Transcript Chapter 4- Linear Programming: Modeling Examples Chapter

Chapter 3: Linear Programming
Formulation and Applications
1.
2.
3.
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6.
7.
Product Mix Problem
Marketing Problem
Transportation Problem
Diet Problem
Portfolio Planning Problem
Human Resource Scheduling Problem
Blend Problem
1 - Chap 03
Product Mix Problem
Quik-Screen is a clothing manufacturing company that specializes in
producing commemorative shirts immediately following major sporting
events like the World Series, Super Bowl, and Final Four. The company has
been contracted to produce a standard set of shirts for the winning team,
either State University or Tech, following a college football bowl game on
New Year's Day. The items produced include two sweatshirts, one with silk
screen printing on the front and one with print on both sides, and two T-shirts
of the same configuration. The company has to complete all production
within 72 hours after the game, at which time a trailer truck will pick up the
shirts. The company will work around the clock. The truck has enough
capacity to accommodate 1,200 standard-size boxes. A standard-size box
holds 12 T-shirts, whereas a box of one dozen sweatshirts is three times the
size of a standard box. The company has budgeted $25,000 for the
production run. They have 500 dozen blank sweatshirts and T-shirts each in
stock ready for production.
The company wants to know how many dozen (boxes) of each type of shirt to
produce in order to maximize profit.
2 - Chap 03
Product Mix Problem
Problem Definition and Data
3 - Chap 03
Product Mix Problem
Problem Definition and Data
- Four-product T-shirt/sweatshirt manufacturing company.
- Must complete production within 72 hours
- Truck capacity = 1,200 standard sized boxes.
- Standard size box holds12 T-shirts.
- One-dozen sweatshirts box is three times size of standard box.
- $25,000 available for a production run.
- 500 dozen blank T-shirts and sweatshirts in stock.
- How many dozens (boxes) of each type of shirt to produce?
4 - Chap 03
Product Mix Problem
Decision Variables and Model Construction
Decision variables:
x1 = dozens of sweatshirts, front printing to produce
x2 = dozens of sweatshirts, back and front printing to produce
x3 = dozens of T-shirts, front printing to produce
x4 = dozens of T-shirts, back and front printing to produce
Objective function:
maximize Z = 90x1 + 125x2 + 45x3 + 65x4
Model constraints:
0.10x1 + 0.25x2+ 0.08x3 + 0.21x4  72
Production time
3x1 + 3x2 + x3 + x4  1,200
Boxes
36x1 + 48x2 + 25x3 + 35x4  $25,000
Production cost
x1 + x2  500
Blank sweatshirts
x3 + x4  500
x1, x2, x3, x4  0 and integers
Blank T-shirts
5 - Chap 03
Product Mix Problem
Computer Solution with Excel Solver
6 - Chap 03
Product Mix Problem
Optimal Solution
Optimal production plan:
The firm should produce
175 dozens of sweatshirts with front printing,
58 dozens of sweatshirts with back and front printing,
500 dozens of T-shirts with front printing and
no T-shirts with back and front printing.
The maximum profit = $45,500
7 - Chap 03
Marketing Problem
The Biggs Department Store chain has hired an advertising firm to
determine the types and amount of advertising it should invest in for
its stores. The three types of advertising available are radio ads.
television commercials and newspaper ads. The retail chain desires to
know the number of each type of advertisement it should purchase in
order to maximize exposure. It is estimated that each ad or
commercial will reach the following potential audience and cost the
following amount.
8 - Chap 03
Marketing Problem
Problem Definition and Data
The company must consider the following resource constraints.
a. The budget limit for advertising is $100,000.
b. The television station has time available for four commercials.
c. The radio station has time available for 10 commercials.
d. The newspaper has space available for seven ads.
e. The advertising agency has time and staff available for producing no
more than a total of 15 commercials and/or ads.
