UNIVERSITY OF CALIFORNIA COLLEGE OF AGRICULTURE AGRICULTURAL EXPERIMENT STATION BERKELEY, CALIFORNIA CIRCULAR 333 January, 1934 METHODS OF STANDARDIZING ICE CREAM MIXES W. C. COLE 1 "Standardizing" an ice cream mix means calculating the relative amounts of ingredients required for a finished product of a desired com- position. The calculations are made for a definite weight of ice cream mix, the exact amount depending upon the size of the standard batch processed in a given plant. For the sake of simplicity, however, the cal- culations made here in most cases will be for a 100-pound batch, since that amount offers a satisfactory basis for making any desired quantity of mix. Errors often occur in making mixes. These are commonly due to care- less weighing of products, or to the fact that the composition of the ingredients are different from the assumed values used during the orig- inal calculations. Provided the errors are great enough, a recalculation is necessary, to ascertain how the original mixture must be modified to assure the desired composition. This process is termed "^standardiza- tion." A satisfactory mix for plain as well as for fruit ice creams contains 12.0 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and sufficient gelatin for proper stabilization. In California, 2 the mini- mum legal fat content of vanilla ice cream is 10.0 per cent; of fruit ice cream, 8.0 per cent. The fat contained in 100 pounds of legal vanilla ice cream would satisfy the requirements for 125 pounds of fruit ice cream — which equals a dilution of 25 per cent. If a higher dilution is desired, it would be necessary for the basic mix to contain more than 10 per cent fat. The standard adopted in many cases is considerably higher than the legal requirements. i Associate in Dairy Industry. 2 Agricultural Code of California, effective August 21, 1933. 2 University of California — Experiment Station This circular is intended to aid commercial ice cream manufacturers in making the computations necessary for the proportioning 1 of ingre- dients in ice cream mixes. Not all the available methods for standardiza- tion and restandardization are considered in this publication; but there are definite reasons for including the methods given as can be seen from the brief advance discussion that follows. 1. Ordinarily the Pearson square method is as simple as any. With this method no formulas need be memorized. The computations, how- ever, are comparatively long, particularly when used to standardize a complex mix such as one requiring the use of evaporated or sweetened condensed milk. 2. The serum point or formula method is shorter than the Pearson square. It can be used in practically all cases and does not become un- duly complicated when used for the solution of complex mixes. Its for- mulas appear at first sight to be rather complicated, but are easily learned. 3. The main advantages of the algebraic method are accuracy and general applicability. In most cases, however, it is longer than the serum point method and, as ordinarily used, is scarcely more accurate; besides, it is often more difficult to check if an error is introduced in the calcu- lations. 4. The graphic method, although not so generally used, is fairly simple and is accurate enough for practical purposes. Its main advantage seems to lie in its easy application to restandardization problems. Those accustomed to the use of graphs should find it convenient and satis- factory. 5. Tabulated recipes (table 2) for ice cream and ice milk 3 on a 100- pound basis, will be sufficient to meet the needs* of many ice cream makers. The particular method chosen for use in restandardization will de- pend upon the type of problem under consideration. Examples are given in a later section to illustrate the application of several methods. 3 lee milk is a frozen dairy product similar in composition to ice cream except that its percentage fat content is usually much lower — the minimum legal standard in California being 4 per cent fat. Cm. 333] Standardizing Ice Cream Mixes METHODS OF MIX STANDARDIZATION PEARSON SQUARE METHOD The Pearson square method, rather generally used in standardizing milk and cream for fat content, will serve equally well in standardiz- ing mixes for milk fat and for milk-solids-not-fat. Here, however, one must make certain arbitrary divisions of the mix to facilitate calcu- lation. In this method, a square or rectangle serves as a guide to the neces- sary steps in calculating the relative amounts of two ingredients re- quired for a mixture of desired composition. Obviously the percentage composition of one ingredient must be above and that of the other must be below the desired value in the final mixture. Before proceeding to a consideration of its application to specific problems, the following working diagram is given to illustrate the basis of this method. A = Percentage of the desired constituent in the final mixture (e.g., milk fat or milk- solids-not-fat) B = Percentage of this constituent in one of the ingredients C = Percentage of this constituent in the other ingredient V = Unit parts of the ingredient in the final mixture containing the B percentage. D is a positive number and equals the difference between C and A E = Unit parts of the ingredient in the final mixture containing the C percentage. E is a positive number and equals the difference between A and B The total unit parts of the final mixture having A percentage of the desired constituent equals the sum of D and E. Consequently, the fraction of the total mixture which will be the ingredient with B per centage the weight of the ingredient with B percentage :l and 1 _ f g "I r total weight of 1 / " \_j) i e J l_the desired mixture J 4 University of California — Experiment Station Likewise, the fraction of the total mixture which ] £ will be the ingredient with C percent- }■ := , and age J D + E the weight of the ingredient with C 1 — |~ & ~| x f" total weight of percentage J |_Z) -4- E J L the desired mixture J The solution to the following problem illustrates the use of this method in standardizing for milk-solids-not-fat and for milk fat. In all problems of this type it is necessary to first standardize for the milk- solids-not-fat in the serum of the mix. Even though there are more than two sources of milk-solids-not-fat the products can ordinarily be grouped so that standardization is accomplished by considering only two sources. The reason for this is the fact that the skim-milk portions of both milk and cream are "normal" skim milk and for practical purposes both may be considered to contain the same percentages of milk-solids-not-fat. Problem: Suppose it is desired to prepare 100 pounds of mix contain- ing 12.0 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.5 per cent gelatin. Consider the following' products to be available : cream testing 40.0 per cent fat; milk testing 4.0 per cent fat; condensed skim milk containing 30.0 per cent milk-solids-not-fat; and the nonmilk products, sugar and gelatin. The steps in the calculations are given below and the results are verified in tabulated check 1 on page 6. Since the nonmilk products can be calculated directly, their weights should first be subtracted from the total weight of the mix. Thus, 100 pounds of mix minus 15.5 pounds of sugar and gelatin leaves 84.5 pounds of milk products, which, in this case, will be standardized by the Pearson