1. Chapter 3 Linear Programming Applications
The process of problem formulation
Production Problem
Marketing and media applications
Financial Applications
Transportation Problem

2. The process of problem formulation
Provide a detailed verbal description of the problem
Determine the overall objective that appears to be relevant.
Determine the factors (constraints) that appear to restrict the attainment of the objective function.
Define the decision variables and state their units of measurement.
Using these decision variables, formulate an objective function.
Formulate a mathematical equations for each of the identified constraints.
Check the entire formulation to ensure linearity.

3. Production Problem
The demand for a particular item for a Cosmopolitan Company is known to be:
The production capacity and the cost per unit in these months is, respectively:
and the storage cost is $5 per month.
At the beginning of January there are 50 items in store, and there should be 80 items left in store at the end of April.
Formulate a linear program to find the best production level for each month.
Month
January
February
March
April
Demand
100
110
120
Month
January
February
March
April
Production
120
150
100
Cost ($) 9 10 8 7

4. Production Problem
Let xi be the amount of production in month i (i=1,2,3,4)
 Production capacity constraints:
x1 ≤ 120
x2 ≤ 150
x3 ≤ 150
x4 ≤ 100
Amount left over at the end of each month (positive)
End of January :50+x1-100 >=0  x1-50 ≥ 0
End of February: x1-50+x2-110 >=0  x1+x2-160 ≥ 0
End of March: x1+x2-160+x3-120 >= 0  x1+x2+x3-280 ≥ 0
End of April: x1+x2+x3-280+x4-110 >= 80  x1+x2+x3+x4-390 ≥ 80

5. Production Problem Objective function:
Production Cost= 9x1+10x2+8x3+7x4
Storage Cost = 5 [(x1-50)+(x1+x2-160)+(x1+x2+x3-280)+(x1+x2+x3+x4-390)]
Minimize Z = Production Cost + Storage Cost
= 29x1+25x2+18x3+13x4-4400
Non-Negativity Constraint : x1, x2, x3, x4 ≥ 0

6. Marketing Applications
One application of linear programming in marketing is media selection.
LP can be used to help marketing managers allocate a fixed budget to various advertising media.
The objective is to maximize reach, frequency, and quality of exposure.
Restrictions on the allowable allocation usually arise during consideration of company policy, contract requirements, and media availability.

7. Media Selection SMM Company recently developed a new instant
salad machine, has $282,000 to spend on advertising. The product is to be initially test marketed in the Dallas
area. The money is to be spent on
a TV advertising blitz during one
weekend (Friday, Saturday, and
Sunday) in November.
The three options available
are: daytime advertising,
evening news advertising, and
Sunday game-time advertising. A mixture of one-minute TV spots is desired.

8. Media Selection Estimated Audience
Ad Type Reached With Each Ad Cost Per Ad
Daytime , $5,000
Evening News , $7,000
Sunday Game , $100,000
SMM wants to take out at least one ad of each type (daytime, evening-news, and game-time). Further, there are only two game-time ad spots available. There are ten daytime spots and six evening news spots available daily. SMM wants to have at least 5 ads per day, but spend no more than $50,000 on Friday and no more than $75,000 on Saturday.

9. Media Selection Define the Decision Variables
DFR = number of daytime ads on Friday
DSA = number of daytime ads on Saturday
DSU = number of daytime ads on Sunday
EFR = number of evening ads on Friday
ESA = number of evening ads on Saturday
ESU = number of evening ads on Sunday
GSU = number of game-time ads on Sunday

10. Media Selection Define the Objective Function
Maximize the total audience reached:
Max (audience reached per ad of each type)
x (number of ads used of each type)
Max 3000DFR +3000DSA +3000DSU +4000EFR
+4000ESA +4000ESU GSU

11. Media Selection Define the Constraints
Take out at least one ad of each type:
(1) DFR + DSA + DSU > 1
(2) EFR + ESA + ESU > 1
(3) GSU > 1
Ten daytime spots available:
(4) DFR < 10
(5) DSA < 10
(6) DSU < 10
Six evening news spots available:
(7) EFR < 6
(8) ESA < 6
(9) ESU < 6

12. Media Selection Define the Constraints (continued)
Only two Sunday game-time ad spots available:
(10) GSU < 2
At least 5 ads per day:
(11) DFR + EFR > 5
(12) DSA + ESA > 5
(13) DSU + ESU + GSU > 5
Spend no more than $50,000 on Friday:
(14) 5000DFR EFR < 50000

13. Media Selection Define the Constraints (continued)
Spend no more than $75,000 on Saturday:
(15) 5000DSA ESA < 75000
Spend no more than $282,000 in total:
(16) 5000DFR DSA DSU EFR
+ 7000ESA ESU GSU7 <
Non-negativity:
DFR, DSA, DSU, EFR, ESA, ESU, GSU > 0

14. Media Selection The Management Scientist Solution
Objective Function Value =
Variable Value Reduced Costs
DFR
DSA
DSU
EFR
ESA
ESU
GSU

