Diet lecture¶
Description: Diet case study
Tags: amplonly, ampllecture
Notebook author: Marcos Dominguez Velad <marcos@ampl.com>
Model author: N/A
# Install dependencies
%pip install q amplpy
# Google Colab & Kaggle integration
from amplpy import AMPL, ampl_notebook
ampl = ampl_notebook(
modules=["coin"], # modules to install
license_uuid="default", # license to use
) # instantiate AMPL object and register magics
This notebook provides the implementation of the production problem described in the book AMPL: A Modeling Language for Mathematical Programming by Robert Fourer, David M. Gay, and Brian W. Kernighan.
Diet problem¶
As an intuitive example of a costminimizing model, we use the wellknown “diet problem”, which finds a mix of foods that satisfies requirements on the amounts of various vitamins. We will construct a small, explicit linear program, and then show how a general model can be formulated for all linear programs of that kind.
After formulating the diet model, we will discuss a few changes that might make it more realistic. The full power of this model, however, derives from its applicability to many situations that have nothing to do with diets. A general model derived from the diet formulation can be also applied to blending, economics, and scheduling.
Solving a diet problem instance¶
Consider the problem of choosing prepared foods to meet certain nutritional requirements. Suppose that precooked dinners of the following kinds are available for the following prices per package:
Name 
Price $ 


BEEF 
beef 
3.19 
CHK 
chicken 
2.59 
FISH 
fish 
2.29 
HAM 
ham 
2.89 
MCH 
macaroni & cheese 
1.89 
MTL 
meat loaf 
1.99 
SPG 
spaghetti 
1.99 
TUR 
turkey 
2.49 
These dinners provide the following percentages, per package, of the minimum daily requirements for vitamins A, C, B1 and B2:
A 
C 
B1 
B2 


BEEF 
60% 
20% 
10% 
15% 
CHK 
8 
0 
20 
20 
FISH 
8 
10 
15 
10 
HAM 
40 
40 
35 
10 
MCH 
15 
35 
15 
15 
MTL 
70 
30 
15 
15 
SPG 
25 
50 
25 
15 
TUR 
60 
20 
15 
10 
The problem is to find the cheapest combination of packages that will meet a week’s requirements  that is, at least 700% of the daily requirement for each nutrient.
Let us write $X_{beef}$ for the number of packages of beef dinner to be purchased, $X_{chk}$ for the number of packages of chicken dinner, and so forth. Then the total cost of the diet will be:
total cost =
3.19 Xbeef + 2.59 Xchk + 2.29 Xfish + 2.89 Xham +
1.89 Xmch + 1.99 Xmtl + 1.99 Xspg + 2.49 Xtur
The total percentage of the vitamin A requirement is given by a similar formula, except that X BEEF , X CHK , and so forth are multiplied by the percentage per package instead of the cost per package:
total percentage of vitamin A daily requirement met =
60 Xbeef + 8 Xchk + 8 Xfish + 40 Xham +
15 Xmch + 70 Xmtl + 25 Xspg + 60 Xtur
This amount needs to be greater than or equal to 700 percent. There is a similar formula for each of the other vitamins, and each of these also needs to be $\geq$ 700.
