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A Linear Programming Model Integrating Resource Allocation and Product Acceptability for Processed Cheese Products
A Linear Programming Model Integrating Resource Allocation and Product Acceptability for Processed Cheese Products KAY L. CRAIG, JOHN P. NORBACK, and MARK E. JOHNSON Department of Food Science University of Wisconsin Madison 53706
Linear programming is a tool of operations research with many well-documented applications in the food industry (3, 8, 15, 16, 18, 29, 30). Smith et al. (30) have described linear programming's food formulation optimization process as "an evolutionary process that allows the manufacturer to rapidly review a large number of formulations and modify ingredients and constraints where necessary." The format of a linear programming model involves an objective function:
ABSTRACT
A linear programming model integrates cheese manufacturing, blending, and aging and flavor components for a processed cheese manufacturing operation. The input resources for the process cheese product are idenified and defined. The constraints define the desired product using legal, quality, and management guidelines. The objective function maximizes the net returns of a pasteurized process cheese batch. Several scenarios are analyzed and compared. The critical factors that have high impact on costs and profits are identified and their economic influence quantified.
Z=CX
subject to a set of contraints: Ax___b
INTRODUCTION
Specialization in dairy product manufacture has produced more easily managed production system, but in the case of processed cheese production the result is not necessarily the most efficient use of milk resources. An integrated approach to managing cheese making resources allows the processor to get the most from incoming materials. In this case "integrated" means having some control of the manufacture or purchase of intermediate products used in processed cheese manufacture, including especially the incoming natural cheeses that may be used in this processing. Controlling the manufacture of these products for overall organization objectives can lead to manufacturing efficiencies and to increased profitability for the organization (14). Received September6, 1988. Accepted May 22, 1989. 1989 J Dairy Sci 72:3098-3108
where: z = the value of the objective function, e.g., maximizing a net return, minimizing costs, and maximizing a cheese yield; c = the vector of net returns, resource costs, or cheese yield associated with one unit of each activity; x = the vector of activities whose component levels in the optimal resource allocation are to be solved (also called the decision variables); A = the matrix of technical coefficients that relate resource use to resource constraints; and b = vector of constraint values that are limits on the amount of resources into and out of the system (19). The emphasis of the published literature on process cheese typically has focused on the manufacturing process. The study of the contribution of the ingredients has generally been limited to their principle effects on melting properties (13, 25), textural (9, 21, 33) and flavor (2) contributions, microorganism contamination (17, 31), and browning problems (4, 5). There are several good reviews of the industry (26) and of the manufacturing pro-
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cess (36). The economies involved in the manufacture of natural cheese and the manufacture of process cheese have traditionally been analyzed independently. A linear programming model is described herein that integrates these two production systems for the purpose of maximizing net returns from natural cheese and processed cheese products. Although the usual restrictions of a linear programming model apply (14), the advantage of developing such a model is that machine optimal solutions provide a place to begin the analysis and adjustment of production plans and methods. Further, the model provides tools that make the economic and resource use consequences of adjusting production plans available to the modeler (28). DEVELOPING THE MODEL
An integrated process cheese manufacturing system combines two components: the conversion of raw materials into the output products of cheese, whey cream, and separated whey and the allocation of these intermediate products with other ingredients to achieve the pasteurized process cheese food product. In the example developed here, Table I lists some of the input resources that could be used in the manufacture of natural cheese. We will assume that Cheddar cheese is to be manufactured and some or all of it will be processed. The Cheddar cheese is to be made with a fat on a dry basis (FDB) of 53.5%. This is accomplished by regulating the casein to fat ratio to be .6925. Moisture of the cheese is assumed to remain at 37%. The values used in the cheese yield formula are 1.09 salt solids retention factor, 93% fat retention, and 96% casein retention. It is assumed that the fatcontent of whey cream removed is 45%. TABLE 1. Several input resources available for use in the manufacture of natural Cheddar cheese. Resource Silo 1 Silo 2 NDM Cream Condensed skim milk
Amount available
Fat %
Casein %
Cost/ pound
300,000 300,000 As needed As needed
3.55 3.57 1.00 45.00
2.44 2.49 28.00 1.39
.1178 .1181 .79 .80
As needed
.37
9.2
.228
TODAY
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TABLE 2. Several input resources available for use in the manufacture of process cheese. Resource
Fat %
Cheese
Determined by LP 3.5
Whey protein concentrate Emulsifiers Fat Whey cream Water
0~ 83.0 45.0 0
Moisture % 37.0 1.5 0~ 17.0 55.0 100.0
Cost/pound Determined by LP .4825 .535 1.53 .782 .01
~Assumed t o be so s m a l l as t o n o t s i g n i f i c a n t l y affect the final product. 2Foregone revenue.
Whey fat is recovered at a rate of 100%. All of the raw milk is used for cheese production. The purchasable quantities of nonmilk resources is assumed to be limitless. Potential inputs for process cheese manufacture are listed in Table 2. All of the manufactured cheese is either sold or processed. It is possible to divide the manufactured cheese, i.e., one-third can be sold and two-thirds processed or half sold and half processed. The value of the output products (cheese, whey cream, and separated whey) and the resources used for process cheese manufacture are related to the value of the final process cheese product. The end product is pasteurized process block cheese valued at a block wholesale price of $1.50/lb. Commercial available software (LINDO, Microsoft Corporation, Copywright 1984), was used for computation with the resulting model. Decision Variables
The decision variables for the natural cheese manufacture are idenified with the resources that may be used and the amount of cream that can be removed from a milk resource during a standardization (18). Other variables are the direct inputs to the process cheese food. This example's decision variables for the process cheese manufacturing operation are listed in Table 3. Constraints
Constraint. In the model presented herein the casein:fat ratio of the cheese mik is .6925. J o u r n a l o f D a i r y Science Vol. 72, N o . 11, 1989
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TABLE 3. The decision variables used in the linear programming model of the process cheese manufacturing operation. Resource
Variable
Milk (silo 1) Milk (silo 2) Cream removed from silo 1 Cream removed from silo 2 NDM Condensed skim milk Cream added to cheese milk Whey cream from cheese mfg ~(used) Whey cream from cheese mfg (sold) Cheddar cheese mfg and processed Cheddar cheese mfg and sold Cheddar cheese purchased 1 mo old Cheddar cheese purchased 3 mo old Cheddar cheese purchased 5 mo old Emulsifiers WPC Fat Water
xl x2 x3 x4 x5 x6 x7 zl z2 M S B1 B3 B5 y3 y4 y5 y6
TABLE 4. Constraints for the model. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11)
1Manufacturing. 12) A s u m m a r y o f c o n t r a i n t s f o r this m o d e l c a n be f o u n d in T a b l e 4. T h e c h o s e n r a t i o s d e p e n d greatly on the company's manufacturing conditions and the type and quality of the natural cheese. S t u d i e s h a v e s h o w n r e l a t i o n s h i p s b e t w e e n t h e c a s e i n : f a t r a t i o in m i l k , m a n u f a c turing conditions, and the resulting FDB, m o i s t u r e in t h e n o n f a t t y s u b s t a n c e ( M N F S ) , a n d fat a n d m o i s t u r e p e r c e n t a g e in t h e c h e e s e (20, 22). T h e g r e a t e r t h e r a n g e a l l o w e d , t h e g r e a t e r t h e c o n t r o l is n e e d e d d u r i n g t h e m a n u f a c t u r i n g o p e r a t i o n to m a i n t a i n a c o n s i s t e n t quality product. T h e g e n e r a l f o r m a t f o r t h e c o n s t r a i n t is: casein percentage of standardized milk fat p e r c e n t a g e o f standardized milk
=.6925
T h i s r a t i o was c h o s e n to e s t a b l i s h t h e F D B o f t h e n a t u r a l c h e e s e at 53.5%, w h i c h is t h e F D B o f u n s t a n d a r d i z e d m i l k in this e x a m p l e . O t h e r r a t i o s c a n be c a l c u l a t e d f r o m t h e c h e e s e yield f o r m u l a : FDB =
FR(F) [FR(F) + CR(C)] SR
w h e r e : F R = fat r e t e n t i o n p e r c e n t a g e d i v i d e d Journal of Dairy Science Vol. 72, No. 11, 1989
13) 14)
.0002xl + .0002x2 + .2977x3 + .2974x4 ÷ .2731x5 + .0894x6 - .2977x7 ~ 0. casein to fat ratio constraint. xl = 300,000. x2 = 300,000 use all the milk constraints. .0789xl - x3_>0 .0793×2 x4 _> 0, limiting the amount of cream removed constraints. .0977xl + .0988x2 - .747x3 - .7472x4 + .4812x5 ÷ .1588x6 + .7472x7 = M + S natural cheese manufactured constraint. .0055xl + .0056x2 - .07x3 - .07x4 + .0013x5 + 0005x6 ÷ .07x7 = Z2 + ZI whey cream yield constraint. M + B1 + B3 + B5_>.70P, where P=(M+BI+B3+B5+y3+y4+y5+y6) natural cheese in the blend constraint. y3 = .03P emulsifier constraint. y4 _< .10P whey protein concentrate constraint. .0y3 + .015y4 + .17y5 + 1.0y6 + .55Z1 + .37B1 + .36B3 + .35B5 --< .43P moisture in the processed cheese constraint. .0y3 + .035y4 + .83y5 + 0y6 + .45Z1 + .33B1 + .33B3 + .33B5 + .3371M >-- .24P fat content constraint. - .005xl .005x2 -.307x3 - .307x4 + .85x5 + .15x6 + .307x7 <-- 0 allowable solids constraint. M + B I > - - . 5 ( M + B I + B 3 +B5) M+B2--<.6(M+B1 +B3+B5) B3_>.2 ( M + B I + B 3 + B 5 ) B3 <: .35 (M + BI + B3 + B5) B5___ .15 (M + B1 + B3 + B5) B5<_.20(M+BI+B3+B5) cheese blend constraints.
