Transcript LP Formulation Practice Set 1 Problem 1. Optimal Product Mix Management is considering devoting some excess capacity to one or more of three.

LP Formulation
Practice Set 1
Problem 1. Optimal Product Mix
Management is considering devoting some excess capacity to one
or more of three products. The hours required from each
resource for each unit of product, the available capacity (hours
per week) of the three resources, as well as the profit of each unit
of product are given below.
Hours used per unit
Product1 Product2 Product3
9
3
5
5
4
0
3
0
2
$50
$20
$25
Total hours avialable
500
350
150
Profit
Sales department indicates that the sales potentials for products
1 and 2 exceeds maximum production rate, but the sales
potential for product 3 is 20 units per week.
Formulate the problem and solve it using excel
LP-Formulation
Ardavan Asef-Vaziri
June-2013
2
Problem Formulation
Decision Variables
x1 : volume of product 1
x2 : volume of product 2
x3 : volume of product 3
Objective Function
Max Z = 50 x1 +20 x2 +25 x3
Constraints
Resources
9 x1 +3 x2 +5 x3  500
5 x1 +4 x2 +
 350
3 x1 +
+2 x3  150
Market
x3  20
Nonnegativity
x1  0, x2  0 , x3 0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
3
Problem 2
An appliance manufacturer produces two models of microwave
ovens: H and W. Both models require fabrication and
assembly work: each H uses four hours fabrication and two
hours of assembly, and each W uses two hours fabrication and
six hours of assembly. There are 600 fabrication hours this
week and 450 hours of assembly. Each H contributes $40 to
profit, and each W contributes $30 to profit.
a) Formulate the problem as a Linear Programming problem.
b) Solve it using excel.
c) What are the final values?
d) What is the optimal value of the objective function?
LP-Formulation
Ardavan Asef-Vaziri
June-2013
4
Problem Formulation
Decision Variables
xH : volume of microwave oven type H
xW : volume of microwave oven type W
Objective Function
Max Z = 40 xH +30 xW
Constraints
Resources
4 xH +2 xW  600
2 xH +6 xW  450
Nonnegativity
xH 0, xW  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
5
Problem 3
A small candy shop is preparing for the holyday season. The owner
must decide how many bags of deluxe mix how many bags of
standard mix of Peanut/Raisin Delite to put up. The deluxe mix has
2/3 pound raisins and 1/3 pounds peanuts, and the standard mix has
1/2 pound raisins and 1/2 pounds peanuts per bag. The shop has 90
pounds of raisins and 60 pounds of peanuts to work with. Peanuts
cost $0.60 per pounds and raisins cost $1.50 per pound. The deluxe
mix will sell for 2.90 per pound and the standard mix will sell for 2.55
per pound. The owner estimates that no more than 110 bags of one
type can be sold.
a) Formulate the problem as a Linear Programming problem.
b) Solve it using excel.
c) What are the final values?
d) What is the optimal value of the objective function?
LP-Formulation
Ardavan Asef-Vaziri
June-2013
6
Problem Formulation
Decision Variables
x1 : volume of deluxe mix
x2 : volume of standard mix
Objective Function
Max Z = [2.9-0.60(1/3)-1.5(2/3)] x1 + [2.55-0.60(1/2)-1.5(1/2)] x2
Max Z = 1.7x1 + 1.5 x2
Constraints
Resources
(2/3) x1 +(1/2) x2  90
(1/3) x1 +(1/2) x2  60
Nonnegativity
x1  0, x2  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
7
Problem 4
The following table summarizes the key facts about two
products, A and B, and the resources, Q, R, and S, required to
produce them.
Resource Usage per Unit Produced
a)
b)
c)
d)
Resource
Product A
Product B
Amount of
resource
available
Q
2
1
2
R
1
2
2
S
3
3
4
Profit/Uni
t
$3000
$2000
Formulate the problem as a Linear Programming problem.
Solve it using excel.
What are the final values?
What is the optimal value of the objective function?
LP-Formulation
Ardavan Asef-Vaziri
June-2013
8
Problem Formulation
Decision Variables
xA : volume of product A
xB : volume of product B
Objective Function
Max Z = 3000 xA +2000 xB
Constraints
Resources
2 xA +1 xB  2
1 xA +2 xB  2
3 xA +3 xB  4
Nonnegativity
xA  0, xB  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
9
Problem 5
The Apex Television Company has to decide on the number of 27”
and 20” sets to be produced at one of its factories. Market research
indicates that at most 40 of the 27” sets and 10 of the 20” sets can be
sold per month. The maximum number of work-hours available is 500
per month. A 27” set requires 20 work-hours and a 20” set requires 10
work-hours. Each 27” set sold produces a profit of $120 and each 20”
set produces a profit of $80. A wholesaler has agreed to purchase all
the television sets produced if the numbers do not exceed the
maximum indicated by the market research.
a)
b)
c)
d)
Formulate the problem as a Linear Programming problem.