square method. By temporarily eliminating from consideration the pounds of fat in the mix, the first part of the problem resolves itself into determining the percentage of milk-solids-not-fat in the serum and then using this value in the Pearson square as indicated below. Thus, 84.5 pounds of milk products minus, 12.0 pounds of milk fat equals 72.5 pounds of serum in the mix. The average percentage of milk-solids-not-fat in the serum of the mix is calculated as follows : Average percentage of ^ng^ of milk-solids-not-fat milk-solids-not-fat in _ m the serum of the mix x 100 . the serum of the mix " weight of serum of the mix Hence, L_ X 100 = 13.79 = the average percentage of milk-solids-not fat in 72.5 the serum of the mix. These calculations obviously indicate that the serum of the mix must Cir. 333] Standardizing Ice Cream Mixes contain 13.79 per cent of milk-solids-not-fat. The products available from which to supply the serum of the mix with its required percentage of serum solids are : condensed skim milk testing 30.0 per cent milk- solids-not-fat, and the skim-milk portion of the milk and cream used. The latter can for all practical purposes be considered to contain 9.0 per cent serum solids. If in any specific case the composition of the skim milk deviates from this value, the correct figure can be substituted in the calculations. Thus, if the skim milk contained 8.8 per cent milk- solids-not-fat use 0.088 instead of 0.09. (For other values, see table 3.) Follow the procedure already outlined on pages 3 and 4, but substi- tute numbers for the letters used in the general example. Thus, 30.0 4.79= unit parts of condensed skim milk Then, 4.79 + 16.21 = 21.00 II = unit parts of skim milk from milk and cream the unit parts of the mixture testing 13.79 per cent serum solids. Hence, 4.79 21.00 X 72.5 = 16.54 = pounds of condensed skim milk required. Similarly, the pounds of skim milk could be calculated, but this is sup- plied by the milk and cream and is called "skim milk" only for purposes of computation. It should be apparent, therefore, that if the milk and cream are correctly proportioned they will of necessity contain the re- quired amount of skim milk. Hence, the next calculations will deal with the proportioning of the milk and cream so as to supply the milk-fat requirements of the mix. This can be accomplished by taking into consideration the milk fat of the mix which was temporarily eliminated in the previous calculations. The average percentage of fat in the milk and cream must be evaluated and used in the rectangle as indicated below. Thus, 84.50 — 16.54 = 67.96 = pounds of milk and cream desired. The average per cent of fat in the milk and cream is calculated as fol- lows: Average per cent fat '_ weight of fat in the milk and cream in the milk and cream weight of milk and cream 17.66 per cent fat in the milk-and-cream mixture. 12.0 Hence, X 100 67.96 6 University of California — Experiment Station Again use the Pearson square method, and calculate the weight of milk and cream required. Thus, 40.0 13.06 = unit parts of cream test- ing 40.0 per cent fat 22.34 =: unit parts of milk testing 4.0 per cent fat Then, 13.66 + 22.34 = 36.00 = unit parts of mixture containing 17.66 per cent fat ; 13.66 X 67.96 = 25.79 = pounds of cream; and, 36.00 22.34 36.00 X 67.96 = 42.17 = pounds of milk. The weight of milk can also be calculated by subtracting the weight of cream already determined from the total weight of milk and cream. Thus, 67.96 — 25.79 = 42.17= pounds of milk. Finally the weights of sugar and gelatin are calculated by multiply- ing the weight of the mix by the fraction of ( , each desired. Thus, 100 X 0.15 = 15.0 =. pounds of sugar; and, 100 X 0.005 = 0.5 = pounds of gelatin. The foregoing calculations are verified in tabulated check 1. TABULATED CHECK 1 Ingredients used Weight Fat supplied Milk-solids- not-fat supplied Total solids supplied Condensed skim milk testing 30.0 per cent milk-solids-not-fat pounds 16.54 pounds pounds 4 96 pounds 4.96 Cream testing 40.0 per cent fat 25.79 10.32 1 39 11 71 Milk testing 4.0 per cent fat 42 17 1 69 3 64 5 33 Sugar 15.00 15 00 Gelatin 50 50 Totals supplied 100.00 12-01 9.99 37.50 Totals required 100.00 12.00 10.00 37 50 ClR - 333 J Standardizing Ice Cream Mixes SERUM POINT OR FORMULA METHOD It is possible by simple algebra to set up an equation for milk-solids- not-f at and one for the milk fat of the mix and from these derive general working formulas for the solution of most standardization problems. A set of such formulas, without showing derivation, is given below. In these formulas, the composition of skim milk, or the skim-milk portion of milk and cream, is taken as 9.0 per cent milk-solids-not-fat, which, as has already been explained, represents a safe average figure. B — 0.09/? (1) z = (2)Z = d — 0.09e (F — cZ) —bV a — c (3) Y = P — X In the above formulas X = pounds of cream with a fraction fat T = pounds of milk with b fraction fat Z 4 = pounds of serum-solids concentrate with c fraction fat, d fraction milk- solids-not-fat, and e fraction serum S — pounds of serum in the mix (milk products minus milk fat) containing E pounds of serum solids F = pounds of milk fat in the mix P = pounds of milk and cream, that is, the quantity (X + Y) It may clarify these formulas somewhat to represent them, respec- tively, in the following manner (with the addition of the formulas for calculating nonmilk products) : ["pounds of milk- (1) Pounds of serum- ^ ie m ^ x solids concentrate = | solids-not-fat in — 0.09 X pounds of mix serum [fraction of milk- solids-not-fat in — q the serum-solids concentrate -J L fraction of serum X in the serum-solids concentrate (2) Pounds of cream " total pounds - of fat _ in mix pounds fat ~ — in serum- solids foncentrate- — ~ fat pounds fraction X milk of and _ milk cream - fraction of fat in cream — fraction of fat in milk (3) Pounds of milk — pounds milk and cream minus pounds cream (4) Pounds of sugar c= pounds of sugar desired minus pounds of sugar added in serum-solids concentrate (5) Pounds of gelatin = weight of mix X fraction of gelatin in mix 4 Z can be condensed skim or whole milk, either sweetened or unsweetened; or whole or skim-milk powder. 8 University of California — Experiment Station Problem: Suppose 100 pounds of mix must contain 13.0 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.4 per cent gelatin. Consider the following products available : cream containing 40.0 per cent fat; milk containing 4.0 per cent fat; sweetened condensed whole milk containing 8.0 per cent fat, 20.0 per cent milk-solids-not-fat, 42.0 per cent sugar; 5 and the nonmilk products, sugar and gelatin. The mix contains 84.6 pounds milk products of which 71.6 pounds is serum. By application of the formulas given above, the following re- sults are obtained : (1) Pounds of condensed milk (2) Pounds of cream 10.00— (0.09 X 71.6) 0.20— (0.09 X 0.50) = 22.94 13.00— (0.08 X 22.94) — (0.04 X 71.29) 0.40 — 0.04 (3) Pounds of milk — 71.29 — 23.09 = 48.20 (4) Pounds of sugar = 15.00 — (22.94 X 0.42) = 5.37 (5) Pounds of gelatin = 100 X 0.004 = 0.40 Tabulated check 2 shows the figures to be correct. 