9 - Chap 03
Marketing Problem
Decision Variables and Model Construction
Decision variables:
X1 = the number of TV Commercials
X2 = the number of Radio Commercials
X3 = the number of Newspaper Ads
Algebraic model:
Maximize Z = 20,000x1 + 12,000x2 + 9,000x3
subject to
15,000x1 + 6,000x 2+ 4,000x3  100,000
1x1  4
1x2  10
1x3  7
1x1 + 1x2 + 1x3  15
x1, x2, x3  0
x1, x2, x3 = integer
Budget
TV commercials
Radio commercials
Newspaper ads
Total
10 - Chap 03
Marketing Problem
Computer Solution with Excel Solver
11 - Chap 03
Marketing Problem
Optimal Solution
Optimal solution:
The company should subscribe
2 TV Commercials
9 Radio Commercials and
4 Newspaper Ads.
The maximum number of people reached = 184,000.
12 - Chap 03
Transportation Problem
The Zephyr Television Company ships televisions from three
warehouses to three retail stores on a monthly basis. Each warehouse
has a fixed supply per month and each store has a fixed demand per
month. The manufacturer wants to know the number of television sets
to ship from each warehouse to each store in order to minimize the
total cost of transportation.
13 - Chap 03
Transportation Problem
Problem Definition and Data
14 - Chap 03
Transportation Problem
Model Summary
As the total amount (700 TV sets) of the supply is greater than the total amount (600 TV
sets) of the demand, we add a dummy node D at demand side. The costs of shipment to
dummy node D are all zero. The decision variables are
xij = number of television sets shipped from warehouse i to store j, where i = 1, 2, 3
and j = A, B, C, D.
Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C
subject to
Node 1: x1A + x1B + x1C + x1D = 300
Supply from Warehouse 1
Node 2: x2A+ x2B + x2C + x2D = 200
Supply from Warehouse 2
Node 3: x3A+ x3B + x3C + x3D = 200
Supply from Warehouse 3
Node A: x1A + x2A + x3A = 150
Demand of Store A
Node B: x1B + x2B + x3B = 250
Demand of Store B
Node C: x1C + x2C + x3C = 200
Demand of Store C
Node D: x1D + x2D + x3D = 100
Demand of Store D
xij  0
15 - Chap 03
Transportation Problem
Network Presentation
300
1
16 X1A
18 X1B
0
11
X1C
X1D
14
200
X2A
2
0
3
Total Supply = 700
150
B
250
C
200
D
100
X2D
X3A
13
200
X2B
12
13 X2C
A
X
15 3B
17 X3C
0 X3D
Total Demand = 700
16 - Chap 03
Transportation Problem
Computer Solution with Excel Solver
17 - Chap 03
Transportation Problem
Optimal Solution
The company should ship
200 TVs from Cincinnati (W1) to Detroit (SC),
200 TVs from Atlanta (W2) to Dallas (SB),
150 TVs from Pittsburgh (W3) to New York (SA) and
50 TVs from Pittsburgh (W3) to Dallas (SB).
100 TVs are kept in Cincinnati (W1) as stock
The minimum shipping cost = $7,300
18 - Chap 03
Diet Problem
Breathtakers, a health and fitness center, operates a morning fitness
program for senior citizens. The program includes aerobic exercise,
either swimming or step exercise, followed by a healthy breakfast in
the dining room. Breathtakers' dietitian wants to develop a breakfast
that will be high in calories, calcium, protein, and fiber, which are
especially important to senior citizens, but low in fat and cholesterol.
She has selected the following possible food items, whose individual
nutrient contributions and cost from which to develop a standard
breakfast menu are shown in the following table.
19 - Chap 03
Diet Problem
Problem Definition and Data
The dietitian wants to minimize the total cost of the breakfast while
satisfying all the requirements as follows. The breakfast has to include at
least 420 calories, 5 milligrams of iron, 400 milligrams of calcium, 20
grams of protein, 12 grams of fiber, and must have no more than 20
grams of fat and 30 milligrams of cholesterol.