Solve it using excel.
What are the final values?
What is the optimal value of the objective function?
LP-Formulation
Ardavan Asef-Vaziri
June-2013
10
Problem Formulation
Decision Variables
x1 : number of 27’ TVs
x2 : number of 20’ TVs
Objective Function
Max Z = 120 x1 +80 x2
Constraints
Resources
20 x1 +10 x2  500
Market
x1
 40
x2  10
Nonnegativity
x1  0, x2  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
11
Problem 6
Ralph Edmund has decided to go on a steady diet of only streak and
potatoes s (plus some liquids and vitamins supplements). He wants to
make sure that he eats the right quantities of the two foods to satisfy
some key nutritional requirements. He has obtained the following
nutritional and cost information. Ralph wishes to determine the number
of daily servings (may be fractional of steak and potatoes that will meet
these requirements at a minimum cost.
Grams of Ingredient per Serving
Ingredient
Steak
Potatoes
Daily Requirements (grams)
Carbohydrates
5
15
≥ 50
Protein
20
5
≥ 40
Fat
15
2
≤ 60
Cost per serving
$4
$2
Formulate the problem as an LP model. Solve it using excel. What are
the final values? What is the optimal value of the objective function?
LP-Formulation
Ardavan Asef-Vaziri
June-2013
12
Problem Formulation
Decision Variables
x1 : serving of steak
x2 : serving of potato
Objective Function
Min Z = 4 x1 +2x2
Constraints
Resources
5 x1 +15 x2 ≥ 50
20 x1 +5 x2 ≥ 40
15 x1 +2 x2 ≤ 60
Nonnegativity
x1  0, x2  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
13
Excel Solution
Steak
Potatoe
LHS
5
15
50
≥
50
Protein
20
5
40
≥
40
Fat
15
2
24.9
≤
60
Cost per serving
$4
$2
10.9
1.27
2.91
Carbohydrates
LP-Formulation
Ardavan Asef-Vaziri
RHS
June-2013
14
Problem 7
A farmer has 10 acres to plant in wheat and rye. He has to plant
at least 7 acres. However, he has only $1200 to spend and each
acre of wheat costs $200 to plant and each acre of rye costs $100
to plant. Moreover, the farmer has to get the planting done in
12 hours and it takes an hour to plant an acre of wheat and 2
hours to plant an acre of rye. If the profit is $500 per acre of
wheat and $300 per acre of rye, how many acres of each should
be planted to maximize profits?
State the decision variables.
x = the number of acres of wheat to plant
y = the number of acres of rye to plant
Write the objective function.
maximize 500x +300y
LP-Formulation
Ardavan Asef-Vaziri
June-2013
15
Problem 7
Write the constraints.
x+y ≤ 10
x+y ≥ 7
200x + 100y ≤ 1200
x + 2y ≤ 12
x ≥ 0, y ≥ 0
LP-Formulation
(max acreage)
(min acreage)
(cost)
(time)
(non-negativity)
Ardavan Asef-Vaziri
June-2013
16
Problem 8
You are given the following linear programming model in
algebraic form, where, X1 and X2 are the decision variables
and Z is the value of the overall measure of performance.
Maximize Z = X1 +2 X2
Subject to
Constraints on resource 1: X1 + X2 ≤ 5 (amount available)
Constraints on resource 2: X1 + 3X2 ≤ 9 (amount available)
And
X1 , X2 ≥ 0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
17
Problem 8
Identify the objective function, the functional constraints, and
the non-negativity constraints in this model.
Objective Function  Maximize Z = X1 +2 X2
Functional constraints  X1 + X2 ≤ 5, X1 + 3X2 ≤ 9
Is (X1 ,X2) = (3,1) a feasible solution?
3 + 1 ≤ 5, 3 + 3(1) ≤ 9  yes; it satisfies both constraints.
Is (X1 ,X2) = (1,3) a feasible solution?
1 + 3 ≤ 5, 1 + 3(9) > 9  no; it violates the second constraint.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
18
Problem 9
You are given the following linear programming model in
algebraic form, where, X1 and X2 are the decision variables
and Z is the value of the overall measure of performance.
Maximize Z = 3X1 +2 X2
Subject to
Constraints on resource 1: 3X1 + X2 ≤ 9 (amount available)
Constraints on resource 2: X1 + 2X2 ≤ 8 (amount available)
And
X1 , X2 ≥ 0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
19
Problem 9
Identify the objective function,
Maximize Z = 3X1 +2 X2
the functional constraints,
3X1 + X2 ≤ 9 and X1 + 2X2 ≤ 8
the non-negativity constraints
X1 , X2 ≥ 0
Is (X1 ,X2) = (2,1) a feasible solution?