Putting these all together, we have the following linear program:
Minimize
3.19 Xbeef + 2.59 Xchk + 2.29 Xfish + 2.89 Xham +
1.89 Xmch + 1.99 Xmtl + 1.99 Xspg + 2.49 Xtur
Subject to
60 Xbeef + 8 Xchk + 8 Xfish + 40 Xham +
15 Xmch + 70 Xmtl + 25 Xspg + 60 Xtur >= 700
20 Xbeef + 0 Xchk + 10 Xfish + 40 Xham +
35 Xmch + 30 Xmtl + 50 Xspg + 20 Xtur >= 700
10 Xbeef + 20 Xchk + 15 Xfish + 35 Xham +
15 Xmch + 15 Xmtl + 25 Xspg + 15 Xtur >= 700
15 Xbeef + 20 Xchk + 10 Xfish + 10 Xham +
15 Xmch + 15 Xmtl + 15 Xspg + 10 Xtur >= 700
Xbeef >= 0, Xchk >= 0, Xfish >= 0, Xham >= 0,
Xmch >= 0, Xmtl >= 0, Xspg >= 0, Xtur >= 0
At the end we have added the commonsense requirement that no fewer than zero packages of a food can be purchased. And now, we can transcribe the model to an AMPL statement of the explicit diet LP:
%%writefile diet0.mod
# Variables
var Xbeef >= 0; var Xchk >= 0; var Xfish >= 0;
var Xham >= 0; var Xmch >= 0; var Xmtl >= 0;
var Xspg >= 0; var Xtur >= 0;
# Objective function (minimizing)
minimize cost:
3.19*Xbeef + 2.59*Xchk + 2.29*Xfish + 2.89*Xham +
1.89*Xmch + 1.99*Xmtl + 1.99*Xspg + 2.49*Xtur;
# Constraints
subject to A:
60*Xbeef + 8*Xchk + 8*Xfish + 40*Xham +
15*Xmch + 70*Xmtl + 25*Xspg + 60*Xtur >= 700;
subject to C:
20*Xbeef + 0*Xchk + 10*Xfish + 40*Xham +
35*Xmch + 30*Xmtl + 50*Xspg + 20*Xtur >= 700;
subject to B1:
10*Xbeef + 20*Xchk + 15*Xfish + 35*Xham +
15*Xmch + 15*Xmtl + 25*Xspg + 15*Xtur >= 700;
subject to B2:
15*Xbeef + 20*Xchk + 10*Xfish + 10*Xham +
15*Xmch + 15*Xmtl + 15*Xspg + 10*Xtur >= 700;
A few AMPL commands then suffice to read the file, send the LP to a solver, and retrieve the results (using CBC solver from coin). AMPL commands within a Jupyter Notebook should be executed in a cell with the %%ampl_eval
header.
%%ampl_eval
model diet0.mod;
option solver cbc;
solve;
display Xbeef,Xchk,Xfish,Xham,Xmch,Xmtl,Xspg,Xtur;
The optimal solution is found quickly, but it is hardly what we might have hoped for. The cost is minimized by a monotonous diet of 46 and 2/3 packages of macaroni and cheese! You can check that this neatly provides $15% \times 46\frac{2}{3} = 700%$ of the requirement for vitamins A, B1 and B2, and a lot more vitamin C than necessary; the cost is only $$1.89 × 46\frac{2}{3} = $88.20.$
You might guess that a better solution would be generated by requiring the amount of each vitamin to equal 700% exactly. Such a requirement can easily be imposed by changing each >= to = in the AMPL constraints. If you go ahead and solve the changed LP, you will find that the diet does indeed become more varied: approximately 19.5 packages of chicken, 16.3 of macaroni and cheese, and 4.3 of meat loaf. But since equalities are more restrictive than inequalities, the cost goes up to $89.99.
An AMPL model for the diet problem¶
Clearly we will have to consider more extensive modifications to our linear program in order to produce a diet that is even remotely acceptable. We will probably want to change the sets of food and nutrients, as well as the nature of the constraints and bounds. As in the production example of the previous chapter, this will be much easier to do if we rely on a general model that can be coupled with a variety of specific data files.
This model deals with two things: nutrients and foods. Thus we begin an AMPL model by declaring sets of each:
set NUTR;
set FOOD;
Next we need to specify the numbers required by the model. Certainly a positive cost should be given for each food:
param cost {FOOD} > 0;
We also specify that for each food there are lower and upper limits on the number of packages in the diet:
param f_min {FOOD} >= 0;
param f_max {j in FOOD} >= f_min[j];
Notice that we need a dummy index j
to run over FOOD
in the declaration of f_max
, in order to say that the maximum for each food must be greater than or equal to the corresponding minimum.