by 100, F = fat p e r c e n t a g e o f s t a n d a r d i z e d milk, CR = casein retention percentage d i v i d e d b y 100, C = c a s e i n p e r c e n t a g e o f s t a n d a r d i z e d m i l k , a n d S R = salt solids r e t e n tion factor. The procedure for determining the final f o r m o f this c o n s t r a i n t is to c o n s i d e r s e p a r ately all a d d i t i o n s a n d s u b t r a c t i o n s t o t h e c h e e s e m i l k o f c a s e i n a n d fat. ( W e m a y remove cream from the batch, for instance.) T h e r e s u l t f o r this c a s e is: . 0 2 4 4 x l + .0249x2 - . 0 1 3 9 x 3 .0142x4 + .28x5+.092x6 + . 0 1 3 9 x 7 . 0 3 5 5 x l + .0357x2 - . 4 5 x 3 - . 4 5 x 4 + .01x5 + .0037x6 + . 4 5 x 7
=.6925.
OUR INDUSTRY TODAY Rearranging the equation yields: -.0002xl + .0002x2 + .2977x3 + .2974x4 + .2731x5 + .0894x6 - . 2 9 7 7 x 7 _ > 0 . The same format would be used for constraining the casein:fat ratio between two limiting values. The o p t i m u m casein:fat ratio is that which maximizes the net return of the process cheese product. Consequently, the cheese milk may or may not be standardized, depending on the profit contributions of the milk components and the natural cheese toward the final process cheese product. The casein:fat ratio has been constrained here to show the difference that can occur when natural cheese is manufactured to maximize the value of the processed cheese product. A manufacturer might test several different casein:fat ratios to discover the impact different fat cheese have on the final processed product. Constraint 2. xl -- 300,000 and x2 = 300,000. These constraints require all the milk to be used. Constraint 3. x3, x4, x5, x6 _> 0. Constraint 4..0789xl - x3 ~ 0 and .0793x2 x4 _> 0. These two constraints limit the amount of cream that can be removed from the two milk resources. Silo 1 has 3.5% fat and 45% fat cream may be removed. In this case there are 3.5/45 = .078 lb of 45% fat cream available from each pound of 3.5% milk from silo 1. Using Kerrigan and Norback's (18) version slightly modified, the constraint to limit the amount of cream removed is: -
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where: M = pounds of cheese manufactured and processed, S = pounds of cheese manufactured and sold, and M + S = total pounds of cheese produced. The coefficients on the input variables are the cheese yield per pound of resource as determined by a modified version (18) of the original formula proposed by Van Slyke and Price (35). This formula is: CY = [FR(F) + CR(C)]SR 1-W where: CY = cheese yield per 100 lb, and W = moisture percentage divided by 100. To get the coefficient of x l , the unknown amount of silo 1 milk, we compute:
CY =
[.93(3.55) + .96(2.44)]1.09 .37 1
= 9.765 lb.
-
9.765/100 lb -- .09765 lb/lb of resource. The same calculation is done for the other input resources to determine the other coefficients. Constraint 6. The whey cream by-product resulting from the cheese manufacture is: WCY =
(F - FR(F)) * W F R WF
where: WCY = whey cream yield per 100 lb., W F R = whey fat recovery percentage divided by 100, and W F = fat percentage of whey cream divided by 100.
. 0 7 8 x i > = x3, An example using silo 1 milk is as follows: and rearranged: WCY.078xl - x3~>0.
(3.55 -.93(3.55)) * 1 .45
Constraint 5. The connecting link between the raw cheese manufacture and the blending of the process cheese ingredients is the cheese yield formula represented by the equation:
Thus, the whey cream yield is .005522 lb/lb of silo 1 milk. Similar calculations are done for the other input resources. The resulting equation is:
.0977xl + .0988x2 -.7472x3 - . 7 4 7 2 x 4 + .4812x5 + .1588x6 + .7472x7 = M + S
.0055xl + .0056x2- .07x3 - . 0 7 x 4 + .0a013x5 ÷ .0005x6 + .07x7 = z2 + zl Journal of Dairy Science Vol. 72, No. 11, 1989
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where: z2 = whey cream extracted and sold, zl = whey cream extracted and used in processing, and z2 + zl = total pounds of whey cream produced from the cheese manufacture. Constraint 7. There must be a constraint to maintain a certain amount of cheese is the process cheese blend. The total process cheese batch is represented by P, where:
product. Because sensory qualities can be adversely affected by too much WPC, limits must be set (24). Management and experience play a major role in deciding the quantity to be used. In this example, W P C is limited to less than or equal to 10% of the total batch, where y4 is pounds of WPC. y4 _< .10P.
P = M + B1 = B3 + B5 + y3 + y4 ÷ y5 + y6. It is necessary to specify a limit in order to retain a product's sensory properties or as in the case of a cheese spread to satisfy the legal standards of identity. For example, a process cheese product that is required to have at least 70% cheese (minimum) will be written as follows: M + B1 + B3 + B5_>.70P. Constraint 8. It is assumed that the process cheese batch size is known and constant. Thus P = batch size of process cheese blend. In our example, P = 100,000 lb. Constraint 9.The amount and type of emulsifiers used for processing must be incorporated into the program in a similar manner as that done for cheese (Constraint 7). Numerous articles, books, and bulletins have been written about the type, quantity, and characteristics of emulsifiers in dairy products (11, 12, 23, 27, 32, 34). Some of the variables that must be considered before choosing the type and quantity of an emulsifier are cheese type, sensory characteristics of the desired finished product, melting properties, and safety considerations. Because of the sensitivity of these factors and the variability of the emulsifiers available for use, only the crudest measure for emulsifier use will be presented, a percentage of total process cheese batch: y3 = .03P As represented here, y3 equals 3% of the total process cheese batch in pounds. Constraint 10. W P C is a supplement to the raw cheese and is generally accepted for its lower cost and the improved sensory qualities it can impart to the final process cheese food Journal of Dairy Science Vol. 72, No. 11, 1989
Constraint 11. The moisture in the process cheese product must be constrained to comply with the legal standards of identity and the characteristics of the product as set by management. Because moisture adds to the yield of the product with next to no cost, the LP model will add as much moisture as it is allowed by the constraints. Consequently, only an upper limit constraint is needed. To ensure the final product has less than or equal to the legal moisture of 43% requires the following constraint: 0y3 + .15y4+ .17y5+ 1.0y6 + .55zl + .37B1 + .36B3 + .35B5 ÷ .37M <_ .43P. Constraint 12. Another constraint is necessary to ensure the fat content of the process cheese food meets the standard of identity for the specific product. Pasteurized process cheese foods must contain at least 23% fat. An extra I% is added for a safety margin. The general inequality is: pounds fat in process cheese ___.24 (total process cheese batch). The fat content of the manufactured cheese can be calculated by knowing the casein:fat ratio and a constant moisture value. Using the cheese yield formula mentioned earlier, we know the F D B is 53.5%. If the moisture is 37%, the fat percentage in the natural cheese is 33.71%. The inequality can thus be written: 0y3 + .035y4 + .83y5 + 0y6 + .45zl + .33B1 + .33B3 + .33B5 + .3371M _> .24P. Constraint 13. A constraint on the allowable solids content of the raw milk used for cheese making has been included in this model [labeled (*)]. This has been done for several reasons. If milk concentrated by reverse osmosis or ultrafiltration is used, it is
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necessary to set a limit on the solids content of the cheese milk. Using reverse osmosis, Barb a n o and B y n u m (1, 7) have shown that at about a 15% reduction in volume (about 14.17% solids content), the increased lactose of the cheese m a y b e c o m e the limiting factor in p r o d u c i n g a g o o d quality aged C h e d a r cheese. T h e y also go on to say that a low moisture barrel cheese used within 60 d m a y tolerate a reduction in volume of greater t h a n 15%. A n o t h e r i m p o r t a n t reason is that the higher cheese yield that results f r o m adding casein and fat may be economically profitable. If this is the case, it is necessary to constrain the solids content to a m a x i m u m level. In this example, the solids content is constrained to less than 12.0%. The m a n u f a c t u r e r would decide on the appropriate milk solids content for their operation. The initial inequality is: . l 1 5 x l +.115x2 - . 4 2 7 x 3 - . 4 2 7 x 4 +.97x5 + .27x6 + .427x7 _<.12(xl + x 2 - x3 - x4 + x5 + x6)
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TABLE 5. The values and costs of the input resources used in the linear programming model (per pound basis). Revenue Cost ($)~ Block processed cheese Purchased price of: Milk (silo 1) Milk (silo 2) NDM Condensed skim milk Emulsifier(s) WPC Butterfat Water Whey cream Sweet cream Direct labor and overhead: per lb natural cheese per lb processed cheese Purchase price of: 1-mo-old Cheddar cheese 3-mo-old Cheddar cheese 5-mo-old Cheddar cheese Revenue from: 1-mo-old mfg. 3 Cheddar
1.50 . ! 178 .1181 .79 .228 .535 .4825 1.53 .01 .781 .822 .25 .35 1.51 1.55 1.59 1.49
~Foregone revenue. value of resource. 3Manufacturing.
