J E Beasley

OR-Notes are a series of introductory notes on topics that fall under the broad heading of the field of operations research (OR). They were originally used by me in an introductory OR course I give at Imperial College. They are now available for use by any students and teachers interested in OR subject to the following conditions.

A full list of the topics available in OR-Notes can be found here.

Integer programming formulation examples

Capital budgeting extension

For the integer programming problem given before related to capital budgeting suppose now that we have the additional condition that either project 1 or project 2 must be chosen (i.e. projects 1 and 2 are mutually exclusive). To cope with this condition we enlarge the IP given above in the following manner.

This condition is an example of an either/or condition "either project 1 or project 2 must be chosen". In terms of the variables we have already defined this condition is "either x1 = 1 or x2 = 1". The standard trick for dealing with either/or conditions relating to two zero-one variables is to add to the IP the constraint

To verify that this mathematical description is equivalent to the verbal description "either x1 = 1 or x2 = 1" we use the fact that x1 and x2 are both zero-one variables.

Choose any one of these two variables (x1, say) and tabulate, for each possible value of the chosen variable, the meaning of the corresponding mathematical and verbal descriptions as below.

Variable value   Mathematical description   Verbal description 
                 x1 + x2 = 1                 "either x1 = 1 or x2 = 1"

x1 = 0           x2 = 1                      x2 = 1 

x1 = 1           1 + x2 = 1                  x2 = 0 
                 i.e. x2 = 0 

It is clear from this tabulation that the mathematical and verbal descriptions are equivalent with the proviso that we have interpreted the condition "either x1 = 1 or x2 = 1" as meaning that the case x1 = 1 and x2 = 1 (both projects 1 and 2 chosen) cannot occur. If this is not so (i.e. both projects 1 and 2 can be chosen) then we replace the constraint x1 + x2 = 1 by the constraint x1 + x2 >= 1 (i.e. at least one of projects 1 and 2 must be chosen).

Note here that, in general, we can regard formulating a problem as translating a verbal description of the problem into an equivalent mathematical description. What we have presented above is a similar process - translating a specific "either ... or ..." verbal description into an equivalent mathematical description. Whilst it is not usual to go to the lengths of a complete tabulation (as shown above) to verify that the verbal and mathematical descriptions are equivalent a complete tabulation is helpful in clarifying thought and detecting incorrect mathematical descriptions.
Other problems

The main areas in which IP is used in practice include:

Integer programming example

Recall the blending problem dealt with before under linear programming. To remind you of it we reproduce it below.

Blending problem

Consider the example of a manufacturer of animal feed who is producing feed mix for dairy cattle. In our simple example the feed mix contains two active ingredients and a filler to provide bulk. One kg of feed mix must contain a minimum quantity of each of four nutrients as below:

Nutrient       A      B     C     D
gram           90     50    20    2

The ingredients have the following nutrient values and cost

                        A      B       C       D      Cost/kg
Ingredient 1 (gram/kg)  100    80      40      10     40
Ingredient 2 (gram/kg)  200    150     20      -      60

What should be the amounts of active ingredients and filler in one kg of feed mix?

Blending problem solution


In order to solve this problem it is best to think in terms of one kilogram of feed mix. That kilogram is made up of three parts - ingredient 1, ingredient 2 and filler so: let
x1 = amount (kg) of ingredient 1 in one kg of feed mix
x2 = amount (kg) of ingredient 2 in one kg of feed mix
x3 = amount (kg) of filler in one kg of feed mix
where x1 >= 0, x2 >= 0 and x3 >= 0.


x1 + x2 + x3 = 1

100x1 + 200x2 >= 90 (nutrient A)
80x1 + 150x2 >= 50 (nutrient B)
40x1 + 20x2 >= 20 (nutrient C)
10x1 >= 2 (nutrient D)

Note the use of an inequality rather than an equality in these constraints, following the rule we put forward in the Two Mines example, where we assume that the nutrient levels we want are lower limits on the amount of nutrient in one kg of feed mix.


Presumably to minimise cost, i.e.

minimise 40x1 + 60x2

which gives us our complete LP model for the blending problem.