To make this model somewhat more general than our examples so far, we also specify similar lower and upper limits on the amount of each nutrient in the diet:
param n_min {NUTR} >= 0;
param n_max {i in NUTR} >= n_min[i];
Finally, for each combination of a nutrient and a food, we need a number that represents the amount of the nutrient in one package of the food. Such a “product” of two sets is written by listing them both:
param amt {NUTR,FOOD} >= 0;
References to this parameter require two indices. For example, amt[i,j]
is the amount of nutrient i
in a package of food j
.
The decision variables for this model are the numbers of packages to buy of the different foods:
var Buy {j in FOOD} >= f_min[j], <= f_max[j];
The number of packages of some food j
to be bought will be called Buy[j]
; in any acceptable solution it will have to lie between f_min[j]
and f_max[j]
.
The total cost of buying a food j
is the cost per package, cost[j]
, times the number of packages, Buy[j]
. The objective to be minimized is the sum of this product over all foods j
:
minimize Total_Cost: sum {j in FOOD} cost[j] * Buy[j];
Similarly, the amount of a nutrient i
supplied by a food j
is the nutrient per package, amt[i,j]
, times the number of packages Buy[j]
. The total amount of nutrient i
supplied is the sum of this product over all foods j
:
sum {j in FOOD} amt[i,j] * Buy[j];
To complete the model, we need only specify that each such sum must lie between the appropriate bounds. Our constraint declaration begins
subject to Diet {i in NUTR}:
to say that a constraint named Diet[i]
must be imposed for each member i
of NUTR
. The rest of the declaration gives the algebraic statement of the constraint for nutrient i
: the variables must satisfy
n_min[i] <= sum {j in FOOD} amt[i,j] * Buy[j] <= n_max[i]
A “double inequality” like this is interpreted in the obvious way: the value of the sum in the middle must lie between n_min[i]
and n_max[i]
. We can write all together into a model file diet.mod
:
%%writefile diet.mod
set NUTR;
set FOOD;
param cost {FOOD} > 0;
param f_min {FOOD} >= 0;
param f_max {j in FOOD} >= f_min[j];
param n_min {NUTR} >= 0;
param n_max {i in NUTR} >= n_min[i];
param amt {NUTR,FOOD} >= 0;
var Buy {j in FOOD} >= f_min[j], <= f_max[j];
minimize Total_Cost: sum {j in FOOD} cost[j] * Buy[j];
subject to Diet {i in NUTR}:
n_min[i] <= sum {j in FOOD} amt[i,j] * Buy[j] <= n_max[i];
Data file¶
By specifying appropriate data, we can solve any of the linear programs that correspond to the above model. Let’s begin by using the data from the previous example.
The values of f_min
and n_min
are as given originally, while f_max
and n_max
are set, for the time being, to large values that won’t affect the optimal solution. In the table for amt
, the notation (tr) indicates that we have “transposed” the table so the columns correspond to the first index (nutrients), and the rows to the second (foods). Alternatively, we could have changed the model to say
param amt {FOOD,NUTR}
in which case we would have had to write amt[j,i] in the constraint.
The data file diet.dat
is created with:
%%writefile diet.dat
data;
set NUTR := A B1 B2 C ;
set FOOD := BEEF CHK FISH HAM MCH MTL SPG TUR ;
param: cost f_min f_max :=
BEEF 3.19 0 100
CHK 2.59 0 100
FISH 2.29 0 100
HAM 2.89 0 100
MCH 1.89 0 100
MTL 1.99 0 100
SPG 1.99 0 100
TUR 2.49 0 100 ;
param: n_min n_max :=
A 700 10000
C 700 10000
B1 700 10000
B2 700 10000 ;
param amt (tr):
A C B1 B2 :=
BEEF 60 20 10 15
CHK 8 0 20 20
FISH 8 10 15 10
HAM 40 40 35 10
MCH 15 35 15 15
MTL 70 30 15 15
SPG 25 50 25 15
TUR 60 20 15 10 ;
Suppose that model and data are stored in the files diet.mod
and diet.dat
, respectively. Then AMPL is used as follows to read these files and to solve the resulting linear program:
%%ampl_eval
reset; # to clean the previous model
model diet.mod;
data diet.dat;
option solver cbc;
solve;
display Buy;
Naturally, the result is the same as before.