2New
Rewritten: (*) -.005x I - .005x2 - .307x3 - .307x4 + .85x5 + .15x6 + .307x7 ___0. Constraint 14. Finally, there are several constraints that must establish the acceptable age blends used in the processed cheese p r o d uct. These age blends m a y be set as absolute or given as a range. F o r example, an absolute age blend could consist of 60% l - m o - o l d cheese, 30% 3-mo-old cheese, and 10% 5-moold cheese. Alternately, an acceptable range could be 50 to 60% 1-mo-old cheese, 20 to 35% 3-mo-old cheese, and 15 to 20% 5-moold cheese. The advantage of a range of values is that it allows some flexibility in the prog r a m depending on cost, availability, grade of cheese, etc. Thus, the 1-mo-old cheese, M and B1, could be greater t h a n 50% but less t h a n 60% of the total cheese used in processing. Rewritten: M + BI ~> .5 (M + BI + B3 + B5) M + B1 <_.6 (M + B1 + B3 + B5)
Likewise, B3_> .2 (M + B1 + B3 + B5) B3_<.35 (M + B1 + B3 + B5)
and B5_> .15 (M + B1 + B3 + B5) B5 <_ .20 (M + B1 + B3 + B5).
Objective function The objective function is to maximize the net returns of a pasteurized process cheese product. Net return is defined as the difference between revenue and cost. Table 5 illustrates an example of the values and costs of various ingredients. The expression for the objective function is pieced together as net profit of process cheese minus the cost to m a n u f a c t u r e natural cheese minus the cost to Journal of Dairy Science Vol. 72, No. i 1, 1989
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buy natural cheese plus revenue from natural cheese sold. Net Process Cheese Profit. The process cheese batch is represented by the following equation: M + B1 + B3 + B5 + Y3 + Y4 + Y5 + Y6 + Z1 If the selling price of block process cheese is $1.50/lb. the expression gives the revenue from its sale: $1.50 ( M + B1 + B3 + B5 + Y3 + Y 4 + Y 5 + Y6 + Z1) The ingredient costs for the processing materials are just their purchase prices. Whey cream produced from the Cheddar cheese manufacturing process has a "cost" value calculated from its foregone fat revenue. If whey cream fat is valued at $1.73/lb and 45% of the cream is fat, then: $1.73/lb. fat * .45 fat in cream -- $.78 (value of fat/lb, cream)
cheese to allow for the option of selling the manufactured cheese or processing it. There are the ingredients costs to consider as well as the direct labor and overhead that go into the manufacturing operation. The sweet cream represented by x3 and x4 have positive "costs" since any cream removed and sold is revenue (18). The sweet cream fat is valued at $1.82/lb with 45% fat in the cream. $1.82/lb fat * .45 fat in cream -- $.82 (value of fat/lb cream) Direct labor and overhead costs are calculated on the total pounds of natural cheese produced, whether sold or processed. .11278xl + . l 1 8 1 x 2 - . 8 x 3 - . 8 x 4 + .79x5 + .228x6 + .82x7 + .25 ( M + S).
Cost to Buy Natural Cheese. Cost values are based on the estimated cost to buy Cheddar cheese of varying ages: 1.51B1 + 1.55B3 + 1.59B5.
The expression representing the cost of the processing ingredients is:
Revenue from Natural Cheese (and Whey Cream).
.535y3 + .4825y4 + 1.535y5 + .01y6 + .78zl.
1.49S + .78z2
The direct labor and overhead cost estimate is based on total pounds of process cheese produced. It is necessary to include these costs because process cheese production and natural cheese production are separate operations requiring different equipment, facilities, and labor. The inclusion of the labor and overhead estimate is written as:
Note that the sold cheese [S] value is $.02 less than the 1-mo-old cheese that could be purchased. It is assumed that it costs more to buy than is gained through sale of similar products. The objective function to be maximized is a combination of all four components:
$.35 (M + B1 + B3 + B5 + Y3 + Y4 + Y5 + Y6 + Z1). Combined, this part of the objective function is written as: 1 . 1 5 M + 1.15B1 + 1.15B3 + 1.15B5 + .615y3 + .6675y4 - .385y5 + 1.14y6 + .37zl.
Cost to Manufacture Natural Cheese. The cost to produce Cheddar cheese was not included directly in the profit from processed Journal of Dairy Science Vol. 72, No. 11, 1989
- . 1 1 7 8 x l - . 1 1 8 1 x 2 + .8x3 + .8x4 -.79x5 - . 2 2 8 x 6 - . 8 2 x 7 +.615y3 +.6675y4 - . 3 8 y 5 + 1.14y6 +.37zl + 1.24S + . 9 M - . 3 6 B 1 - . 4 B 3 -.45B5. RESULTS
Results of applying linear p r o g r a m m i n g software to the model described are presented in Tables 5-8. There are several underlying assumptions c o m m o n to all of the scenarios. First is that all raw milk resources are used in their entirety to make natural Cheddar
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T A B L E 6. Comparison of the o p t i m u m formulation and profit potential that results from modeling the natural cheese and process cheese manufacture as individual process systems versus an integrated process cheese operation. Pounds Scenario 1 (separate)
Scenario 2 (integrated)
300,000 300,000 0 0 0 0 0
300,000 300,000 0 0 0 12,386 3719
3000 9382 329 17,288 0 3,322 58,935 16,935 0 17,500 10,500
3000 9382 329 17,288 0 3582 63,680 21,680 0 17,500 10,500
Total processed batch size, lb Cheddar cheese yield, lb Total Cheddar cheese cost Cost/lb Cheddar cheese
100,000 58,935 $ 85,522 $ 1.4511
100,000 63,680 $ 92,564 $ 1.4536
Objective function value
$
$
Natural cheese input resources xl = Silo 1 milk x2 = Silo 2 milk x3 = Remove cream x4 -- Remove cream x5 -- N D M x6 = Condition skim x7 = Cream Process cheese input resources y3 = Emulsifier y4 -- W P C y5 -- Fat y6 = Water zl = Whey cream used z2 -- Whey cream sold M = Manufacturing cheese S = Cheese sold B1 -- Purchased cheese, 1 mo. B3 -- Purchased cheese, 3 mo. B5 = Purchased cheese, 5 mo.