3(2) + 1 ≤ 9 and 2 + 2(1) ≤ 8 yes; it satisfies both constraints
Is (X1 ,X2) = (2,3) a feasible solution?
3(2) + 3 ≤ 9 and 2 + 2(3) ≤ 8 yes; it satisfies both constraints
Is (X1 ,X2) = (0,5) a feasible solution?
3(0) + 5 ≤ 9 and 0 + 2(5) > 8 no; it violates the second constraint
LP-Formulation
Ardavan Asef-Vaziri
June-2013
20
Problem 10. Product mix problem : Narrative representation
The Quality Furniture Corporation produces
benches and tables.
The firm has two main resources
Resources
labor and redwood for use in the furniture.
During the next production period
1200 labor hours are available under a union agreement.
A stock of 5000 pounds of quality redwood is also available.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
21
Problem 10. Product mix problem : Narrative representation
Consumption and profit
Each bench that Quality Furniture produces requires
4 labor hours and 10 pounds of redwood
Each picnic table takes 7 labor hours and 35 pounds of
redwood.
Total available 1200, 5000
Completed benches yield a profit of $9 each,
and tables a profit of $20 each.
Formulate the problem to maximize the total profit.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
22
Problem 10. Product Mix : Formulation
x1
= number of benches to produce
x2
= number of tables to produce
Maximize Profit = ($9) x1 +($20) x2
subject to
Labor: 4 x1 + 7 x2
 1200 hours
Wood:10 x1 + 35 x2  5000 pounds
and
x1  0, x2  0.
We will now solve this LP model using the Excel Solver.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
23
Problem 10. Product Mix : Excel solution
Labor
Wood
Contribution Margin
Changing Cells
LP-Formulation
Bench
4
10
9
100
Table Required
7
1200
35
5000
20
3185.714
114.2857
Ardavan Asef-Vaziri
June-2013
Available
1200
5000
24
Problem 11. Make / buy decision : Narrative representation
Electro-Poly is a leading maker of slip-rings.
A new order has just been received.
Model 1
Model 2
Model 3
3,000
2,000
900
Hours of wiring/unit
2
1.5
3
Hours of harnessing/unit
1
2
1
Cost to Make
$50
$83
$130
Cost to Buy
$61
$97
$145
Number ordered
The company has 10,000 hours of wiring capacity and 5,000
hours of harnessing capacity.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
25
Problem 11. Make / buy decision : decision variables
x1 = Number of model 1 slip rings to make
x2 = Number of model 2 slip rings to make
x3 = Number of model 3 slip rings to make
y1 = Number of model 1 slip rings to buy
y2 = Number of model 2 slip rings to buy
y3 = Number of model 3 slip rings to buy
The Objective Function
Minimize the total cost of filling the order.
MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3
LP-Formulation
Ardavan Asef-Vaziri
June-2013
26
Problem 11. Make / buy decision : Constraints
Demand Constraints
x1 + y1 = 3,000
} model 1
x2 + y2 = 2,000
} model 2
x3 + y3 = 900 } model 3
Resource Constraints
2x1 + 1.5x2 + 3x3 <= 10,000 } wiring
1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing
Nonnegativity Conditions
x1, x2, x3, y1, y2, y3 >= 0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
27
Problem 11. Make / buy decision : Excel
Make
Buy
Make
Buy
Produced
Required
Capacity
Wiring
Harnessing
LP-Formulation
Model 1 Model 2 Model 3
50
83
130
61
97
145
Model 1 Model 2 Model 3
3000
550
900
0
1450
0
3000
2000
900
3000
2000
900
2
1
1.5
2
3
1
453300
Required Available
9525
10000
5000
5000
Ardavan Asef-Vaziri
June-2013
28
Problem 11. Make / buy decision : Constraints
Do we really need 6 variables?