23.09 TABULATED CHECK 2 Ingredients used Weight Milk fat supplied Milk-solids- not-fat supplied Sugar supplied Total solids supplied Condensed whole milk testing 8.0 per cent fat, 20.0 per cent milk-solids-not- fat, 42.0 per cent sugar pounds 22.94 pounds 1 84 pounds 4 59 pounds 9.63 pounds 16 06 Cream testing 40.0 per cent lat 23.09 9.24 1 25 10.49 Milk testing 4.0 per cent fat 48.20 1 93 4 16 6.09 Sugar 5.37 5 37 5 37 Gelatin 40 40 Totals supplied 100 00 13 01 10.00 15 00 38.41 Totals required 100 00 13 00 10 00 15 00 38.40 TABULATION AS A SHORT-CUT FOR ROUTINE CALCULATIONS In many plants a mix of uniform composition is made from day to day; in addition the composition of the serum-solids concentrate used does not fluctuate. When such is the case, the amount of serum-solids concentrate required for a given-sized batch will be constant, as also will 5 The fraction of serum in the serum-solids concentrate is calculated by subtract- ing from the whole, the fraction of fat and the fraction of added sugar. Thus, in this case 1.00 minus (0.08 fat plus 0.42 sugar) equals 0.50 serum. This equals the value of e in formula (1). Cm. 333] Standardizing Ice Cream Mixes be the amounts of nonmilk products, no matter by which method they are calculated. A table posted in the work room giving the quanti- ties of these products, along with the total weight and the average fat test of the milk and cream mixture required for the remainder of the standardization, will make unnecessary the repeated calculations for these values. Table 1 illustrates such a tabulation for specified quantities of a mix to contain 12.0 per cent fat, 10.5 per cent milk-solids-not-fat, 15.0 per cent sugar and 0.35 per cent gelatin making 37.85 per cent total solids; TABLE 1 Ingredients Kequired for Batches of Ice Cream Mix Containing 12.0 Per Cent Fat, 10.5 Per Cent Mllk-Solids-Not-Fat, 15.0 Per Cent Sugar, and 0.35 Per Cent Gelatin Mix Standardized milk-and- cream mix- ture testing 17.8 per cent fat* Condensed skim milk testing 32.0 per cent milk-solids- not-fat Sugar Gelatin pounds 100 pounds 67.43 pounds 17.22 pounds 15.00 pounds 35 200 134.86 34.44 30 00 0.70 300 202.29 51.66 45.00 1.05 400 269.72 68.88 60.00 1.40 500 337.15 86.10 75.00 1.75 GOO 404 58 103.32 90.00 2 10 700 472.01 120.54 105 00 2.45 800 539 44 137.76 120.00 2 80 900 606.87 154.98 135 00 3 15 * A milk-and-cream mixture testing 17.8 per cent fat can easily be standardized by the Pearson square method. where condensed skim milk testing 32.0 per cent milk-solids-not-fat is used as the serum-solids concentrate. Since the percentage of fat in milk and cream, respectively, can be expected to vary considerably from day to day, it usually is not practical to tabulate these for specified fat con- tents. The exact weight of each, however, can easily be calculated by the Pearson square method each time a mix is made. These are the only cal- culations required for routine standardization of such mixes. Thus, the final step in proportioning the ingredients for the problem illustrated in table 1, consists in calculating by this method the exact quantities of milk and cream with known fat contents respectively, to supply the weights of the mixture testing 17.80 per cent fat, which are indicated in column 2 of table 1 for various-sized batches. The recipes given in table 2 likewise illustrate this method. 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Suppose it is desired to make 100 pounds of mix to contain 14.0 per cent fat, 9.0 per cent milk-solids-not-fat, 14.5 per cent sugar, 0.5 per TABLE 3 Kelation of Percentage of Fat to Percentage of Other Solids in Milk* Milk fat Milk-solids- not-fat Total milk Ratio solids solids of fat to -not-fat Calculated niilk- solids-not-fat in skim-milk portion (serum) per cent 3 per cent 8 33 per cent 11.33 1 2.77 per cent 8 59 3 1 8 40 11.50 1 2.71 8 67 3 2 8 46 11 66 1 2 64 8 74 3.3 8.52 11 82 1 2.58 8 81 3.4 8 55 11 95 1 2.52 8.85 3 5 8.60 12 10 1 2 46 8 91 3.6 8 65 12.25 1 2 40 8.97 3 7 8 69 12 39 1 2.35 9 02 3.8 8.72 12 52 1 2 30 9 06 3.9 8.76 12 66 1 2 25 9.12 4 8.79 12 79 1 2.20 9 16 4 1 8.82 12.92 1 2.15 9 20 4.2 8 86 13 06 1 2.11 9 25 4.3 8.89 13.19 1 2.07 9 29 4 4 8.92 13.32 1 2 03 9.33 4 5 8.95 13 45 1 1 99 9 37 4.6 8.98 13 58 1 1 95 9 41 4.7 9 01 13.71 1 1.92 9 45 4.8 9 04 13 84 1 1.88 9.50 4.9 9.07 13 97 1 1.85 9 54 5.0 9.10 14 10 1 1.82 9.58 * The first four columns of this table are taken from: Kelley, E., and C. E. Clements. Market milk, 2nd ed., p. 42. John Wiley and Sons, New York, 1931. The values given in this table represent averages compiled from more than 250,000 milk samples. cent gelatin, and 0.5 per cent egg-yolk solids. Consider the following in- gredients available : cream testing 30.0 per cent fat; milk testing 3.6 per cent fat; evaporated whole milk containing 8.0 per cent fat and 20.0 per cent milk-solids-not-fat; egg yolk containing 50.0 per cent egg-yolk solids; sugar; and gelatin. In some cases analyses of milk and cream for both fat and total solids are made, but usually only the percentage of fat is determined. Tables Cir. 333] Standardizing Ice Cream Mixes 13 3 and 4 show the relations between fat and serum solids that are likely to occur, and the values for milk-solids-not-fat given can be used where TABLE 4 Percentages of Milk-Solids-Not-Fat in Cream Where Its Serum Contains the Percentages of Milk-Solids-Not-Fat Given in the Headings of the Eespective Columns Fat in Milk-solids-not-fat in cream where the serum contains cream 8.7 8.8 8.9 9.0* 9.1 9.2 15.0 7.40 7.48 7.57 7.65 7.74 7.82 1(3 7.31 7.39 7.48 7.56 7.64 7.73 17 7.22 7 30 7.39 7.47 7.55 7.64 18.0 7.13 7 21 7 30 7 38 7.46 7 54 19 7 05 7 13 7.21 7.29 7.37 7 45 20.0 6.96 7 04 7.12 7 20 7.28 7.36 21.0 6.87 6.95 7.03 7.11 7.19 7.27 22 6.79 6.86 6.94 7.02 7.10 7 18 23.0 0.70 6.78 6 85 6 93 7 01 7.08 24 6.61 6.69 6.76 6.84 6.92 6.99 2.5 6.53 6.60 6.68 6.75 6.83 6.90 26.0 6.44 6 51 6.59 6 66 6.73 6.81 27 6 35 6.42 6 50 6.57 6.64 6.72 28 6.26 6 34 6 41 6.48 6 55 6.62 29 6 18 6 25 6 32 6.39 6.46 6.53 30.0 6.09 6.16 6.23 6.30 6.37 6.44 31 6.00 6.07 6.14 6.21 6.28 6.35 32 5.92 5.98 6.05 6.12 6.19 6.26 33 5 83 5 90 5 96 6.03 6 10 6.16 34 5.74 5 81 5.87 5 94 6 01 6.07 35 5.66 5.72 5.79 5.85 5 92 5 98 36 5 57 5 63 5 70 5.76 5.82 5.89 37 5 48 5 54 5.61 5.67 5.73 5.80 38 5 39 5 46 5 52 5 58 5.64 5.70 39 5 31 5 37 5 43 5 49 5 55 5 61 40 5 22 5.28 5.34 5 40 5.46 5 52 41 5 13 5.19 5.25 5 .31 5 37 5 43 42 5.05 5 10 5.16 5.22 5.28 5.34 43 4 96 5 02 5.07 5 13 5.19 5.24 44 4 87 4 93 4.98 5 04 5 10 5 15 45 4 79 4.84 4.90 4.95 5 01 5 06 46 4 70 4.75 4.81 4.86 4.91 4.97 47 4 61 4.66 4 72 4.77 4.82 4.88 48 4 52 4 58 4 63 4.68 4 73 4.78 49 4 44 4 49 4.54 4 59 4.64 4.69 50 4 35 4 40 4 45 4.50 4 55 4 60 55 3.92 3 96 4 01 4 05 4 10 4 14 60 3.48 3 52 3.56 3.60 3.64 3.68 * A safe average for practical purposes is 9.0 per cent milk-solids-not-fat in the serum of milk and cream. Hence, the values in this column should be used unless it is known that some other value more nearly represents the composition of the sample under con- sideration. analytical data are lacking. If in any case it is known that some other values more nearly represent the compositions of the products, ob- viously they should be used. Id University of California — Experiment Station The amounts of nonmilk products are found as follows : 0.145 X 100 pounds = 14.5 pounds sugar 0.005 X 100 pounds = 0.5 pounds gelatin 0.005 X 100 pounds — = 1.0 pounds egg yolk (contains 0.5 pound egg yolk solids) Hence, (14.5 + 0.5 + 1.0) pounds = 16.0 pounds of nonmilk products; and (100.0 — 16.0) pounds = 84.0 pounds of milk products. In order to solve for the amount of each of the milk