cheese. A second assumption is that there is no waste. Third, the process cheese batch size is known and constant. Fourth, all ingredients are of known quality and composition and there is no problem with availability. Fifth, marginal revenue and costs are not applicable over the stated batch sizes. The optimum input resources to maximize the profits of the process cheese manufacturing operation, their amounts and some costs are listed in Table 6 (scenario 2). This is compared to a run using a decision support program in the unpublished program manual mentioned earlier designed by Kerrigan (scenario 1). Kerrigan's program is computer based and gives an economic evaluation of standardizing milk for cheese making. The objective is to maximize the profit of the nat-
4,365
6,805
ural cheese. This cheese was then used in the current model to formulate process cheese under the same constraints and objective function as for scenario 2. The constantly changing economic environment regularly influences the optimal solution to the model. A small change in the cost of an input resoure can dramatically change the optimal solution. An important question is how sensitive the solution is to these changes. It is important to know how close the unused decision variables (those held a value 0) came to be included in the optimal solution. If using a resource corresponding to a decision variable is obviously not profitable, then little extra effort is needed to accurately estimate its cost. Conversely, if a small change in a decision variaJournal of Dairy Science Vol. 72, No. 11, 1989
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T A B L E 7. Reduced costs associated with the decision variables from Scenario 2. Decision variables
Resource
Old coefficient
Reduced cost
New coefficient
x1 x2 x3 x4 x5 x6 y3 y4 y5 y6 x7 zl z2 S M B1 B3 B5
Silo 1 milk Silo 2 milk Removed cream Removed cream NDM Conditioned skim Emulsifiers WPC Fat Water Cream Cream used C r e a m sold Cheese sold Cheese processed l-too-old cheese 3-mo-old cheese 5-mo-old cheese
-. 11782 -.11814 .800 .800 .79 -.228 .615 .6675 -.38 1.14 .82 .37 .78 1.24 .90 .36 -.40 .45
0 0 0
- 11782 .11814 .800 .8201 -6638 .228 .615 .6675 -.38 1.14 -.82 1.0744 .78 1.24 .90 .33 -.40 -.45
.0201 .1268 0 0 0 0 0 0 .7044 0 0 0 .03 0 0
T A B L E 8. Dual prices associated with the constraints from Scenario 2. Constraint
Slack
Dual Prices
x3>0
0
- .02
x4>0 x6>O x5>0 Cream removed from x l milk Cream removed from x2 milk xl = 300,000 x2 = 300,000 Cheese yield equation
0 12,386 0 23,670 23,790 0 0 0
0 0 0 0 0
Fat in process cheese > 24%
0
Moisture in process cheese _< 43% Whey cream yield
0 0
1 -.401427 1 .429899 -.78
Solids in cheese milk _< 12% Process cheese batch variables
0 0
.072705 '-.71010
Process batch = I00,000 y4<10,000 y3 = 3000
0 617 0
Cheese = 70% of batch
0
-.769859
cheese cheese cheese cheese cheese
7000 0 3500 7000 0
0
5-mo-old cheese <_ 20% of cheese Fat in process cheese ~ 28% Casein:fat ratio = .6925
3500 4298 0
0 0
1-mo-old 1-mo-old 3-mo-old 3-mo-old 5-mo-old
cheese cheese cheese cheese cheese
>_ 50% _< 60% _> 20% _ 35% _> 15%
of of of of of
Journal of Dairy Science Vol. 72, No. 11, 1989
.007863 .009200 .2400
.68
.710101 0 -.095101
.065651 0 0 -.045701
.466183
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ble results in a new optimal solution, that decision variable and the constraints limiting its use should be analyzed. A reduced cost of a decision variable is the change in the value of the objective function for each unit increase in the value of the decision variable. Table 7 shows the reduced cost for the decision variables for scenario 2. Any decision variable that is part of the solution has a reduced cost of zero. This interpretation of a reduced cost is only valid for small changes in variable values. Most linear programming computer models supplement the solution report with a range or sensitivity analysis report. This report indicates the amounts by which unilateral changes in objective function coefficients or constraints can be altered without affecting which variables are nonzero in the optimal solution. For more detailed information, see Hillier (14) and Schrage (28). The reduced cost gives the cost, or price, as in the case of the removed cream, the resource must change before it will become a nonzero component in the optimal solution. Nonfat dry milk reduced cost of .1268 can be interpreted to mean the cost of N D M must decrease from $.79 to $.6638/lb before it would replace a resource that is already being used in the optimal formultion. The cream price would have to increase $.0201/lb, or up to $.8201, before it would be economically practical to remove it from the cheese milk. Associated with each constraint is a quantity known as the dual price. Its value is the rate at which the objective function value will improve as the right-hand side or constant term of the constraint is increased a small amount. Another way of stating this is that the dual price of a constraint is the amount by which profits will decrease if the availability of the resource associated with this constraint is reduced by one unit. Because many of the constraints have been rewritten to have the right-hand side term greater than, less than, or equal to zero, interpretations are complicated. Table 7 shows the dual prices for the constraints used in this example. The slack column lets the user know how much " r o o m " there is between the constraint and what is actually occurring in the optimum solution. For example, the constraint limiting
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the 5-mo-old cheese to less than 20% has a slack value of 3500. Looking closer, we see the constraint before forces the solution to contain at least 15% (10,500 lb) 5-mo-old cheese. There is no slack associated with this constraint. Because we are allowing a range of values, the solution could allow the 5-mo-old Cheddar to increase to 20% (or 14,000 lb). The slack is the difference between what is used and what could be used, 3500. The model is a guideline to assist in decision making. Its assumptions and solutions must be regularly tested as economic and manufacturing conditions change. H u m a n judment is needed to evaluate the proposed solution and adjust it to the specific situation,
ACKNOWLEDGMENTS
This research was supported by the Center for Dairy Research at the University of Wisconsin, and the College of Agriculture and Life Sciences, University of Wisconsin, Madison.
REFERENCES
1 Barbano, D. M. and D. G. Bynum. 1984. Wholemilk reverse osmosis retentates for Cheddar cheese manufacture: cheese composition and yield. J. Dairy Sci. 67:2839. 2 Barker, C. R. 1947. Practical suggestionson the manufacture of process cheese. Natl. Butter Cheese J. 38(1):42. 3 Bender, F. E., A. Kramer, and G. Kahan. 1982. Linear programming and its applications in the food industry. Food Technol. 36(7):94. 4 Bley, M. E., M. E. Johnson, and N. F. Olson. 1984. Factors affectingnonenzymaticbrowning of process cheese. J. Dairy Sci. 68:555. 5 Bley,M., M. E. Johnson, and N. R. Olson. 1985Predictive test for the tendency of Cheddar cheese to brown after processing. J. Dairy Sci. 68:2517. 6 Bradley, S. P., A. C. Hax, and T. L. Magnanti. 1977. Applied mathematical programming. Reading, MA. [Cited in Kerrigan, G. L. and Norback, J. P. 1986. Linear programmingin the allocation of milk resources for cheese making. J. Dairy Sci. 69:1432]. 7 Bynum,D. G.. and D. M. Barbano. 1985.Wholemilk reverse osmosis retentates for Cheddar cheese manufacture: chemical changes during aging. J. Dairy Sci. 68:1. 8 Avins,J. F., G. E. Inglett, and J. S. Wall. 1972.Linear programming controls amino acid balance in food formulation. Food Technol. 26(6):46. 9 Chen, A. H., J. W. Larkin, C. J. Clark, and W. E. Journal of Dairy Science Vol. 72, No. 11, 1989
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14 15 16 17
18 19 20 21 22 23
