Calculations for Ice Cream and Frozen Dairy Desserts

# 19 Mix Calculations

The general objective in calculating ice cream mixes is to turn your formula into a recipe based on the ingredients you intend to use and the amount of mix you desire. The formula is given as percentages of fat, milk solids-not-fat, sugar, corn syrup solids (glucose solids), stabilizers, emulsifiers or other mix components. The ingredients to supply these components are chosen on the basis of availability, quality and cost, for example cream or butter as a fat source, skim milk powder or condensed skim milk as SNF sources. The complexity is that several ingredients will supply more than one componet, for example cream contains all of milkfat, milk SNF and water. In this section, I include several examples of how to calculate mix recipes based on desired formulations.

The following table illustrates the relationship between the major components, the main ingredients that supply the major components, and the minor components that are supplied with the major ones for each ingredient.

Component and Ingredients to supply that component (but note that each of these ingredients also supplies the following other components):

Milkfat, supplied by Cream (which also supplies SNF and water) or Butter (which also supplies SNF, water);

Milk solids-not-fat (SNF, or sometimes also called serum solids, S.S.), supplied by any of the following:

• Skim powder (which also supplies water, about 3%)
• Condensed skim (which also supplies water)
• Condensed milk (which also supplies water and fat)
• Sweetened condensed (which also supplies water and sugar)
• Whey powder (which also supplies water)

Water, supplied by Skim milk (which also supplies msnf), or milk (which also supplies fat and msnf), or pure water.

Sweetener, supplied by dry or liquid (which also then supplies water) sucrose or corn syrup solids.

The first step in a mix calculation is to identify for each ingredient we intend to use its components. If there is only one source of the component we need for the formula, for example the stabilizer or the sugar, we determine it directly by multiplying the percentage we need by the amount we need, e.g., 100 kg of mix @ 10% sugar would require 10 kg sugar. If there are two or more sources, for example we need 10 % fat and it is coming from both cream and milk, then we need to utilize an algebraic method.

Computer programs developed for mix calculations generally solve a simultaneous equation based on mass and component balances. To solve simultaneous equations, you need as many independent equations as you have unknowns. A free on-line simultaneous equation solver is available at https://www.idomaths.com/simeq.php. For an example of a on-line mix calculator by subscription, see https://www.dairyscience.info/calculator1/index.php . Other commercial options are available from http://www.articagel.it/frame_freezerlabpro3.htm or https://owlsoft.com

For manual calculations, a method known as the “Serum Point” method has been derived. This method has solved the simultaneous equations in a general way so that only the equations need to be known and not resolved each time.

In standardizing mixes, the composition of the various ingredients used must be known. In some cases the percentage of solids contained in a product is taken as constant, while in others the composition must be obtained by analysis. Information on the various ingredients is given below:

(a) Skim milk – can be determined by analysis or assumed at 9 percent serum solids. Fat (0.01% – 0.10%) should be taken into account if significant.

(b) Dried products, e.g. skim milk powder, whey powder, WPC, milk powder blends, usually taken to be 97 percent solids as they retain some moisture.

(c) Cream – Percent fat usually measured by an acceptable method.

Percent MSNF found by formula as follows: (100 – percent fat) x .09 = % snf (assuming that the “skim milk” contains 9% total solids). Example: In cream testing 30% fat, the percent snf would be (100 – 30) x .09 = 6.3% snf

(d) Milk – Percent fat measured by an acceptable method.

Percent snf may be found same as for cream or by making a total solids test and deducting the percent fat.

(e) Condensed Milk Products – Composition of these products should be obtained by the supplier.

(f) Sweeteners – Sucrose – Dry 100% solids
Sucrose – Liquid 66% solids
Dextrose – Dry 100% solids
Corn Syrup Solids 100% solids
Corn Syrup Liquid 80% solids
Glucose 80% solids
Honey 80% solids

(g) Stabilizers and Emulsifiers (if solid) – Because of the small percentage used may be figured as 100 percent solids.

(h) Egg Products – Fresh whole eggs: 10% fat, 25% solids
Fresh egg yolk: 33% fat, 50% solids
Frozen egg yolk: 33% fat, 50% solids
Dried egg yolk: 60% fat, 100% solids

Please follow along the several examples below, solved by both algebraic methods (including use of the simultaneous equation solver) and by the srum point method (which is explained in Example Problem 2).