products, they will be designated as follows : X = pounds of 30.0 per cent cream required Y = pounds of 3.6 per cent milk required Z = pounds of evaporated milk required Three equations using' these three unknowns follow : (a) Equation for total milk products: X + Y + Z = 84.0 pounds (b) Equation for fat in mix: 0.30Z + 0.036F + 0.08Z= 14.0 pounds (c) Equation for serum solids in mix: 0.063X -f 0.0865F -f- 0.20Z = 9.0 pounds These equations can be solved by eliminating one or more unknowns at a time; either by combining two equations by addition or subtraction, or by substituting the value for one unknown in another equation. These are commonly called the addition or subtraction method, and the sub- stitution method, respectively. Usually with three unknowns the first method is preferable, although sometimes a combination of the two meth- ods makes the calculations easier. Only the first method will be illus- trated here. In this case, one of the equations will have to be multiplied by a factor such that the coefficients of one of the unknowns in two equa- tions are the same. Thus, if equation (c) above is multiplied by 5, the coefficient of Z becomes the same as the coefficient of Z in equation (a) . Subtracting, (a) Z+ Y 4- 2 = 84.00 (d) 0.315X + 0.4325Y -f Z = 45.00 (Equation c X 5) (e) 0.685Z + 0.5675F = 39.00 (Equation a — equation d) Similarly, Z can be eliminated from equations (a) and (6) if equation (a) is multiplied by 8, and equation (ft) by 100. Subtracting, (/) 30Z + 3.6Y + 8Z= 1,400.00 (Equation b X 100) (g) 8X + 8.0Y 4- 8Z — 672.00 (Equation a X 8) (h) 22X — 4.4F =,728.00 (Equation/ — equation #) 6 If the condensed milk used contains sugar, proper allowance should be made f or this in the equations. Thus if it contains 42.0 per cent sucrose, equation (a) above would become X + Y + 0.58Z = 84.0. Cm. 333] Standardizing Ice Cream Mixes 15 One more unknown can be eliminated from equations (e) and (h) by dividing each equation by its respective coefficient of X. Subtracting, (i) X + 0.8285Y = 56.934 (Equation e -r- 0.685) (j) X—0.2000r = 33.091 (Equation h -r- 22.0) (Jc) 1.0285F = 23.843 (Equation i — equation j) Solving equation (Jo) for Y : Y — 23.182 Substitute this value of Y in equation (7i) and solve for X; 22X — 4.4 X 23.182 = 728.000 22X= 830.001 X— 37.727 Substitute values for X and Y in equation (a) and solve for Z : 37.727 + 23.182 + Z — 84.0 Z — 23.091 The results are verified as shown in tabulated check 3. TABULATED CHECK 3 Ingredients Weight Milk fat supplied Milk-solids- not-fat supplied Total solids supplied Sugar pounds 14.50 pounds pounds pounds 14.50 Gelatin 50 50 Egg yolk 1 00 50 Milk testing 3.6 per cent fat 23.182 835 2.005 2.84 Evaporated milk testing 8.0 per cent fat, 20.0 per cent milk-solids-not-fat 23.091 1.847 4.618 6.465 Cream testing 30.0 per cent fat 37.727 11.318 2.377 13 695 Totals supplied 100 000 14 000 9.000 38.500 Totals required 100 000 14 000 9.000 38 500 GRAPHIC METHOD OF STANDARDIZATION The graphic method of standardizing ice cream mixes, introduced by J. A. Cross 7 and later reported by W. V. Price 8 as applicable for re- standardization as well as for original standardization problems, should prove satisfactory for those accustomed to the use of graphs. The method is simple, and the calculations required are relatively few. Inaccuracies 7 Mojonnier, T., and H. C. Troy. Technical control of dairy products. Mojonnier Brothers Company, Chicago, p. 416-418. 1922. s Price, W. V. A graphical method of proportioning and standardizing ice cream mixes. Journal Dairy Science 10:292-299. 1927. 16 University of California — Experiment Station introduced by plotting and reading of the graph need not be great enough to make the method impracticable for general use. In fact, its simplicity and fair degree of accuracy and especially its adaptability for restandardizing off-batches recommend its more general use. The method can perhaps best be explained by an illustration. Sup- pose it is desired to make 100 pounds of mix to test 13.0 per cent fat, 10.5 per cent milk-solids-not-fat, 14.6 per cent sugar, and 0.4 per cent gela- tin; and the products available are sugar, gelatin, cream testing 33.0 per cent fat, milk testing 3.6 per cent fat, and evaporated milk testing 8.0 per cent fat and 20.0 per cent milk-solids-not-fat. The calculations for the nonmilk products are the same as in any of the other methods previously considered. Thus, 0.146 X 100 pounds = 14.6 pounds sugar; and, 0.004 X 100 pounds — 0.4 pound gelatin. Hence, (14.6 + 0.4) pounds i= 15.0 pounds of nonmilk products; and, (100.0 — 15.0) pounds — 85.0 pounds of milk products. In order to solve the problem by this method, first determine the per- centage of fat and the percentage of milk-solids-not-fat in the milk products. Thus, 13.0 X 100 = 15.29 per cent fat; and 85 10.5 X 100 — 12.35 per cent milk-solids-not-fat. 85 Using coordinate paper as indicated in figure 1, let the horizontal axis represent the percentage of milk-solids-not-fat, and let the vertical axis represent the percentage of fat. Now, on this graph, plot point A, using the values just calculated to represent the composition of the milk prod- ucts portion of the mix to be standardized. In addition to using the data given in the problem, refer to tables 3 and 4 for the percentages of milk- solids-not-fat in the milk and cream; then using these values, plot points representing the composition of the three milk products available for standardization: B representing the evaporated milk; C the milk; and D the cream. Obviously, the line segment CD represents the composition of all pos- sible mixtures of milk and cream represented by the points C and D. Now the line BA, when extended, will intersect the line CD at E, repre- senting the composition of the milk-and-cream mixture, which when combined in the correct proportion with the evaporated milk of B com- position, will result in a mixture of A composition. 9 The other desig- s When used for routine standardization the work can be simplified if a large sheet of coordinate paper is used and the necessary points represented by pins. Lines can be indicated by thread stretched between the pins. Cm. 333] Standardizing Ice Cream Mixes 17 >J^> r> P r 1 30 es <£ eo V 5 n •K) 1 c j i pf- $" -A y o 1 / / V f /o % c 4 r 1 > 8 - s bC n o s/o/5 eo as Per cent n? ilk - solids - not- fat f serum so/ids) Fig. 1. — Graph for standardizing mixes and restandardizing off -batches. A, Composition of dairy products portion of desired mix. B, Composition of condensed milk available. C, Composition of milk available. D, Composition of cream available. E, and G, Composition of two milk-and-cream mixtures determined graphically. F 1} F 2 , and F 3 , Compositions of the dairy products portions of three off -batches. 