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Irwin. 1979. Textural analysis of cheese. J. Dairy Sci. 62:901. Cornell, G. K. 1981. Page 258 in Specifications for procurement of cheese for processing. 2nd Bienn. Marschall Int. Cheese Conference, Madison, WI. Ellinger, R. H. 1984. Fate ofpolymerized phosphates used as emulsifying salts in pasteurized process cheese products. Cheese Inst., Northbrook, IL. Gupta, S. K., C. Karahadian, and R. C. Lindsay. 1984. Effect of emulsifier salts on textural and flavor properties of processed cheese. J. Dairy Sci. 67:764. Hanrez-LaGrange, V. 1987. An optimization procedure with sensory analysis data in the development of cheese toppings. MS Thesis. Univ. Wisconsin, Madison. Hillier, F. S., G. J. Lieberman. 1980. Pages 2-4 in Introduction to operations research. 3rd ed. HoldenDay, Inc., San Francisco, CA. Hsu, H. W., L. D. Satterlee, andj. G. Kendrick. 1977. Computer blending predetermines properties of protein foods - Part I. Food Prod. Dev. 11(9):52. Hsu, H. W., L. D. Satterlee, and J. G. Kendrick. 1977. Computer blending predetermines properties of protein foods - Part II. Food Prod. Dev. 11(10):70. Karahadian, C., R. C. Lindsay, L. L. Dillman, and R. L. Deibel. 1985. Evaluation of the potential for botulinal toxigenesis in reduced-sodium processed American cheese foods and spreads. J. Food Prot. 48:63. Kerrigan, G. L., and J. P. Norback. 1986. Linear programming in the allocation of milk resourses for cheese making. J. Dairy Sci. 69:1432. Krigsman, J. G. 1985. Phosphates in food processing. Food Technol. Aust. 37:414. Lawrence, R. C., and J. Gilles. 1980. The assessment of the potential quality of young Cheddar cheese. N. Z. J. Dairy Sci. Technol. 15(1):1. Lee, C., E. M. Imoto, and C. Rha. 1978. Evaluation of cheese texture. J. Food Sci. 43:1600. Lelievre, J. 1983. Influence of the casein/fat ratio in milk on the moisture in the non-fat substance in Cheddar cheese. J. Soc. Dairy Technol. 36(4): 119. Melnchyn, P., and J. M. Wolcott. 1971. Interactions between phosphates and individual food com-
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24 25 26
27 28 29 30
31
32 33
34
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ponents. Page 49 in Phosphates in food processing. J. M. Deman, ed. AVI Pub. Co., Westport, CT. Meyer, A. 1973. Processed cheese manufacture. Food Trade Press Ltd., London, UK. Olson, N. F., and W. V. Price. 1958. A meltingtest for pasteurized process cheese spreads. J. Dairy Sci. 41:999. Price, W. V., and M. G. Bush. 1974. The process cheese industry in the United States: a review - - II Research and Development. J. Milk Food Technol. 37(4):179. Scharpf Jr., L. G. 1971. The use of phosphates in cheese processing Page 120 in in food processing. J. M. Deman, ed. AVI Publ. Co., Westport, CT. Schrage, L. 1984. Linear, integer, and quadratic programming with LINDO. The Scientific Press, Palo Alto, CA. Skinner, R. H., and G. B. Debling. 1969. Food industry applications of linear programming. Food Manuf. 44(10):35. Smith, J. P., B. Ooraikul, and E. D. Jackson. 1984. Linear programming: a tool in reformulation studies to extend the shelf life of English-style crumpets. Food Technol. Aust. 36:454. Tanaka, N., E. Traisman, P. Plantinga, L. Finn, W. Flom, L. Meske, and J. Guggisberg. 1986. Evaluation of factors involved in antibotulinal properties of pasteurized process cheese spreads. J. Food Prot. 49:526. Templeton, H. L., and H. H. Sommer. 1936. Studies on the emulsifying salts used in processed cheese. J. Dairy Sci. 19:561. Thomas, M. A., F. B. Brown, and A. T. Hanly. 1970. Studies in processed cheese manufacture - - the evaluation of physical properties. Aust. J. Dairy Technol. 25(1):46. Thomas, M. A., G. Newell, G. A. Abad, and A D. Turner. 1980. Effect of emulsifying salts on objective and subjective properties of processed cheese. J. Food Sci. 45:458. VAn Slyke, L. L., and W. V. Price. 1979. Cheese. Ridgeview PubL Co. Independence, OH. Wearmouth, W. G. 1954. Problems in the manufacture of processed cheese. Dairy Ind. 19:1016.
Linear programming is a tool of operations research with many well-documented applications in the food industry (3, 8, 15, 16, 18, 29, 30). Smith et al. (30) have described linear programming's food formulation optimization process as "an evolutionary process that allows the manufacturer to rapidly review a large number of formulations and modify ingredients and constraints where necessary." The format of a linear programming model involves an objective function:
ABSTRACT
A linear programming model integrates cheese manufacturing, blending, and aging and flavor components for a processed cheese manufacturing operation. The input resources for the process cheese product are idenified and defined. The constraints define the desired product using legal, quality, and management guidelines. The objective function maximizes the net returns of a pasteurized process cheese batch. Several scenarios are analyzed and compared. The critical factors that have high impact on costs and profits are identified and their economic influence quantified.
Z=CX
subject to a set of contraints: Ax___b
INTRODUCTION
Specialization in dairy product manufacture has produced more easily managed production system, but in the case of processed cheese production the result is not necessarily the most efficient use of milk resources. An integrated approach to managing cheese making resources allows the processor to get the most from incoming materials. In this case "integrated" means having some control of the manufacture or purchase of intermediate products used in processed cheese manufacture, including especially the incoming natural cheeses that may be used in this processing. Controlling the manufacture of these products for overall organization objectives can lead to manufacturing efficiencies and to increased profitability for the organization (14). Received September6, 1988. Accepted May 22, 1989. 1989 J Dairy Sci 72:3098-3108
where: z = the value of the objective function, e.g., maximizing a net return, minimizing costs, and maximizing a cheese yield; c = the vector of net returns, resource costs, or cheese yield associated with one unit of each activity; x = the vector of activities whose component levels in the optimal resource allocation are to be solved (also called the decision variables); A = the matrix of technical coefficients that relate resource use to resource constraints; and b = vector of constraint values that are limits on the amount of resources into and out of the system (19). The emphasis of the published literature on process cheese typically has focused on the manufacturing process. The study of the contribution of the ingredients has generally been limited to their principle effects on melting properties (13, 25), textural (9, 21, 33) and flavor (2) contributions, microorganism contamination (17, 31), and browning problems (4, 5). There are several good reviews of the industry (26) and of the manufacturing pro-
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cess (36). The economies involved in the manufacture of natural cheese and the manufacture of process cheese have traditionally been analyzed independently. A linear programming model is described herein that integrates these two production systems for the purpose of maximizing net returns from natural cheese and processed cheese products. Although the usual restrictions of a linear programming model apply (14), the advantage of developing such a model is that machine optimal solutions provide a place to begin the analysis and adjustment of production plans and methods. Further, the model provides tools that make the economic and resource use consequences of adjusting production plans available to the modeler (28). DEVELOPING THE MODEL
An integrated process cheese manufacturing system combines two components: the conversion of raw materials into the output products of cheese, whey cream, and separated whey and the allocation of these intermediate products with other ingredients to achieve the pasteurized process cheese food product. In the example developed here, Table I lists some of the input resources that could be used in the manufacture of natural cheese. We will assume that Cheddar cheese is to be manufactured and some or all of it will be processed. The Cheddar cheese is to be made with a fat on a dry basis (FDB) of 53.5%. This is accomplished by regulating the casein to fat ratio to be .6925. Moisture of the cheese is assumed to remain at 37%. The values used in the cheese yield formula are 1.09 salt solids retention factor, 93% fat retention, and 96% casein retention. It is assumed that the fatcontent of whey cream removed is 45%. TABLE 1. Several input resources available for use in the manufacture of natural Cheddar cheese. Resource Silo 1 Silo 2 NDM Cream Condensed skim milk
Amount available
Fat %
Casein %
Cost/ pound
300,000 300,000 As needed As needed
3.55 3.57 1.00 45.00
2.44 2.49 28.00 1.39
.1178 .1181 .79 .80
As needed
.37
9.2
.228
TODAY
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TABLE 2. Several input resources available for use in the manufacture of process cheese. Resource
Fat %
Cheese
Determined by LP 3.5
Whey protein concentrate Emulsifiers Fat Whey cream Water
0~ 83.0 45.0 0
Moisture % 37.0 1.5 0~ 17.0 55.0 100.0
Cost/pound Determined by LP .4825 .535 1.53 .782 .01
~Assumed t o be so s m a l l as t o n o t s i g n i f i c a n t l y affect the final product. 2Foregone revenue.
