Exam II review - student version

# The machine is available 6 hours a day 25 days a

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The machine is available 6 hours a day, 25 days a month. Setups take .5 hours. Both products must be produced every day How many units of X can be produced in a month? 687.5 units How many units of Y can be produced in a month? 2062.5 What is production in standard units? 1375

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Effect of set up time 20 There is one machine in the shop to produce two products : X, Y. Product X (standard product): 6 min/unit. Product Y: 2 min/unit We must produce X and Y in the ratio 1 : 3 Then how to allocate production time to each product? Questions: Suppose there is .5 hour set-up time when you change products. How should we schedule, and what will be monthly production if Part (a) We need to maximize monthly production. Part (b) We must produce both products daily . Production time division between X and Y: Production time of one unit of X : 1 * 6 = 6 minutes Production time of three units of Y: 3 * 2 = 6 minutes. Therefore, the production time should be divided in the ratio 6:6 or 50% to 50% . 6/(6+6) and 6/(6+6)
Time [Min. /unit] Production [units] Production in standard units X 6.0 747.5 747.5 * (6/6) = 747.5 Y 2.0 2242.5 2242.5 * (2 / 6) = 747.5 Total 1495 21 Example: Effect of set-up time Part (a) We need to maximize monthly production. X Y S Production time per month: 149.5 hours, set up time .5 hour. 149.5 hours = 149.5 * 60 = 8970 minutes. 50% of 8970 = 4485 min. We can produce 4485 / 6 = 747.5 units of X. 50% of 8970 = 4485 min. We can produce 4485 / 2 = 2242.5 units of Y. Did we maintain our desired ratio? 747.5:12242.5 = 1:3 Machine available: 6 hrs/day, 25 days/month or 150 hrs/month. Machine Capacity in term of standard product (with considering set up time): = 150 [hrs/month] * 60 [ min/hr] / 6 [min/unit] = 1500 [units/month] Small loss in capacity! (1495 units vs. 1500 units)

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Time [Min. /unit] Production [units] Production in standard units X 6.0 687.5 687.5 * (6/6) = 687.5 Y 2.0 2062.5 2062.5 * (2 / 6) = 687.5 Total 1375 22 Example: cont.. Part (b) We must produce both products daily . Production time per day: 5.5 hours, set up time .5 hour. Production time per month = 5.5 * 25 = 137.5 hours 137.5 hours = 137.5 * 60 = 8250 minutes. 50% of 8250 = 4125 min. We can produce 4125 / 6 = 687.5 units of X. 50% of 8250 = 4125 min. We can produce 4125 / 2 = 2062.5 units of Y. 687.5:2062.5 = 1:3 X Y S Day 1 Y S X Day 2 Capacity = 1500 In aggregate planning we may want to reduce aggregate capacity by 10% to 15% to allow for set ups . Machine available: 6 hrs/day, 25 days/month or 150 hrs/month.
23 EPQ Q” Time Inventory t 0 t’’ Q Only consumption Only production Production & consumption Annual Demand: D, Usage Rate : u, Production rate: p Average inventory = ½ Q” = ½ Q . (p u) / p Annual number of runs = D / Q AIC(Q) = S . D + H. Q. (p - u) Q 2 p 2 * S * D H Q O = p (p - u) What if p is very large compared to u? In EPQ model, for the optimal Q formula, the first term is similar to EOQ, but the second term sqrt(p/p- u) is interesting. When p is very large (infinite) this becomes 1. Intuitively, in the graph of EPQ, when production rate is very high (going to infinity) (while demand rate is not high), the production time is going to be zero and EPQ becomes EOQ.

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A.) 18000 / 360 = 50 [units / day] B.) 18000/4.2 = 4285.71 [units / day] C.) 18000 / 90 = 2000 [units / day] D.) 175 * 4.2 = 735 [units/ year] Using the EPQ model D = 18,000 units / year, S = \$175, 1 year = 360 days H = \$4.2 / unit / year, p = 90 units / day What is u?
Ex. 2.1 Run time = 1800 [units] / 90 [units/day] = 20 days Cycle time = 1800 [units] / 50 [units/day] = 36 days Q” = maximum inventory AIC (1800) = S.D/Q + ½H.Q.(p-u)/p Q O = [2 * D * S / H] *

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• Spring '10
• panwalker
• inventory position

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