18 University of California — Experiment Station nated points in figure 1 can be ignored for the present but will be used later in a problem of restandardization presented on pages 27 and 28. The problem can be finished in either of two ways : Procedure 1 : Using the Pearson square, standardize in each case for either the milk fat or the milk-solids-not-fat. Usually it is preferable to standardize on the basis of the fat as illustrated in the problem below. Determine the pounds of evaporated milk by taking into considera- tion the fact that evaporated milk of B composition contains 8.0 per cent fat and the milk-and-cream mixture of E composition contains 20.2 per cent fat, as can be seen by referring to the graph. Tims, 20.2 7.29 = unit parts of milk and cream mixture (E) 4.91 = unit parts of evaporated milk (B) Hence, (7.29 + 4.91) = 12.20 = total unit parts of mixture containing 15.29 per cent fat; and, (A) 4.91 the pounds of evaporated milk = X 85.0 = 34.21. Therefore, the pounds of milk-and-cream mixture = (85.00 — 34.21) = 50.79. Next determine the pounds of milk and cream required to make 50.79 pounds of the mixture containing 20.20 per cent fat. Thus, 33.0 K — — s, 16.60 = unit parts of cream (D) 12.80 = unit parts of milk (C) Hence, (16.60 -f- 12.80) = 29.40 = total unit parts of milk-and-cream mixture; and, (E) 16.60 the pounds of cream = X 50.79 = 28.68. 29.40 Therefore, the pounds of milk = 50.79 — 28.68 = 22.11. Cm. 333] Standardizing Ice Cream Mixes 19 Procedure 2: The lengths of line segments on the graph can be used to determine the proportion of the ingredients required. The unit of length chosen is immaterial as long as it is uniform throughout the cal- culations, but usually the most convenient for this purpose is the unit of space on the graph. The length in any case can be obtained by using dividers and reading the length off the base line or by simply subtract- ing on one of the scales the difference between two points; for example, length of line segment CD = (33.00 — 3.60) =29.40; length of line segment AB = (15.29 — 8.00) = 7.29. By this last method of determin- ing the length, the pounds of each ingredient can be calculated as follows : length of line EA Pounds of evaporated milk = ; X pounds of milk products; length of line EB 4.91 12.20 X 85.0 == 34.21 Pounds of milk-and-cream mixture = (85.00 — 34.21) = 50.79 length of line CE Pounds of cream e= X pounds of milk and cream length of line CD 16.60 X 50.79 = 28.68 29.40 Pounds of milk == (50.79 — 28.68) = 22.11 When this method of determining the length of the line segment is used, the calculations are identical with those by the Pearson square method, except that ordinarily the measurements cannot be scaled so ac- curately as they can be computed. Tabulated check 4 shows that the calculations are correct for prac- tical purposes. TABULATED CHECK 4 Ingredients Weight Milk fat Milk-solids- not-fat Total solids Sugar pounds 14.60 pounds pounds pounds 14.60 Gelatin 0.40 40 Evaporated milk testing 8.0 per cent fat and 20.0 per cent milk-solids-not-fat 34.21 2.74 6.84 9 58 Cream testing 33.0 per cent fat 28.68 9 46 1.73 11 19 Milk testing 3.6 per cent fat 22 11 0.80 1.91 2.71 Totals supplied 100 00 13.00 10 48 38 48 Totals required 100.00 13.00 10 50 38 50 20 University of California — Experiment Station STANDARDIZING MIXES TO BE MADE IN A VACUUM PAN The calculations for standardizing mixes to be prepared in a vacuum pan are fewer than in the more usual process where a serum-solids con- centrate is used. The weights of nonmilk products are calculated for a definite finished weight of mix. Then the composition of a mixture of fat milk and cream having the correct ratio of ~ ttz ~ is calcu- milk-solids-not-iat lated, or determined by reference to a table such as table 5 or a graph such as figure 2; and finally the weight of this mixture required is calcu- lated by the Pearson square method. The following example will illus- trate this procedure. Problem: Determine the amounts of ingredients required to prepare 1 ,000 pounds of mix, the latter to be made in a vacuum pan. The desired composition of the mix is 14.0 per cent fat, 10.0 per cent milk-solids-not- fat, 14.7 per cent sugar, and 0.3 per cent gelatin. Milk testing 3.7 per cent fat and cream testing 40.0 per cent fat are available in addition to sugar and gelatin. The order in which the ingredients are determined is of no conse- quence in this case, but for convenience the calculations are listed in steps. Calculate the amounts of nonmilk products required : 0.147 X 1,000 pounds — 147.0 pounds sugar; and 0.003 X 1,000 pounds — 3.0 pounds gelatin. fat Calculate the ratio of — ttz j . j desired in the final mix. milk-sohds-not-iat Thus, 14.0 =1.40. 10.0 Determine the percentage of fat in the mixture of milk and cream having the ratio 1.40 by referring to either table 5 or figure 2. This is found to be 11.2 per cent fat. It may also be calculated in the following manner : Let F = fat fraction of the desired mix S = milk-solids-not-f at fraction of the desired mix A = milk-solids-not-f at fraction of the serum (skim-milk portion) of tlie mixture of milk and cream X = fat fraction of the desired milk-and-cream mixture Hence, (1 — X)A — milk-solids-not-fat fraction of the milk-and-cream mixture; and F _ X T~~ (1 — X)A Solving this equation for X, AF X— S + AF VNDARDIZING IcE CREAM MlXES 21 0600 0600 Rot io of fot to /77 //A- so/ids- not -fot ( /77 , J -. /7 . / -/ LEGEND ......... w/) en the skim mi/k portion tests 6 6% sn-s-n-f. " • * » » ■' 90% - * " 9a% - Fig. 2. — Grapli showing the percentage of fat in milk-and-cream mixtures as re- r — - — i Lmilk -solids-not-fat J lated to the ratio of fat to milk solids-not-fat such mixtures. Assuming that the serum tests 9.0 per cent milk-solids-not-fat, the value of A becomes 0.09. When substituted in the preceding equation, it is as follows : _ 0.09F ~~ S +0.09^ 22 University of California — Experiment Station TABLE 5 Ratios of Fat to Milk-Solids-Not-Fat in Mixtures of Milk and Cream Per cent fat in mixture Per cent milk-solids-not-fat in serum of mixture 8.7 8.8 9.0t 9.2 Ratio of fat to milk-solids-not-fat in mixture (milk-solids-not-fat= 1) 6 0. 6.1. 6.2. 6.3. 6.4 6.5. 6.6. 6.7. 6.8. 6.9. 7.0. 7.1. 7.2. 7.3. 7.4. 7.5. 7.6. 7.7. 7.8. 7.9. 8.0 8.1 8.2. 8.3. 8.4. 8.5. 8.6. 8.7. 8.8. 8.9. 9 9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 733 747 760 773 780 800 812 825 838 852 865 879 892 932 945 959 973 986 000 013 026 040 054 068 082 096 110 122 136 150 165 179 193 207 221 234 248 263 726 0.738 0.752 0.765 0.777 0.790 0.803 0.816 0.829 842 0.856 0.868 0.881 0.895 0.908 921 0.935 948 0.962 0.975 0.988 1.001 1.015 1.029 1.042 1.056 1.070 1.083 1.096 1.110 1 124 1.138 1.151 1 165 1.179 1 193 1.206 1.220 1.234 1.248 0.717 0.730 0.743 0.755 0.768 0.781 0.794 0.807 0.820 0.832 845 0.859 0.872 0.885 0.898 0.911 0.925 938 0.950 0.963 0.977 0.990 1.004 1.017 1 031 1 044 1.058 1 070 1.084 1 097 1.111 1.125 1.139 1.152 1.166 1.180 1.193 1.206 1 220 1 234 0.709 0.722 0.735 747 760 772 785 798 810 823 836 0.849 0.862 0.875 0.888 900 0.913 927 940 953 966 979 993 1.006 1 019 1.032 1.045 1.058 1.072 1.085 1.099 1.112 1.126 1.140 1.153 1.166 1.179 1.193 1.207 1.221 702 0.714 0.726 0.739 751 0.764 0.776 0.789 0.802 0.815 0.827 840 0.853 0.865 .878 891 917 930 943 950 982 995 1.007 1 020 1.034 1.047 1.060 1.074 1 087 1.100 1.114 1.127 1.141 1.153 1.166 1.180 1.194 1.207 694 0.706 718 731 743 0.756 0.768 781 793 805 0.818 0.830 0.843 0.856 0.869 0.881 0.894 907 0.920 0.933 0.946 0.959 970 0.983 0.996 1.010 1 023 1.036 1.049 1.062 1.075 1.089 1.102 1.115 1.127 1.140 1.154 1.167 1.181 1 194 * The ratios in this table were derived by dividing the percentage of fat by the percentage of milk- solids-not-fat for each mixture. Thus the first value in the table is obtained as follows: A mixture contain- ing 6 per cent fat will contain 94 per cent serum. As indicated in the column heading, the serum contains 8.7 per cent milk-solids-not-fat. Hence, the mixture contains(0.94 X 8.7) or 8.178 