Whey fat is recovered at a rate of 100%. All of the raw milk is used for cheese production. The purchasable quantities of nonmilk resources is assumed to be limitless. Potential inputs for process cheese manufacture are listed in Table 2. All of the manufactured cheese is either sold or processed. It is possible to divide the manufactured cheese, i.e., one-third can be sold and two-thirds processed or half sold and half processed. The value of the output products (cheese, whey cream, and separated whey) and the resources used for process cheese manufacture are related to the value of the final process cheese product. The end product is pasteurized process block cheese valued at a block wholesale price of $1.50/lb. Commercial available software (LINDO, Microsoft Corporation, Copywright 1984), was used for computation with the resulting model. Decision Variables
The decision variables for the natural cheese manufacture are idenified with the resources that may be used and the amount of cream that can be removed from a milk resource during a standardization (18). Other variables are the direct inputs to the process cheese food. This example's decision variables for the process cheese manufacturing operation are listed in Table 3. Constraints
Constraint. In the model presented herein the casein:fat ratio of the cheese mik is .6925. J o u r n a l o f D a i r y Science Vol. 72, N o . 11, 1989
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TABLE 3. The decision variables used in the linear programming model of the process cheese manufacturing operation. Resource
Variable
Milk (silo 1) Milk (silo 2) Cream removed from silo 1 Cream removed from silo 2 NDM Condensed skim milk Cream added to cheese milk Whey cream from cheese mfg ~(used) Whey cream from cheese mfg (sold) Cheddar cheese mfg and processed Cheddar cheese mfg and sold Cheddar cheese purchased 1 mo old Cheddar cheese purchased 3 mo old Cheddar cheese purchased 5 mo old Emulsifiers WPC Fat Water
xl x2 x3 x4 x5 x6 x7 zl z2 M S B1 B3 B5 y3 y4 y5 y6
TABLE 4. Constraints for the model. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11)
1Manufacturing. 12) A s u m m a r y o f c o n t r a i n t s f o r this m o d e l c a n be f o u n d in T a b l e 4. T h e c h o s e n r a t i o s d e p e n d greatly on the company's manufacturing conditions and the type and quality of the natural cheese. S t u d i e s h a v e s h o w n r e l a t i o n s h i p s b e t w e e n t h e c a s e i n : f a t r a t i o in m i l k , m a n u f a c turing conditions, and the resulting FDB, m o i s t u r e in t h e n o n f a t t y s u b s t a n c e ( M N F S ) , a n d fat a n d m o i s t u r e p e r c e n t a g e in t h e c h e e s e (20, 22). T h e g r e a t e r t h e r a n g e a l l o w e d , t h e g r e a t e r t h e c o n t r o l is n e e d e d d u r i n g t h e m a n u f a c t u r i n g o p e r a t i o n to m a i n t a i n a c o n s i s t e n t quality product. T h e g e n e r a l f o r m a t f o r t h e c o n s t r a i n t is: casein percentage of standardized milk fat p e r c e n t a g e o f standardized milk
=.6925
T h i s r a t i o was c h o s e n to e s t a b l i s h t h e F D B o f t h e n a t u r a l c h e e s e at 53.5%, w h i c h is t h e F D B o f u n s t a n d a r d i z e d m i l k in this e x a m p l e . O t h e r r a t i o s c a n be c a l c u l a t e d f r o m t h e c h e e s e yield f o r m u l a : FDB =
FR(F) [FR(F) + CR(C)] SR
w h e r e : F R = fat r e t e n t i o n p e r c e n t a g e d i v i d e d Journal of Dairy Science Vol. 72, No. 11, 1989
13) 14)
.0002xl + .0002x2 + .2977x3 + .2974x4 ÷ .2731x5 + .0894x6 - .2977x7 ~ 0. casein to fat ratio constraint. xl = 300,000. x2 = 300,000 use all the milk constraints. .0789xl - x3_>0 .0793×2 x4 _> 0, limiting the amount of cream removed constraints. .0977xl + .0988x2 - .747x3 - .7472x4 + .4812x5 ÷ .1588x6 + .7472x7 = M + S natural cheese manufactured constraint. .0055xl + .0056x2 - .07x3 - .07x4 + .0013x5 + 0005x6 ÷ .07x7 = Z2 + ZI whey cream yield constraint. M + B1 + B3 + B5_>.70P, where P=(M+BI+B3+B5+y3+y4+y5+y6) natural cheese in the blend constraint. y3 = .03P emulsifier constraint. y4 _< .10P whey protein concentrate constraint. .0y3 + .015y4 + .17y5 + 1.0y6 + .55Z1 + .37B1 + .36B3 + .35B5 --< .43P moisture in the processed cheese constraint. .0y3 + .035y4 + .83y5 + 0y6 + .45Z1 + .33B1 + .33B3 + .33B5 + .3371M >-- .24P fat content constraint. - .005xl .005x2 -.307x3 - .307x4 + .85x5 + .15x6 + .307x7 <-- 0 allowable solids constraint. M + B I > - - . 5 ( M + B I + B 3 +B5) M+B2--<.6(M+B1 +B3+B5) B3_>.2 ( M + B I + B 3 + B 5 ) B3 <: .35 (M + BI + B3 + B5) B5___ .15 (M + B1 + B3 + B5) B5<_.20(M+BI+B3+B5) cheese blend constraints.
by 100, F = fat p e r c e n t a g e o f s t a n d a r d i z e d milk, CR = casein retention percentage d i v i d e d b y 100, C = c a s e i n p e r c e n t a g e o f s t a n d a r d i z e d m i l k , a n d S R = salt solids r e t e n tion factor. The procedure for determining the final f o r m o f this c o n s t r a i n t is to c o n s i d e r s e p a r ately all a d d i t i o n s a n d s u b t r a c t i o n s t o t h e c h e e s e m i l k o f c a s e i n a n d fat. ( W e m a y remove cream from the batch, for instance.) T h e r e s u l t f o r this c a s e is: . 0 2 4 4 x l + .0249x2 - . 0 1 3 9 x 3 .0142x4 + .28x5+.092x6 + . 0 1 3 9 x 7 . 0 3 5 5 x l + .0357x2 - . 4 5 x 3 - . 4 5 x 4 + .01x5 + .0037x6 + . 4 5 x 7
=.6925.
OUR INDUSTRY TODAY Rearranging the equation yields: -.0002xl + .0002x2 + .2977x3 + .2974x4 + .2731x5 + .0894x6 - . 2 9 7 7 x 7 _ > 0 . The same format would be used for constraining the casein:fat ratio between two limiting values. The o p t i m u m casein:fat ratio is that which maximizes the net return of the process cheese product. Consequently, the cheese milk may or may not be standardized, depending on the profit contributions of the milk components and the natural cheese toward the final process cheese product. The casein:fat ratio has been constrained here to show the difference that can occur when natural cheese is manufactured to maximize the value of the processed cheese product. A manufacturer might test several different casein:fat ratios to discover the impact different fat cheese have on the final processed product. Constraint 2. xl -- 300,000 and x2 = 300,000. These constraints require all the milk to be used. Constraint 3. x3, x4, x5, x6 _> 0. Constraint 4..0789xl - x3 ~ 0 and .0793x2 x4 _> 0. These two constraints limit the amount of cream that can be removed from the two milk resources. Silo 1 has 3.5% fat and 45% fat cream may be removed. In this case there are 3.5/45 = .078 lb of 45% fat cream available from each pound of 3.5% milk from silo 1. Using Kerrigan and Norback's (18) version slightly modified, the constraint to limit the amount of cream removed is: -
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where: M = pounds of cheese manufactured and processed, S = pounds of cheese manufactured and sold, and M + S = total pounds of cheese produced. The coefficients on the input variables are the cheese yield per pound of resource as determined by a modified version (18) of the original formula proposed by Van Slyke and Price (35). This formula is: CY = [FR(F) + CR(C)]SR 1-W where: CY = cheese yield per 100 lb, and W = moisture percentage divided by 100. To get the coefficient of x l , the unknown amount of silo 1 milk, we compute:
CY =
[.93(3.55) + .96(2.44)]1.09 .37 1
= 9.765 lb.
-
9.765/100 lb -- .09765 lb/lb of resource. The same calculation is done for the other input resources to determine the other coefficients. Constraint 6. The whey cream by-product resulting from the cheese manufacture is: WCY =
(F - FR(F)) * W F R WF
where: WCY = whey cream yield per 100 lb., W F R = whey fat recovery percentage divided by 100, and W F = fat percentage of whey cream divided by 100.
. 0 7 8 x i > = x3, An example using silo 1 milk is as follows: and rearranged: WCY.078xl - x3~>0.
(3.55 -.93(3.55)) * 1 .45
Constraint 5. The connecting link between the raw cheese manufacture and the blending of the process cheese ingredients is the cheese yield formula represented by the equation:
Thus, the whey cream yield is .005522 lb/lb of silo 1 milk. Similar calculations are done for the other input resources. The resulting equation is:
.0977xl + .0988x2 -.7472x3 - . 7 4 7 2 x 4 + .4812x5 + .1588x6 + .7472x7 = M + S
.0055xl + .0056x2- .07x3 - . 0 7 x 4 + .0a013x5 ÷ .0005x6 + .07x7 = z2 + zl Journal of Dairy Science Vol. 72, No. 11, 1989
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CRAIG ET AL.
where: z2 = whey cream extracted and sold, zl = whey cream extracted and used in processing, and z2 + zl = total pounds of whey cream produced from the cheese manufacture. Constraint 7. There must be a constraint to maintain a certain amount of cheese is the process cheese blend. The total process cheese batch is represented by P, where:
product. Because sensory qualities can be adversely affected by too much WPC, limits must be set (24). Management and experience play a major role in deciding the quantity to be used. In this example, W P C is limited to less than or equal to 10% of the total batch, where y4 is pounds of WPC. y4 _< .10P.
