6-1
Chapter 6 Integer Programming
Section 6.1 Introduction to Integer Programming
Integer programming problem:
a linear programming problem in which the domains of all (or some) of the variables
are restricted to be integral.
For example
1, Transportation problem
The units to be shipped are automobiles, refrigerators and so on.
2, Production problem
The units to be produced are homes, swimming pools and so on.
Difficulties
A most obvious possible solution technique:
Initially ignore the integral restriction, solve the linear programming problem first,
then round off the calculated solution to integral values.
This method fails in the following example.
Example
Maximize f ( x1 , x2 )  3x1  13x2
Subject to
2 x1  9 x2  40
11x1  8 x2  82
x1 , x2  0 and integral
Ignoring the integral restriction, we apply the simplex algorithm.
6-2
Table 6.1
x3
x4
x2
x4
x1
x2
x3
x4
2
11
-3
2
9
9
-8
-13
1
0
0
1
9
8
9
13
9
11
115
8
115
167
115
0
1
0
1
115
9
0
1
9
0

x2
0
1
x1
1
0
0
0
0
1
0
2
115
9
115
1
115

Slack
variables:
x 3 , x4
40
82
0
4
4
9
5
9
7
57
9
2
2
5
1
9
5
4
58
5
117
4
Maximum value of f for the simple linear programming problem = 58 ,
5
1
2
attained at (9 , 2 )
5
5
round off to
(9, 2), (10, 2), (10, 3) or (9, 3).
all are not feasible
The simplex algorithm has provided us with no useful information.
optimal solution
infeasible
6-3
For the problem with integral restriction, the actual solution (only two variables) can
be obtained graphically (see Figure 6.1).
All of feasible solutions:
36 lattice points (both coordinates are integrals).
The coefficients (in f):
3 > 0, 13 > 0
 possible optimal solution points:
(2,4), (6,3), (8,2)
f (2,4) = 58
f (6,3) = 57
f (8,2) = 50
 optimal solution point = (2,4).
Problem Set 6.1: 2; 4.
6-4
Section 6.2 Models with Integer Programming Formulations
The Allocation of Resource in Discrete Amounts
Problem:
to utilize a given amount of a resource in a way that maximizes profit
or gain.
Supposition: to each of alternative methods of utilizing the resource,
only multiples of fixed sized lots can be allocated.
Integrally restricted variables have to be introduced to represent the number of lots.
Example 1
An investment firm has $100,000 to invest:
A construction project
Portfolio of stocks
Shares of a stock
Question:
Investment
$48,000
$19,000/unit
$1750/share
How would the firm invest to maximize profit?
Variables:
1, if the constructi on project is invested
x1  
0, if not
x2  number of units of the portfolio bought
x3  number of shares of the stock bought
Integer programming problem:
2900 x1  1100 x2  95x3
Maximize
Subject to
48,000 x1  19,000 x2  1,750 x3  100,000
0  x1  1, x2 , x3  0 and x1, x2 , x3 integral .
Profit
$2,900
$1,100/unit
$95/share
6-5
Example 2 (cargo loading problem)
An airplane can carry up to w lb of cargo.
There are n different items:
Item i: weight ai lb, profit c i dollars.
Question:
What items should be shipped to maximize profit?
Variables
1, if the ith item is to be shipped
xi  
0, if not
Integer programming problem
n
c x
Maximize:
i 1
i
i
Subject to:
n
a x
i 1
i
i
w
0  xi  1 and integral, i  1,..., n .
Two more restrictions:
1. Item 2 cannot be shipped unless Item 1 is also shipped. Add a constraint: x2  x1 .
2. Item 3 and Item 4 cannot be shipped together. Add a constraint: x3  x4  1 .
Modification of the Objective Function
An example – The Fixed Charge Problem
In an operation, there is a fixed charge or cost for a process only if that process
is used.
For Example
If
A machine is to be used
A new market is to be opened
Warehouses are to be used
there could be
A setup cost
An initial market development cost
A rental cost
To formulate models, introduce for each fixed charge
1, if the fixed charge is to be assessed
an auxiliary variable= 
0, if not
Then
(the fixed charge)(the auxiliary variable)
is added to
the objective function,and
an inequality is added to
