INTEGER PROGRAMMING AND LOCATION PROBLEMS
T: Welcome to the next exercise. I hope you enjoyed the previous exercise.
S: It was easy to begin with. Towards the end, it was tough. I request you to make it simple today.
T: I will do so. Can you plot points on a graph sheet if I give you the coordinates?
S: Of course I can. I have not forgotten the mathematics I learnt in high school.
T: I will give you five points. We can call them P1 to P5. The coordinates are P1 = (1, 4), P2 = (1, 5), P3 =
(2, 5), P4 = (3, 2) and P5 = (4, 1). Please plot them on a graph sheet.
S: Yes. They are shown in Figure.4.1
P3
P2
P1
P4
P5
Figure 4.1 – Location of five points
T: You have plotted five points. How many groups are there?
S: I think there are two groups.P1, P2 and P3 form a group while P4 and P5 form the second group.
T: Can you give me a solution with one group?
S: Yes. I can put all the points into a single group. That is the solution. This is easy and trivial.
T: Can you think of 5 groups?
S: Yes, I can. Each point becomes a group. This is also easy and trivial. Surprisingly, you have made it
very easy this time. I also have a feeling that you are going to make it complicated very soon.
T: Don’t worry too much about how easy or tough the questions are. You have given me solutions for 1
group, 2 groups and 5 groups. Can you give me solutions for 3 groups and 4 groups?
S: I will try. It is not as easy as the other cases. If you need 4 groups, I can think of P1 and P2 in one
group and the remaining three points forming a group each. For three groups I can think of P1 and P2,
P3, P4 and P5 as three groups.
T: Let me denote your 3 group solution as {P1, P2}, {P3}, {P4, P5}. Will you accept if I give you another
solution {P1, P4}, {P2, P5}, {P3}.
S: I won’t. My solution is better.
T: On what basis?
S: The distance between the points in a group is lesser in my case.
T: That is exactly what I wanted to hear from you. A group is made up of points that are close to each
other. Groups should be as far away as possible.
S: This principle adequately explains my solution with 2 groups as well as with 3 groups. For example,
the solution {P1, P3, P5}, {P2, P4} is not a good solution for 2 groups.
T: The within group distance should be small and between group distance should be large.
S: How do I measure the within group distance and between group distance? If there are two points in a
group, there is one distance measurement. If there are 3 points in a group there are 3 distances.
T: If there are four points in a group, how many distances are there among the points? Is it six?
S: I had that question and I thought if requesting you to answer that question.
T: One of the ways to handle this issue is to define a representative point for a group. The within group
distance is the distance between the representative point and the points in the group. The distance
between groups is the distance between the representative points.
S: How is the representative point determined?
T: It is either one of the points themselves or it can be the centroid.
S: I remember the centroid of a triangle.
T: The same centroid. You can compute the centroid for any figure that is made up of points.
S: I do not know to compute the centroid if there are four points. This is getting complicated.
T: Do not worry about the centroid. We will instead use one of the points to act as the representative
point. Consider the solution with 5 groups. What are the representative points?
S: The five points themselves are the representative points.
T: What is the within group distance?
S: Zero. The distance between a point and itself is zero. For each group the distance is zero.
T: Now can you give me a solution with four groups?
S: {P1}, {P2}, {P3}, {P4, P5}
T: Is this grouping unique?
S: I could have given multiple solutions. In fact I could have given {P1, P2}, {P3}, {P4}, {P5} also.
T: Can you identify a representative point for each group in solution 1?
S: P1, P2, P3 and P4 can be the representative points. Either P4 or P5 can be the point.
T: Will you now compute the total distance between the given points and the representative points?
S: The distance is zero for the first three points and will be the distance between P4 and P5. I remember
how to compute distance between points on a graph sheet. The distance is √2 = 1.414.
T: Let us avoid square roots and we can assume that the distance is 1.4.
S: The total distance is 1.4
T: What about solution 2?
S: For P3, P4 and P5 the distance is zero. Between P1 and P2 the distance is 1. The total distance is 1. I
would now prefer solution 2 to solution 1 because the within group distances is less.
T: Can you provide the best solution for 3 groups.
S: I will use the solution with four groups and I can merge two groups to get a solution with three
groups. I should identify two groups with the least distance and merge them. The distance between P4
and P5 is 1.4 and if i assume that P2 is the representative point for the group, the distance between {P1,
P2} and P3 is 1. I will merge P3 with {P1, P2}. The solution is {P1, P2, P3}, {P4} and {P5}.
