AACIMP 2009 Summer School lecture by Boris Goldengorin. "Logistics" course. 3rd - 4th hour. Part 2.

1. Operations Research 1
Chapter 3
Introduction to Operations Research
Hillier & Liebermann
Eighth Edition
McGrawHill
Operations Research 1 – 06/07 – Chapter 3 1

2. Contents
• Linear Programming
– Example
– Formulations
– Graphical solution
– The linear programming model
– Additional examples
Operations Research 1 – 06/07 – Chapter 3 2

3. The Wyndor Glass Co. Example
• Three plants for glass products (windows and doors):
– Aluminium frames and hardware are made in Plant 1.
– Wood frames are made in Plant 2.
– Plant 3 produces the glass and assembles the products.
• Capacity is released voor new products:
– Two products are an option: product 1 and product 2.
• Product 1 requires capacity in Plants 1 and 3 only.
• Product 2 requires capacity in Plants 2 and 3 only.
• Both products need capacity in plant 3. What is the most
profitable mix to produce?
Operations Research 1 – 06/07 – Chapter 3 3

4. Formulation as a Linear Programming Problem
X1 = number of batches of product 1 produced per week
X2 = number of batches of product 2 produced per week
Z = total profit per week (in thousands of dollars) from producing these two products
Table 3.1 Production Time per
batch, Hours
Product Production Time
1 2 Available per Week,
Hours
Plant
1 1 0 4
2 0 2 12
X1 and X2 are the decision variables for the model. Using the bottom row of
table 3.1 we obtain Z = 3X1 + 5X3
3 2
2 18
Operations Research batch – Chapter 3
Profit per 1 – 06/07 $ 3,000 $ 5,000 4

5. Mathematical formulation of the LP problem
Maximize Z = 3X1 + 5X2
Subject to restrictions
X1 ≤4
2X2 ≤ 12
3X1 + 2X2 ≤ 18
And
X1 ≥ 0, X2 ≥ 0.
Operations Research 1 – 06/07 – Chapter 3 5

6. Graphical Solution figure 3.1
Figure 3.1
Shaded area shows the
X1 ≤ 4
values of (X1,X2) allowed
by:
X2 ≥ 0
X1 ≥ 0,
X2 ≥ 0,
X1 ≤ 4
X1 ≥ 0
Operations Research 1 – 06/07 – Chapter 3 6

7. Graphical Solution figure 3.2
Figure 3.2
3X1 + 2X2 = 18 Shaded area shows the
X1 =4 set of permissible values
of (X1,X2), called the
2X2 = 12
feasible region.
Operations Research 1 – 06/07 – Chapter 3 7

8. Graphical Solution figure 3.3
Z = 36 = 3X1 + 5X2 (2,6)
Z = 20 = 3X1 + 5 X2
Z = 10 = 3X1 + 5X2
Operations Research 1 – 06/07 – Chapter 3 8

9. Terminology for Linear Programming
TABLE 3.2
Prototype Example General Problem
Production capacities of plants Resources
3 plants m resources
Production of products Activities
2 products n activities
Production rate of product j, xj Level of activity j, Xj
Profit Z Objective function (Overall measure
of performance Z)
Operations Research 1 – 06/07 – Chapter 3 9

10. Data needed for a LP Model
TABLE 3.3 Data needed for a linear programming model involving the allocation of
resources to activities
Resource Usage per Unit of
Activity
Activity Amount of Resource
Resource 1 2 … n available
1 a11 a12 … a1n b1
2 a21 a22 … a2n b2
. .
. … … … … .
m am1 am2 … amn bm
Contribution to 1 – 06/071 – Chapter23
Operations Research c c … cn 10
Z per unit of

11. Symbols used as notation of the various
components commonly used for a LP Model
• Below certain Symbols are listed, along with their interpretation of
the general problem.
– Z = Value of overall measure of performance.
– xj = level of activity j (for j = 1, 2, …, n).
– cj = increase in Z that would result from each unit increase in level of
activity j.
– bi = amount of resource i that is available for allocation to activities (for i
= 1, 2, …, m).
– aij = amount of resource i consumed by each other unit of acivity j.
• The model poses the problem in terms of making decisions about
the levels of the activitys, so x1, x2, …, xn are called the decision
variables. In Table 3.3 the values of ci, bi, and aij (for i = 1, 2, …, m
and j = 1, 2, … , n) are the input constants also referred to as the
parameters of the model.
Operations Research 1 – 06/07 – Chapter 3 11

12. A standard form of the model
• We can now formulate the mathematical model for this
general problem of allocation resources to activities. The
model is to select the values of x1, x2, …, xn so as to
Maximize Z = c1x1 + c2x2 + … + cnxn,
subject to restrictions
a11x1 + a12x2 + … + a1nxn ≤ b1
a21x1 + a22x2 + … + a2nxn ≤ b2
…………………………
am1x1 + am2x2 + … + amnxn ≤ bm,
and
x1 ≥ 0, x2 ≥ 0 , …, xn ≥ 0
Operations Research 1 – 06/07 – Chapter 3 12

13. Summary of terminology in LP 1
objective function = the function being maximized: c1x1 +
c2x2 + … + cnxn.
constraints = restrictions for the objective function.
functional constraints = the first m constraints with a
function of all the variables (ai1x1 + ...+ ainxn) on the left-
hand side.
nonnegativity constraints = the xj ≥ 0 restrictions
Feasible solution = a solution for which all the constraints
are satisfied.
Infeasible solution = a solution for which at least one
constraint is violated.
Feasible region = the collection of all feasible solutions.
Operations Research 1 – 06/07 – Chapter 3 13