Suppose now we have the additional conditions:

Give the complete MIP formulation of the problem with these two new conditions added.


To cope with the condition that if x2>=0 we have a fixed cost of 15 incurred we have the standard trick of introducing a zero-one variable y defined by

y = 1 if x2>=0 
  = 0 otherwise 


and add the additional constraint

In this case it is easy to see that x2 can never be greater than one and hence our additional constraint is x2<= y.

To cope with condition that need only satisfy three of the four nutrient constraints we introduce four zero-one variables zi (i=1,2,3,4) where

zi = 1 if nutrient constraint i (i=1,2,3,4) is satisfied
   = 0 otherwise 

and add the constraint

and alter the nutrient constraints to be

The logic behind this change is that if a zi=1 then the constraint becomes the original nutrient constraint which needs to be satisfied. However if a zi=0 then the original nutrient constraint becomes

which (for the four left-hand sides dealt with above) is always true and so can be neglected - meaning the original nutrient constraint need not be satisfied. Hence the complete (MIP) formulation of the problem is given by

minimise 40x1 + 60x2 + 15y 
subject to 
x1 + x2 + x3 = 1
100x1 + 200x2 >= 90z1 
80x1 + 150x2 >= 50z2 
40x1 + 20x2 >= 20z3 
10x1 >= 2z4 
z1 + z2 + z3 + z4 >= 3
x2 <= y
zi = 0 or 1    i=1,2,3,4 
y = 0 or 1
xi >= 0    i=1,2,3 

Integer programming example

In the planning of the monthly production for the next six months a company must, in each month, operate either a normal shift or an extended shift (if it produces at all). A normal shift costs £100,000 per month and can produce up to 5,000 units per month. An extended shift costs £180,000 per month and can produce up to 7,500 units per month. Note here that, for either type of shift, the cost incurred is fixed by a union guarantee agreement and so is independent of the amount produced.

It is estimated that changing from a normal shift in one month to an extended shift in the next month costs an extra £15,000. No extra cost is incurred in changing from an extended shift in one month to a normal shift in the next month.

The cost of holding stock is estimated to be £2 per unit per month (based on the stock held at the end of each month) and the initial stock is 3,000 units (produced by a normal shift). The amount in stock at the end of month 6 should be at least 2,000 units. The demand for the company's product in each of the next six months is estimated to be as shown below:

Month   1     2     3     4     5     6 
Demand  6,000 6,500 7,500 7,000 6,000 6,000 

Production constraints are such that if the company produces anything in a particular month it must produce at least 2,000 units. If the company wants a production plan for the next six months that avoids stockouts, formulate their problem as an integer program.

Hint: first formulate the problem allowing non-linear constraints and then attempt to make all the constraints linear.



The decisions that have to be made relate to:

Hence let:

xt = 1 if we operate a normal shift in month t (t=1,2,...,6) 
   = 0 otherwise 
yt = 1 if we operate an extended shift in month t (t=1,2,...,6) 
   = 0 otherwise 
Pt (>= 0) be the amount produced in month t (t=1,2,...,6) 

In fact, for this problem, we can ease the formulation by defining three additional variables - namely let:

zt = 1 if we switch from a normal shift in month t-1 to an extended 
       shift in month t (t=1,2,...,6) 
   = 0 otherwise 
It be the closing inventory (amount of stock left) at the end 
   of month t (t=1,2,...,6)

wt = 1 if we produce in month t, and hence from the production 
       constraints Pt >= 2000 (t=1,2,...,6) 
   = 0 otherwise (i.e. Pt = 0) 

The motivation behind introducing the first two of these variables (zt, It) is that in the objective function we will need terms relating to shift change cost and inventory holding cost. The motivation behind introducing the third of these variables (wt) is the production constraint "either Pt = 0 or Pt >= 2000", which needs a zero-one variable so that it can be dealt with using the standard trick for "either/or" constraints.

In any event formulating an IP tends to be an iterative process and if we have made a mistake in defining variables we will encounter difficulties in formulating the constraints/objective. At that point we can redefine our variables and reformulate.