# Problems

Below are some example problems to look at, if you are interested in the mathematics of mix calculations.

• Example 1. Basic mix using butter, skim powder, and water (only one source of each component). (Algebraic Method)
• Example 2. Mix using cream, skim, and skim powder (three sources of milk SNF, three sources of water). (Algebraic and Serum Point Methods)
• Example 3. Mix using cream, milk, and skim powder (three sources of milk SNF, three sources of water, and two source of fat). (Algebraic and Serum Point Methods)
• Example 4. Mix using cream, milk, and sweetened, condensed skim (three sources of milk SNF, three sources of water, two sources of fat, and two sources of sugar). (Serum Point Method)
• Example 5. Mix using cream, milk, and sweetened, condensed milk (three sources of milk SNF, three sources of water, three sources of fat, and two sources of sugar). (Serum Point Method)
• Example 6. Mix using a given amount of cream and skim, with the balance coming from butter, milk, and skim powder. (Serum Point Method)
• Example 7. Mix using cream, milk, condensed skim, and liquid sweeteners (water needs to be accounted for). (Serum Point Method)

After completing a problem, you should do a proof of your calculation, by ensuring that the mass sums to the desired value, and that the mass fraction of all components also sum to the desired value – see all the examples below. There is only one unique solution, so you know by calculation if you have it right or not!

Note: these are all solved on the basis of 100 kg. If you are making more or less than that, you can still solve on the basis of 100 kg and then scale up or down your answer accordingly (by dividing your answers per 100 kg by 100 and multipying by the actual wt., e.g. 45 kg cream/100 kg mix, if you are making 4000 kg, it would be 45/100*4000=180 kg cream) , or you can use the desired weight directly in the calculations, but be careful with the serum point method equations – see Example 3 below. If you have scaled up or down from 100 kg, you should do the proof total on the desired weight, and ensure it meets the desired percentage – see Example 3 or 6.

I am sometimes asked how to incorporate whole milk powder into mix calculations. This is common in some parts of the world. I will attach a pdf file below that shows such a calculation.

# Problem 1

Desired: 100 kg mix testing 14% fat, 10% MSNF, 15% sucrose, 0.4% stabilizer/ emulsifier.

Ingredients on hand: Butter 80% fat, skim milk powder 97% solids, water sucrose, stabilizer/emulsifier.

Solution:

1. Find the amount of butter required to supply 14 kg of fat/ 100 kg mix,

14 kg fat x 100 kg butter/80 kg fat = 17.5 kg butter

2. Find the amount of skim milk powder needed to supply a total of 10 kg of snf/ 100 kg mix.

The butter contributes 17.5 kg butter x 1.8 kg s.s / 100 kg butter = 0.315 kg s.s.

Powder must contribute 10 kg snf – 0.315 kg = 9.685 kg s.s.

9.685 kg snf x 100 kg powder/97 kg snf = 9.98 kg powder

3. Sucrose required will be 15.0 kg/ 100 kg mix.

4. Stabilizer/ emulsifier required will be 0.4 kg/ 100 kg mix.

5. The amount of water required will be equal to 100 minus the sum of the weights of the other ingredients, thus,

100 – (17.5 + 9.98 + 15 + 0.4) = 57.12 kg water

Note: In Problem 1, the serum solids content of the butter was calculated as follows: butter at 80% fat, remaining 20% skim milk at 9% milk solids-not-fat, therefore msnf in butter = 20% x 9% = 1.8%.

In the manufacture of butter, fat is churned from cream (which can be thought of as a mixture of fat and skim milk). If no washing of the butter is performed after churning, the above assumption of 1.8% msnf in 80% fat butter is correct. However, the 20% skim milk could be substituted wholly or in part with wash water, which would reduce the msnf level to anywhere between 1.8% and 0%. Each butter sample either needs to be analyzed for solids or an assumption of no msnf should be made to assure that at least the required msnf is supplied from other ingredients.

## Problem 1 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Total Solids (kg0
Butter 17.50 14.00 0.32 14.32
Skim powder 9.98 -- 9.68 9.68
Sucrose 15.00 -- -- 15.00
Stabilizer 0.40 -- -- 0.40
Water 57.12 -- -- --
Totals 100.00 14.00 10.00 39.40

Note: 14% fat + 10% msnf + 15% sucrose + 0.4% stab. = 39.4% TS

# Problem 2

Desired : 100 kg mix @ 13% fat, 11% MSNF, 15% sucrose, 0.5% stabilizer, 0.15% emulsifier

On hand: Cream @ 40% fat, 5.4% msnf; skimmilk @ 9% msnf; skimmilk powder @ 97% msnf; sugar; stabilizer; emulsifier.