14. Summary of terminology in LP 2
Optimal solution = a feasible solution that has the most
favorable value of the objective function.
Most favorable value = the largest value if the objective
function is to be maximized or the smallest value if the
objective function is to be minimized.
It is possible for a problem to have no feasible solutions,
multiple optimal solutions. If there are no optimal
solutions we speak of an unbounded Z or an
unbounded objective
Operations Research 1 – 06/07 – Chapter 3 14

15. Figure 3.4
Maximize Z = 3x1 + 5x2,
Subject to X1 ≤ 4
3x1 + 5x2 ≥ 50
2x2 ≤ 12
3x1 + 2x2 ≤ 18
3x1 + 5x2 ≥ 50
And x1 ≥ 0, x2 ≥ 0
2x2 ≤ 12
3x1 + 2x2 ≤ 18
x1 ≤ 0 Figure 3.4
X1 ≤ 4 The Wyndor Glass Co.
x2 ≥ 0 Problem would have no
solutions if the constraint
3x1 + 5x2 ≥ 50 were
added to the problem.
Operations Research 1 – 06/07 – Chapter 3 15

16. Figure 3.5
Maximize Z = 3x1 + 2x2,
Subject to X1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
And x1 ≥ 0, x2 ≥ 0
Every point on this darker line
segment is optimal, each with Z = 18
Feasible Figure 3.5
region
The Wyndor Glass Co.
Problem would have
multiple solutions if the
objective function were
changed to Z = 3x1 + 2x2
Operations Research 1 – 06/07 – Chapter 3 16

17. Figure 3.6
(4,10), Z = 62
(4,8), Z = 52 Maximize
Z = 3x1 + 5x2 Figure 3.6
Subject to x1 ≤ 4
(4,6), Z = 42 The Wyndor Glass Co.
And x1 ≥ 0, x2 ≥ 0 Problem would have no
Feasible optimal solutions if the
region
(4,4), Z = 32 only functional
constraint were x1 ≤ 4,
because x2 then could
(4,2), Z = 22 be increased indefinitely
in the feasible region
without ever reaching
the maximum value of Z
= 3x1 + 5x2
Operations Research 1 – 06/07 – Chapter 3 17

18. Figure 3.7
(0,6) (2,6)
(4,3)
Figure 3.7
The five dots are the five
(4,0) CPF solutions for the
Wyndor Glass Co.
(0,0) Problem.
Operations Research 1 – 06/07 – Chapter 3 18

19. Assumptions of LP
Proportionality assumption: The contribution of each
activity to the value of the objective function Z is
proportional to the level of the activity xj, as represented
by the cjxj term in the objective function. Similarly, the
contribution of each activity to the left-hand side of each
functional constraint is proportional to the level of activity
xj, as represented by the aijxj term in the constraint.
Consequently, this assumption rules out any exponent
other than 1 for any variable in any term of any function
(whether the objective function or the function on the left-
hand side of a functional constraint) in a linear
programming model.
Operations Research 1 – 06/07 – Chapter 3 19

20. Graphs of Table 3.4
x2
20
18 18
16
14 Proportionality
Satisfied
12 12
11 Case 1
10
8 8 Case 2
7
6 6 6
5 Case 3
4
3
2 2
0 0
0 1 2 3 4 x1
In case 1 through 3 proportionality is violated.
Operations Research 1 – 06/07 – Chapter 3 20

21. Additivity assumption & Table 3.5
Additivity Assumption: Every function in a linear programming
model (whether the objective function or the function on the left-hand
side of a functional constraint) is the sum of the individual
contributions of the respective activities.
• TABLE 3.5 Examples of satisfying or violating additivity for the objective function
Value of Z
Additivity Violated
(x1,x2) Additivity Satisfied Case 1 Case 2
(1,0) 3 3 3
(0,1) 5 5 5
(1,1) 8 9 7
Operations Research 1 – 06/07 – Chapter 3 21

22. Table 3.6
• TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint
Amount of Resource Used
Additivity Violated
(x1,x2) Additivity Satisfied Case 3 Case 4
(2,0) 6 6 6
(0,3) 6 6 6
(2,3) 12 15 10.8
Operations Research 1 – 06/07 – Chapter 3 22

23. Divisibility assumption
Divisibility Assumption: Decision variables in a linear
programming model are allowed to have any values,
including noninteger values, that satisfy the functional
and nonnegativity constraints. Thus, these variables are
not restricted to just integer values. Since each decision
variable presents the level of some activity, it is being
assumed that the activities can be run at fractional
levels.
Operations Research 1 – 06/07 – Chapter 3 23

24. Certainty assumption
Certainty assumption: The value assigned to each
parameter of a linear programming model is assumed to
be a known constant.
Operations Research 1 – 06/07 – Chapter 3 24

25. Regional Planning: The southern confederation of
kibbutzim
• TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim
Kibbutz Usable Land (Acres) Water Allocation (Acre Feet)
1 400 600
2 600 800
3 300 375
• TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim
Crop Maximum Water Consumption Net Return ($ / Acre)
Quota (Acres) (Acre Feet/ Acre)
Sugar beets 600 3 1,000
Cotton 500 2 750
Sorghum 325 1 250
Operations Research 1 – 06/07 – Chapter 3 25

26. Regional Planning (continued): Table 3.10
• TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem
Allocation (Acres)
Kibbutz
Crop 1 2 3
Sugar beets X1 X2 X3
Cotton X4 X5 X6
Sorghum X7 X8 X9
Operations Research 1 – 06/07 – Chapter 3 26