PTA Chemical manufactures 4 products (SKUs) and needs to determine the optimal production levels to maximize profit for the upcoming month. The company needs to produce at least 10,000 tons of SKU1, between 4,000-5,000 tons of SKU2, no more than 1,500 tons of SKU3, and between 25-2,000 tons of SKU4. Raw materials, reactor availability, and sales prices are provided. An optimization model is formulated to determine the production mix that results in maximum profit of $40,243 by producing 10,300 tons of SKU1, 4,467 tons of SKU2, 1,500 tons of SKU3, and 25 tons

1. Case 1: How much to manufacture
PTA Chemical LLC manufactures 4 SKUs. The company is struggling to keep up with the increasing
demand for these products; however, increasing raw material prices are forcing PTA Chemical to watch
its margin to ensure it is operating at the most optimum level.
To meet next month’s the sales forecast, PTA Chemical needs to produce:
1. SKU 1: At least 10,000 tons
2. SKU2 – between 4,000 to 5,000 tons
3. SKU3 – not more than 1,500 tons
4. SKU 4 – between 25 to 2,000 tons
To manufacture a ton of any SKUs, Chemical 1 & Chemical 2 required as a raw material in the
following proportions:
Final Product Input - Chemical 1 Input - Chemical 2
SKU 1 60% 40%
SKU 2 40% 60%
SKU 3 20% 80%
SKU4 100% 0%
The warehouse is having 11,000 Tons of Chemical 1 and 8,000 tons of Chemical 2 in the stock.
Each SKU needs processing in different reactors for different amount of time as shown below. For
instance, SKU1 needs 1 minutes of mixing, 2 minutes of heating, 1 minutes in a pressure chamber and
2.5 minutes of cooling chamber.
Processing Minutes required per ton Maximum Time available in
a month (hours)
SKU1 SKU2 SKU3 SKU4
Mixing 1.00 1.00 1.00 1.00 480
Heating 2.00 1.50 1.00 1.75 480
Pressure chamber 1.00 0.70 0.20 0.00 240
Cooling 2.50 1.60 1.25 1.00 720
The objective is to maximize the Profit by producing optimum mix of SKUs in a month for following sales
and cost price matrix.
$ /Ton SKU1 SKU2 SKU3 SKU4
Selling price 7.5 6.00 4.8 6.75
Cost 4.73 3.90 3.24 4.65

2. Solution:
1. PTA Chemical will get maximum profit of $ 40,243 by optimum utilization of available resources
and manufacturing products in following volume. Find below formulation of the model and excel
file for calculation.
SKU 1 SKU2 SKU3 SKU4
10,300 tons 4467 tons 1500 tons 25 tons
Refer attached Excel file “case1 _calculation” for detailed calculations.
2. Formulation of Optimization model:
1. Maximize profit
2. Variables
1. 𝑥 = Tons of SKU 1 to be manufactured
2. 𝑦 = Tons of SKU 2 to be manufactured
3. 𝑧 = Tons of SKU 3 to be manufactured
4. 𝑢 = Tons of SKU 4 to be manufactured
3. Constraints
1. 𝑥 >=10,000
2. 4000 <= 𝑦 <= 5000
3. 0 < 𝑧 < 1500
4. 25 < 𝑢 < 2000

3. 5. Only 11,000 tons of Chemical 1 is available
6. Only 8,000 tons of Chemical 1 is available
7. Mixing Reactor time <= 480 hours/month
8. Heating Reactor time <= 480 hours/month
9. Pressure Chamber time <= 240 hours/month
10. Cooling time <= 720 hours/month
4. Objective function: 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 (2.77 𝑥 + 2.1 𝑦 + 1.56 𝑧 + 2.1 𝑢)
5. Subject to
Reactor Time:
a. Mixing Reactor Time: (1𝑥 + 1𝑦 + 1 𝑧 + 1𝑢 ≤ 28,800 𝑚𝑖𝑛𝑢𝑡𝑒𝑠)
b. Heating Reactor Time: (2𝑥 + 1.5𝑦 + 1 𝑧 + 1.75 𝑢 ≤ 28,800 𝑚𝑖𝑛𝑢𝑡𝑒𝑠)
c. Pressure Chamber Time: (1𝑥 + 0.7𝑦 + 0.2𝑧 + 0𝑢 ≤ 14,400 𝑚𝑖𝑛𝑢𝑡𝑒𝑠)
d. Cooling Time: (2.5𝑥 + 1.6𝑦 + 1.25 𝑧 + 1𝑢 ≤ 43,200 𝑚𝑖𝑛𝑢𝑡𝑒𝑠)
Direct material:
e. Chemical 1: (0.6𝑥 + 0.4𝑦 + 0.2 𝑧 + 1𝑢 ≤ 10,0000 𝑡𝑜𝑛𝑠)
f. Chemical 2: (0.4𝑥 + 0.6𝑦 + 0.8 𝑧 ≤ 8,0000 𝑡𝑜𝑛𝑠)
c. Production constraints
g. SKU 1: 10,000 ≤ 𝑥 ≤ 99,9999 𝑡𝑜𝑛𝑠)
h. SKU 2: 4,000 ≤ 𝑦 ≤ 5,000 𝑡𝑜𝑛𝑠)
i. SKU 3: 0 ≤ 𝑧 ≤ 1500 𝑡𝑜𝑛𝑠)
j. SKU 4: 25 ≤ 𝑢 ≤ 2000 𝑡𝑜𝑛𝑠)

4. 3. Solver Screen Shot