the constraints
in order to maintain the “on or off” role of the auxiliary variable.
6-6
Example 3
A machine shop has three processes to produce two parts. There are
weekly limits on
raw material, and
labor;
weekly fixed demands for two parts;
operation cost (per hour) for Process i : c i ;
setup and maintenance charge (fixed charge) d i for Process i if Process i is
utilized.
Question:
What is the weekly operation to minimize the total cost which meets the weekly
demands?
Variables:
xi = number of hours per week that Process i is used.
Auxiliary variables:
0, xi  0
yi  
1, xi  0
Objective function:
c1 x1  c2 x2  c3 x3  d1 y1  d 2 y2  d 3 y3
Constraints
for restrictions on raw material and labor, impose
“  ” type inequalities in the xi ' s ;
for weekly demands for two parts, impose
“  ” type inequalities in the xi ' s ;
in order to regulate the yi ' s , suppose
xi  40 ,
estimate an obvious upper bound for xi
then we add to constraints:
xi
40
0  yi  1 and integral, i  1,2,3
yi 
Note
If Process i is utilized in the optimal operation
x
 i  0  yi  1 .
40
If Process i is not used
 yi is permitted to be 0 or 1.
Minimization of the objective function
 yi = 0
 the “on or off” role of yi is achieved.
6-7
Example 4
A distributor supplies n retail outlets, and he may consider to use some warehouses.
Known data:
d j (units) = monthly requirement of Outlet j
s i (units) = storage capacity of Warehouse i
ri (dollars) = monthly rental fee of Warehouse i
cij (dollars) = cost to ship 1 unit from Warehouse i to Outlet j
(it is known that the total capacity of warehouses is larger than the total demand of
retail outlets.)
Question:
What warehouse should be utilized to minimize the total cost?
Variables:
x ij = number of units shipped monthly from Warehouse i to Outlet j.
Auxiliary variables:
0, warehouse i is not utilized
yi  
1, warehouse i is utilized
And we know that warehouse i is used if and only if
x
ij
 0.
j
Integer programming problem:
Minimize:
 cij xij   ri yi
i, j
i
Subject to:
x
ij
 si , i  1,..., m
j
x
ij
i
 d j , j  1,..., n
x
ij
yi 
j
si
, i  1,..., m
0  xij , 0  yi  1 and integral.
(1)
Example 4 (continued)
In Example 4, if both Warehouses 1 and 2 are utilized, then the monthly rent can be
reduced by $R.
Introduce another “on or off” variable
0, y1  y2  2
z
1, y1  y2  2
Then
-Rz
is added to
the objective function
6-8
y1  y2
2
0  z  1 and integral
z
are added to
the constraints.
Constraints.
Example 5a
A company produces products A&B
A
B
M(material)
(units)
5
8
Machine Time
(min)
2
4
weekly available 110min of machine time at no cost
weekly purchase price of material M
$2/unit for 1st 150 units
$3/unit for additional 100 units
Problem:
Formulate a model for maximizing profit.
Variables:
x1 = number of A’s to be produced
x2 = number of B’s to be produced
x3 = number of units of M purchased at $2/unit
x4 = number of units of M purchased at $3/unit
Linear programming problem
25 x1  45 x2  2 x3  3x4
Maximize:
Subject to:
5 x1  8 x2  x3  x4
2 x1  4 x2  110
x3  150
x4  100
x1 , x2 , x3 , x4  0
Note
(2)
Here we do not use auxiliary discrete variables.
The purchase price for the first 150 units is cheaper
 in any optimal solution, x4  0 only if x3  150 .
Selling Price
(dollars)
25
45
6-9
Example 5b
The purchase price of M is changed to be
$2/unit for 1st 150 units
$1.50/unit for next 100 units
Variables:
x1 , x2 , x3 as before
x4 = number of units of M purchased at $1.50/unit
Objective function in (2) is changed to be
25x1  45x2  2 x3  1.50 x4
In order to force the use of the first 150 units of M before the use of the less expensive
last 100 units. Introduce an auxiliary variable
0, x3  150 (cannot use the cheap material)
y1  
1, x3  150 (can use the cheap material)
and x4  0 only if y1  1 .
Integer programming problem:
25x1  45x2  2 x3  1.5x4
Maximize:
Subject to:
5 x1
2 x1
 8 x2
 4 x2
x3
y1
x4

