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Alligation is a method of solving arithmetic problems related to mixtures of ingredients. There are two types of alligation, Alligation medial and Alligation alternate. Alligation medial involves finding the a weighted mean concentration, or mean price of a mixture of ingredients by mass. Alligation alternate is a procedure where the proportions of ingredients needed to achieve a target mean concentration or price are found.

There are two further variations on Alligation alternate, Alligation Partial and Alligation Total, These involve finding the precise quantities of ingredients needed to achieve a target mean concentration or price.

Suppose you make a cocktail drink combination out of 1/2 Coke, 1/4 Sprite, and 1/4 orange soda. The Coke has 120 grams of sugar per liter, the Sprite has 100 grams of sugar per liter, and the orange soda has 150 grams of sugar per liter. How much sugar does the drink have? This is an example of alligation medial because you want to find the amount of sugar in the mixture given the amounts of sugar in its ingredients. The solution is just to find the weighted average by composition:

${\displaystyle {1 \over 2}\times 120+{1 \over 4}\times 100+{1 \over 4}\times 150=122.5}$ grams per liter

Alligation alternate may be thought of as the inverse of alligation medial. Given several numbers, to find multiples of those numbers that would result in a given mean value. Such a problem does not have a unique answer, and there are several solutions that give valid answers. There are several methods by which such problems may be solved.

Suppose you like 1% milk, but you have only 3% whole milk and ½% low fat milk. How much of each should you mix to make an 8-ounce cup of 1% milk? This is an example of alligation alternate because you want to find the amount of two ingredients to mix to form a mixture with a given amount of fat. Since there are only two ingredients, there is only one possible way to form a pair. The difference of 3% from the desired 1%, is assigned to the low fat milk, and the difference of ½% from the desired 1%, is assigned alternately to the whole milk. The total amount, 8 ounces, is then divided by the sum  to yield , and the amounts of the two ingredients are

ounces whole milk and  ounces low fat milk.

This article incorporates text from a publication now in the public domain:

A general formula that works for both alligation "alternate" and alligation "medial" is the following: Aa + Bb = Cc.

In this formula, A is the volume of ingredient A and a is its mixture coefficient (i.e. a= 3%); B is volume of ingredient B and b is its mixture coefficient; and C is the desired volume C, and c is its mixture coefficient. So in the above example we get: A(0.03) + B(0.005) = 8oz(0.01). We know B = (8oz-A), and so can easily solve for A and B to get 1.6 and 6.4oz, respectively. Using this formula you can solve for any of the 6 variables A,a,B,b,C,c, regardless of whether you're dealing with medial, alternate, etc.

This method was described by several early modern arithmetic books such as Hodder's Arithmetick and Cocker's Arithmetick. This method allowed one to solve such problems without knowledge of algebra.

1. Draw a vertical line. Write the mean on one side, and and each of the numbers from largest to smallest on the other side.
2. Draw a line connecting a number greater than the mean, with one or more numbers that is less than the mean.
3. Find the difference between the mean and each of the numbers. Write the difference beside the numbers that are linked to the subtrahend, in a third column. (Do not write this difference beside the subtrahend itself)
4. The numbers found in step (3) are the relative proportions of the quantities to be taken, to generate the mean. If more than one number results from step 3, add the numbers together.^[1]

Wheat costs 60 cents a bushel, Rye costs 36 cents a Bushel, Barley costs 24 cents a bushel, and Oats costs 18 cents a Bushel. How much of each kind of grain must be combined to create a mixture that costs 32 cents per bushel?^[2]

1. The prices of each kind of grain are written from largest to smallest, along with the mean.

  | 60
32| 36
  | 24
  | 18

2. Lines are drawn, connecting one quantity that's greater than the mean, with one greater than the mean.

  | 60───┐         
32| 36──┐│
  | 24──┘│   
  | 18───┘   

3. The numbers written down in step (1) are one by one, subtracted from the mean (or, if greater than the mean, the mean subtracted from them). The result of the subtraction is written beside the number that was linked to the minuend in step 2, and not the minuend itself.

In this example, 32 is subtracted from 60. The difference, 28 is written beside the number 18, which was linked to 20 in step 2. Likewise, the difference between 18 and 32, 14, is written beside 60, the number that it was linked to.

              (explanation)
  | 60───┐ 14 (32-18=14) 
32| 36──┐│  8 (24-32=8)
  | 24──┘│  4 (36-32=4) 
  | 18───┘ 28 (60-32=28)

The numbers that are produced by in this step are the relative proportions of the four grains that must be mixed: 14 parts of wheat, 8 parts of rye, 4 parts of barley and 28 parts of oats^[3]

Linking the numbers in the following manner will also yield a correct result: 8 parts of wheat, 14 parts of rye, 28 parts of barley and 4 parts of oats

  | 60───┐  8  
32| 36──┐│ 14
  | 24──┼┘ 28
  | 18──┘   4

It is also possible for one number above the mean to be linked to two or bore below the mean, and still yield a valid result. In this case 60 to be linked to both 24 and 18, and 36 to be linked to 24 and 18.^[4]

  | 60───┬──┐   
32| 36─┐─┤┐ │ 
  | 24─┴─┘│ │
  | 18────┴─┘

The subtractions are done as above, except in this case both differences are written against each number

                    (explanation)
  | 60───┬──┐  8,14 (32-18),(32-24)
32| 36─┐─┤┐ │  8,14 (32-18),(32-24)
  | 24─┴─┘│ │  28,4 (60-32),(36-32)
  | 18────┴─┘  28,4 (60-32),(36-32)

Each of the differences is is added together to yield a the final answer: 22 parts of wheat, 22 parts of rye, 32 parts of barley and 32 parts of oats

