## Abstract

One of the first practical problems to be formulated and solved by linear programming methods was the so-called *diet problem*, which is concerned with planning a diet from a given set of foods which will satisfy certain nutritive requirements while keeping the cost at a minimum. For each food the nutritional values in terms of vitamins, calories, etc. per unit of food are known constants and these are the a’s of the problem, aii being the amount of the *i*th nutritional factor contained in a unit of the *j*th food. If it is required that there shall be at least b_{ i } units of the *i*th nutrient in the diet the nutritional requirements will take the form of a set of linear inequalities’ in the variables *x1*, which represent the amounts of the respective foods which shall be present in the diet. These restrictions will in general be satisfied by a large number of combinations of ingredients (foods) and we want to select a combination which minimizes the total cost of ingredients, i.e., a linear function in the *x* _{ j } where the coefficients c_{ j } are the prices per unit of the respective foods.

## Keywords

Setup Time Linear Programming Problem Transportation Problem Slack Variable Total Profit## Preview

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## References

- 1.Nutritional requirements will usually have the form of inequalities since the organism can dispose of excess amounts of the nutrients.Google Scholar
- 2.Cf. Ch. IV.Google Scholar
- 3.The term “process” is here used in a broader sense than elsewhere in this book, meaning any transformation of inputs into outputs.Google Scholar
- 1.Cf. I. KATZMAN (1956).Google Scholar
- 2.Cf. requirements (5) and (6) in the ice cream blending problem.Google Scholar
- 1.Exercise: In an (xi, x2) diagram, draw the geometric picture of the set of feasible solutions defined by (1). Vary the constants on the right-hand sides in turn to show that not only lower limits but upper limits as well (inequalities of type , such as the second inequality) can be effective restrictions in a minimization problem in the sense that they affect the position of the optimal point. (Consider this as a problem in xi and x2 only, i.e., as a problem of minimizing a linear form in these two variables subject to (1), thus neglecting the other variables and restrictions of the complete problem.)Google Scholar
- 2.Many of the best known examples are found in recent volumes of the Journal of Farm Economics.Google Scholar
- 3.Exercise: Show this by constructing an example.Google Scholar
- 1.This case is formally analogous to the occurrence of setup costs in problems of planning production subject to machine capacity limitations.Google Scholar
- 2.Cf. Ch. VI, C.Google Scholar
- 3.An example of such a problem (only with additional restrictions on the quantities of crudes available and with two alternate processes for one of the crudes) is given in G. H. SYMONDS (1955), Ch. 3.Google Scholar
- 4.Cf. N. V. RErNFELD and W. R. VOGEL (1958), pp. 122–5.Google Scholar
- 5.Thus the three crude gasolines are intermediate products rather than primary raw materials.Google Scholar
- 1.Exercise: What statistical information is needed for calculating the coefficients in the linear preference function (i.e., the unit gross profit coefficients 3 and 2)Y Cf. Ch. IV, B above.Google Scholar
- 2.Exercise: Solve the problem geometrically.Google Scholar
- 3.When the company’s ultimate objective is to maximize profits, the avoidance of material waste, idle machine hours, etc. becomes a secondary purpose. It may very well happen that a solution in which some of the slack variables are positive will be optimal even though there exist positive solutions in structural variables only. In such cases intuition and common-sense considerations may lead to a solution which is irrational from the point of view of profit maximization.Google Scholar
- 4.The highest prices which the company can afford to pay for these additional supplies will be equal to the simplex coefficients of xi’ and x2’, a result which follows from the Duality Theorem. (Cf. Ch. VII below.)Google Scholar
- 1.Borrowed from A. BoxDiN (1954), Ch. I II: Un esempio di programmazione lineare nell’industria, by A. BARGOrri, B. GIARDINA and S. RicossA. To preserve company confidentiality, and partly for didactic reasons, the authors have adjusted some of the technological and economic data used in the following.Google Scholar
- 1.The units of measurement used are those of the metric system.Google Scholar
- 1.Exercise: Explain why an equation whose slack variable is known to be positive can always be disregarded.Google Scholar