We first express each constraint in words and then in terms of the variables defined above.

xt + yt <= 1    t=1,2,...,6 

Note here that we could not have made do with just one variable (xt say) and defined that variable to be one for a normal shift and zero for an extended shift (since in that case what if we decide not to produce in a particular month?).

Although we could have introduced a variable indicating no shift (normal or extended) operated in a particular month this is not necessary as such a variable is equivalent to 1-xt-yt.

Pt <= 5000xt + 7500yt    t=1,2,...,6 

Note here the use of addition in the right-hand side of the above equation where we are making use of the fact that at most one of xt and yt can be one and the other must be zero.

It >= 0    t=1,2,...,6 

closing stock = opening stock + production - demand

where I0 = 3000. Hence letting Dt = demand in month t (t=1,2,...,6) (a known constant) and assuming

we have that

It = It-1 + Pt - Dt    t=1,2,...,6 

As noted above this equation assumes that we can meet demand in the current month from goods produced that month. Any time lag between goods being produced and becoming available to meet demand is easily incorporated into the above equation. For example for a 2 month time lag we replace Pt in the above equation by Pt-2 and interpret It as the number of goods in stock at the end of month t which are available to meet demand i.e. goods are not regarded as being in stock until they are available to meet demand. Inventory continuity equations of the type shown are common in production planning problems.

I6 >= 2000 

Here we make use of the standard trick we presented for "either/or" constraints. We have already defined appropriate zero-one variables wt (t=1,2,...,6) and so we merely need the constraints

Pt <= Mwt       t=1,2,...,6 
Pt >= 2000wt    t=1,2,...,6 

Here M is a positive constant and represents the most we can produce in any period t (t=1,2,...,6). A convenient value for M for this example is M = 7500 (the most we can produce irrespective of the shift operated).

The obvious constraint is

zt = xt-1yt    t=1,2,...,6 

since as both xt-1 and yt are zero-one variables zt can only take the value one if both xt-1 and yt are one (i.e. we operate a normal shift in period t-1 and an extended shift in period t). Looking back to the verbal description of zt it is clear that the mathematical description given above is equivalent to that verbal description. (Note here that we define x0 = 1 (y0 = 0)).

This constraint is non-linear. However we are told that we can first formulate the problem with non-linear constraints and so we proceed. We shall see later how to linearise (generate equivalent linear constraints for) this equation.


We wish to minimise total cost and this is given by

SUM{t=1,...,6}(100000xt + 180000yt + 15000zt + 2It)

Hence our formulation is complete.

Note that, in practise, we would probably regard It and Pt as taking fractional values and round to get integer values (since they are both large this should be acceptable). Note too here that this is a non-linear integer program.


In practice a model of this kind would be used on a "rolling horizon" basis whereby every month or so it would be updated and resolved to give a new production plan.

The inventory continuity equation presented is quite flexible, being able to accommodate both time lags (as discussed previously) and wastage. For example if 2% of the stock is wasted each month due to deterioration/pilfering etc then the inventory continuity equation becomes It = 0.98It-1 + Pt - Dt. Note that, if necessary, the objective function can include a term related to 0.02It-1 to account for the loss in financial terms.

In order to linearise (generate equivalent linear constraints) for our non-linear constraint we again use a standard trick. Note that that equation is of the form

where A, B and C are zero-one variables. The standard trick is that a non-linear constraint of this type can be replaced by the two linear constraints

To see this we use the fact that as B and C take only zero-one values there are only four possible cases to consider:

B   C   A = BC     A <= (B+C)/2    A >= B+C-1 
        becomes    becomes         becomes 
0   0   A = 0      A <= 0          A >= -1 
0   1   A = 0      A <= 0.5        A >= 0 
1   0   A = 0      A <= 0.5        A >= 0 
1   1   A = 1      A <= 1          A >= 1 

Then, recalling that A can also only take zero-one values, it is clear that in each of the four possible cases the two linear constraints (A <= (B+C)/2 and A >= B+C-1) are equivalent to the single non-linear constraint (A=BC).