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Diet Problem
Decision Variables and Model Construction
x1 = Number of cups of bran cereal
x2 = Number of cups of dry cereal
x3 = Number of cups of oatmeal
x4 = Number of cups of oat bran
x5 = Number of eggs
x6 = Number of slices of bacon
x7 = Number of oranges
x8 = Number of cups of milk
x9 = Number of cups of orange juice
x10 = Number of slices of wheat toast
21 - Chap 03
Diet Problem
Decision Variables and Model Construction
minimize Z = 0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6+ 0.40x7 + 0.16x8 + 0.50x9 + 0.07x10
subject to
90x1 + 110x2 + 100x3 + 90x4 + 75x5 + 35x6 + 65x7 + 100x8 + 120x9 + 65x10  420 Calories
2x2 + 2x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10  20
Fat
270x5 + 8x6 + 12x8  30
Cholesterol
6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10  5
Iron
20x1 + 48x2 + 12x3 + 8x4+ 30x5 + 52x7 + 250x8 + 3x9 + 26x10  400
Calcium
3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7+ 9x8+ x9 + 3x10  20
Protein
5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10  12
Fiber
xi  0
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Diet Problem
Computer Solution with Excel Solver
23 - Chap 03
Diet Problem
Optimal Solution
Optimal solution:
The breakfast should include
1.02 cups of oatmeal,
1.24 cups of milk, and
2.98 slices of wheat toast.
The minimum cost = $0.51.
24 - Chap 03
Portfolio Planning Problem
Winslow Savings has $20 million available for investment. It wishes to
invest over the next four months in such a way that it will maximize the total
interest earned over the four month period as well as have at least $10
million available at the start of the fifth month for a high rise building
venture in which it will be participating.
For the time being, Winslow wishes to invest only in 2-month government
bonds (earning 2% over the 2-month period) and 3-month construction
loans (earning 6% over the 3-month period). Each of these is available each
month for investment. Funds not invested in these two investments are
liquid and earn 3/4 of 1% per month when invested locally.
Formulate a linear program that will help Winslow Savings determine how
to invest over the next four months if at no time does it wish to have more
than $8 million in either government bonds or construction loans.
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Portfolio Planning Problem
Decision Variables and Model Construction
• Define the decision variables
Gj = amount of new investment in government bonds at the beginning
of month j
Cj = amount of new investment in construction loans at the beginning
of month j
Lj = amount invested locally at the beginning of month j,
where j = 1,2,3,4
• Define the objective function
Maximize total interest earned over the 4-month period.
MAX (interest rate on investment) (amount invested)
MAX .02G1 + .02G2 + .02G3 + .02G4 + .06C1 + .06C2 + .06C3 + .06C4 +
.0075L1 + .0075L2 + .0075L3 + .0075L4
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Portfolio Planning Problem
Decision Variables and Model Construction
• Define the constraints
Balance equations (total investment = total funds available) at the
beginning of each month:
(1) At the beginning of Month 1: G1 + C1 + L1 = 20
(2) At the beginning of Month 2: G2 + C2 + L2 = 1.0075L1
 G2 + C2 - 1.0075L1 + L2 = 0
(3) At the beginning of Month 3: G3 + C3 + L3 = 1.02G1 + 1.0075L2
 - 1.02G1 + G3 + C3 - 1.0075L2 + L3 = 0
(4) At the beginning of Month 4: G4 + C4 + L4 = 1.06C1 + 1.02G2 +
1.0075L3
 - 1.02G2 + G4 - 1.06C1 + C4 - 1.0075L3 + L4 = 0
(5) At the beginning of Month 5: 1.06C2 + 1.02G3 + 1.0075L4 >= 10
27 - Chap 03
Portfolio Planning Problem
Decision Variables and Model Construction
• Define the constraints (continued)
No more than $8 million in government bonds at any time:
(6) G1 <= 8
(7) G1 + G2 <= 8
(8) G2 + G3 <= 8
(9) G3 + G4 <= 8
No more than $8 million in construction loans at any time:
(10) C1 <= 8
(11) C1 + C2 <= 8
(12) C1 + C2 + C3 <= 8
(13) C2 + C3 + C4 <= 8
(14) Gj, Cj, Lj > 0 for j = 1,2,3,4
28 - Chap 03
Portfolio Planning Problem
Computer Solution with Excel Solver
29 - Chap 03
Portfolio Planning Problem
Optimal Solution
Optimal investment portfolio:
Winslow Savings should invest
$8.0 million in government bonds, $8.0 million in construction
loan and $4.0 million in local investment in month 1;
$4.03 million in local investment in month 2;
$5.11 million in government bonds and $7.11 million in local
investment in month 3; and
$2.89 million in government bonds, $8.0 million in construction
loan and $4.75 million in local investment in month 4.