- 2.Cf. the similar procedure by which the ice cream blending problem of Ch. III was partially solved.Google Scholar
- 3.Using the simplex method, with x2, x11, x12, xis, and x14 as a starting basis, the problem is solved in three steps and the optimal solution turns out to be x2=288.8, x7=37.52, x8= 29.0909, x12=50.0, x13=81.6 (all in tons per day), all other variables =0. Total profit is 1-= 19,403,380 (lire per day).Google Scholar
- 4.Exercise: What price and cost data are needed to compute the coefficients of the profit function? Should the costs of the cracking and electrolysis processes be taken into account?Google Scholar
- 5.Exercise: In the gasoline blending problem above, the introduction of sales restrictions implies the additional side conditions xi xl and x2 Y2. Show geometrically how this may or may not affect the optimal solution, depending on the values of xl and 12.Google Scholar
- 1.It will serve no useful purpose to store them for later use since they will be in abundant supply also in the following periods so long as the data of the problem remain the same.Google Scholar
- These slack variables will have zero coefficients in the profit function, except for chlorine (v4) which is poisonous and cannot be disposed of costlessly. Excess amounts of chlorine are rendered innocuous by means of calcium hydroxide, i.e., the chlorine is transformed into calcium hypochlorite which is harmless and can be thrown away. The costs of this process give rise to the negative profit coefficient —20.Google Scholar
- 2.Exercise: Write out the new system of equations, or its coefficient matrix.Google Scholar
- 3.Exercise: Solve the problem numerically, using the procedure indicated.Google Scholar
- i This assumption of additivity—in chemical engineering known as the “Mixture Law”, cf. J. H. PERRY (1941), p. 616—is not always satisfied even in cases of physical blending. For example, if alcohol and water are mixed the volume of the mixture will not be the sum of the volumes of the ingredients although there is no chemical reaction.Google Scholar
- 2.Exercise: Explain why. (Hint: Interpret the numerator and the denominator of the fraction.)Google Scholar
- .3Exercise: Would there have been a problem at all if there had been no restrictions other than (2), and if so what would the solution have been?Google Scholar
- 1.Exercise: Use the data given above to set up a table showing the coefficients of the problem. How many restrictions are there ? Would there have been any problem if unlimited amounts of the raw materials could have been purchased at normal prices?Google Scholar
- 1.The prices have been adjusted for reasons of confidentiality.Google Scholar
- 1.Specifying the heaviness of a beer is not equivalent to specifying its alcoholic strength since the latter depends not only on the content of extract (as measured by gravity) but on the degree of fermentation as well. The following empirical relation holds approximately for gravities in the neighborhood of 1.045:Google Scholar
- 2.Exercise: What is the precise interpretation of x6 and x7?Google Scholar
- 1.One way of tackling the problem might be to construct a set of grading scales—based upon experiments with a team of beer-tasters—by which the various beer types could be classified with respect to each taste component separately. The gradings would be based on subjective judgment but they would nevertheless be quantitative. Only experiment can show whether it is possible to establish, for each quality component, an order of preference on which most of the tasters agree, and whether a numerical grading scale based upon it will blend linearly.Google Scholar
- 2.Only such ingredients are considered as might be expected to give a satisfactory product.Google Scholar
- 3.The procedure used for computing second-best solutions is shown in Ch. VI, C, below.Google Scholar
- 2.Exercise: Carry out the calculations.Google Scholar
- 3.They are not only cheaper per hl. than nos. 4 and 5. Dividing the coefficients in each column by the corresponding price we will find that ingredient no. 3 is the cheapest source of alcohol while no. 2 gives the largest contribution to gravity, color, and hop resin content per crown’s worth. (Cf. the ice cream problem of Ch. III.)Google Scholar
- 4.This is a “degenerate” solution with fewer than eight positive variables.Google Scholar
- 1.Although there are three structural variables, geometrical illustration in two dimensions is possible if xi is treated as a slack variable in (4a) and replaced by (100 - xa - x3) in (4d). For x4 = x5 = 0, (4a) represents a half plane in an (xa, x3) diagram, whereas (4b) corresponds to a straight line. The three pairs of upper and lower limits, cf. (4 c-h), are represented by three belts in the (x2, x3) plane. These restrictions together define a set of feasible solutions.Google Scholar