Returning now to our original non-linear constraint

this involves the three zero-one variables zt, xt-1 and yt and so we can use our general rule and replace this non-linear constraint by the two linear constraints

    zt <= (xt-1 + yt)/2    t=1,2,...,6
and zt >= xt-1 + yt - 1    t=1,2,...,6 

Making this change transforms the non-linear integer program given before into an equivalent linear integer program.

Integer programming example 1996 MBA exam

A toy manufacturer is planning to produce new toys. The setup cost of the production facilities and the unit profit for each toy are given below. :

Toy   Setup cost (£)   Profit (£)
1     45000            12
2     76000            16

The company has two factories that are capable of producing these toys. In order to avoid doubling the setup cost only one factory could be used.

The production rates of each toy are given below (in units/hour):

            Toy 1   Toy 2
Factory 1   52      38
Factory 2   42      23

Factories 1 and 2, respectively, have 480 and 720 hours of production time available for the production of these toys. The manufacturer wants to know which of the new toys to produce, where and how many of each (if any) should be produced so as to maximise the total profit.



We need to decide whether to setup a factory to produce a toy or not so let fij = 1 if factory i (i=1,2) is setup to produce toys of type j (j=1,2), 0 otherwise

We need to decide how many of each toy should be produced in each factory so let xij be the number of toys of type j (j=1,2) produced in factory i (i=1,2) where xij >=0 and integer.


x11/52 + x12/38 <= 480
x21/42 + x22/23 <= 720

x11 <= 52(480)f11
x12 <= 38(480)f12
x21 <= 42(720)f21
x22 <= 23(720)f22


The objective is to maximise total profit, i.e.

maximise 12(x11+ x21) + 16(x12+ x22) - 45000(f11 + f21) - 76000(f12 + f22)

Note here that the question says that in order to avoid doubling the setup costs only one factory could be used. This is not a constraint. We can argue that if it is only cost considerations that prevent us using more than one factory these cost considerations have already been incorporated into the model given above and the model can decide for us how many factories to use, rather than we artificially imposing a limit via an explicit constraint on the number of factories that can be used.

The above mathematical model could be used in production planning in the following way:

Integer programming example 1995 MBA exam

A project manager in a company is considering a portfolio of 10 large project investments. These investments differ in the estimated long-run profit (net present value) they will generate as well as in the amount of capital required.

Let Pj and Cj denote the estimated profit and capital required (both given in units of millions of £) for investment opportunity j (j=1,...,10) respectively. The total amount of capital available for these investments is Q (in units of millions of £)

Investment opportunities 3 and 4 are mutually exclusive and so are 5 and 6. Furthermore, neither 5 nor 6 can be undertaken unless either 3 or 4 is undertaken. At least two and at most four investment opportunities have to be undertaken from the set {1,2,7,8,9,10}.

The project manager wishes to select the combination of capital investments that will maximise the total estimated long-run profit subject to the restrictions described above.

Formulate this problem using an integer programming model and comment on the difficulties of solving this model. (Do not actually solve it).

What are the advantages and disadvantages of using this model for portfolio selection?



We need to decide whether to use an investment opportunity or not so let xj = 1 if we use investment opportunity j (j=1,...,10), 0 otherwise


SUM{j=1,...,10}Cjxj <= Q

x3 + x4 <= 1
x5 + x6 <= 1

x5 <= x3+ x4
x6 <= x3+ x4

x1 + x2 + x7 + x8 + x9 + x10 >= 2
x1 + x2 + x7 + x8 + x9 + x10 <= 4


The objective is to maximise the total estimated long-run profit i.e.

maximise SUM{j=1,...,10}Pjxj

The model given above is a very small zero-one integer programming problem with just 10 variables and 7 constraints and should be very easy to solve. For example even by complete (total) enumeration there are just 210 = 1024 possible solutions to be examined.

