1. MATH143 Lec 4 Quantitative Methods
Charles W Jackson
Andrew Nunekpeku
MATH143 – Fall 2011
6/3/2018

2. Agenda for today Translation of Word Problems into LP mathematical
statements
Another Two Examples, solved in class
Translation to LP math
Organizing our numbers in Excel
Graphical Technique
Identifying Corner Points and Evaluating There
Discovering the Optimum Program
MATH143 – Fall 2011
6/3/2018

3. Linear Programing Allocate some limited resources
One sentence description:
Allocate some limited resources
Between competing activities
In a best possible way
You have to know your business very well in order to formulate these problems. Running by the seat of your pants is insufficient!
MATH143 – Fall 2011
6/3/2018

4. Word Problem Translation
Step One: What are the Decision Variables?
These are the minimum info required to describe what you want done during the period. What are the competing activities?
Step Two: What are the Constraints?
Which linear combinations of activities is limited? May be more or less than number of decision variables. Beware of implied non-negativity constraints.
MATH143 – Fall 2011
6/3/2018

5. Word Problem Translation
Step Three: What is the Objective Function? Describe mathematically how you will decide which of two plans (levels of the activities) you would prefer. Often, this can be cast in the notion of maximizing profit for profit-making enterprises, but other situations might not be based on that metric.
MATH143 – Fall 2011
6/3/2018

6. Word Problem #1 Take this“word problem” (business situation):
George is a painter who also knows plumbing. He gets paid 1 GHS per hour painting, but 3 GHS per hour plumbing. He is willing to work up to 40 hours per week, no more than 20 of them being spent painting (paint fumes make him sick). He can get 10 jerrycans of water from the borehole at his house each week, and he can fill two cans each hour he is working as a plumber, but must use one can each hour to clean brushes when he is painting. How many hours should he work each job so that he makes the most money?
MATH143 – Fall 2011
6/3/2018

7. Word Problem #1 Identify the Objective (what is “best”)
Identify the Decision Variables (what defines the “program”)
Identify the Constraints (what resources are limited?)
Maximize the Profit for the week’s work.
The number of hours of painting (x) and the number of hours of plumbing (y)
Water, total hours, maximum painting hours.
Also some non-negativity constraints on x and y are assumed.
MATH143 – Fall 2011
6/3/2018

8. In Algebraic Form Maximize: x+3y (Total Wages)
Subject to: x + y ≤ 40 (Max hours worked each week)
x -2y ≤ 10 (Jerrycans water used)
x ≤ (Fumes limit painting time)
Also assume non-negativity on activities:
x ≥ 0 (non-negative painting hrs)
y ≥ 0 (non-negative plumbing hrs)
(Notice that the coefficient of y in the second equation is negative, since George gets water each hour he is plumbing, while the constraint is on the total consumption of water.)
MATH143 – Fall 2011
6/3/2018

9. In Spreadsheet Form Here, the decision variables have been highlighted
in yellow, while the objective is highlighted in green.
MATH143 – Fall 2011
6/3/2018

10. 2-D Solutions - Graphical Method
Add in constraints one at a time.
The constraints form straight lines.
The inequalities include the line and the points
on one side of each line.
As we add constraints,
the feasible area contracts. Lastly, draw constant-profit lines,
find solution at a corner or along an edge.
MATH143 – Fall 2011
6/3/2018

11. MATH143 – Fall 2011
6/3/2018

12. Pay at the corners Painting hours Plumbing hours Total Pay 1*0+3*0=0
1*0+3*0=0 10
1*10+3*0=10 30
1*30+3*10=60 40
1*0+3*40=120  WINNER
Footer Text
6/3/2018

13. Word Problem #2 Take this“word problem” (business situation):
Forestry Commission can sell seedlings (5GHS/acre) or charcoal (4GHS/acre). They have 8 carbon credits, seedlings credit 4 per acre, while charcoal uses 2 per acre. They have fertilizer for only seven acres worth of seedlings. They have 29 trucks, each acre of seedlings takes 3 trucks, and each acre of charcoal fill 2 trucks. The forest has 18 acres total. What should they do to maximize profits without exceeding their credits?
MATH143 – Fall 2011
6/3/2018

14. Word Problem #2 Identify the Objective (what is “best”)
Identify the Decision Variables (what defines the “program”)
Identify the Constraints (what resources are limited?)
Maximize the Profit for the Forestry Commission.
The number of acres of charcoal (x) and the number of acres of seedlings (y)
Carbon credits, truck capacity, fertilizer, total land.
Also some non-negativity constraints on x and y are assumed.
MATH143 – Fall 2011
6/3/2018

15. In Algebraic Form Maximize: 5x+4y (Total Sales)
Subject to: -4x + 2y ≤ 8 (Available Carbon Credits)
x ≤ 7 (Fertilizer)
3x + 2y ≤ (Trucks)
x + y ≤ (Total Forest acreage)
Also assume non-negativity on activities:
x ≥ 0 (non-negative charcoal)
y ≥ 0 (non-negative seedlings)
(Notice that the coefficient of x in the first constraint is negative, since Forestry Commission earns carbon credits on each acre of seedlings.)
MATH143 – Fall 2011
6/3/2018

16. In Spreadsheet Form Here, the decision variables have been highlighted
in yellow, while the objective is highlighted in green.
MATH143 – Fall 2011
6/3/2018

17. MATH143 – Fall 2011
6/3/2018

18. Sales at the corners Seedlings Charcoal Total Sales 5*0+4*0=0 4
5*0+4*0=0 4
5*10+4*0=10 3 10
5*3+4*10=55  MAX 7
5*7+4*4=51
5*7+4*0=35
Footer Text
6/3/2018

19. Tutorial Question One
MATH 143 Fall Semester 2011
6/3/2018

20. Tutorial Question Two
MATH 143 Fall Semester 2011
6/3/2018

21. Tutorial Question Two
LP Statement Let x = number of generators, y = number of alternators to be produced. Maximize P = 250x + 150y Subject to: Wiring: 2x + 3y ≤ 260 Testing: x + 2y ≤ 140 Non-Neg: x ≥ 0, y ≥ 0
MATH 143 Fall Semester 2011
6/3/2018

22. Tutorial Question Two
MATH 143 Fall Semester 2011
6/3/2018

23. Tutorial Question Two…
MATH 143 Fall Semester 2011
6/3/2018

24. Tutorial Question Two
LP Statement Let x = number of generators, y = number of alternators to be produced. Maximize P = 250x + 150y Subject to: Wiring: 2x + 3y ≤ 260 Testing: x + 2y ≤ 140 Policy: x ≥ 20, y ≥ 20
MATH 143 Fall Semester 2011
6/3/2018

25. Tutorial Question Two MATH 143 Fall Semester 2011 – Monday Week 2
6/3/2018

26. Tutorial Question Two…
Since the Wiring constraint is just now binding, and the Testing and policy constraints will keep us pegged to the same corner, even free wiring would not improve our solution.
MATH 143 Fall Semester 2011
6/3/2018