To fulfill an order for tinted plastic lenses, the plastics manufacturing company must first produce ceramic powder made from a mixture of three raw materials — A, B and C — in the ratio 10:3:1 by weight. The raw materials are mixed, baked and ground to a powder. The powder then is screened to remove lumps. The screening removes on average 15 percent, but the lumps can be returned to the raw mixture and reprocessed so they are not wasted.

The production manager consults with an industrial engineer, who must specify the ratios of raw materials needed to meet quality specifications of the job. The production manager then assigns a production and planning clerk to calculate the order and ensure the timely delivery of the appropriate quantities of the raw materials needed to produce 1,000 blocks, each weighing 2.5 kg on each of seven consecutive production days.

Ignoring the waste for a moment, the total amount of powder needed for one day is

.

Since the powder is produced from the raw materials in the ratio of 10:3:1, the total of 2,500 kg needs to be divided into 14 parts: 2500/14 kg = 178.6 kg per part. Thus the after-waste weights of the three ingredients A, B and C will be

WA = after-waste weight of A = 10 x 178.6 kg = 1786 kg.
WB = after-waste weight of B = 3 x 178.6 kg = 536 kg.
WC = after-waste weight of C = 1 x 178.6 kg = 179 kg.

If the production clerk started with these amounts on the first day, the output would be 15 percent short because of the screening loss in grinding.

To meet demand at the end of the day, the machine operator, following the written instructions from the production manager, must add additional raw material. The necessary amount for each raw material can be calculated most easily from the ratio:

Before-waste weight -----1.00 -------1.00
––––––––––––––––– = ––––––––– = ––––––– = 1.1765.
After-waste weight----1.00 – .15----- .85

This means that the before-waste weight of each ingredient must be 17.65 percent larger than the desired after-waste weight. So the extra material that must be ordered is

XA = extra amount of A = .1765 x 1786 kg = 315 kg.
XB = extra amount of B = .1765 x 536 kg = 95 kg.
XC = extra amount of C = .1765 x 179 kg = 32 kg.

The operator must use an extra 315 kg of A, 95 kg of B and 32 kg of C to complete the order on the first day. This will result in 2,500 kg of ceramic at the end of the day, with 442 kg to be recycled on the next day.

For each of the next six days, the operator needs to use 1,786 kg of A, 536 kg of B and 179 kg of C. The recycled material from the previous day takes care of the 15 percent waste. The total order for each of the raw materials for the week's production is (rounded up):

• Material A: (7 x 12,500 kg) + 315 kg = 87,815 kg.
• Material B: (7 x 3,750 kg) + 95 kg = 26,345 kg.
• Material C: (7 x 1,250 kg) + 32 kg = 8,782 kg.

This work is just a starting point in the analysis of the scheduling problem. In the workplace, it often is necessary to ask how the solution will change if some of the assumptions change.

• The level of waste (15 percent) actually is not fixed but will change from day to day. The industrial engineer has collected data to show that the level of waste fits a normal distribution with a mean of 15 percent and a standard deviation of 2 percent. How should the production manager take this new information into account?
• The production and planning clerk realizes that product is not shipped every day but only once every seven days. It is necessary only to average 1,000 blocks per day for seven days. Does this added flexibility allow the company to order lower quantities of the raw materials?