T: Very good. Now you have formed groups hierarchically meaning that you started with five groups,
merged two of them to get 4 groups and again merged two groups to get three groups and so on. The
groups chosen for merger were based on minimum distance between them.
S: Using the above principle my solution for two groups will be {P1, P2, P3}, {P4, P5}. This is because the
distance of 1.4 is lesser than the distance between P4 and the bigger group and between P5 and the
bigger group.
T: You get the same solution that you provided earlier because you grouped them based on your
perception of their closeness (or distance) using the naked eye.
S: Yes. I have learnt a systematic way to get groups. I can start with each point as a group and
progressively reduce the number of groups by merging groups that are nearest to each other.
T: The drawback of this approach is that once points join in the same group, they cannot belong to
different groups subsequently. Many times this is not an undesirable thing also.
S: Are there ways to overcome this aspect? Can points go to different groups as the groups reduce?
T: We revisit the idea of the representative point. Instead of progressively defining the representative
points, we can define as many representative points as the number of groups that we wish to. Can you
think of two representative points?
S: Yes. P2 and P4 are good representative points.
T: By defining the two representative points, you have created two groups, Group 1 that has P2 and
Group 2 has P4. You have to assign the remaining three points to the two groups. Can you assign them
based on minimum distance with the representative point?
S: P1 has distance = 1 with P2 and distance = √8 = 2.8 with P4. P1 will go to group 1. P3 has distance =
1 with P2 and distance = √10 = 3.16 with P4. P3 goes to group 1. P5 has distance of 5 with P2 and
distance = √2 = 1.4 with P4. Therefore P5 will go to group 2. The groups are {P1, P2, P3} and {P4 and
P5}.
T: Can you compute the sum of distances?
S: 1 + 1 + 1.4 = 3.4
T: Instead of choosing P2 and P4 as the representative points, if you had chosen P4 and P5 as the
representative points?
S: That is a bad choice of representative points. I will not choose this pair. I would want the
representative points to be far away.
T: It is a very good observation that the representative points should be as far away as they can be.
However, can you provide the groups for the given representative points?
S: Group 1 will have P4 and group 2 will have P5. P1, P2 and P3 will go to group 1. The groups are {P1,
P2, P3, P4} and {P5}.
T: What is the total distance between the given points and the representative points?
S: The answer is the sum of the distance between P1 and P4, between P2 and P4 and between P3 and
P4. This is equal to √8 + √13 + √10 = 2.8 + 3.6 + 3.16 = 9.56.
T: This solution is not better than the previous one where the distance was 3.4
S: I told you immediately that P4 and P5 provide a bad choice for the representative points.
T: This was validated by the fact that the sum of distances was higher. In other words the best solution is
given by choosing that pair which results in minimum sum of distances.
S: That is not difficult. We can plot the points and from the graph either choose the groups or choose the
representative points.
T: It can get difficult if the points have more than two coordinates or if the distance matrix is given and
the location of points is not given.
S: In such cases should i evaluate the sum of distances for all possible pairs of representative points?
T: That is one way to do this. There are optimization based ways of solving this problem. You will see
some aspects in this chapter.
S: Thank you. This exercise was not very difficult but was interesting.
We begin this chapter by introducing Integer Programming problems. We learn to solve IP
problems using the solver. Few applications that result in IP formulations are presented.
Some of these are discussed in detail in subsequent chapters. In this chapter we also discuss
Location and layout problems after the initial discussion on Integer Programming problems.
In many applications it is necessary to define the variables as integers rather than as
continuous variables which the LP formulations assumed. These lead to a different class of
problems called Integer programming (IP) problems which are solved very differently from
LP problems. IP is used to model several situations where the variables can be integer or
binary (zero-one). Formulations of several real life instances result in Mixed linear integer
programming (MILP) where some variables take integer values while some can take
continuous variables.
In this chapter we study IP problems. We begin with an illustration to explain the concepts.
Illustration 4.1
“Fresh and tasty” bakers make cakes and pastries. On a Sunday morning they have to make
‘plum cake” and “fruit pastries”. They have 12 units of flour and they use 3 units of flour per
cake and 4 units per pastry. They also have 15 units of time in the oven that day and each
cake requires 5 units in the oven and each pastry requires 3 units of time in the oven. They
sell each cake for Rs 25 and each pastry for Rs 20. How many cakes and pastries they should
make to maximize their sale?