x1 + y1 = 3,000 ===> y1 = 3,000 - x1
x2 + y2 = 2,000 ===> y2 = 2,000 - x2
x3 + y3 = 900 ===> y3 = 900 - x3
The objective function was
MIN: 50x1 + 83x2 + 130x3 + 61y1 + 97y2 + 145y3
Just replace the values
MIN: 50x1 + 83x2 + 130x3 + 61 (3,000 - x1 ) + 97 ( 2,000 - x2) +
145 (900 - x3 )
MIN: 507500 - 11x1 -14x2 -15x3
We can even forget 507500, and change the the O.F. into
MIN - 11x1 -14x2 -15x3 or
MAX + 11x1 +14x2 +15x3
LP-Formulation
Ardavan Asef-Vaziri
June-2013
29
Problem 11. Make / buy decision : Constraints
MAX + 11x1 +14x2 +15x3
Resource Constraints
2x1 + 1.5x2 + 3x3 <= 10,000 } wiring
1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing
Demand Constraints
x1 <= 3,000
} model 1
x2 <= 2,000
} model 2
x3 <= 900
} model 3
Nonnegativity Conditions
x1, x2, x3 >= 0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
30
Problem 11. Make / buy decision : Constraints
y1 = 3,000- x1
MIN: 50x1 + 83x2 + 130x3
y2 = 2,000-x2
+ 61y1 + 97y2 + 145y3
y3 =
Demand Constraints
x1 + y1 = 3,000
} model 1
x2 + y2 = 2,000
} model 2
MIN: 50x1 + 83x2 + 130x3
+ 61(3,000- x1)
+ 97(2,000-x2)
+ 145(900-x3)
x3 + y3 = 900 } model 3
Resource Constraints
2x1 + 1.5x2 + 3x3 <= 10,000 } wiring
1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing
Nonnegativity Conditions
y1 = 3,000- x1>=0
y2 = 2,000-x2>=0
y3 = 900-x3>=0
x1 <= 3,000
x2 <= 2,000
x3 <= 900
x1, x2, x3, y1, y2, y3 >= 0
LP-Formulation
900-x3
Ardavan Asef-Vaziri
June-2013
31
Problem 11. Make / buy decision : Constraints
Model1
Wiring
2
Harnessing
1
Marginal Profit
11
Demand
3000
3000
LP-Formulation
Model2
1.5
2
14
2000
550
Model3
3
1
15
900
900
Ardavan Asef-Vaziri
9525
5000
54200
10000
5000
453300
June-2013
32
Problem 12. Marketing : narrative
A department store want to maximize exposure.
There are 3 media; TV, Radio, Newspaper
each ad will have the following impact
Media
Exposure (people / ad)
Cost
TV
20000
15000
Radio
12000
6000
News paper
9000
4000
Additional information
1-Total budget is $100,000.
2-The maximum number of ads in T, R, and N are limited to
4, 10, 7 ads respectively.
3-The total number of ads is limited to 15.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
33
Problem 12. Marketing : formulation
Decision variables
x1 = Number of ads in TV
x2 = Number of ads in R
x3 = Number of ads in N
Max Z = 20 x1 + 12x2 +9x3
15x1 + 6x2 + 4x3 
x1

x2

x3 
x1 + x2 + x3 
100
4
10
7
15
x1, x2, x3  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
34
Problem 13. ( From Hillier and Hillier)
Men, women, and children gloves.
Material and labor requirements for each type and the
corresponding profit are given below.
Glove
Material (sq-feet) Labor (hrs) Profit
Men
2
0.5
8
Women
1.5
0.75
10
Children
1
0.67
6
Total available material is 5000 sq-feet.
We can have full time and part time workers.
Full time workers work 40 hrs/w and are paid $13/hr
Part time workers work 20 hrs/w and are paid $10/hr
We should have at least 20 full time workers.
The number of full time workers must be at least twice of that of
part times.
LP-Formulation
Ardavan Asef-Vaziri
June-2013
35
Problem 13. Decision variables
X1 : Volume of production of Men’s gloves
X2 : Volume of production of Women’s gloves
X3 : Volume of production of Children’s gloves
Y1 : Number of full time employees
Y2 : Number of part time employees
LP-Formulation
Ardavan Asef-Vaziri
June-2013
36
Problem 13. Constraints
Row material constraint
2X1 + 1.5X2 + X3  5000
Full time employees
Y1  20
Relationship between the number of Full and Part time employees
Y1  2 Y2
Labor Required
.5X1 + .75X2 + .67X3  40 Y1 + 20Y2
Objective Function
Max Z = 8X1 + 10X2 + 6X3 - 520 Y1 - 200 Y2
Non-negativity
X1 , X2 , X3 , Y1 , Y2  0
LP-Formulation
Ardavan Asef-Vaziri
June-2013
37
Problem 13. Excel Solution
2
1.5
1
0.5
8
0.75
10
0.67
6
1
1
-40
-520
X1
X2
X3
Y1
2
1.5
1
0.5
8
2500
X1
LP-Formulation
0.75
10
0
X2
0.67
6
0
X3
1
1
-40
-520
25
Y1
-2
-20
-200
Y2
-2
-20
-200
12.5
Y2
Ardavan Asef-Vaziri
0
0
0
0
0
<=
>=
>=
<=
5000
20
0
0
5000
25
0
0
4500
<=
>=
>=
<=
5000
20
0
0
June-2013
38