Now suppose that we want to make the following enhancements. To promote variety, the weekly diet must contain between 2 and 10 packages of each food. The amount of sodium and calories in each package is also given; total sodium must not exceed 40,000 mg, and total calories must be between 16,000 and 24,000. All of these changes can be made through a few modifications to the data. Putting this new data in file diet2.dat
, we can run AMPL again:
%%writefile diet2.dat
set NUTR := A B1 B2 C NA CAL ;
set FOOD := BEEF CHK FISH HAM MCH MTL SPG TUR ;
param: cost f_min f_max :=
BEEF 3.19 2 10
CHK 2.59 2 10
FISH 2.29 2 10
HAM 2.89 2 10
MCH 1.89 2 10
MTL 1.99 2 10
SPG 1.99 2 10
TUR 2.49 2 10 ;
param: n_min n_max :=
A 700 20000
C 700 20000
B1 700 20000
B2 700 20000
NA 0 40000
CAL 16000 24000 ;
param amt (tr):
A C B1 B2 NA CAL :=
BEEF 60 20 10 15 938 295
CHK 8 0 20 20 2180 770
FISH 8 10 15 10 945 440
HAM 40 40 35 10 278 430
MCH 15 35 15 15 1182 315
MTL 70 30 15 15 896 400
SPG 25 50 25 15 1329 370
TUR 60 20 15 10 1397 450 ;
%%ampl_eval
reset data; # to reset the data not the model
data diet2.dat;
solve;
display Buy;
The message infeasible problem tells us that we have constrained the diet too tightly; there is no way that all of the restrictions can be satisfied.
AMPL lets us examine a variety of values produced by a solver as it attempts to find a solution. We can use marginal (or dual) values to investigate the sensitivity of an optimum solution to changes in the constraints. Here there is no optimum, but the solver does return the last solution that it found while attempting to satisfy the constraints. We can look for the source of the infeasibility by displaying some values associated with this solution:
%%ampl_eval
display Diet.lb, Diet.body, Diet.ub;
For each nutrient, Diet.body
is the sum of the terms amt[i,j] * Buy[j]
in the constraint Diet[i]
. The Diet.lb
and Diet.ub
values are the “lower bounds” and “upper bounds” on the sum in Diet[i]  in this case, just the values
n_min[i]and
n_max[i]`. We can see that the diet returned by the solver does not supply enough vitamin B2, while the amount of sodium (NA) has reached its upper bound.
At this point, there are two obvious choices: we could require less B2 or we could allow more sodium. If we try the latter, and relax the sodium limit to 50,000 mg, a feasible solution becomes possible (let statement permits modifications of data):
%%ampl_eval
let n_max["NA"] := 50000;
solve;
display Buy;
This is at least a start toward a palatable diet, although we have to spend $118.06, compared to $88.20 for the original, less restricted case. Clearly it would be easy, now that the model is set up, to try many other possibilities. Section 11.3 of the AMPL book describes ways to quickly change the data and resolve.
One still disappointing aspect of the solution is the need to buy 5.36061 packages of beef, and 9.30605 of spaghetti. How can we find the best possible solution in terms of whole packages? You might think that we could simply round the optimal values to whole numbers  or integers, as they’re often called in the context of optimization  but it is not so easy to do so in a feasible way. Using AMPL to modify the reported solution , we can observe that rounding up to 6 packages of beef and 10 of spaghetti, for example, will violate the sodium limit:
%%ampl_eval
let Buy["BEEF"] := 6;
let Buy["SPG"] := 10;
display Diet.lb, Diet.body, Diet.ub;
You can similarly check that rounding the solution down to 5 of beef and 9 of spaghetti will provide insufficient vitamin B2. Rounding one up and the other down doesn’t work either. With enough experimenting you can find a nearby allinteger solution that does satisfy the constraints, but still you will have no guarantee that it is the leastcost allinteger solution.