The formulation is familiar and can be written as follows
Let us define X1 as the number of cakes made and X2 as the number of pastries made.
Maximize 25X1 + 20X2
Subject to
3 X1  4 X 2  12
5 X1  3 X 2  15
X1 , X 2  0 .
We solve the LP using the solver to get the solution X1 = 2.1818, X2 = 1.3636 with Z = Rs.
81.81
Unlike in the earlier example, we obtain fractional solutions to the LP. We cannot make 2.18
plum cakes. We can make only an integer number of the plum cakes. We temporarily try
and resolve the issue by checking if a rounded solution is feasible. We consider the four
possible rounded solutions
1.
2.
3.
4.
(3, 2) is infeasible because it violates both the constraints.
(3, 1) violates both the constraints.
(2, 2) violates both the constraints.
(2, 1) is feasible and satisfies both the constraints. The value of the objective
function is 70.
Many times we will be tempted to believe that this is the best possible solution and
implement it. Is this the optimum solution to the situation when the variables have to be
integer valued?
When we have a LP problem where the decision variables are further constrained to the
integers is called an Integer Programming problem or Linear Integer Programming problem.
It is customary to call it Integer Programming and we use an abbreviation IP to represent
Integer Programming.
How to solve IPs using the solver?
We know that the LP becomes an IP when we add an integer restriction on the variables.
We formulate the problem as a LP in the solver. We go to the options and choose both
1. Assume Linear model
2. Assume non negative.
After the two constraints are added, click ‘add’ again and the window to add a constraint
pops up. Click on the position B2 (where we have defined values for X 1) on the LHS of the
constraint. Use the drop box to choose integer (int) and it appears on the RHS. Press OK and
add another constraint to restrict B3 (variable X2) to integer. Now solve the problem.
The solver gives the optimum solution to the IP which is X1 = 3 with Z = 75.
If we had followed the earlier procedure by choosing the best rounded solution we would
have used (2,1) with Z = 70.
Few observations and questions
1. Would at least one of the rounded solutions be feasible always?
The answer is No. We can construct examples to show that all rounded solutions are
infeasible but the IP has an optimum.
Example:
Maximize X1 + X2
Subject to
7X1 - 5X2  7
-12X1 + 15X2  7
X1, X2  0 and integer.
The LP solution after relaxing the integer restrictions is X1 = 28/9, X2 = 133/45 Z = 273/45.
There are four rounded integer solutions given by (4,3), (3,3), (4,2) and (3,2). Points (4,3)
and (4,2) violate the first constraint while points (3,3) and (3,2) violate the second
constraint. All the rounded integer solutions are infeasible. The IP optimum to this problem
is (2,2) with Z = 4.
2. It is also to be noted that if the problem has m constraints and if all the m variables
in the LP optimum have fractional values, we can get 2m rounded solutions because
each variable can take two values. This creates evaluation of a large number of
solutions for feasibility.
3. In example 7.1 the optimum solution to the LP was X1 = 2.1818, X2 = 1.3636 with Z =
81.81. The IP optimum was X1 = 3 with Z = 75. Is it necessary that at least one
variable will take a rounded value (upper or lower integer value)?
No.
Example
Maximize 30X1 + 25X2
Subject to
4X1 + 3X2  12
3X1 + 4X2  15
X1, X2  0 and integer.
The LP optimum is X1 = 3/7, X2 = 24/7 with Z = 690/7. The IP optimum was X1 = 3 with
Z = 90.
4. For a maximization problem, the IP optimum Z*IP ≤ Z*LP. The LP optimum is an upper
estimate (or upper bound for the IP optimum. For a minimization problem, the IP
optimum Z*IP ≥ Z*LP. The LP optimum is a lower estimate (or lower bound for the IP
optimum.
Let us consider some examples that involve the decision variables to be binaries or integers.
Illustration 4.2
Consider a shuttle service that send car to pick up and drop customers. There are six orders
and these require 7, 6, 5, 4, 3 and 3 hours. Each car is available for 10 hours. What is the
minimum cars to which we can assign the tasks?