The advantages and disadvantages of using this model for portfolio selection are:

Integer programming example 1994 MBA exam

A food is manufactured by refining raw oils and blending them together. The raw oils come in two categories:

The prices for buying each oil are given below (in £/tonne)

115   128   132   109   114

The final product sells at £180 per tonne. Vegetable oils and non-vegetable oils require different production lines for refining. It is not possible to refine more than 210 tonnes of vegetable oils and more than 260 tonnes of non-vegetable oils. There is no loss of weight in the refining process and the cost of refining may be ignored.

There is a technical restriction relating to the hardness of the final product. In the units in which hardness is measured this must lie between 3.5 and 6.2. It is assumed that hardness blends linearly and that the hardness of the raw oils is:

8.8   6.2   1.9   4.3   5.1

It is required to determine what to buy and how to blend the raw oils so that the company maximises its profit.

The following extra conditions are imposed on the food manufacture problem stated above as a result of the production process involved:

Introducing 0-1 integer variables extend the linear programming model you have developed to encompass these new extra conditions.



We need to decide how much of each oil to use so let xi be the number of tonnes of oil of type i used (i=1,...,5) where i=1 corresponds to VEG1, i=2 corresponds to VEG2, i=3 corresponds to OIL1, i=4 corresponds to OIL2 and i=5 corresponds to OIL3 and where xi >=0 i=1,...,5


x1 + x2 <= 210
x3 + x4 + x5 <= 260

(8.8x1 + 6.2x2 + 1.9x3 + 4.3x4 + 5.1x5)/(x1 + x2 + x3 + x4 + x5) >= 3.5
(8.8x1 + 6.2x2 + 1.9x3 + 4.3x4 + 5.1x5)/(x1 + x2 + x3 + x4 + x5) <= 6.2


The objective is to maximise total profit, i.e.

maximise 180(x1 + x2 + x3 + x4 + x5) - 115x1 - 128x2 - 132x3 - 109x4 - 114x5

The assumptions we make in solving this problem by linear programming are:

Integer program


In order to deal with the extra conditions we need to decide whether to use an oil or not so let yi = 1 if we use any of oil i (i=1,...,5), 0 otherwise


x1 <= 210y1
x2 <= 210y2
x3 <= 260y3
x4 <= 260y4
x5 <= 260y5

y1 + y2 + y3 + y4 + y5 <= 3

xi >= 30yi           i=1,...,5

y4 >= y1
y4 >= y2


The objective is unchanged by the addition of these extra constraints and variables.

Integer programming example 1985 UG exam

A factory works a 24 hour day, 7 day week in producing four products. Since only one product can be produced at a time the factory operates a system where, throughout one day, the same product is produced (and then the next day either the same product is produced or the factory produces a different product). The rate of production is:

Product                                 1   2   3   4 
No. of units produced per hour worked   100 250 190 150 

The only complication is that in changing from producing product 1 one day to producing product 2 the next day five working hours are lost (from the 24 hours available to produce product 2 that day) due to the necessity of cleaning certain oil tanks.

To assist in planning the production for the next week the following data is available:

         Current   Demand (units) for each day of the week 
Product  stock     1     2     3     4     5     6     7 
1        5000      1500  1700  1900  1000  2000  500   500 
2        7000      4000  500   1000  3000  500   1000  2000 
3        9000      2000  2000  3000  2000  2000  2000  500
4        8000      3000  2000  2000  1000  1000  500   500 

Product 3 was produced on day 0. The factory is not allowed to be idle (i.e. one of the four products must be produced each day). Stockouts are not allowed. At the end of day 7 there must be (for each product) at least 1750 units in stock.

If the cost of holding stock is £1.50 per unit for products 1 and 2 but £2.50 per unit for products 3 and 4 (based on the stock held at the end of each day) formulate the problem of planning the production for the next week as an integer program in which all the constraints are linear.



The decisions that have to be made relate to the type of product to produce each day. Hence let:

In fact, for this problem, we can ease the formulation by defining two additional variables - namely let:


x1t + x2t + x3t + x4t = 1    t=1,2,...,7 
Iit >= 0    i=1,...,4 t=1,...,7 

closing stock = opening stock + production - demand

Letting Dit represent the demand for product i (i=1,2,3,4) on day t (t=1,2,...,7) we have

I10 = 5000 
I20 = 7000 
I30 = 9000 
I40 = 8000 

representing the initial stock situation and

Iit = Iit-1 + Pit - Dit    i=1,...,4 t=1,...,7 

for the inventory continuity equation.