per cent milk-solids-not- fat and ^=0.733. t Values given in this column should be used unless it is known for a specific case that a value given in one of the other columns more nearly represents the relation which exists between the fat and milk- solids-not-fat. Gir.333] Standardizing Ice Cream Mixes TABLE 5— (Concluded) 23 Per cent milk-solids-not-fat in serum of mixture Per cent fat in mixture 8.7 8.8 8.9 9 Of 9 1 9 2 Ratio of fat to milk-solids-not-fat in mixture ( milk-solids-not-fat = 1) 10.0 1 277 1.292 1.306 1.321 1.333 1.348 1.362 1.377 1.392 1 406 1.421 1 436 1.449 1 464 1.479 1.494 1.508 1.523 1 538 1 554 1.567 1.582 1 597 1.612 1.627 1.643 1.658 1 671 1.686 1.702 1 717 1.263 1 277 1.291 1 248 1 235 1.248 1.262 1.276 1.290 1.303 1 317 1 331 1 345 1 359 1 373 1.388 1 402 1 416 1 430 1 443 1 457 1 472 1 486 1 501 1 515 1.530 1 544 1 559 1 574 1.586 1 601 1 616 1 631 1.645 1 660 1 221 1 208 10.1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 263 277 291 305 317 332 346 360 375 389 403 418 432 445 459 474 489 503 518 533 547 562 575 590 605 620 634 649 665 680 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 235 248 262 276 290 302 316 330 344 358 372 386 400 414 429 443 455 469 484 498 513 527 541 556 570 585 599 612 627 641 1 1 1 1 1 1 1 1 I 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 221 10.2 235 10.3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 305 320 332 347 361 376 390 405 419 434 447 462 476 491 506 521 535 550 563 578 593 608 623 638 654 669 684 697 248 10.4 262 10.5 276 10.6 290 10.7 302 10.8 315 10.9 329 11.0 .. 343 11.1 . 357 11.2 371 11.3 385 11.4 399 11.5 413 11.6 .. 427 11.7 441 11.8 455 11.9 .. 467 12.0 .. 481 12.1 496 12.2 510 12.3 524 12.4 538 12.5 553 12.6 567 12.7 582 12.8 596 12.9. .. 610 13.0 625 t Values given in this column should be used unless it is known for a specific case that a value given in one of the other columns more nearly represents the relation which exists between the fat and milk-solids-not-fat. In this problem F = 0.14 and 8 = 0.10. Hence, 0.09 X 0.14 X — =0.1119 0.10 + 0.09 X 0.14 This equals 11.19 per cent, or, for practical purposes, 11.2 per cent. Then calculating the pounds of this milk-and-cream mixture required for the finished mix : 0.14 X 1,000 = 140.0 pounds fat required, or 140.00 ■= 1,250 pounds of mixture testing 11.2 per cent fat required. 0.112 Finally the pounds of milk and cream, respectively, required to pre- pare the above mixture are calculated by the Pearson square method. 24 University of California — Experiment Station 7.5 = unit parts of 40 per cent croam 28.8 = unit parts of 3.7 per cent milk (7.5 + 28.8) = 3(3.3 = the total unit parts of the mixture testing 11.2 per cent fat. 7.5 Hence, the pounds of cream testing 40.0 per cent fat := X 1,250 — 258.3, 36.3 and the pounds of milk testing 3.7 per cent f at = (1,250.0 — 258.3) = 991.7. Verification of these calculations will be found in tabulated check 5. TABULATED CHECK 5 Ingredients Weight Milk fat Milk-solids- not-fat Total solids Sugar pounds 147.0 pounds pounds pounds 147.00 Gelatin 3 3 00 Cream testing 40.0 per cent fat 258.3 103.32 13.95 117.27 Milk testing 3.7 per cent fat 991.7 36.69 86.18 122 87 Totals supplied 1,400 140.01 100.13 390.14 Totals required 1,000 140 00 100 00 390 00 As shown in tabulated check 5 the required solids for 1 ,000 pounds of mix of the desired composition are supplied by a weight of 1,400 pounds. Hence, all that remains to be done is to evaporate the amount of water represented by the difference between that desired and the mixture above — that is, 1,400 — 1,000 = 400 pounds of water to be evaporated. The correct composition is determined by using' a hydrometer that gives the specific gravity corresponding to a given total-solids content at the temperature of operation. ClR - 333 I Standardizing Ice Cream Mixes 25 CORRECTING OFF-BATCHES BY RESTANDARDIZATION Often an ice cream mix does not have the desired composition. If it deviates appreciably from the standard adopted the mix should he "re- standardized" — that is, its composition should he adjusted to the desired values. Slight variations from the desired composition ordinarily need not be corrected. The composition of the mix is usually determined by analyzing it for fat and total solids, then assuming that the nonmilk products are present in the correct proportion and computing the milk-solids-not-fat by dif- ference. To illustrate this point, suppose a mix was found by analysis to contain 13.0 per cent fat and 39.5 per cent total solids, whereas it would have contained 12.5 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.5 per cent gelatin, provided no mistake had oc- curred in the original standardization or processing. The sugar and gela- tin contents are assumed to be correct, since, generally speaking, errors are less likely to occur in the sugar and gelatin contents of the mix than they are elsewhere. Hence the percentages of fat in the off -batch and the desired percentages of sugar and gelatin will all be subtracted from the total solids in the off-batch, as a basis of calculating the milk-solids-not- fat contained in the batch under consideration. Thus, 39.5 — (13.0 + 15.0 + 0.5) — 11.0 —per cent of milk-solids-not-fat in the off -batch. Mixes can be restandardized by adding the minimum weight of in- gredients to correct the off-batch or by simply adding a convenient weight of ingredients, provided the latter is large enough to permit the desired correction in composition to be made. In some cases there is a decided advantage in a minimum-sized final batch; in others, there is little or no choice between the two methods of deciding the weight of the final mix, except for convenience in calculating. RESTANDARDIZING ON ARBITRARY-WEIGHT BASIS In case the mix is restandardized by the use of an arbitrary quantity equaling or exceeding the minimum required, almost any method used for the original standardization can be employed. In such a case the off- batch is considered one of the ingredients that must be completely util- ized. The remainder of the mix can be adjusted, and calculations for its standardization made, so that when the two portions are combined the final mixture will have the desired composition. The following problem will illustrate restandardization where the mix is increased by an arbitrary quantity. 