The maximum interest return is $1.43 million.
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Human Resource Scheduling Problem
Mr. Chan is the shop manager of a fast food restaurant. He now needs to
schedule the staffing of the restaurant which is open from 8 AM until
midnight. Mr. Chan has monitored the catering demand at various times of
the day and determined that the following number of waiters are required:
Time of Day
Minimum Number of Waiters Required
8 am – noon
6
Noon – 4 pm
8
4 pm – 8 pm
12
8 pm – midnight
6
Two types of waiters can be hired: full-time and part-time. The full-time
waiters work for eight consecutive hours in any of the following shifts:
morning (8 AM - 4 PM), afternoon (noon - 8 PM), and evening (4 PM midnight).
31 - Chap 03
Human Resource Scheduling Problem
Full-time waiters are paid $60 per hour. Part-time waiters can be hired to work
any of the four shifts listed in the table. Part-time waiters are paid $35 per
hour. An additional requirement is that during every time period, there must
be at least two full-time waiters on duty for every part-time waiter on duty. Mr.
Chan would like to determine how many full-time and part-time waiters
should work each shift to meet the above requirements at the minimum
possible cost.
32 - Chap 03
Human Resource Scheduling Problem
Decision Variables and Model Construction
Decision variables.
F1 = number of full-time waiters working the morning shift (8 a.m.-4 p.m.),
F2 = number of full-time waiters working the afternoon shift (12 p.m.-8 p.m.),
F3 = number of full-time waiters working the evening shift (4 p.m.-midnight),
P1 = number of part-time waiters working the first shift (8 a.m.-12 p.m.),
P2 = number of part-time waiters working the second shift (12 p.m.-4 p.m.),
P3 = number of part-time waiters working the third shift (4 p.m.-8 p.m.),
P4 = number of part-time waiters working the fourth shift (8 p.m.-midnight).
33 - Chap 03
Human Resource Scheduling Problem
Decision Variables and Model Construction
Model Summary:
Minimize C = ($60 / hour)(8 hours)(F1 + F2 + F3) + ($35 / hour)(4 hours)(P1 + P2 + P3 + P4)
= $(480F1 + 480F2 + 480 F3 + 140P1 + 140P2 + 140P3 + 140P4)
subject to
F1 + P1 ≥ 6
Staff required, 8 am – noon
F1 + F2 + P2 ≥ 8
Staff required, Noon – 4 pm
F2 + F3 + P3 ≥ 12
Staff required, 4 pm – 8 pm
F3 + P4 ≥ 6
Staff required, 8 pm – midnight
F1 ≥ 2P1
Staff ratio, 8 am – noon
F1 + F2 ≥ 2P2
Staff ratio, Noon – 4 pm
F2 + F3 ≥ 2P3
Staff ratio, 4 pm – 8 pm
F3 ≥ 2P4
Staff ratio, 8 pm – midnight
F1, F2, F3, P1, P2, P3, P4 ≥ 0
F1, F2, F3, P1, P2, P3, P4 are integers
34 - Chap 03
Human Resource Scheduling Problem
Computer Solution with Excel Solver
Decision Variables
F1 F2 F3 P1 P2 P3 P4
No. of Waiters
4
2
6
2
2
4
0
Cost per Shift
$480 $480 $480 $140 $140 $140 $140 $6,880 <= Min. Cost
S.T.
LHS
RHS
Staff required, 8 am - noon
1
1
6 >=
6
Staff required, Noon - 4 pm
1
1
1
8 >=
8
Staff required, 4 pm - 8 pm
1
1
1
12 >=
12
Staff required, 8 pm - midnight
1
1
6 >=
6
Staff Ratio, 8 am - noon
1
-2
0 >=
0
Staff Ratio, noon - 4 pm
1
1
-2
2 >=
0
Staff Ratio, 4 pm - 8 pm
1
1
-2
0 >=
0
Staff Ratio, 8 pm - midnight
1
-2
6 >=
0
35 - Chap 03
Human Resource Scheduling Problem
Optimal Solution
Mr. Chan should employ:
4 full-time waiters working the morning shift (8 a.m.-4 p.m.),
2 full-time waiters working the afternoon shift (12 p.m.-8 p.m.),
6 full-time waiters working the evening shift (4 p.m.-midnight),
2 part-time waiters working the first shift (8 a.m.-12 p.m.),
2 part-time waiters working the second shift (12 p.m.-4 p.m.),
4 part-time waiters working the third shift (4 p.m.-8 p.m.) and
no part-time waiters working the fourth shift (8 p.m.-midnight).