- 2.Exercise: Draw the diagram on graph paper and show the positions of the basic solutions that exist for x2 and x3 0 and x4 =x5 = 0. How many are there and how many of them are feasible ?Google Scholar
- 2.For example, (4b), (4g), and (4h) can be solved for x2 and x3; (4a) then gives xi, and so forth.Google Scholar
- 1.This includes the decision as to which products are to be made at all.Google Scholar
- 2.Borrowed from R. DORFMAN (1953).Google Scholar
- 3.Alternatively, the capacity restriction may be written with machine hours per truck and per automobile as coefficients on the left-hand side and machine hours available per month on the right-hand side. This form of the restriction is obtained from (7) by multiplying by the number of machine hours available per month.Google Scholar
- 4.Exercise: Solve the program geometrically to find the optimal basis, and compute the exact numerical solution.Google Scholar
- 1.The ctual number of products was far greater than six, but they were arranged in six major groups in order to simplify the problem. Similarly, each of the eleven machine groups of the problem represents a number of different machine types which have been reduced to one type by applying suitable correction factors (efficiency factors).Google Scholar
- 2.This aggregation procedure has since been abandoned as it proved possible to solve the complete problem numerically without first having to simplify it in this fashion Cf. op. cit., p. 117.Google Scholar
- 1.Cf. A. HENDERSON and R. SCHLAIFER, op. cit., p. 86.Google Scholar
- 2.Cf. Ch. IV, p. 31.Google Scholar
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- 4.I. e., setup time on a particular machine for a particular activity.Google Scholar
- 5.See A. HENDERSON and R. SCHLAIFER, op. cit., pp. 86–89, where the procedure is illustrated by a numerical example.Google Scholar
- 1.The data were kindly provided by the tobacco factory of the Danish Cooperative Union. The model was developed in collaboration with Mr. E. LYxx i JENSEN, University of 1 In actual practice, a positive inventory will usually be left over from the previous period, even if it was intended to close that period with zero stock. However, the general nature of the model will not be changed if a positive inventory is entered at the beginning and at the end of the planning period.Google Scholar
- 2.These are the capacities of the roll-making machines which are the bottleneck factors of the planning problem.Google Scholar
- 3.Exercise: Interpret the slack variables in all of the 11 inequalities.Google Scholar
- 2.Storing here implies that unutilized capacity in the first shift is transferred to following quarters.Google Scholar
- 2.In a graph showing cumulated production (according to the solution) and cumulated sales as functions of t, storing time and inventory are represented respectively by the horizontal and vertical distances between the curves at time t, and we can then examine whether the solution respects the additional inequalities. If this is not the case, the inspection methods breaks down and the production program will have to be recomputed by the simplex method, taking explicit account of all restrictions on the variables. Exercise: Indicate the general form of the additional constraints.Google Scholar
- 3.For an example, see S. DAN and E. L. JENSEN (1958).Google Scholar
- 1.Exercise: Why are the variables required to be non-negative?Google Scholar
- 2.This fact makes it possible to solve such problems by a considerably simplified variant of the simplex method, see Ch. VI, F.Google Scholar
- 8.Solutions of transportation problems will often turn out to be degenerate, i.e., fewer than m +n —1 variables are positive in the optimal combination of routes.Google Scholar
- 1.Quite apart from the question of number of routes, it is difficult to solve transportation problems by common-sense methods. It is no use assuming that each warehouse shall be supplied from the nearest factory, for this assumption will only in exceptional cases be compatible with the restrictions.Google Scholar
- 2.Cf. A. HENDERSON and R. SCHLAIFER, op. cit., pp. 77 f.Google Scholar
- 3.See, for example, A. HENDERSON and R. SCHLAIFER, op. cit., pp. 79 ff.Google Scholar
- 1.Exercise: Construct a simple numerical example.Google Scholar
- 2.Exercise: Indicate, for each example in the present chapter, the dimensions of the coefficients and of the variables (e g, machine hours per unit produced, units to be produced per month, etc.).Google Scholar