                  (explanation)
  | 60───┬──┐  22 (8+14)
32| 36─┐─┤┐ │  22 (8+14)
  | 24─┴─┘│ │  32 (28+4)
  | 18────┴─┘  32 (28+4)

There are 4 sorts of wine, costing $10, $8, $6 and $4 per gallon. How much of each must be combined to get a mixture worth $5 per gallon?^[5]

1. Since there is only one term ($4) that's less than the mean price of $5, the remaining three terms must be connected to it thus:

  | 10─────┐
 5|  8───┐ │
  |  6─┐ │ │
  |  4─┴─┴─┘

2. As all three terms are linked to '4', three subtractions are written beside it (10-5, 8-5 and 6-5)

                   (explanation) 
  | 10─────┐ 1     (5-4)
 5|  8───┐ │ 1     (5-4)
  |  6─┐ │ │ 1     (5-4)
  |  4─┴─┴─┘ 5,3,1 (10-5),(8-5),(3-5)

3. The multiple differences agains '4' are added together, to yield the final answer: 1 gallon each of the $10,$8, and $6 wines, and 9 gallons of the $4 wine.

             (answer)
  | 10─────┐ 1    
 5|  8───┐ │ 1 
  |  6─┐ │ │ 1   
  |  4─┴─┴─┘ 9

There five types of tea. Each costs $20, $12, $3, $2 and $1 per kilogram. In what proportions must the teas be mixed to give a mixture with an average cost of $8 per kilogram?

Owing to the odd number of terms, three terms are be linked to one another. Here 20 is linked to 1 and 2

  | 20───┐─┐ 
  | 12──┐│ │ 
 8|  3──┘│ │  
  |  2───┘ │ 
  |  1─────┘             

The subtractions are performed as in the previous example. As two terms are linked to "20", so two differences are written beside it.

  | 20───┐─┐ 7, 6     (8-1),(8-2) 
  | 12──┐│ │ 5        (8-3)
 8|  3──┘│ │ 14       (20-8)
  |  2───┘ │ 12       (20-8)
  |  1─────┘ 12       (20-8)

There are two differences beside "20".These are added up. The final sum is the answer.

                 (answer)
  | 20───┐─┐ 7,6 │13
  | 12──┐│ │ 5   │5
 8|  3──┘│ │ 14  │4
  |  2───┘ │ 12  │12
  |  1─────┘ 12  │2

Another answer may be obtained, by linking 20 to 3 and 2, as follows:

  | 20─────┐ 11    |
  | 12──┐  │ 7     |
 8|  3──│── 12   8|
  |  2──│──┘ 12    |
  |  1──┘    4     |

Alligation partial finds the exact quantities of ingredients needed to generate a mixture of a given concentration, when the quantity of one of the ingredients is already fixed. This is a matter of proportion.

Alligation total is a variation on alligation alternate, when the total mass of the final mixture is fixed.

Suppose you like 1% milk, but you have only 3% whole milk and 0.5% low fat milk. How much of each should you mix to make an 8-ounce cup of 1% milk?

By performing alligation alternate, the respective proportions of milk are 0.5 parts of 3%, and 2 parts of 0.5%.

  | 3.0──┐0.5
 1|      │
  | 0.5──┘2

As the total amount of milk in the final mixture is fixed at 8 ounces, finding the respective masses of milk, is simply a matter of finding the appopriate proportion of eight ounces.

The total amount, 8 ounces, is then divided by the sum {\displaystyle 2+{1 \over 2}={5 \over 2}} {\displaystyle 2+{1 \over 2}={5 \over 2}} to yield {\displaystyle 16 \over 5} {\displaystyle 16 \over 5}, and the amounts of the two ingredients are {\displaystyle {16 \over 5}\times {1 \over 2}={8 \over 5}} {\displaystyle {16 \over 5}\times {1 \over 2}={8 \over 5}} ounces whole milk and {\displaystyle {16 \over 5}\times 2={32 \over 5}} {\displaystyle {16 \over 5}\times 2={32 \over 5}} ounces low fat milk.

A general formula that works for both alligation "alternate" and alligation "medial" is the following:

${\displaystyle Aa+Bb=Cc.}$

In this formula, A is the volume of ingredient A and a is its mixture coefficient (i.e. a= 3%); B is volume of ingredient B and b is its mixture coefficient; and C is the desired volume C, and c is its mixture coefficient. Returning to example 4,

A(0.03) + B(0.005) = 8oz(0.01). We know B = (8oz-A), and so can easily solve for A and B to get 1.6 and 6.4oz, respectively. Using this formula you can solve for any of the 6 variables A,a,B,b,C,c, regardless of whether you're dealing with medial, alternate, etc.

Alligation, Forerunner of Linear Programming, Frederick V. Waugh, Journal of Farm Economics Vol. 40, No. 1 (Feb., 1958), pp. 89–103 http://www.jstor.org/stable/1235348

• Cocker's Arithmetick ... Perused and published, by John Hawkins ... (22nd edition, 1702) https://www.google.com/books/edition/Cockers_Arithmetick_perused_by_J_Hawkins/GWcFAAAAQAAJ?hl=en&gbpv=1&dq=cocker's%20arithmetick&pg=PA156&printsec=frontcover
• Hodder's Arithmetick ... The six and twentieth edition, https://books.google.com/books?id=2-9ZAAAAcAAJ&newbks=1&newbks_redir=0&pg=PA185#v=onepage&q&f=false

• Alligation alterne et medial: www.formatp.ca/alligation.php
• Alligation Alternate and the Composition of Medicines: Arithmetic and Medicine in Early Modern England
• Robinson's Progressive Practical Arithmetic