26 University of California — Experiment Station Problem: Restandardize 100 pounds of mix testing 12.5 per cent fat and 39.0 per cent total solids so that the final mixture will contain 14.0 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.5 per cent gelatin. The milk products available are cream testing 30.0 per cent fat, milk testing 3.6 per ceut fat, and condensed skim milk test- ing 32.0 per cent milk-solids-not-fat. Arbitrarily assume that a total of 30 pounds will be added to correct the off-batch. Tabulated check 6 summarizes the problem. TABULATED CHECK 6 Fat Milk-solids- not-fat Sugar Gelatin Total solids Amount required for 130 pounds cor- rected mix pounds 18.20 pounds 13.00 pounds 19 50 pounds 0.65 pounds 51.35 Amount supplied by 100 pounds of the off-batch 12.50 11.00 15.00 50 39 00 Difference to be supplied in 30 pounds added for restandardization 5 70 2 00 4 50 0.15 12 35 Any one of the first four methods already illustrated in the section de- voted to standardization can be used in solving this problem. Only the solution by the serum point method will be given here. Tabulated check 6 shows that the correcting batch of 30.0 pounds must contain 4.65 pounds of nonmilk products and 25.35 pounds of milk products. The fat present in these added milk products amounts to 5.70 pounds; consequently, they will contain 19.65 pounds of serum which is the difference between the weight of milk products and the fat contained therein. Proceed by substituting the correct values in the serum point formu- las given on page 7 : 2.00 — 0.09 X 19.65 Pounds condensed skim milk required = Pounds milk and cream 0.32 25.35 — 1.01 = 24.34 0.09 1.01 5.70 — 0.036 X 24.34 Pounds cream s= = 18.27 0.30—0.036 Pounds milk == 24.34 — 18.27 = 6.07 Tabulated check 7 verifies these calculations. Where this method is employed, the procedure is the same whether the milk fat or milk-solids-not-fat in the off -batch are both too high, or both too low, or whether one is too high and the other too low. The only pre- requisite is that the correcting batch equal or exceed the minimum re- quired in any case. Cir. 333] Standardizing Ice Cream Mixes TABULATED CHECK 7 27 Ingredients Weight Fat Milk-solids- not-fat Total solids Off-batch pounds 100 00 pounds 12.50 pounds 11.00 pounds 39 00 Sugar 4 50 4 50 Gelatin 15 0.15 Condensed skim milk testing 32.0 per cent serum solids 1 01 32 0.32 Cream testing 30.0 per cent fat 18.27 5.48 1.15 6.63 Milk testing 3.6 per cent fat 6.07 0.22 53 75 Totals supplied 130.00 18.20 13 00 51 35 Totals required 130 00 18.20 13.00 51.35 RESTANDARDIZING ON MINIMUM- WEIGHT BASIS Ordinarily, though not always, it is an advantage to correct an off- batch by increasing the weight as little as possible. Whether or not this makes the problem more difficult will depend upon the correction to be made and upon the method employed for restandardization. The graphic method permits the worker to restandardize on the basis of a minimum weight more easily than on the basis of any larger quan- tity, and to determine readily the ingredients required. The algebraic method ordinarily involves no additional calculations to determine the minimum weight of the correcting batch, although in some cases it is not readily apparent which ingredients will be required for restandardization. Formulas derived algebraical^ can likewise be used. A few of these are given later, and the methods of solution by their use illustrated. Graphic Method of Restandardization. — In order to illustrate this method, reference will be made to figure 1. Suppose that F 1 represents the composition of the dairy-products por- tion of the mix, where A is the desired composition. If a line drawn through the points F x and A is extended to intersect line CD at G, then G represents the composition of the milk-and-cream mixture which, when combined with F 1 in the proper proportion, will give a mix of the desired composition A. If F 2 represented the composition of the off- batch, then a mixture of cream and evaporated milk would be required for restandardization; whereas if the off-batch were of F 3 composition, a mixture of milk and evaporated milk would be necessary to adjust it to 28 University of California— Experiment Station the desired composition. The method of restandardization in all three cases would be the same; hence only the solution for F^ will he given. By referring to figure 1 and reading from the graph the values repre- senting the compositions of the ingredients designated by the respective points under consideration, it will be seen that : F lf the composition of the dairy products portion of the off-batch, contains 16.50 per cent fat and 13.50 per cent milk-solids-not-fat, which is equivalent to 14.03 per cent fat and 11.48 per cent milk-solids-not-fat for the off -batch as a whole. A, the desired composition of the dairy products portion of the mix, contains 15.29 per cent fat and 12.35 per cent milk-solids-not-fat, which is equivalent to 13.00 per cent fat and 10.50 per cent milk-solids-not-fat in the final batch. G, the composition of the milk and cream mixture which is to be com- bined with the dairy products portion of the off -batch (F x ), contains 10.67 per cent fat and 8.00 per cent milk-solids-not-fat, as can be seen by referring to the graph. unit parts of off-batch 10.67 1.21 = unit parts of mixture of milk and cream Then, 4.62 + 1.21 = 5.83 = total unit parts of mixture testing 15.29 per cent fat, 4.62 _ Hence, _ 5.83 100.00 pounds 0.79245 0.79245 == fraction that the off -batch will be of the final weight of the mix; and, := 126.19 pounds = weight of the final batch. It is more convenient to combine the two previous steps as follows : 5.83 — — X 100 pounds — 126.19 pounds — weight of final batch. Consequently, (126.19 — 100.00) pounds = 26.19 pounds == weight of minimum cor- recting batch. Then, 0.146 X 26.19 pounds = 3.82 pounds of sugar required; and 0.004 X 26.19 pounds = 0.10 pound of gelatin required. Hence, 26.19 pounds — (3.82 + 0.10) pounds = 22.27 pounds — weight of milk-and- cream mixture re- quired. Cir. 333] Standardizing Ice Cream Mixes 29 Next determine the pounds of milk and cream respectively required to make 22.27 pounds of the mixture containing 10.67 per cent fat. Thus, 33.0 k „ 7.07 = unit parts of cream (D) J2.33 = unit parts of milk (C) Then, (7.07 + 22.33) = 29.40 = total unit parts of milk-and-cream mixture testing 10.67 per cent fat; and 7.07 the pounds of cream = X 22.27 = 5.36. 29.40 Therefore, the pounds of milk = 22.27 — 5.36 = 16.91. Tabulated check 8 verifies the above calculations. TABULATED CHECK 8 Ingredients Weight Fat Milk-solids- not-fat Total solids Off-batch pounds 100 00 pounds 14 03 pounds 11.48 pounds 40.51 Sugar 3.82 3 82 Gelatin 0.10 0.10 Cream testing 33 percent fat 5 36 1.77 32 2 09 Milk testing 3.6 per cent fat 16 91 61 1 47 2 08 Totals supplied 126 19 16 41 13 27 48 60 Totals required 126.19 16 40 13 25 48 57 On a minimum-weight basis, unless the graphic method is employed, it is ordinarily somewhat easier to restandardize a mix containing the fat and milk-solids-not-fat in excess of that desired, than to restandard- ize one with the fat and serum solids, both too low, or with one of them too low and the other too high. Here the problem consists in getting the fat and milk-solids-not-fat in the desired ratio and then diluting with water to the correct final weight, after considering the required amount of nonmilk products. This can be done by using only one dairy product, 30 University of California — Experiment Station either a fat concentrate, such as cream, or a serum-solids concentrate, such as condensed milk, according to the problem at hand. If the worker desires to use more than one dairy product in any one case, he can then restandardize without the addition of water. Both types of problem will be illustrated in the following sections. Formula Method of Restandardization. — The following formulas may be used for calculating the amount of cream or condensed milk required for restandardization where the fat and milk-solids-not-fat are both too high and where water will be used for part of the dilution. First case : Where the fat is not proportionately as high as the milk- solids-not-fat, obviously cream or some other fat concentrate will be necessary for restandardization. per cent fat Let -Kj = in the desired batch; per cent milk-solids-not-fat then, pounds of cream [7?, X pounds r pounds milk-solids-not-fat — fat in in off -batch |_off-bateli J [per cent ~j R t X per cent fat in — milk-solids-not-fat cream J |_ in cream J X 100. Problem: Suppose 100 pounds of a mix testing 13.0 per cent fat and 41.0 per cent total solids is to be restandardized on a minimum-weight basis to contain 12.0 per cent fat, 10.0 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.5 per cent gelatin. The products available for re- standardization are cream testing 30.0 per cent fat, sugar, gelatin, and water. The gelatin and sugar contents in the original mix are assumed to be correct. The steps in calculating the solution to this problem are given below : Calculate the per cent of milk-solids-not-fat in the off-batch by sub- tracting from the per cent total solids the percentages of fat, sugar, and gelatin. Thus, 41.0— (13.0 + 15.0 + 0.5) =12.5 = per cent milk-solids-not-fat in the off-batch. Calculate the ratio of fat to serum solids in the desired mix. Thus, 12.0 E t = — —=1.2. 