. The machine will produce the item at a of 200 units a day. Eighty units will be used daily in assembling the final product. Assembly wll take place five days a week, 50 weeks a year. The manager estimates that it will take almost a full day to get the machine ready for a production run, at a cost of $300. Inventory holding costs will be $10 a year. rate

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Demand rate, d = 80 per day
Production rate, p = 200 per day
Setup cost, K = $300 per setup
Carrying cost, h = $10 per unit per year

(a)

Annual demand, D = 80 per day * 5 days per week * 50 weeks per year = 20,000 units

Optimal run quantity that minimizes the total cost, Q_p = [2.D.K.p/{h.(p - d)}]^1/2 = (2*20000*300*200/(10*(200-80)))^0.5 = 1,414 units

(b)

Number of days to produce of lot = Q_p / p = 1414/200 = 7.07 days

(c)

Average inventory = (Q_p/2) * (1 - d/p) = (1414/2)*(1 - 80/200) = 424.2 units

(d)

Cycle time = Q_p / d = 1414/80 = 17.7 days
Number of days to produce of lot = 7.07 days
The setup time is given as 1 day

So, the idle time remaining in a cycle = 17.7 - 7.07 - 1 = 9.6 days which is more than 10 days. So, it cannot accommodate the job requiring 10 days' time.

(e)

1. Increase the production rate by in-house modification of the machine
2. Split the new job into two parts and do it in two consecutive cycles
3. Deviate from the optimal production run size and produce a little less in a particular cycle to accommodate the new job

(f)

The required time including the setup of 1 days should be 10+1 = 11 days

Q x (1/d - 1/p) = 11 days

or, Q x (1/80 - 1/200) = 11

or, Q = 11/(1/80 - 1/200) = 1,467. So, the run size must at least be 1,467 units to accommodate the job.