TWO HYPOTHETICAL EXAMPLES
1.
A furniture manufacturer produces two types of desks: Standard and Executive.
These desks are sold to an office furniture wholesaler, and for all practical
purposes, there is an unlimited market for any mix of these desks, at least within the
manufacturer’s production capacity. Each desk has to go through four basic
operations: cutting of the lumber, joining of the pieces, prefinishing, and final
finish. Each unit of the Standard desk produced takes 48 minutes of cutting time, 2
hours of joining, 40 minutes of prefinishing, and 5 hours and 20 minutes of final
finishing time. Each unit of the Executive desk requires 72 minutes of cutting, 3
hours of joining, 2 hours of prefinishing, and 4 hours of final finishing time. The
daily capacity for each operation amounts to 16 hours of cutting, 30 hours of
joining, 16 hours of prefinishing, and 64 hours of final finishing time. The profit
per unit produced is $40 for the Standard desk and $50 for the Executive desk.
What product mix is optimal?
2.
A firm produces 3 types of refined chemicals: A, B, and C. At least 4 tons of A, 2
tons of B, and 1 ton of C have to be produced per day. The inputs used are
compounds X and Y. Each ton of X yields 1/4 ton of A, 1/4 ton of B, and 1/12 ton
of C. Each ton of Y yields 1/2 ton of A, 1/10 ton of B, and 1/12 ton of C.
Compound X costs $250 per ton, compound Y $400 per ton. The cost of processing
is $250 per ton of X and $200 per ton of Y. Amounts produced in excess of daily
requirements have no value as the products undergo chemical changes if not used
immediately. The problem is to find the minimum cost input mix.
15.066J
21
Summer 2003
E
(0, 40/3)
4/5 S + 6/5 E <= 16
(20, 0)
S
E
2 S + 3 E <= 30
(0, 10)
(15, 0)
15.066J
S
22
Summer 2003
E
2/3 S + 2 E <= 16
(0, 8)
(24, 0)
S
E
(0, 16)
16/3 S + 4 E <= 64
(12, 0)
15.066J
S
23
Summer 2003
E
(0, 16)
(0, 40/3)
(0, 10)
(0, 8)
(12, 0)
(15, 0)
(20, 0)
(24, 0)
S
15.066J
24
Summer 2003
E
(0, 8)
(6, 6)
(9, 4)
(12, 0)
(0, 0)
15.066J
25
S
Summer 2003
E
Z = 800
Z = 560
(0, 8)
(6, 6)
(9, 4)
Z = 400
(0, 0)
(12, 0)
S
15.066J
26
Summer 2003
Y
(0, 8)
X/4 + Y/2 >= 4
(16, 0)
15.066J
27
X
Summer 2003
(0, 20)
Y
X/4 + Y/10 >= 2
X
(8, 0)
Y
(0, 12)
X/12 + Y/12 >= 1
(12, 0)
15.066J
28
X
Summer 2003
Y
(0, 20)
Z=9K
Z = 6.4K
(16/3, 20/3)
(8, 4)
Z=4K
(16, 0)
X
15.066J
29
Summer 2003
OBSERVATIONS ON THE GRAPHICAL METHOD
FOR SOLVING A LINEAR PROGRAM
1.
The set of feasible plans is called the feasible set or feasible region. It is always
polygonal or polyhedral.
2.
A constraint is said to be binding or active if it is satisfied at equality at the
optimal solution.
3.
Every equality (=) constraint is binding.
4.
A constraint is said to be inactive or non binding if it is satisfied with strict
inequality at the optimal solution.
5.
Geometrically, an active constraint is one that passes through the optimal
solution.
6.
Geometrically, an inactive constraint is one that does not pass through the
optimal solution.
7.
There may be more than one optimal solution to a linear program, i.e., there
may be multiple optima.
8.
A linear program is unbounded if there exist feasible plans with arbitrarily
large (if a max problem) or arbitrarily small (if a min problem) objective values.
If this happens in practice, the formulation has been mis-specified.
9.
A linear program is infeasible if there is no plan that satisfies all of the
constraints.
15.066J
30
Summer 2003