Note here that we assume that we can meet demand in month t from goods produced in month t and also that the opening stock in month t = the closing stock in month t-1.

Let Ri represent the work rate (units/hour) for product i (i=1,2,3,4) then the production constraint is

Pit = xit[24Ri]    i=1,3,4 t=1,...,7 

which covers the production for all except product 2 and

P2t = [24R2]x2t - [5R2]x2tx1t-1    t=1,...,7 

i.e. for product 2 we lose 5 hours production if we are producing product 2 in period t and we produced product 1 the previous period. Note here that we initialise by

x10 = 0 

since we know we were not producing product 1 on day 0. Plainly the constraint involving P2t is non-linear as it involves a term which is the product of two variables. However we can linearise it by using the trick that given three zero-one variables (A,B,C say) the non-linear constraint

can be replaced by the two linear constraints

Hence introduce a new variable Zt defined by the verbal description

Zt = 1 if produce product 2 on day t and product 1 on day t-1
   = 0 otherwise 


Zt = x2tx1t-1    t=1,...,7 

and our non-linear equation becomes

P2t = [24R2]x2t - [5R2]Zt    t=1,...,7 

and applying our trick the non-linear equation for Zt can be replaced by the two linear equations

Zt <= (x2t + x1t-1)/2    t=1,...,7  
Zt >= x2t + x1t-1 - 1    t=1,...,7
Ii7 >= 1750    i=1,...,4 

Note that, in practise, we would probably regard (Iit) and (Pit) as taking fractional values and round to get integer values (since they are both quite large this should be acceptable).


We wish to minimise total cost and this is given by

SUM{t=1,...,7}(1.50I1t + 1.50I2t + 2.50I3t + 2.50I4t) 

Note here that this program may not have a feasible solution, i.e. it may simply not be possible to satisfy all the constraints. This is irrelevant to the process of constructing the model however. Indeed one advantage of the model may be that it will tell us (once a computational solution technique is applied) that the problem is infeasible.

Integer programming example 1987 UG exam

A company is attempting to decide the mix of products which it should produce next week. It has seven products, each with a profit (£) per unit and a production time (man-hours) per unit as shown below:

Product   Profit (£ per unit)   Production time (man-hours per unit)
1         10                    1.0 
2         22                    2.0 
3         35                    3.7
4         19                    2.4 
5         55                    4.5 
6         10                    0.7 
7         115                   9.5  

The company has 720 man-hours available next week.

Incorporate the following additional restrictions into your integer program (retaining linear constraints and a linear objective):


Let xi (integer >=0) be the number of units of product i produced then the integer program is


10x1 + 22x2 + 35x3 + 19x4 + 55x5 + 10x6 + 115x7 

subject to

1.0x1 + 2.0x2 + 3.7x3 + 2.4x4 + 4.5x5 + 0.7x6 + 9.5x7 <= 720 
xi >= 0 integer i=1,2,...,7 


z7 = 1 if produce product 7 (x7 >= 1) 
   = 0 otherwise 



     x7 <= (720/9.5)z7
i.e. x7 <= 75.8z7
so   x7 <= 75z7 (since x7 is integer) 

Let y2 = number of units of product 2 produced in excess of 100 units then add the constraints

This will work because x2 and y2 have the same objective function coefficient but y2 requires longer to produce so will always get more flexibility by producing x2 first (up to the 100 limit) before producing y2.


Z = 1 if produce both product 3 and product 4 (x3 >= 1 and x4 >= 1) 
  = 0 otherwise 

z3 = 1 if produce product 3 (x3 >= 1) 
   = 0 otherwise 

z4 = 1 if produce product 4 (x4 >= 1) 
   = 0 otherwise 


and add the two constraints


x3 <= 194z3 and x4 <= 300z4 

and relate Z to z3 and z4 with the non-linear constraint

which we linearise by replacing the non-linear constraint by the two linear constraints