10.0 Calculate the pounds of cream required for restandardization by using the formula given above for this case. Thus, (1.2 X 12.5) —13.0 pounds cream = X 100 = 8.91 pounds cream. 30— (1.2 X 6.3) Cir. 333] Standardizing Ice Cream Mixes 31 Calculate the final weight to which the mix must be diluted in either of the following ways : Divide the total pounds of fat supplied, by the fraction of fat desired in the corrected mix; or, divide the total pounds of milk-solids-not-fat by the fraction of milk-solids-not-fat desired in the corrected batch. Inasmuch as the method of calculation is the same in both cases, only the first way will be used here. Thus, 100 pounds X 0.13 = 13.00 pounds fat 8.91 pounds X 0.30 = 2.67 pounds fat Hence, Total supplied == 15.67 pounds fat 15.67 pounds 0.12 = 130.58 pounds = final weight of mix. The additional weights of sugar and gelatin required are calculated in the usual manner. Thus, (130.58 — 100.00) pounds = 30.58 pounds = weight of correcting batch. Then, 30.58 pounds X 0.15 == 4.59 pounds sugar; and 30.58 pounds X 0.005 = 0.15 pound gelatin. The weight of water for the final dilution equals the difference be- tween the total weight desired and the weight of the ingredients re- quired to supply the necessary solids in the mix. Thus, 30.58 pounds — (8.91 + 4.59 + 0.15) pounds :== 16.93 pounds of water. These calculations are verified in tabulated check 9. TABULATED CHECK 9 Ingredients Weight Fat Milk-solids- not-fat Total solids Off-batch testing 13.0 per cent fat and 41.0 per cent total solids pounds 100 00 pounds 13 00 pounds 12 50 pounds 41 00 Cream testing 30.0 per cent fat 8 91 2.87 56 3 23 Sugar 4 59 4 59 Gelatin 15 15 Water 16.93 Totals supplied 130 58 15 67 13.06 48 97 Totals required 130 58 15 67 13.06 48.97 32 University of California — Experiment Station Second case: Where the milk-solids-not-fat are not proportionally as high as the fat, condensed milk or some other serum-solids concen- trate will be necessary to adjust the mix to the desired composition. per cent milk-solids-not-fat Lot En — in the dosired batch. per cent fat [B 2 X pounds " |~ pounds milk- ~j fat in — solids-not-fat off -batch J |_ in off -batch J [~por cent milk- - ! [~ solids-not-fat in condensed L milk J L _. ■ .. , oft-batch I m oft -batch J ihen the pounds of condensed milk = X 100. "~ Bo X per cent "* fat in condensed milk J L milk Problem: Suppose 100 pounds of a mix testing 15.0 per cent fat and 41.0 per cent total solids is to be restandardized on a minimum-weight basis to contain 12.5 per cent fat, 10.0 per cent milk-solids-not-fat, 14.5 per cent sugar, and 0.5 per cent gelatin. The gelatin and sugar contents in the original batch are assumed to be correct. The milk product avail- able for restandardization is evaporated milk testing 8.0 per cent fat and 20.0 per cent milk-solids-not-fat. The steps in the restandardization are as follows : Calculate the per cent milk-solids-not-fat in the off-batch by differ- ence as in the first case (page 30) . Thus, 41.0 per cent — (15.0 4- 14.5 4- 0.5) per cent= 11.0 per cent. The ratio of milk-solids-not-fat to fat is calculated for the desired mix. Thus, 10.0 K. = = 0.80. " 12.5 Calculate the pounds of evaporated milk required by using the for- mula given for this case. Thus, (0.80 X 15.0) —11.0 pounds evaporated milk = X 100 = 7.35 pounds. 20.0 — (0.80 X 8.0) Ascertain the final weight to which the mix must be diluted by either of the methods suggested in the first case. The calculations here will be made on the basis of the fat content. Thus, 100 pounds X 0.15 = 15.00 pounds fat 7.35 pounds X 0.08 = 0.59 pound fat Total fat in mix ;= 15.59 pounds fat 15.59 pounds Hence, final weight of mix = = 124.72 pounds. S 0.125 Cm. 333] Standardizing Ice Cream Mixes 33 The additional weights of sugar and gelatine required are calculated as usual. Thus, (124.72 — 100.00) pounds = 24.72 pounds — weight of correcting batch. Then, 24.72 pounds X 0.145 = 3.58 pounds sugar; and 24.72 pounds X 0.005 = 0.12 pound gelatin. The weight of water for the final dilution is calculated by difference, as in the first case. Thus, 24.72 pounds — (7.35 + 3.58 -f 0.12) pounds = 13.67 pounds = weight of water required. Tabulated check 10 verifies the preceding calculations. TABULATED CHECK 10 Ingredients Weight Fat Serum solids Total solids Off-batch testing 15.0 per cent fat and 41.0 per cent total solids pounds 100.00 pounds 15 00 pounds 11 00 pounds 41 00 Evaporated milk testing 8.0 per cent fat and 20.0 per cent milk-solids-not-fat 7.35 59 1 47 2.06 Sugar 3 58 3.58 Gelatin 12 12 Water 13.67 Totals supplied 124.72 15.59 12.47 46.76 Totals required 124.72 15 59 12.47 46.77 Algebraic Method of Re standardization. — When the algebraic method is used for restandardization on a minimum-weight basis without dilu- tion with water, the principle is the same whether both fat and serum solids in the off-batch are too high or too low, or whether one is too high and the other too low. The first step is to determine which of the dairy products available will be required for restandardization. Then set up simultaneous equations using the unknowns and solve as previously indicated. As an illustration of this method, suppose it is desired to restandard- ize 100 pounds of mix testing 11.0 per cent fat, 10.0 per cent milk-solids- not-fat, 15.0 per cent sugar, and 0.4 per cent gelatin so that the final batch will contain 13.0 per cent fat, 10.5 per cent milk-solids-not-fat, 15.0 per cent sugar, and 0.4 per cent gelatin. The dairy products avail- able are cream testing 40.0 per cent fat, milk testing 4.0 per cent fat, and condensed skim milk testing 32.0 per cent serum solids. 34 University of California — Experiment Station Obviously, in this case, only cream and condensed skim milk will be required for the restandardization, as far as dairy products are con- cerned. Just which products will be required for the restandardization is not always so apparent, although by use of the graph illustrated in figure 1, this can easily be determined in all cases. The unknowns are as follows : X = weight of cream required Y = weight of condensed skim milk required Z = minimum weight of the final batch The equations will then be (a) 0.846Z = X + Y + 84.60 (milk products equation) (ft) 0.105Z = 0.054X + 0.32Y + 10.00 (milk-solids-not-fat equation) (c) 0.130Z = 0.40X + 11.00 (fat equation) The solution follows; subtracting, (d) 27.072Z = 32X + 32Y 4- 2,707.2 (Equation a X 32) (