x3  x 4

110

150
 x3 / 150
 100 y1
x4  0 only if y1  1
x1 , x2 , x3 , x4  0
0  y1  1 and integral
only if x3  150
Example 5c
Suppose the amount of M is unlimited, and the price at
$2/unit
for
1st 150 units
$1.50/unit
for
next 100 units
$1/unit
for
any number of units purchased over the first 250 units
Variables:
x1 , x2 as before
x3 = number of units of M purchased at $2/unit
x4 = number of units of M purchased at $1.5/unit
x5 = number of units of M purchased at $1/unit
6-10
Auxiliary variables:
0,
y1  
1,
x3  150
x3  150
0,
y2  
1,
,
x4  100
x4  100
Integer programming problem:
Maximize:
25 x1  45 x2  2 x3  1.5 x4  1x5
Subject to:
5 x1  8 x2  x3  x4  x5
2 x1  4 x2  110
x3
 150
y1  x3 / 150,
y 2  x4 / 100
x4  100 y1 ,
x5  245 y 2
0  x1 , x2 , x3 , x4 , x5
0  y1 , y 2  1 and integral
Note
Induction of the constraint: x5  245y2
We need to have a upper bound for x5 .
1st constraint
x5  5 x1  8 x2  x3  x4

nd
2 constraint
x1  55,
x2  27.5

x1 , x2  0

x5
 5(55)  8(27.5)  x3

495
 x3
On the other hand
x5  0 only if x3  150 and x4  100

x5  495  150  100  245
Notice
y2  1 only if x4  100
then the constraint
x5  245y2
allows x5 to be a positive value (  245 ).
 x4
 x4
245 is an upper bound of x5
6-11
The Problem of Alternative Constraints
Situation:
variables must satisfy one (or more) set (sets) from two (or more)
alternative sets of constraints
For example,
In
the manufacture of a certain product
the bidding for various contracts
the utilization of some resources
Possibilities
to produce either none or an amount
exceeding some minimal batch size
the bids are either 0 or  some lower
bound
One of several methods may be
employed, each method corresponds to a
system of constraints reflecting the
limited supplies and required demands
To formulate models, introduce auxiliary variables equal to either 0 or 1, use these
variables to modify the system of constraints.
Example 6
Suppose a variable xi must be either 0 or  bi (lower bound).
Introduce an “on or off” variable yi , and ask that
xi  0 if yi  0, and xi  bi if yi  1 .
This requirement can be realized by adding constraints:
xi  mi yi
Suppose in any feasible solution xi  mi
xi  bi yi
0  yi  1 and integral
Note yi can equal only 0 or 1.
If yi  0  xi  0
xi is properly restricted
If yi  1  xi  bi
(when yi  1 , the 1st inequality xi  mi has no significance)
In fact if only one variable, say x1 , has such restriction, then we may not need to
introduce the on and off variable y1 , because we can first solve the problem with the
request x1  0 , then solve the problem again by replacing x1  0 with the constraint
x1  b1 , and at last compare the optimal values of the two cases and choose the better
solution. However, if many variables, say x1 , x2 ,..., x10 , have such restriction, then we
should introduce 0-1 variables y1 , y2 ,..., y10 by the above method, for otherwise there
would be 210  1024 possible cases to compare.
Problem Set 6.2 : 6; 11.
6-12
Section 6.3 Gomory’s Cutting Plane Algorithm
Pure integer programming problem:
a linear programming problem with integral restrictions on all of the variables.
Fundamental idea underlying Gemory’s Cutting Plane Algorithm:
add one constraint to the problem at a time if in the solution some variables do
not have integer value;
eventually have a linear programming problem with an optimal solution with
integral coordinates.
Process of the algorithm:
First, solve the original linear programming problem ignoring the integral
restrictions.
Second, if this solution has all integral coordinates, it is also an optimal
solution to the integer programming problem. If not, generate a new constraint with
two fundamental properties:
(1)
(2)
the non-integral optimal solution will not satisfy this constraint.
all integral feasible solution will satisfy this constraint.
Note
The new constraint essentially cuts off a subset of the feasible region for the
case of continuous variables.
Contains no feasible integral solutions
Then, add the new constraint to the problem and proceed to solve the expanded
problem as before. (again ignore the integral requirement)
Example 1
Minimize:
x1  3x2
Subject to:
x1  x2  2
2 x1  4 x2  15
x1 , x2  0 and integral
Introduce integrally restricted slack variables:
x3 , x4 .
Use the simplex algorithm to solve the problem ignoring the integral restrictions.
6-13
Table 6.2
x3
x4
x3
x2
x1
x2
x3
x4
1
2
1
3
2
1
2
5
2
-1
4
-3
1
0
0
0
1
1
0
0
0
0
1
0
1
4
1
4
3
4
2
15
0
23
4
15
4
45
4
Minimal value is attained at
15
3
3
x1  0, x2 
 3 , ( x3  5 , x4  0)
4
4
4
Non-integral
Generate a new constraint:
Select a constraint equation with non-integral constant term, say the first one:
3
1
23
x1  x3  x4 
.
2
4
4
Separate all constants into their integral and fractional parts:
1
1
3
(1  ) x1  x3  x4  5 
2
4
4
1
1
3