P = M + B1 = B3 + B5 + y3 + y4 ÷ y5 + y6. It is necessary to specify a limit in order to retain a product's sensory properties or as in the case of a cheese spread to satisfy the legal standards of identity. For example, a process cheese product that is required to have at least 70% cheese (minimum) will be written as follows: M + B1 + B3 + B5_>.70P. Constraint 8. It is assumed that the process cheese batch size is known and constant. Thus P = batch size of process cheese blend. In our example, P = 100,000 lb. Constraint 9.The amount and type of emulsifiers used for processing must be incorporated into the program in a similar manner as that done for cheese (Constraint 7). Numerous articles, books, and bulletins have been written about the type, quantity, and characteristics of emulsifiers in dairy products (11, 12, 23, 27, 32, 34). Some of the variables that must be considered before choosing the type and quantity of an emulsifier are cheese type, sensory characteristics of the desired finished product, melting properties, and safety considerations. Because of the sensitivity of these factors and the variability of the emulsifiers available for use, only the crudest measure for emulsifier use will be presented, a percentage of total process cheese batch: y3 = .03P As represented here, y3 equals 3% of the total process cheese batch in pounds. Constraint 10. W P C is a supplement to the raw cheese and is generally accepted for its lower cost and the improved sensory qualities it can impart to the final process cheese food Journal of Dairy Science Vol. 72, No. 11, 1989
Constraint 11. The moisture in the process cheese product must be constrained to comply with the legal standards of identity and the characteristics of the product as set by management. Because moisture adds to the yield of the product with next to no cost, the LP model will add as much moisture as it is allowed by the constraints. Consequently, only an upper limit constraint is needed. To ensure the final product has less than or equal to the legal moisture of 43% requires the following constraint: 0y3 + .15y4+ .17y5+ 1.0y6 + .55zl + .37B1 + .36B3 + .35B5 ÷ .37M <_ .43P. Constraint 12. Another constraint is necessary to ensure the fat content of the process cheese food meets the standard of identity for the specific product. Pasteurized process cheese foods must contain at least 23% fat. An extra I% is added for a safety margin. The general inequality is: pounds fat in process cheese ___.24 (total process cheese batch). The fat content of the manufactured cheese can be calculated by knowing the casein:fat ratio and a constant moisture value. Using the cheese yield formula mentioned earlier, we know the F D B is 53.5%. If the moisture is 37%, the fat percentage in the natural cheese is 33.71%. The inequality can thus be written: 0y3 + .035y4 + .83y5 + 0y6 + .45zl + .33B1 + .33B3 + .33B5 + .3371M _> .24P. Constraint 13. A constraint on the allowable solids content of the raw milk used for cheese making has been included in this model [labeled (*)]. This has been done for several reasons. If milk concentrated by reverse osmosis or ultrafiltration is used, it is
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necessary to set a limit on the solids content of the cheese milk. Using reverse osmosis, Barb a n o and B y n u m (1, 7) have shown that at about a 15% reduction in volume (about 14.17% solids content), the increased lactose of the cheese m a y b e c o m e the limiting factor in p r o d u c i n g a g o o d quality aged C h e d a r cheese. T h e y also go on to say that a low moisture barrel cheese used within 60 d m a y tolerate a reduction in volume of greater t h a n 15%. A n o t h e r i m p o r t a n t reason is that the higher cheese yield that results f r o m adding casein and fat may be economically profitable. If this is the case, it is necessary to constrain the solids content to a m a x i m u m level. In this example, the solids content is constrained to less than 12.0%. The m a n u f a c t u r e r would decide on the appropriate milk solids content for their operation. The initial inequality is: . l 1 5 x l +.115x2 - . 4 2 7 x 3 - . 4 2 7 x 4 +.97x5 + .27x6 + .427x7 _<.12(xl + x 2 - x3 - x4 + x5 + x6)
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TABLE 5. The values and costs of the input resources used in the linear programming model (per pound basis). Revenue Cost ($)~ Block processed cheese Purchased price of: Milk (silo 1) Milk (silo 2) NDM Condensed skim milk Emulsifier(s) WPC Butterfat Water Whey cream Sweet cream Direct labor and overhead: per lb natural cheese per lb processed cheese Purchase price of: 1-mo-old Cheddar cheese 3-mo-old Cheddar cheese 5-mo-old Cheddar cheese Revenue from: 1-mo-old mfg. 3 Cheddar
1.50 . ! 178 .1181 .79 .228 .535 .4825 1.53 .01 .781 .822 .25 .35 1.51 1.55 1.59 1.49
~Foregone revenue. value of resource. 3Manufacturing.
2New
Rewritten: (*) -.005x I - .005x2 - .307x3 - .307x4 + .85x5 + .15x6 + .307x7 ___0. Constraint 14. Finally, there are several constraints that must establish the acceptable age blends used in the processed cheese p r o d uct. These age blends m a y be set as absolute or given as a range. F o r example, an absolute age blend could consist of 60% l - m o - o l d cheese, 30% 3-mo-old cheese, and 10% 5-moold cheese. Alternately, an acceptable range could be 50 to 60% 1-mo-old cheese, 20 to 35% 3-mo-old cheese, and 15 to 20% 5-moold cheese. The advantage of a range of values is that it allows some flexibility in the prog r a m depending on cost, availability, grade of cheese, etc. Thus, the 1-mo-old cheese, M and B1, could be greater t h a n 50% but less t h a n 60% of the total cheese used in processing. Rewritten: M + BI ~> .5 (M + BI + B3 + B5) M + B1 <_.6 (M + B1 + B3 + B5)
Likewise, B3_> .2 (M + B1 + B3 + B5) B3_<.35 (M + B1 + B3 + B5)
and B5_> .15 (M + B1 + B3 + B5) B5 <_ .20 (M + B1 + B3 + B5).
Objective function The objective function is to maximize the net returns of a pasteurized process cheese product. Net return is defined as the difference between revenue and cost. Table 5 illustrates an example of the values and costs of various ingredients. The expression for the objective function is pieced together as net profit of process cheese minus the cost to m a n u f a c t u r e natural cheese minus the cost to Journal of Dairy Science Vol. 72, No. i 1, 1989
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CRAIG ET AL.
buy natural cheese plus revenue from natural cheese sold. Net Process Cheese Profit. The process cheese batch is represented by the following equation: M + B1 + B3 + B5 + Y3 + Y4 + Y5 + Y6 + Z1 If the selling price of block process cheese is $1.50/lb. the expression gives the revenue from its sale: $1.50 ( M + B1 + B3 + B5 + Y3 + Y 4 + Y 5 + Y6 + Z1) The ingredient costs for the processing materials are just their purchase prices. Whey cream produced from the Cheddar cheese manufacturing process has a "cost" value calculated from its foregone fat revenue. If whey cream fat is valued at $1.73/lb and 45% of the cream is fat, then: $1.73/lb. fat * .45 fat in cream -- $.78 (value of fat/lb, cream)
cheese to allow for the option of selling the manufactured cheese or processing it. There are the ingredients costs to consider as well as the direct labor and overhead that go into the manufacturing operation. The sweet cream represented by x3 and x4 have positive "costs" since any cream removed and sold is revenue (18). The sweet cream fat is valued at $1.82/lb with 45% fat in the cream. $1.82/lb fat * .45 fat in cream -- $.82 (value of fat/lb cream) Direct labor and overhead costs are calculated on the total pounds of natural cheese produced, whether sold or processed. .11278xl + . l 1 8 1 x 2 - . 8 x 3 - . 8 x 4 + .79x5 + .228x6 + .82x7 + .25 ( M + S).
Cost to Buy Natural Cheese. Cost values are based on the estimated cost to buy Cheddar cheese of varying ages: 1.51B1 + 1.55B3 + 1.59B5.
The expression representing the cost of the processing ingredients is:
Revenue from Natural Cheese (and Whey Cream).
.535y3 + .4825y4 + 1.535y5 + .01y6 + .78zl.