Let Xij = 1 if the ith order is assigned to the jth car (Xij = 0 otherwise). We have the following
constraints:
Each order has to go to only one car. This gives us the constraint  X ij  1 for all i. Each car
j
has handle orders such that the time required for the orders does not exceed 10. This gives
us the constraint
t X
i
ij
 10 for all j.
i
The objective function is to minimize the number of cars used. We know that the maximum
number of cars that we would use is 6 because we can give each car an order. This gives us j
= 1 to 6 and 36 variables and 12 constraints. We now define Yj = 1 if car j is chosen and we
6
Minimize  Y j . The final formulation has 42 variables and 12 constraints. The theoretical
j 1
minimum number required is 28/10 = 3. If we can get a solution using three cars, the
optimum solution would show three of the six Yj values to 1 and will allot the order to the
three chosen cars (Yj). We also need to link the Xij and Yj values and modify the constraint to
t X
i
ij
 10Y j .
i
1. One way to solve the problem is to solve it for 3 cars. This will give us 18 X ij variables
and 3 Yj variables indicating 21 variables. There are 9 constraints. If there is a
solution with three cars, we will get this solution.
2. Solving the problem using the excel solver we get a message that solver is unable to
find a solution.
3. We now verify if we have a solution with 4 cars. We now have 24 Xij variables and 4
Yj variables indicating 28 variables. There are 10 constraints. Solving the problem
using the excel solver we get the following optimal solution given in Table 4.1
Table 4.1 – Optimum solution from solver
x11 x12 x13 x21 x22 x23 x31 x32 x33 y1 y2 y3 y4
x41 x42 x43 x51 x52 x53 x61 x62 x63 x14 x24 x34 x44 x54 x64
-0
-0
obj function
constraints
0
0
1
0
-0
1
4
-1
-3
<=
<=
0
0
-2
1
<=
=
0
1
1
1
1
=
=
=
1
1
1
1
1
-2
=
=
<=
1
1
0
1
0
0
0
1
0
0
0
0
0
1
0
1
0
1
0
1
1
0
0
There are four cars as given in the objective function. The optimum solution has variables
X13 = X22 = X31 = X44 = X51 = X64 = 1. Orders 3 and 5 go to car 1 (load = 8), order 2 goes to car 2
(load = 6), order 1 goes to car 3 (load = 7) and orders 4 and 6 go to car 4 (load = 7).
If we had solved with 6 cars (Yj = 1 to 6) with 30 variables and 12 constraints, we would have
got the optimum solution with 4 cars in one trial with the solver.
If we had a closer look at the data and try to fit the orders to a total of 10 hours, we could fit
the first three orders to three cars and the last three orders to a fourth car. We can
formulate the problem now considering 4 cars and this would give us a formulation with 28
variables and 10 constraints (fewer variables and constraints when compared to the 6 car
formulation).
0
1
The best approach therefore is to find a feasible solution with minimum number of cars and
try to solve it for that number rather than to approach the problem from a lower limit on
the number of cars.
If the orders were 7, 6, 5, 4, 3, 2 and we started with 4 cars, the optimum solution is given in
Table 4.2
Table 4.2 – Optimum solution
x11
x12 x13
x41
-7E-12
1
x42
1
0
x43
0
0
x21 x22 x23 x31 x32 x33
y1
y2
y3
y4
x51 x52 x53 x61 x62 x63 x14 x24 x34 x44 x54 x64
0
0 1
1
0
0
1
1
1
0
0
0
0
0 1
0
1
0
0
0
0
0
0
0
obj function
3
constraints
-3E-11
<=
0
-1E-11
-1E-11
1
<=
<=
=
0
0
1
1
1
1
=
=
=
1
1
1
1
1
=
=
1
1
7E-12
<=
0
There are three cars as given in the objective function. The optimum solution has variables
X12 = X23 = X31 = X41 = X53 = X62 = 1. Orders 3 and 4 go to car 1 (load = 9), orders 1 and 6 go to
car 2 (load = 9), orders 2 and 5 go to car 3 (load = 9). Here, we started with four cars and
since a solution with 3 cars was possible, the solver gave the solution.
The problem formulated above is called the one dimensional bin packing problem (Coffman
et al, 1978). We discuss this problem and its variants in detail subsequently in this course.