x1  x4   5  x1  x3
2
4
4
For integral solutions, the right side and therefore the left side must be integral.
Notice that all variables  0
1
1
3
3

x1  x4   
2
4
4
4
1
1
3

x1  x4   0 and integral.
2
4
4
Let this be the new constraint, and change it to an equation by introducing a slack
variable x5 , i.e.,
1
1
3
x5  x1  x4  ,
2
4
4
then x5  0 and integral.
6-14
The expanded problem is:
5
3
45
z  x1  x4 
Minimize:
2
4
4
Subject to
3
1
x1
 x3  x4
2
4
1
1
x1  x2
 x4
2
4
1
1
x1
 x4  x5
2
4
x1 , x2 , x3 , x4 , x5  0 and integral
23
4
15

4
3

4

former optimal solution
15
23
x1  x4  0, x2  , x3 
4
4
does not satisfy
Table 6.3
x3
x2
x5
x3
x2
x4
x1
3
2
1
2
1

2
5
2
1
0
2
1
x2
x3
0
1
1
0
0
0
0
0
0
1
0
0
1
0
0
0
x5
x4
1
4
1
4
1

4
3
4
0
0
1
0
0
0
1
0
1
1
-4
3
23
4
15
4
3

4
45
4
5
3
3
9
Proceed as above:
Solve the expanded problem ignoring integral restriction with the Dual Simplex
Algorithm (Table 6.3).
Multiplying the new constraint by (-1), the basic variables are
x3 , x2 , x5 .
After one pivot operation, we obtain minimal value = -9, attained at x1  0, x2  3 .
Geometrically,
x2
integral solution
x1
6-15
The added constraint
2 x1  x4  3  0
1
1
3
x1  x4   0
2
4
4
x4  15  2 x1  4 x2 (2nd row in Table 6.2)
2 x1  (15  2 x1  4 x2 )  3

4 x2  12

x2  3.
The new constraint x2  3 cuts off
3
(0, 3 )
the original non-integral optimal solution
4
from the feasible set, but does not exclude any feasible integral points.
We now consider general case.
Generation of new constraints:
Suppose bi is not integral. The ith constraint is
a x
ij
j
 bi
(1)
j
Let [a] be the greatest integer in a (i.e., not exceeding a), then
a-[a] is the fractional part of a:
0  a  [a]  1 .
6-16
Denote
(1) 
f ij  aij  [aij ]
f i  bi  [bi ]
 [a
ij
]  f ij x j  bi   f i
j
or
f
ij
x j  f i  bi   [aij ]x j
j
(2)
j
For any integral solution to the original constraints
the right side of (2) is integral. So,
the left side of (2) is also integral.
Notice that all x j  0, 0  f ij  1, and 0  f i  1 , then
any integral feasible solution must satisfy:
 fij x j  f i   f i  1
j

f
ij
x j  f i  0 and integral
(3)
j
~this is the new constraint
Note 1
Note 2
Note 3
Adding (3), we still have a pure integer programming problem.
Any feasible integral solution to the original constraints satisfies (3).
The optimal basic feasible solution from the final tableau does not
satisfy (3). In fact f i  0 (since bi is not integral).
For any basic variable x j , in column j, aij  0 or 1, hence f ij  0 ;
For any nonbasic variable x j , x j  0 ;