The minimum cost = $6,880
36 - Chap 03
Blend Problem
A chemical company produces three grades of motor oil – super, premium,
and extra – from three components. The company wants to determine the
optimal mix of the three components in each grade of motor oil that will
maximize profit. The maximum quantities available of each component and
their cost per barrel are as follows. Determine the optimal mix of the three
components in each grade of motor oil that will maximize profit. Company
wants to produce at least 3,000 barrels of each grade of motor oil.
37 - Chap 03
Blend Problem
Decision Variables and Model Construction
Decision variables.
The quantity of each of the three components used in each grade of motor oil (9 decision
variables):
xij = barrels of component i used in motor oil grade j per day,
where i = 1, 2, 3 and j = s (super), p (premium), and e (extra).
Model Summary:
maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e
subject to
x1s + x1p + x1e  4,500
C1 available
x2s + x2p + x2e  2,700
C2 available
x3s + x3p + x3e  3,500
C3 available
0.50x1s - 0.50x2s - 0.50x3s  0
At least 50% of C1 in S
0.70x2s - 0.30x1s - 0.30x3s  0
No more than 30% of C2 in S
0.60x1p - 0.40x2p - 0.40x3p  0
At least 40% of C1 in P
0.75x3p - 0.25x1p - 0.25x2p  0
No more than 25% of C3 in P
0.40x1e- 0.60x2e- - 0.60x3e  0
At least 60% of C1 in E
0.90x2e - 0.10x1e - 0.10x3e  0
At least 10% of C2 in E
x1s + x2s + x3s  3,000
Min. requirement of S
x1p+ x2p + x3p  3,000
Min. requirement of P
x1e+ x2e + x3e  3,000
Min requirement of E
xij  0
38 - Chap 03
Blend Problem
Computer Solution with Excel Solver
Decision Variables
Barrels used
Profot($/bbl)
S.T.
C1 available
C2 available
C3 available
No more than 30% of C2 in S
No more than 25% of C3 in P
At least 50% of C1 in S
At least 40% of C1 in P
At least 60% of C1 in E
At least 10% of C2 in E
Min. Requirement of S
Min. Requirement of P
Min. Requirement of E
X1S X2S X3S X1P X2P X3P X1E X2E X3E
1,500 600 900 1,200 1,800
0 1,800 300 900
$11 $13
$9
$8 $10
$6
$6
$8
$4 $76,800.00
LHS
1.00
1.00
1.00
4500.00
1.00
1.00
1.00
2700.00
1.00
1.00
1.00
1800.00
-0.30 0.70 -0.30
-300.00
-0.25 -0.25 0.75
-750.00
0.50 -0.50 -0.50
0.00
0.60 -0.40 -0.40
0.00
0.40 -0.60 -0.60
0.00
-0.10 0.90 -0.10
0.00
1.00 1.00 1.00
3000.00
1.00 1.00 1.00
3000.00
1.00 1.00 1.00
3000.00
<=Max. Profit
RHS
<=
4,500.00
<=
2,700.00
<=
3,500.00
<=
0.00
<=
0.00
>=
0.00
>=
0.00
>=
0.00
>=
0.00
>=
3,000.00
>=
3,000.00
>=
3,000.00
39 - Chap 03
Blend Problem
Optimal Solution
The company should use:
1,500 barrels of component 1, 600 barrels of component 2 and 900
barrels of component 3 to blend 3,000 barrels of Super motor oil;
1,200 barrels of component 1, and 1,800 barrels of component 2 to
blend 3,000 barrels of Premium motor oil; and
1,800 barrels of component 1, 300 barrels of component 2 and 900
barrels of component 3 to blend 3,000 barrels of Extra motor oil;
The maximum profit = $76,800.
40 - Chap 03