Solution: (Note: only one source of fat, sugar, stabilizer, and emulsifier, but two sources of serum solids)

Cream:
100 kg mix x 13 kg fat/100 kg mix x 100 kg cream/40 kg fat = 32.5 kg cream

Sucrose:
100 kg mix x 15 kg sucrose/100 kg mix = 15 kg sucrose

Stabilizer:
100 kg mix x 0.5 kg stabilizer/100 kg mix = 0.5 kg stabilizer

Emulsifier:
100 kg mix x 0.15 kg emulsifier/100 kg mix = 0.15 kg emulsifier

Algebraic method:

Skim milk and Skim powder, Note: two sources of the MSNF

Now, let x = skim powder, y = skim milk

MASS BALANCE (All the components add up to 100 kg)

x + y = 100 – (32.5 + 15 + 0.5 + 0.15) (1)

MSNF BALANCE (Equal to 11% of the mix and coming from the skim milk, the skim powder, and the cream)

0.97 x + 0.09 y = .11(100) – (.054 x 32.5) (2)
x + y = 51.85 so y = 51.85 – x from (1)
.97 x + .09 y = 9.245 from (2)
.97 x + .09 (51.85 – x) = 9.245 substituting
.97 x – .09 x + 4.67 = 9.245
.88 x = 4.58

x = 5.20 kg skim powder

y = 46.65 kg skim milk

The above shows the solution of a 2-unknown simultaneous equation. Likewise, if there were 3 unknowns, e.g., fat, msnf, and the total weight, then three equations could be written, one for each of fat, msnf, and weight. However, the above problem could also be solved with the Serum Point method, and the example and that solution along with the derivation of the equations follows. The Serum Point calculation assumes 9% msnf in skimmilk and the skim portion of all dairy ingredients. It then solves the calculation beginning with the most concentrated source of serum solids first.

Solution via the serum point method:

1. Amount of powdered skim milk needed is found by the following formula:

(SNF needed – (serum of mix X .09))/(% SNF in powder – 9) X 100 = kg skim powder

This is a generalized equation solved from a mas balance, that works in all situations where milk powder is the most concentrated source of serum solids, and all of the serum products are milk products (i.e., there is no water used in the recipe). An assumption is that the serum fraction of all dairy ingredients contains 9% solids-not-fat, e.g., 40% cream contains 60% skim (100-40), which contains 9% snf, so the snf content of the cream is .60 x .09 = 5.4%

The serum of the mix is found by adding the desired percentages of fat, sucrose, stabilizer and emulsifier together and subtracting from 100. In the present problem then,

100 – (13 + 15 + 0.5 + 0.15) = 71.35 kg serum.

Substituting in the formula we have:

(11 – (71.35 x .09))/(97 – 9) x 100 = 4.58/88 x 100 = 5.20 kg skim powder

2. The weight of cream will be 13 kg x 100 kg cream/40 kg fat = 32.5 kg cream

3. The sucrose will be 15 kg/ 100 kg mix.

4. The stabilizer will be 0.5 kg/ 100 kg mix.

5. The emulsifier will be 0.15 kg/ 100 kg mix.

6. The weight of mix supplied so far is,

Cream 32.50 kg
Skim powder 5.20 kg
Sucrose 15.00 kg
Stabilizer .50 kg
Emulsifier .15 kg
Total 53.35 kg

The skim milk needed therefore is 100 – 53.35 = 46.65 kg.

## Problem 2 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Total Solids (kg)
Cream 32.50 13.00 1.75 14.75
Skim milk 46.65 -- 4.20 4.20
Skim milk powder 5.20 -- 5.04 5.04
Sucrose 15.00 -- -- 15.00
Stabilizer 0.50 -- -- 0.50
Emulsifier 0.15 -- -- 0.15
Totals 100.00 13.00 11.00 39.64

Note: 13% fat + 11% snf + 15% sucrose + 0.50% stab. + 0.15% emul. = 39.65% TS

DERIVATION OF THE SERUM POINT EQUATIONS: Let’s resolve problem 2 again using simultaneous equations in a general way to show where the serum point equations come from.