f
ij
x j  fi   fi  0 .
j
Note 4
If several of bi are not integral, in practice, choosing bi with the
largest fractional value is easy to apply and usually quite effective.
Gomory’s Cutting Plane Algorithm
for solving the pure integer programming problem
Minimize c x
(or Maximize c x )
Subject to
Ax  b
x  0 and integral
Algorithm
1.
Solve the linear programming problem ignoring the integral restriction on x.
If this solution has all integral coordinates
 it is an optimal solution to the original problem.
6-17
2.
Otherwise a new constraint is added to the problem.
a. Select a non-integral bi with the largest fractional value.
n
b.
Suppose bi is selected and the corresponding equation is
a x
j 1
   f ij x j  x   f i
ij
j
 bi
j
f ij  aij  [aij ] , j  1,2,..., n
where
f i  bi  bi 
x is a new slack variable, x  0 and integral.
Add this constraint to the problem and return to step 1. Use the Dual
Simplex Algorithm to solve the corresponding linear programming problem.
c.
Example 2
Minimize:
x1  2x2
Subject to:
2 x1  x2  5
 4 x1  4 x2  5
x1 , x2  0 and integral
Integrally restricted slack variables: x3 , x4
Apply the Simplex Algorithm
Table 6.4
x1
x2
x3
x4
x4
2
-4
1
1
4
○
-2
1
0
0
0
1
0
x3
3
0
1
x2
-1
1
0
-1
0
0
x1
1
0
x2
0
1
0
0
x3
Add a new constraint
1
1
1
 x3  x4  x5   .
3
6
2
slack variable
1
3
1
3
1
3
5
5
0
1
4
15
4
1
4
1
2
1

12
1
6
5
12
5
4
5
2
5
4
5
2
15
4

Non-integral
solution
(select the second row)
6-18
Apply the Dual Simplex Algorithm
detailed steps:
1
1
5
2nd equation :
1x2  x3  x4 
3
6
2
1
1
1
fractional

0 x2  x3  x4   0
3
6
2
part of the
1
1
1
coefficients

x3  x4   x5 ,
3
6
2
& integral
x5  0, integral
Table 6.5
x3
1
3
1
3
1

3
1
3
x1
x2
x1
1
0
x2
0
1
x5
0
0
0
0
x1
1
0
0
x2
0
1
0
x3
0
0
1
0
0
0
Add a new constraint (select the first row)
3
3
 x4  x6  
4
4
slack variable
 1
 4   1
1  1
    
4  4
3

4
Apply the Dual Simplex Algorithm.
x4
1

12
1
6
1

6
5
12
1

4
0
1
2
1
4
x5
0
0
1
0
1
1
-3
1
5
4
5
2
1

2
15
4
3
4
2
3
2
13
4
Non-integral
solution
6-19
Table 6.6
x1
x2
x3
x1
1
0
0
x2
0
1
0
x.3
0
0
1
x6
0
0
0
0
0
0
x1
1
0
x2
0
x3
x4
x5
x6
1
0
1
0
-3
0
0
1
1
0
x4
1

4
0
1
2
3

4
1
4
0
0
1
1
0
0
1
0
0
1
0
-3
0
0
0
1
0
0
0
0
0
1
Minimal value = -3
attained at x1  1, x2  2
1
3
0
2
3
4

3
1
3

3
4
2
3
2
3

4
13
4
1
2
1
1
3
Optimal solution to the original problem
all are integers
Geometrically
1st added constraint
1
1
1
x3  x4   0
3
6
2
1
1
1
(5  2 x1  x2 )  (5  4 x1  4 x2 )   0
3
6
2

x2  2
second point
first point
optimal integral solution (third point)
first cutting plane
second cutting plane
6-20
2nd added constraint
3 3
  x4  0
4 4

3 3
  (5  4 x1  4 x2 )  0
4 4
 x1  x2  1
Problem Set 6.3 : 1; 4; 5.
6-21
Section 6.4 A Branch and Bound Algorithm
Basic process of the Branch and Bound Algorithm
(for solving integer programming problem):
First, solve the problem ignoring the integral restriction.
If there is no feasible solution, the original problem also has no feasible
solution. Stop.
If the optimal solution is integral, it is also optimal for the original problem.
Stop.
If the optimal solution has non-integral coordinates, then proceed the
following iterations:
1.
Generate two new problems (branching process):
each is obtained by adding a new constraint to the previous problem.
arbitrarily select a variable, say x j ,
x j  bi
non-integral
two new constraints:
x j  [bi ]
x j  [bi ]  1
Note 1
Note 2
Note 3
2.
The previous non-integral optimal solution is not feasible for
both new problems.
Any previous integral feasible solution is feasible for one of
these two new problems.
The integral optimal solution to the previous problem could be
contained in either problem.
Solve the new problems ignoring the integral restriction by using the Dual
Simplex Algorithm.

