Mixing three gravel quarries to specification

Aim: find the proportions $a$, $b$, $c$ (in %) of quarries A, B and C, with $a + b + c = 100$, so that the combined % passing at every sieve falls within the specification. The mix passing at a sieve is

$$P_{mix} = \frac{a\,P_A + b\,P_B + c\,P_C}{100}$$

Assumption: the "Specification" column gives the allowable range of % passing; the target is near the middle of each range.

Step 1: Decide the main fraction by control sieves

• Quarry A is a fine material (100% passes even 0.30 mm, and 40% passes 0.075 mm). Quarry B is a sand (all of it passes 4.76 mm, with few fines). Quarry C is the coarsest (gravel).
• 4.76 mm sieve: only C has coarse material (55% passing), so $P = 100 - 45\,c/100$. The range 60-75 gives $c = 55.6$ to $88.9\%$, so C is the main component (about 65%).
• 0.075 mm sieve: $P = [40a + 3(b + c)]/100 = 3 + 0.37a$ (since $b + c = 100 - a$). For 5 to 10, $a = 5.4$ to $18.9\%$, so A is a small proportion (about 10%).
• The remainder is B: $b = 100 - a - c$.

Step 2: Trial mix A : B : C = 10 : 25 : 65

All sieves are inside the specification. (A check of all combinations in 1% steps gives the feasible range as A 6-15%, B 0-38%, C 56-85%; the least-squares fit to the mid-points is about 8 : 25 : 67, which is also acceptable. The trial mix 10 : 25 : 65 is rounded and convenient for site batching.)

Answer: Quarry A : Quarry B : Quarry C = 10% : 25% : 65%, giving a blend passing 100, 96.8, 70.8, 45.8, 31.2, 17.9 and 6.7% on the 25.4, 12.7, 4.76, 1.18, 0.30, 0.15 and 0.075 mm sieves.