On hand: cream @ 40% fat
(supplies fat, water, and serum solids, therefore can be thought of as a mixture of fat and skim milk)
skim milk @ 9% solids not fat
(supplies water and serum solids)
skim milk powder @ 97% solids not fat
(supplies water and serum solids)
sucrose
stabilizer
emulsifier

Solution

– Only one source of fat, sucrose, stabilizer, and emulsifier

kg fat = 100 kg mix x 13 kg fat/100 kg mix = 13 kg fat (The explanation for this assumption becomes clearer in a moment!)

kg sucrose = 100 kg mix x 15 kg sucrose/100 kg mix = 15 kg sucrose

kg stabilizer = 100 kg mix x 0.5 kg stab./100 kg mix = 0.5 kg stabilizer

kg emulsifier = 100 kg mix x 0.15 kg emul./100 kg mix = 0.15 kg emulsifier

– Two sources of serum solids

Let X = skim powder (kg)
Let Y = skim milk (kg) + skim milk in cream (kg)

MASS BALANCE X + Y = Total mix – components already added
X + Y = 100 – (13 + 15 + 0.5 + 0.15), (the “Serum of the Mix”)
X + Y = 71.35
(so Y = 71.35 – X)

MSNF BALANCE 0.97X + 0.09Y = (0.11 x 100)
“Serum “Serum “Serum fraction
fraction fraction in mix”
in powder” in skim”

0.97 X + 0.09 (71.35 – X) = 11
0.97 X + (0.09 x 71.35) – 0.09 X = 11
0.97 X – 0.09 X = 11 – (0.09 x 71.35)
X = 11 – (.09 x 71.35)/ 0.97 – 0.09

Which is equal to:

kg skim powder = S.S. needed – (0.09 x serum of mix) x 100 % S.S. in powder – 9

X = 4.58/0.88 = 5.20 kg powder

kg cream = 13 kg fat x 100 kg cream/40 kg fat = 32.5 kg cream

kg skim = 100 – 32.5 – 15 – 0.5 – 0.15 – 5.2 = 46.65 kg

# Problem 3

Desired: 100 kg mix containing 18% fat, 9.5% SNF, 15% sucrose, 0.4% stabilizer, 1% frozen egg yolk.

On hand: Cream 30% fat, milk 3.5% fat, skim milk powder 97% solids, sucrose, stabilizer, and egg yolk.

The solution to this problem will be shown by simultaneous equations, since there are three sources of milk SNF, three sources of water, and two source of fat, which require three equations, and by the serum point method. Both produce the same results. Follow whichever method you prefer. Computer programs exist that solve simultaneous equations.

Solution via the algebraic method:

Sucrose: 100 kg mix x 15 kg sucrose/100 kg mix = 15 kg sucrose

Stabilizer: 100 kg mix x 0.4 kg stabilizer/100 kg mix = 0.4 kg stabilizer

Egg yolk: 100 kg mix x 1 kg egg yolk/100 kg mix = 1 kg egg yolk

Now, let x = skim powder, y = milk, z = cream.

MASS BALANCE All the components add up to 100 kg

x + y + z = 100 – (15 + 0.4 + 1) (1)

MSNF BALANCE Equal to 9.5% of the mix and coming from the milk, the skim powder, and the cream; assume 9% in the skim portion of the milk and cream so that the msnf of the milk = .09 x (100 – 3.5) and of the cream = .09 x (100-30)

0.97 x + 0.08685 y + 0.063 z = .095 (100) (2)

FAT BALANCE Equal to 18% of the mix and coming from the milk and cream

.035 y + .3 z = .18 (100) (3)

Solution via the serum point method:

1. Find the amount of skim milk powder required by the following formula:

(SNF needed – (serum of mix x .09))/(% SNF in powder – 9) x 100 = skim powder

Substituting we have,

(9.5 – ( 65.6 x .09 ))/(97-9) x 100 = 3.596/88 x 100 = 4.08 kg powder

2. Amount of sucrose required is 15.0 kg.

3. Amount of stabilizer required is .4 kg.

4. Amount of egg required is 1.0 kg.

5. Find weight of milk and cream needed.

Materials supplied so far are 4.08 kg powder, 15 kg sucrose, 0.4 kg stabilizer, and 1 kg egg yolk, a total of 20.48 kg. 100 – 20.48 = 79.52 kg milk and cream needed.