AMPL does provide for putting the integrality restriction directly into the declaration of the variables:
var Buy {j in FOOD} integer >= f_min[j], <= f_max[j];
This will only help, however, if you use a solver that can deal with problems whose variables must be integers. CBC solver can handle these so called integer programs. If we add integer to the declaration of variable Buy
as above, save the resulting model in the file dieti.mod
, and add the higher sodium limit to diet2a.dat
, then we can resolve as follows:
%%writefile dieti.mod
set NUTR;
set FOOD;
param cost {FOOD} > 0;
param f_min {FOOD} >= 0;
param f_max {j in FOOD} >= f_min[j];
param n_min {NUTR} >= 0;
param n_max {i in NUTR} >= n_min[i];
param amt {NUTR,FOOD} >= 0;
var Buy {j in FOOD} integer >= f_min[j], <= f_max[j];
minimize Total_Cost: sum {j in FOOD} cost[j] * Buy[j];
subject to Diet {i in NUTR}:
n_min[i] <= sum {j in FOOD} amt[i,j] * Buy[j] <= n_max[i];
%%writefile diet2a.dat
set NUTR := A B1 B2 C NA CAL ;
set FOOD := BEEF CHK FISH HAM MCH MTL SPG TUR ;
param: cost f_min f_max :=
BEEF 3.19 2 10
CHK 2.59 2 10
FISH 2.29 2 10
HAM 2.89 2 10
MCH 1.89 2 10
MTL 1.99 2 10
SPG 1.99 2 10
TUR 2.49 2 10 ;
param: n_min n_max :=
A 700 20000
C 700 20000
B1 700 20000
B2 700 20000
NA 0 50000
CAL 16000 24000 ;
param amt (tr):
A C B1 B2 NA CAL :=
BEEF 60 20 10 15 938 295
CHK 8 0 20 20 2180 770
FISH 8 10 15 10 945 440
HAM 40 40 35 10 278 430
MCH 15 35 15 15 1182 315
MTL 70 30 15 15 896 400
SPG 25 50 25 15 1329 370
TUR 60 20 15 10 1397 450 ;
%%ampl_eval
model dieti.mod;
data diet2a.dat;
option solver cbc;
solve;
display Buy;
Since integrality is an added constraint, it is no surprise that the best integer solution costs about $1.24 more than the best “continuous” one. But the difference between the diets is unexpected; the amounts of 3 foods change, each by two or more packages. In general, integrality and other “discrete” restrictions make solutions for a model much harder to find.
Generalizations to blending, economics and scheduling¶
Your personal experience probably suggests that diet models are not widely used by people to choose their dinners. These models would be much better suited to situations in which packaging and personal preferences don’t play such a prominent role — for example , the blending of animal feed or perhaps food for college dining halls.
The diet model is a convenient, intuitive example of a linear programming formulation that appears in many contexts. Suppose that we rewrite the model in a more general way:
set INPUT; # inputs
set OUTPUT; # outputs
param cost {INPUT} > 0;
param in_min {INPUT} >= 0;
param in_max {j in INPUT} >= in_min[j];
param out_min {OUTPUT} >= 0;
param out_max {i in OUTPUT} >= out_min[i];
param io {OUTPUT,INPUT} >= 0;
var X {j in INPUT} >= in_min[j], <= in_max[j];
minimize Total_Cost: sum {j in INPUT} cost[j] * X[j];
subject to Outputs {i in OUTPUT}:
out_min[i] <= sum {j in INPUT} io[i,j] * X[j] <= out_max[i];
The objects that were called foods and nutrients in the diet model are now referred to more generically as inputs and outputs. For each input j
, we must decide to use a quantity X[j]
that lies between in_min[j]
and in_max[j]
; as a result we incur a cost equal to cost[j]·X[j]
, and we create io[i,j]·X[j]
units of each output i
. Our goal is to find the leastcost combination of inputs that yields, for each output i
, an amount between out_min[i]
and out_max[i]
.