Illustration 4.3
You want to pick up 3 friends from their homes and bring them to your house. The distances
among the four places A to D (where A is your home and B, C and D are the homes of your
three friends) is given in Table 4.3
Table 4.3 – Distance matrix
A
B
C
A
0
10
8
B
10
0
7
C
8
7
0
D
6
5
9
D
6
5
9
0
We formulate this problem as a binary IP problem (with zero – one variables). Let Xij = 1 if
you visit j immediately after visiting i. Here both i and j take values from 1 to 4 (the four
homes). The constraints are:
1. From a given home i you have to visit another home j. This is given by
X
ij
 1; j  i .
j
2. You have to reach a given home j from some other home i. This is given by
X
ij
 1; j  i
i
3. Assuming that you start from A (your home) you should come back home after
visiting all the other homes. You should not visit only A-C-A or A-C-B-A. These routes
where we come home without completing the visit of all the other homes are called
a sub tour. The complete route (for example, A-C-B-D-A) where we come back after
visiting all other homes is called a tour. Sub tours are to be eliminated and only tours
have to be considered.
Sub tours can be of length 1 (example, A-A), length 2 (example, A-C-A) and length 3
(example, A-D-B-A). X11 = 1 means that we visit home 1 immediately after visiting home 1,
indicating a sub tour of length 1. The best way to eliminate sub tours of length 1 is by
leaving out the Xjj variables. Sub tours of length 2 can be eliminated by the constraint
X ij  X ji  1; j  i .
Sub tours of length 3 are automatically eliminated if we eliminate sub tours of length 1
because for every sub tour of length 3 there has to be one sub tour of length 1.
The objective function is to minimize the total distance travelled and is given by
n
Minimize
n
 d
i 1 j 1
ij
X ij .
The formulation has 12 variables (there are 16 Xij variables and we have eliminated the 4 Xjj
variables) and 14 constraints. The first two set of constraints have three constraints each
while there are 6 sub tour elimination constraints. All the variables are binaries given by X ij =
0,1.
We solve the binary IP and the optimum solution is given in Table 4.4
Table 4.4 – Optimum solution
x12
x13
x14
x21
x23
x24
x31
0
x32
0
x34
1
x41
0
x42
1
x43
0
1
0
0
0
1
0
obj fn
26
constraints
1
=
1
1
1
=
=
1
1
1
=
1
1
1
=
=
1
1
1
1
0
=
=
<=
1
1
1
1
1
1
<=
<=
<=
1
1
1
1
<=
1
0
<=
1
From the table we observe that the minimum distance to be travelled is 26. The solution is
X14 = X23 = X31 = X42 = 1 indicating that the route is 1-4-2-3-1 which is A-D-B-C-A.
The formulation presented above is that of the Travelling Salesman problem (Flood, 1956).
We discuss this problem and its variants in detail in subsequent portions of this course.
Illustration 4.4
Tasty Tea has four outlets in the busy areas of the city. The demand for “Special east tea” in
these areas is 200, 400, 500 and 700 units per day. They wish to locate production units that
send the material to the four outlets. They have identified three potential locations (A to C).
The cost of locating these is 5000, 6000 and 3000. The capacities of these locations are
1100, 1000 and 800 units per day. The cost of transportation from the potential locations to
the outlets per unit is given in Table 4.5
Table 4.5 – Cost of transportation
1
2
3
4
A
3
3
4
5
B
2
3
4
2
C
1
3
4
5
How many production units should they locate and where? How many units are to be
transported from these units to the outlets?
Let Yi = 1 if they locate a production facility at i. Let Xij be the quantities transported from i
to outlet j.
The objective function is to Minimize the sum of location costs and the transportation costs.
3
Minimize
3
4
 fiYi   cij X ij
i 1
i 1 j 1
The constraints are
3
X
i 1
ij
 D j . This ensures that the demands of the outlets are met.
ij
 BiYi . This constraint ensures that the material is sent from the facilities that are
4
X
j 1
created and that the amount sent from a facility does not exceed the capacity.
Yi = 0,1.
There are 3 binary Yi variables and 12 Xij variables. There are 7 constraints.
We solve the problem optimally using the solver. The output is shown in Table 4.6
Table 4.6 – Optimum solution
x21
0
y1
x22
2.89E-17
y2
x23
1
y3
x24
1
x11
x31
0
x12
x32
3.26E-14
x13
x33
0
x14
x34
0
0
obj fn
300
13800
0
700
200
100
500
0
cons
200
>=
200
400
>=
400
500
700
>=
>=
500
700
8.55E-16
3.32E-10
-3.2E-10
<=
<=
<=
0
0
0
The optimum solution suggests that the company start facilities in locations 2 and 3. The
available capacities are 1000 and 800 units. The cost of location is 9000. The demands are
200, 400, 500 and 700 adding to 1800. The optimum solution has variables X31 = 200, X22 =
300, X32 = 100, X33 = 500, X24 = 700. Facility 2 supplies to meet the entire demand of shop 4
and meets part demand of shop 2. Facility 3 meets all the demand of shops 1 and 3 and
meets part demand of shop 2. The total cost of location and transportation is 13800.