6. Find the amount of cream by following formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

substituting we have,

(18 – ( 79.52 x 3.5/100 ))/(30-3.5) x 100 = 15.217/26.5 x 100 = 57.42 kg cream.

7. Amount of milk needed = 79.52 – 57.42 = 22.10 kg of milk.

## Problem 3 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Total Solids (kg)
Cream 57.42 17.23 3.62 20.85
Milk 22.10 0.77 1.92 2.69
Skim milk powder 4.08 -- 3.96 3.96
Sucrose 15.00 -- -- 15.00
Stabilizer 0.40 -- -- 0.40
Egg Yolk 1.00 -- -- 0.50
Totals 100.00 18.00 9.50 43.40

Note: 18% fat + 9.5% snf + 15% sucrose + 0.40% stab. + 0.50% egg yolk solids (half the egg yolk) = 43.4% TS

If you wanted to make 3000 kg (for example) instead of 100 kg, you could multiply all of the numbers above by 30, or you could set up the equation to solve directly for 3000 kg, as shown below.

Solution via the serum point method for 3000 kg:

1. Find the amount of skim milk powder required by the following formula:

(MSNF needed – (serum of mix x .09))/(% snf in powder – 9) x 100 = skim powder

MSNF needed = 3000 x 9.5% = 285 kg; Serum of the mix = 3000 – 540 (fat) – 450 (sugar) – 12 (stab.) – 30 (egg yolk) = 1968 kg. Substituting we have,

(285 – ( 1968 x .09 ))/(97-9) x 100 = 107.88/88 x 100 = 122.59 kg powder

2. Amount of sucrose required is 3000 x 15% = 450.0 kg.

3. Amount of stabilizer required is 3000 x.4% = 12.0 kg.

4. Amount of egg required is 3000 x 1.0% = 30 kg.

5. Find weight of milk and cream needed.

Materials supplied so far are 122.59 kg powder, 450.0 kg sucrose, 12.0 kg stabilizer, and 30.0 kg egg yolk, a total of 614.59 kg. 3000 – 614.59 = 2385.41 kg milk and cream needed.

6. Find the amount of cream by following formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

substituting we have,

(540 – ( 2385.41 x 3.5/100 ))/(30-3.5) x 100 = 456.51/26.5 x 100 = 1722.68 kg cream.

7. Amount of milk needed = 2385.41 – 1722.68 = 662.73 kg of milk.

## Problem 3 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Total Solids (kg)
Cream 1722.68 516.8 108.53 625.33
Milk 662.73 23.2 57.56 80.76
Skim milk powder 122.59 -- 118.91 118.91
Sucrose 450.00 -- -- 450.00
Stabilizer 12.0 -- -- 12.0
Egg Yolk 30.0 -- -- 15.0
Totals 3000.0 540.0 285.0 1302.0

Note: 540/3000 = 18% fat; 285/3000 = 9.5% SNF; 1302/3000 = 43.4% Total solids

# Problem 4

Desired: 100 kgs. mix testing 14% fat, 10% MSNF, 15% sucrose, 0.5% stabilizer/emulsifier.

On hand: Cream 32% fat, milk 3.5% fat, sweetened condensed skim milk 28% serum solids and 40% sugar, sucrose and stabilizer.

Solution via the Serum Point Method :

1. Find amount of condensed skim milk required by the following formula:

(SNF needed – (serum of mix x .09))/( % SNF in cond. – (serum of cond. x .09)) x 100 = sweet cond. skim milk

Note: Serum of condensed is calculated the same as serum of mix, i.e., 100 – (Fat + Sugar + Stab.)

Substituting we have:

(10 – ( 70.5 x .09 ))/(28 – ( 60 x .09 )) x 100 = 16.17 kg cond. skim milk

2. Find amount of sucrose needed:

16.17 x .40 = 6.47 kg of sucrose in the condensed milk.
15 – 6.47 = 8.53 kg of sucrose still needed.

Note: If you added too much sugar by using sweetened condensed skim to supply the desired serum solids, then scale back the sweetened condensed skim to supply all the sugar you need and make up the deficiency in serum solids with skim powder.