15. Media Selection Solution Summary Total new audience reached = 199,000
Number of daytime ads on Friday = 8
Number of daytime ads on Saturday = 5
Number of daytime ads on Sunday = 2
Number of evening ads on Friday = 0
Number of evening ads on Saturday = 0
Number of evening ads on Sunday = 1
Number of game-time ads on Sunday = 2

16. Financial Planning
A bank makes four kinds of loans to its personal customers and these loans yield the following annual interest rates to the bank:
•First mortgage 14%
•Second mortgage 20%
•Home improvement 20%
•Personal overdraft 10%
The bank has a maximum foreseeable lending capability of 250 BD million and is further constrained by the policies:
1. First mortgages must be at least 55% of all mortgages issued and at least 25% of all loans issued (in BD terms)
2. Second mortgages cannot exceed 28% of all loans issued (in BD terms)
3. To avoid public displeasure and the introduction of a new windfall tax the average interest rate on all loans must not exceed 15%.
Formulate the bank's loan problem as an LP so as to maximize interest income whilst satisfying the policy limitations.

17. Financial Planning Define the Decision Variables
Essentially we are interested in the amount (in BD) the bank has loaned to customers in each of the four different areas (not in the actual number of such loans). Hence let
xi = amount loaned in area i in million of BD (where i=1 corresponds to first mortgages, i=2 to second mortgages i=3 to home improvement, and i=4 to personal overdraft)

18. Financial Planning Define the Objective Function
To maximize interest income (which is given above)
Maximize Z=0.14x x x x4
Define the Constraints
(a) Limit on amount lent
x1 + x2 + x3 + x4 ≤ 250
(b) Policy condition 1
x1 ≥ 0.55(x1 + x2)
i.e. first mortgages >= 0.55(total mortgage lending) and also
x1 ≥ 0.25(x1 + x2 + x3 + x4)
i.e. first mortgages >= 0.25(total loans)

19. Financial Planning Define the Constraints (continued)
c) policy condition 2
x2 ≤ 0.28(x1 + x2 + x3 + x4)
d) policy condition 3
we know that the total annual interest is 0.14x x x x4 on total loans of (x1 + x2 + x3 + x4). Hence the constraint relating to policy condition (3) is:
0.14x x x x4 ≤ 0.15(x1 + x2 + x3 + x4)
Non-Negativity Constraint:
xi ≥ 0 (i=1,2,3,4)

20. Transportation Problem
The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si) to n destinations (each with a demand dj), when the unit shipping cost from an origin, i, to a destination, j, is cij.
The network representation for a transportation problem with two sources and three destinations is given on the next slide.

21. Transportation Problem
Network Representation 1 d1
c11 1
c12 s1
c13 2 d2
c21 2
c22 s2
c23 3 d3
Sources
Destinations

22. Transportation Problem
Steel Mills in three cities produce the following amounts of steel:
These mills supply steel to four cities, where manufacturing plants have the following demands:
The cost of sending 1 tons of steel from a production location to a plant depends on the distance the truck must travel.
The problem is to determine how many tons of steel to transport from each production mill to each manufacturing plant on a weekly basis to minimize the total cost of transportation.
Formulate the above problem as a linear programming model.
Location A B C
Production (tons)
150
210
320
Plants 1 2 3 4
Demand (tons)
190 70
180
240

23. Transportation tableau
A transportation problem is specified by the supply, the demand, and the shipping costs. So the relevant data can be summarized in a transportation tableau. The transportation tableau implicitly expresses the supply and demand constraints and the shipping cost between each demand and supply point.
From
To Plant 1 2 3 4
Supply (tons) A
$14 $9
$16
$18
150 B
$11 $8 $7
210 C
$12
$10
$22
320
Demand (tons)
190 70
180
240

24. Transportation Problem
Decision Variable:
Let Xij = number of tons of steel shipped from location i to plant j (i = A, B, C; j = 1, 2, 3, 4)
Objective function
Transportation costs of steel shipped from location A
= 14 XA1 + 9 XA2 +16 XA3 +18 XA4
Transportation costs of steel shipped from location B = 11 XB1 + 8 XB2 + 7 XB3 +16 XB4
Transportation costs of steel shipped from location C = 16 XC XC2+ 10 XC3+22 XC4
Therefore, the objective function is:
Minimize XA1 + 9 XA2 +16 XA3 +18 XA XB1 + 8 XB2 + 7 XB XB XC XC2+ 10 XC3+22 XC4

25. Transportation Problem
3. Supply Constraints
Since each supply point has a limited production capacity;
XA1 + XA2 + XA3 + XA4 ≤ 150
XB1 + XB2 + XB3 + XB4 ≤ 210
XC1 + XC2+ XC3+ XC4 ≤ 320

26. Transportation Problem
4. Demand Constraints
Since each supply point has a limited production capacity;
XA1 + XB1 + XC1 ≥ 190
XA2 + XB2 + XC2 ≥ 70
XA3 + XB3 + XC3 ≥ 180
XA4 + XB4 + XC4 ≥ 240
5. Sign Constraints
Xij ≥ 0 (i= A,B,C; j= 1,2,3,4)