We modify the problem such that the entire demand of a shop has to be met from only one
facility. Now, Xij = 1 if facility i meets the entire demand of shop j.
The formulation becomes
3
3
4
Minimize  fiYi   cij D j X ij
i 1
i 1 j 1
3
The constraints are
 X ij  1 and
i 1
4
D X
j 1
j
ij
 BY
i i and Yi, Xij = 0, 1
The optimum solution locates facilities in locations 1 and 3. Facility 1 supplies to shops 2 and
4 while facility 3 supplies to shops 1 and 3. The total cost of location and transportation is
14900. The optimum solution from the solver is shown in Table 4.7
Table 4.7 – Optimum solution
y1
y2
y3
x11
x12
x13
x14
x21
0
x22
1
x23
0
x24
1
x31
0
x32
1
x33
0
x34
1
0
obj fn
0
14900
0
0
1
0
1
0
cons
1
1
=
=
1
1
1
1
=
=
1
1
0
0
-100
<=
<=
<=
0
0
0
The formulation discussed in this section is that of the fixed charge problem (Hirsch and
Dantzig, 1954). This problem and its variants are discussed in detail in subsequent portions
of this chapter.
Illustration 4.5
Four patients are waiting for their appointment with the dentist. The times they are
expected to spend with the dentist are 6, 10, 8 and 6 minutes respectively. They would like
to leave by 10, 15, 20 and 22 minutes respectively. The secretary to the dentist wishes to
send them in the order such that the sum of the extra time (more than their desired leaving
time) they spend in the clinic is minimized. The dentist can attend to only one patient at a
time.
Let the start time of the treatment of patient j be Sj. The processing time is pj for patient j
(given). The completion time of the treatment is therefore Sj + pj. The delay for patient j is Tj
= Sj + pj – Lj (where Lj is the expected leaving time for patient j) if Sj + pj > Lj and 0 if Sj + pj ≤
Lj.
The objective is to
4
Minimize
T
j 1
j
Subject to
Tj  S j  p j  L j
Since the dentist attends to only one patient at a time, for every pair of patients i and j,
either j visits after i or i visits after j. This means that either Si  S j  p j or S j  Si  pi . This


is written as two constraints S j  p j  Si  M  ij and Si  pi  S j  M 1   ij where M is a
large positive constant and δij is a binary variable. Also Tj ≥ 0.
The formulation has 4 Sj variables, four Tj variables and 6 binary variables. There are 4
constraints to compute the delay and 12 constraints (two for each pair of patients).
The formulation was solved optimally using the solver and the optimum solution is shown in
Table 4.8
Table 4.8 – Optimum solution
s1
s2
s3
s4
t1
t2
d12
d13
d14
d23
d24
d34
0
0
6
0
22
0
16
0
0
0
1
1
obj fn
11
cons
0
5
<=
<=
4
5
12
16
<=
<=
12
16
6
>=
6
6
22
22
<=
>=
<=
90
6
92
16
16
>=
<=
6
94
16
16
10
>=
<=
>=
10
92
10
10
<=
94
94
>=
8
94
<=
94
t3
t4
10
0
0
0
In the formulation we used the value M = 100. The optimum solution has S 1 = 0, S2 = 6, S3 =
22 and S4 =16 indicating that the patients are seen in the order 1-2-4-3. The completion
times (or time of leaving) are 6, 16, 22 and 30. The delays are 0, 1, 0, 10 respectively with a
total delay of 11.
(It is customary to use a Earliest due date (EDD) schedule to get a feasible solution. The due
dates are 10, 15, 20 and 22 and the EDD sequence would be 1-2-3-4. The completion times
are 6, 16, 24, 30. The delays are 0, 1, 4 and 8 and the total delay is 13)
The above formulation comes under the broad category of sequencing and scheduling
problems (Baker 1974). Scheduling and sequencing problems are discussed in detail
subsequently in this course.