3. Amount of stabilizer required is 0.5 kg.

4. Find weight of milk and cream needed.

Material so far supplied is, 16.17 kg condensed milk, 8.53 kg sugar and .5 kg stabilizer, a total of 25.2 kg.
100 – 25.2 = 74.8 kg milk and cream required.

5. Find amount of cream by the following formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

Substituting we have:
(14 – ( 74.8 x .035 ))/(32 – 3.5) x 100 = 39.93

6. Find milk required:

74.8 – 39.93 = 34.87 kgs. milk.

## Problem 4 Proof:

Ingredients Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Sugar (kg) Wt. of Total Solids (kg)
Cream 39.93 12.78 2.44 -- 15.22
Milk 34.87 1.22 3.03 -- 4.25
Swt. Cond. Milk 16.17 -- 4.53 6.47 11.00
Sucrose 8.53 -- -- 8.53 8.53
Stabilizer 0.50 -- -- -- 0.50
Totals 100.00 14.00 10.00 15.00 39.50

Note: 14% fat + 10% snf + 15% sucrose + 0.5% stab = 39.5% TS

# Problem 5

Desired: 100 kg mix testing 14% fat, 10% MSNF, 15% sucrose, 0.5% stabilizer/ emulsifier.

On hand: Cream 30% fat; milk 4% fat; sweetened condensed whole milk 8% fat, 20% snf, 42% sugar; stabilizer/ emulsifier; and sucrose.

Solution via the Serum Point Method:

1. Find the amount of sweetened condensed milk by formula:

(SNF needed – (serum of mix X .09))/(% SNF in cond. – (% serum in cond. X .09)) X 100 = kg cond. milk

Substituting we have:

(10 – (70.5 X .09))/(20 – (50 X .09)) X 100 = 23.58 kg sweetened cond. milk

2. Stabilizer/ emulsifier required will be 0.5 kg.

3. Find amount of sucrose needed.

23.58 X .42 = 9.90 kg sucrose in cond. milk.
15 – 9.90 = 5.1 kg sucrose still required.

Note: If you added too much sugar by using sweetened condensed skim to supply the desired serum solids, then scale back the sweetened condensed skim to supply all the sugar you need and make up the deficiency in serum solids with skim powder.

4. Find amount of milk and cream needed.

100 – 29.18 (cond. milk, sugar, and stabilizer) = 70.82 kg.

5. Find amount of cream required.

23.58 X .08 = 1.89 kg fat in the condensed milk.
14 – 1.89 = 12.11 kg fat still needed.

Use formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

Substituting we have:

(12.11 – (70.82 X .04)) / (30 – 4) X 100 = 35.69 kg cream

6. Find amount of milk required:

70.82 – 35.69 = 35.13 kg milk.

## Problem 5 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Sugar (kg) Wt. of Total Solids (kg)
Cream 35.69 10.71 2.25 -- 12.96
Milk 35.13 1.40 3.03 -- 4.43
Swt. Cond. Milk 23.58 1.89 4.72 9.90 16.51
Sucrose 5.10 -- -- 5.10 5.10
Stabilizer 0.50 -- -- -- 0.50
Totals 100.00 14.00 10.00 15.00 39.50

Note: Note: 14% fat + 10% snf + 15% sucrose + 0.5% stab = 39.5% TS

# Problem 6

Desired: Make 2000 kg of mix testing 15% fat, 10.5% MSNF, 15% sucrose, 0.5% stabilizer, 1% egg yolk

From the following: 450 kgs. 30% cream; 300 kgs. skim milk; Get balance from butter @ 84% fat, milk @ 4% fat, skim milk powder, sucrose, stabilizer, and egg yolk.

Solution via the Serum Point Method:

1. Find skim milk powder needed.

Use formula:

(SNF needed – (serum of mix X .09)) / (% snf in powder – 9 ) X 100

Substituting we have:

(210 – (1370 X .09)) / (97 – 9) X 100 = 98.5 kg powder

2. Find sugar, stabilizer, and egg needed.

2000 X .15 = 300 kg sucrose
2000 X .005 = 10 kg stabilizer
2000 X .01 = 20 kg egg yolk

3. List the materials supplied so far:

Cream               450.00 kg
Skim Milk          300.00 kg
Skim Powder       98.50 kg
Sucrose             300.00 kg
Stabilizer           10.00 kg
Egg Yolk           20.00 kg

Total                1,178.50 kg

4. Find amount of butter and milk needed.

2000 – 1178.5 = 821.5 kg butter and milk required.