If either problem has no feasible solution
 eliminate from considering this problem.

If either problem has an integral optimal solution
 this solution is feasible for the original integer programming
problem, and it provides a bound for the optimal value of the original
problem (a lower bound for Max problem, an upper bound for Min
problem).

Eliminate from consideration any problem generated through the
branching process with an optimal value no better than this bound.

If there is no other problem generated through the branching process
with optimal value better than this bound,
6-22
 this solution is optimal for the original integer programming
problem.
3.
Go to the next iteration with the remaining problems.
Notes 4
If two or more branches remain open, two frequently used rules
to select the problem are
(1)
(2)
the problem with the most favorable optimal value.
the problem most recently generated.
Example 1 Example 2, Section 6.3
Minimize:
z  x1  2x2
Subject to:
2 x1  x2  5
 4 x1  4 x2  5
x1 , x2  0 and integral
Ignoring the integral restriction, we have
3
Min z  3
4
1
1
attained at x1  1 , x2  2
4
2
Non-integral
Min
x1  2, x2  1 is an
integral feasible
solution to Problem 1
an upper bound for
the optimal value of
Problem 1 is 0
6-23
Arbitrarily select x1 and use
x1  1
x1  2
to form two new problems. Use the Dual Simplex Algorithm to solve these two new
problems ignoring the integral restrictions (see Figure 6.7).
1
Min z in problem 2 A  3  0  Min z in problem 2 B
2
better integral solutions may be contained in Problem 2A, branch again off

Problem 2A using x2 , we get problems 3A and 3B:
Continuing by branching at Problem 3A leads to Figure 6.9.
6-24
0  3
eliminate
this problem
1
 3
2
eliminate
this problem
2
x1  1, x2  2
is an integral feasible solution
upper bound = -3

the optimal value of the original integer programming problem = -3,
attained at x1  1, x2  2 .
Example 2
Maximize:
Subject to:
10 x1
2 x1
z  8x1  15x2
 21x2
 x2
 156
 22
x1 , x2  0 and integral
Use the Branch and Bound Algorithm to solve this problem. See Figure 6.10
6-25
Max
110  117
eliminate
this problem
lower bound
 117
3
Max z in problem 3B  117  117  Max z in problem 3 A, but only by a fractional
5
number
3

value of z at any feasible solution to Problem 3B  117 .
5
As all coefficients of z are integrals
value of z at any integral feasible solution to Problem 3B  117 .


Algorithm terminates.
The optimal value for the original integer programming problem = 117,
attained at x1  9, x2  3 .
Example 3
Maximize:
Subject to:
2 x1
11x1
The example in Section 6.1
z  3x1  13x2
 9 x2
 8 x2
 40
 82
x1 , x2  0 and integral
Use the Branch and Bound Algorithm to solve this problem. See Figure 6.11.
6-26
Max
1
8
 52
2
11
Hold 2A in abeyance
for future consideration
58
57  58
Eliminate
4A
8
 Max z in problem 2 A
11
further consideration of problem 2A is unnecessary
terminate at x1  2, x2  4 with optimal value =58.
Max z in problem 4 B  58  52


Advantages of the Branch and Bound Algorithm
1.
The algorithm generates feasible integral solutions as it proceeds, along with
estimates on how close to the optimal value these solutions might be

for a large and complicated problem, if we are unable to complete the
computation (maybe because of limited computer capacities), the algorithm may still
provide an integral feasible solution, which is not necessarily optimal but adequate for
our purposes.
2.
The method can be used to handle many different cases, for example it can
solve some problems in which only a part of variables should have integral values, or
some variables are 0-1 binary variables.
6-27
When we stop further branching at a particular problem?
1. the problem has no feasible solution;
2. the problem has an optimal integral solution;
3. the bounds so far we obtained show that this problem cannot contain the
optimal solution of the original problem.
Problem Set 6.4: 1(b), (d), (e).
(use LINGO to solve each sub-problem)