In one common class of applications for this model, the inputs are raw materials to be mixed together. The outputs are qualities of the resulting blend. The raw materials could be the components of an animal feed, but they could equally well be the crude oil derivatives that are blended to make gasoline, or the different kinds of coal that are mixed as input to a coke oven. The qualities can be amounts of something (sodium or calories for animal feed), or more complex measures (vapor pressure or octane rating for gasoline), or even physical properties such as weight and volume.
In another wellknown application, the inputs are production activities of some sector of an economy, and the outputs are various products. The in_min
and in_max
parameters are limits on the levels of the activities, while out_min
and out_max
are regulated by demands. Thus the goal is to find levels of the activities that meet demand at the lowest cost. This interpretation is related to the concept of an economic equilibrium.
In still another, quite different application, the inputs are work schedules, and the outputs correspond to hours worked on certain days of a month. For a particular work schedule j
, io[i,j]
is the number of hours that a person following schedule j
will work on day i
(zero if none), cost[j]
is the monthly salary for a person following schedule j
, and X[j]
is the number of workers assigned that schedule. Under this interpretation, the objective becomes the total cost of the monthly payroll, while the constraints say that for each day i
, the total number of workers assigned to work that day must lie between the limits out_min[i]
and out_max[i]
. The same approach can be used in a variety of other scheduling contexts, where the hours, days or months are replaced by other periods of time.
Although linear programming can be very useful in applications like these, we need to keep in mind the assumptions that underlie the LP model. We have already mentioned the “continuity” assumption whereby X[j]
is allowed to take on any value between in_min[j]
and in_max[j]
. This may be a lot more reasonable for blending than for scheduling.
As another example, in writing the objective as
sum {j in INPUT} cost[j] * X[j]
we are assuming “linearity of costs”, that is, that the cost of an input is proportional to the amount of the input used, and that the total cost is the sum of the inputs’ individual costs.
In writing the constraints as
out_min[i] <= sum {j in INPUT} io[i,j] * X[j] <= out_max[i]
we are also assuming that the yield of an output i
from a particular input is proportional to the amount of the input used, and that the total yield of an output i
is the sum of the yields from the individual inputs. This “linearity of yield” assumption poses no problem when the inputs are schedules, and the outputs are hours worked. But in the blending example, linearity is a physical assumption about the nature of the raw materials and the qualities, which may or may not hold. In early applications to refineries, for example, it was recognized that the addition of lead as an input had a nonlinear effect on the quality known as octane rating in the resulting blend.
AMPL makes it easy to express discrete or nonlinear models, but any departure from continuity or linearity is likely to make an optimal solution much harder to obtain. At the least, it takes a more powerful solver to optimize the resulting mathematical programs.
Bibliography¶
George B. Dantzig, “The Diet Problem.” Interfaces 20, 4 (1990) pp. 4347. An entertaining account of the origins of the diet problem.
Robert Fourer, David M. Gay, and Brian W. Kernighan, “AMPL: A Modeling Language for Mathematical Programming (2nd edition).” Cengage Learning (2002).
Susan Garner Garille and Saul I. Gass, “Stigler’s Diet Problem Revisited.” Operations Research 49, 1 (2001) pp. 113. A review of the diet problem’s origins and its influence over the years on linear programming and on nutritionists.
Said S. Hilal and Warren Erikson, “Matching Supplies to Save Lives: Linear Programming the Production of Heart Valves.” Interfaces 11, 6 (1981) pp. 4856. A less appetizing equivalent of the diet problem, involving the choice of pig heart suppliers.