5. Find amount of fat that still has to be made up:

300 – 135 (Fat in 450 kg 30% cream) = 165 kg

6. Find amount of butter needed by following formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

Substituting we have:

(165 – (821.5 X .04)) / (84 – 4 ) X 100 = 165.17 kg butter

7. Find amount of milk needed.

821.5 – 165.17 = 656.33 kg milk

## Problem 6 Proof:

Ingredient Total wt. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Total Solids (kg)
Cream 450.00 135.00 28.35 163.35
Milk 656.33 26.25 56.71 82.96
Butter 165.17 138.74 2.37 140.39
Skim Milk 300.00 -- 27.00 27.00
Skim milk powder 98.50 -- 95.55 95.55
Sucrose 300.00 -- -- 300.00
Stabilizer 10.00 -- -- 10.00
Egg Yolk 20.00 -- -- 10.00
Totals 2000.00 300.00 210.00 829.25

Note: 300/2000 = 15% fat; 210/2000 = 10.5% SNF; 829.25/2000 = 41.46% Total solids (= 15% fat + 10.5% snf + 15% sucrose + 0.5% stab + 0.5% egg yolk solids (half of the egg yolk))

# Problem 7 (Using Liquid Sweeteners)

Desired: 100 kgs. of mix testing 12% fat, 11% MSNF, 14% sucrose, 3% corn syrup solids, 0.35% stabilizer, 0.15% emulsifier.

On hand: Cream 40% fat; milk 3.5% fat; condensed skim milk 35% solids; liquid sucrose 66% solids; regular conversion corn syrup 80% solids; stabilizer; emulsifier.

Solution via the Serum Point Method:

1. Calculate the pounds of condensed skim first, but determine serum of the mix as follows:

(a) Find the amount of liquid sucrose that must be added to provide 14 kg of sucrose solids:

14 kg sucrose x 100 kg liq. sucrose/66 kg sucrose = 21.21 kg.

(b) Find the amount of corn syrup that must be added to provide 3 kgs. of corn syrup solids:

3 kg solids x 100 kg liq. css/80 kg solids = 3.75 kg.

Serum of the mix is found by adding together the percentage of fat, liquid sucrose, liquid corn syrup, stabilizer and emulsifier and subtracting from 100:

• 12.00 kg fat
• 21.21 kg liquid sucrose
• 3.75 kg corn syrup
• 0.35 kg stabilizer
• 0.15 kg emulsifier

Total 37.46 kg

100 – 37.46 = 62.54, the serum of the mix

Use formula:

(SNF needed – ( serum of mix x .09 )) / (% SNF in Cond. skim – 9) x 100

Substituting we have:

(11 – ( 62.54 x .09 )) / (35 – 9) x 100 = 20.65 kg of condensed skimmilk

2. Liquid sucrose required = 21.21 kg.

3. Liquid corn syrup required = 3.75 kg.

4. Stabilizer required = 0.35 kg.

5. Emulsifier required = 0.15 kg.

6. Find the amount of milk and cream needed:

100 – (20.65 + 21.21 + 3.75 + 0.35 + 0.15) = 53.89 kg.

7. Find the amount of cream needed by formula:

((kg fat needed – (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream – % fat in milk)) x 100

Substituting we have:

(12 – ( 53.89 x 3.5/100 ))/(40 – 3.5) x 100 = 27.69 kg of cream.

8. Find the amount of milk required:

53.89 – 27.69 = 26.20 kgs. of milk.

## Problem 7 Proof:

Ingredients Total wk. (kg) Wt. of Fat (kg) Wt. of SNF (kg) Wt. of Sugar (kg) Wt. of Total Solids (kg)
Cream 27.69 11.08 1.50 -- 12.58
Milk 26.20 0.92 2.27 -- 3.19
Cond. Skim 20.65 -- 7.23 -- 7.23
Sucrose 21.21 -- -- 14.00 14.00
Corn syrup solids 3.75 -- -- 3.00 3.00
Stabilizer 0.35 -- -- -- 0.35
Emulsifier 0.15 -- -- -- 0.15
Totals 100.00 12.00 11.00 17.00 40.50

Note: 12% fat + 11% snf + 14% sucrose + 3.0% css + 0.35% stab + 0.15% emul. = 40.5% TS