1.49S + .78z2
The direct labor and overhead cost estimate is based on total pounds of process cheese produced. It is necessary to include these costs because process cheese production and natural cheese production are separate operations requiring different equipment, facilities, and labor. The inclusion of the labor and overhead estimate is written as:
Note that the sold cheese [S] value is $.02 less than the 1-mo-old cheese that could be purchased. It is assumed that it costs more to buy than is gained through sale of similar products. The objective function to be maximized is a combination of all four components:
$.35 (M + B1 + B3 + B5 + Y3 + Y4 + Y5 + Y6 + Z1). Combined, this part of the objective function is written as: 1 . 1 5 M + 1.15B1 + 1.15B3 + 1.15B5 + .615y3 + .6675y4 - .385y5 + 1.14y6 + .37zl.
Cost to Manufacture Natural Cheese. The cost to produce Cheddar cheese was not included directly in the profit from processed Journal of Dairy Science Vol. 72, No. 11, 1989
- . 1 1 7 8 x l - . 1 1 8 1 x 2 + .8x3 + .8x4 -.79x5 - . 2 2 8 x 6 - . 8 2 x 7 +.615y3 +.6675y4 - . 3 8 y 5 + 1.14y6 +.37zl + 1.24S + . 9 M - . 3 6 B 1 - . 4 B 3 -.45B5. RESULTS
Results of applying linear p r o g r a m m i n g software to the model described are presented in Tables 5-8. There are several underlying assumptions c o m m o n to all of the scenarios. First is that all raw milk resources are used in their entirety to make natural Cheddar
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T A B L E 6. Comparison of the o p t i m u m formulation and profit potential that results from modeling the natural cheese and process cheese manufacture as individual process systems versus an integrated process cheese operation. Pounds Scenario 1 (separate)
Scenario 2 (integrated)
300,000 300,000 0 0 0 0 0
300,000 300,000 0 0 0 12,386 3719
3000 9382 329 17,288 0 3,322 58,935 16,935 0 17,500 10,500
3000 9382 329 17,288 0 3582 63,680 21,680 0 17,500 10,500
Total processed batch size, lb Cheddar cheese yield, lb Total Cheddar cheese cost Cost/lb Cheddar cheese
100,000 58,935 $ 85,522 $ 1.4511
100,000 63,680 $ 92,564 $ 1.4536
Objective function value
$
$
Natural cheese input resources xl = Silo 1 milk x2 = Silo 2 milk x3 = Remove cream x4 -- Remove cream x5 -- N D M x6 = Condition skim x7 = Cream Process cheese input resources y3 = Emulsifier y4 -- W P C y5 -- Fat y6 = Water zl = Whey cream used z2 -- Whey cream sold M = Manufacturing cheese S = Cheese sold B1 -- Purchased cheese, 1 mo. B3 -- Purchased cheese, 3 mo. B5 = Purchased cheese, 5 mo.
cheese. A second assumption is that there is no waste. Third, the process cheese batch size is known and constant. Fourth, all ingredients are of known quality and composition and there is no problem with availability. Fifth, marginal revenue and costs are not applicable over the stated batch sizes. The optimum input resources to maximize the profits of the process cheese manufacturing operation, their amounts and some costs are listed in Table 6 (scenario 2). This is compared to a run using a decision support program in the unpublished program manual mentioned earlier designed by Kerrigan (scenario 1). Kerrigan's program is computer based and gives an economic evaluation of standardizing milk for cheese making. The objective is to maximize the profit of the nat-
4,365
6,805
ural cheese. This cheese was then used in the current model to formulate process cheese under the same constraints and objective function as for scenario 2. The constantly changing economic environment regularly influences the optimal solution to the model. A small change in the cost of an input resoure can dramatically change the optimal solution. An important question is how sensitive the solution is to these changes. It is important to know how close the unused decision variables (those held a value 0) came to be included in the optimal solution. If using a resource corresponding to a decision variable is obviously not profitable, then little extra effort is needed to accurately estimate its cost. Conversely, if a small change in a decision variaJournal of Dairy Science Vol. 72, No. 11, 1989
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T A B L E 7. Reduced costs associated with the decision variables from Scenario 2. Decision variables
Resource
Old coefficient
Reduced cost
New coefficient
x1 x2 x3 x4 x5 x6 y3 y4 y5 y6 x7 zl z2 S M B1 B3 B5
Silo 1 milk Silo 2 milk Removed cream Removed cream NDM Conditioned skim Emulsifiers WPC Fat Water Cream Cream used C r e a m sold Cheese sold Cheese processed l-too-old cheese 3-mo-old cheese 5-mo-old cheese
-. 11782 -.11814 .800 .800 .79 -.228 .615 .6675 -.38 1.14 .82 .37 .78 1.24 .90 .36 -.40 .45
0 0 0
- 11782 .11814 .800 .8201 -6638 .228 .615 .6675 -.38 1.14 -.82 1.0744 .78 1.24 .90 .33 -.40 -.45
.0201 .1268 0 0 0 0 0 0 .7044 0 0 0 .03 0 0
T A B L E 8. Dual prices associated with the constraints from Scenario 2. Constraint
Slack
Dual Prices
x3>0
0
- .02
x4>0 x6>O x5>0 Cream removed from x l milk Cream removed from x2 milk xl = 300,000 x2 = 300,000 Cheese yield equation
0 12,386 0 23,670 23,790 0 0 0
0 0 0 0 0
Fat in process cheese > 24%
0
Moisture in process cheese _< 43% Whey cream yield
0 0
1 -.401427 1 .429899 -.78
Solids in cheese milk _< 12% Process cheese batch variables
0 0
.072705 '-.71010
Process batch = I00,000 y4<10,000 y3 = 3000
0 617 0
Cheese = 70% of batch
0
-.769859
cheese cheese cheese cheese cheese
7000 0 3500 7000 0
0
5-mo-old cheese <_ 20% of cheese Fat in process cheese ~ 28% Casein:fat ratio = .6925
3500 4298 0
0 0
1-mo-old 1-mo-old 3-mo-old 3-mo-old 5-mo-old
cheese cheese cheese cheese cheese
>_ 50% _< 60% _> 20% _ 35% _> 15%
of of of of of
Journal of Dairy Science Vol. 72, No. 11, 1989
.007863 .009200 .2400
.68
.710101 0 -.095101
.065651 0 0 -.045701
.466183
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ble results in a new optimal solution, that decision variable and the constraints limiting its use should be analyzed. A reduced cost of a decision variable is the change in the value of the objective function for each unit increase in the value of the decision variable. Table 7 shows the reduced cost for the decision variables for scenario 2. Any decision variable that is part of the solution has a reduced cost of zero. This interpretation of a reduced cost is only valid for small changes in variable values. Most linear programming computer models supplement the solution report with a range or sensitivity analysis report. This report indicates the amounts by which unilateral changes in objective function coefficients or constraints can be altered without affecting which variables are nonzero in the optimal solution. For more detailed information, see Hillier (14) and Schrage (28). The reduced cost gives the cost, or price, as in the case of the removed cream, the resource must change before it will become a nonzero component in the optimal solution. Nonfat dry milk reduced cost of .1268 can be interpreted to mean the cost of N D M must decrease from $.79 to $.6638/lb before it would replace a resource that is already being used in the optimal formultion. The cream price would have to increase $.0201/lb, or up to $.8201, before it would be economically practical to remove it from the cheese milk. Associated with each constraint is a quantity known as the dual price. Its value is the rate at which the objective function value will improve as the right-hand side or constant term of the constraint is increased a small amount. Another way of stating this is that the dual price of a constraint is the amount by which profits will decrease if the availability of the resource associated with this constraint is reduced by one unit. Because many of the constraints have been rewritten to have the right-hand side term greater than, less than, or equal to zero, interpretations are complicated. Table 7 shows the dual prices for the constraints used in this example. The slack column lets the user know how much " r o o m " there is between the constraint and what is actually occurring in the optimum solution. For example, the constraint limiting
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the 5-mo-old cheese to less than 20% has a slack value of 3500. Looking closer, we see the constraint before forces the solution to contain at least 15% (10,500 lb) 5-mo-old cheese. There is no slack associated with this constraint. Because we are allowing a range of values, the solution could allow the 5-mo-old Cheddar to increase to 20% (or 14,000 lb). The slack is the difference between what is used and what could be used, 3500. The model is a guideline to assist in decision making. Its assumptions and solutions must be regularly tested as economic and manufacturing conditions change. H u m a n judment is needed to evaluate the proposed solution and adjust it to the specific situation,
ACKNOWLEDGMENTS
This research was supported by the Center for Dairy Research at the University of Wisconsin, and the College of Agriculture and Life Sciences, University of Wisconsin, Madison.
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