6407 ISM Fall 2009
LINEAR PROGRAMMING FORMULATION PROBLEMS AND SOLUTIONS
7-14
The Electrocomp Corporation manufactures two electrical products: air conditioners and large
fans. The assembly process for each is similar in that both require a certain amount of wiring and
drilling. Each air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go
through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of
wiring time are available and up to 140 hours of drilling time maybe used. Each air conditioner
sold yields a profit of $25. Each fan assembled may be sold for a $15 profit. Formulate and solve
this LP production mix situation to find the best combination of air conditioners and fans that
yields the highest profit. Use the corner point graphical approach.
Let X1 = the number of air conditioners scheduled to be produced
X2 = the number of fans scheduled to be produced
Maximize
Subject to:
25X1 +
3X1 +
2X1 +
15X2
(maximize profit)
2X2
240 (wiring capacity constraint)
X2
140 (drilling capacity constraint)
X1, X2 0
(non-negativity constraints)
Optimal Solution: X1 = 40 X2 = 60 Profit = $1,900
7-15
Electrocomps management realizes that it forgot to include two critical constraints (see Problem
7-14). In particular, management decides that to ensure an adequate supply of air conditioners for
a contract, at least 20 air conditioners should be manufactured. Because Electrocomp incurred an
oversupply of fans in the preceding period, management also insists that no more than 80 fans be
produced during this production period. Resolve this product mix problem to find the new
optimal solution.
Let X1 = the number of air conditioners scheduled to be produced
X2 = the number of fans scheduled to be produced
Maximize
Subject to:
25X1 +
3X1 +
2X1 +
X1
15X2
(maximize profit)
2X2
240 (wiring capacity constraint)
X2
140 (drilling capacity constraint)
20 (a/c contract constraint)
X2
80 (maximum # of fans constraint)
X1, X2 0
(non-negativity constraints)
Optimal Solution: X1 = 40 X2 = 60 Profit = $1,900
ISM 6407 Fall 2009
7-16
A candidate for mayor in a small town has allocated $40,000 for last-minute advertising in the
days preceding the election. Two types of ads will be used: radio and television. Each radio ad
costs $200 and reaches an estimated 3,000 people. Each television ad costs $500 and reaches an
estimated 7,000 people. In planning the advertising campaign, the campaign manager would like
to reach as many people as possible, but she has stipulated that at least 10 ads of each type must
be used. Also, the number of radio ads must be at least as great as the number of television ads.
How many ads of each type should be used? How many people will this reach?
Let X1 = the number of radio ads purchased
X2 = the number of television ads purchased
Maximize 3,000X1 + 7,000X2
Subject to:
200X1 +
500X2
X1
X2
X1
X1, X2
(maximize exposure)
40,000 (budget constraint)
10 (at least 10 radio ads purchased)
10 (at least 10 television ads purchased)
X2 (# of radio ads # of television ads)
0
(non-negativity constraints)
For solution purposes, the fourth constraint would be rewritten as:
X1 X2 0
Optimal Solution: X1 = 175
7-17
X2 = 10
Exposure = 595,000 people
The Outdoor Furniture Corporation manufactures two products, benches and picnic tables, for
use in yards and parks. The firm has two main resources: its carpenters (labor force) and a supply
of redwood for use in the furniture. During the next production cycle, 1,200 hours of labor are
available under a union agreement. The firm also has a stock of 3500 feet of good-quality
redwood. Each bench that Outdoor Furniture produces requires 4 labor hours and 10 feet of
redwood; each picnic table takes 6 labor hours and 35 feet of redwood. Completed benches will
yield a profit of $9 each, and tables will result in a profit of $20 each. How many benches and
tables should Outdoor Furniture produce to obtain the largest possible profit? Use the graphical
LP approach.
Let X1 = the number of benches produced
X2 = the number of tables produced
Maximize
Subject to:
9X1 +
4X1 +
10X1 +
20X2
6X2
1,200
35X2
3,500
X1, X2 0
Optimal Solution: X1 = 262.5
X2 = 25
(maximize profit)
(labor hours constraint)
(redwood capacity constraint)
(non-negativity constraints)
Profit = $2,862.50
ISM 6407 Fall 2009
7-18
The dean of the Western College of Business must plan the schools course offerings for the fall
semester. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate
courses in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each
undergraduate course taught costs the college an average of $2,500 in faculty wages, and each
graduate course costs $3,000. How many undergraduate and graduate courses should be taught in
the fall so that total faculty salaries are kept to a minimum?
Let X1 = the number of undergraduate courses scheduled
X2 = the number of graduate courses scheduled
Minimize
Subject to:
2,500X1 + 3,000X2
X1
X2
X1 +
X2
X1, X2
Optimal Solution: X1 = 40
7-19
0
X2 = 20
(minimize faculty salaries)
30 (schedule at least 30 undergrad courses)
20 (schedule at least 20 grad courses)
60 (schedule at least 60 total courses)
(non-negativity constraints)
Cost = $160,000
MSA Computer Corporation manufactures two models of minicomputers, the Alpha 4 and the
Beta 5. The firm employs five technicians, working 160 hours each per month, on its assembly
line. Management insists that full employment (i.e., all 160 hours of time) be maintained for each
worker during next months operations. It requires 20 labor hours to assemble each Alpha 4
computer and 25 labor hours to assemble each Beta 5 model. MSA wants to see at least 10 Alpha
4s and at least 15 Beta 5s produced during the production period. Alpha 4s generate $1,200 profit
per unit, and Beta 5s yield $1,800 each. Determine the most profitable number of each model of
minicomputer to produce during the coming month.
Let X1 = the number of Alpha 4 computers scheduled for production next month
X2 = the number of Beta 5 computers scheduled for production next month
Maximize
Subject to:
1,200X1 + 1,800X2
20X1 +
25X2
X1
X2
X1, X2
Optimal Solution: X1 = 10
(maximize profit)
=
800 (full employment, 5 workers x 160 hours)
10 (make at least 10 Alpha 4 computers)
15 (make at least 15 Beta 5 computers)
0
(non-negativity constraints)
X2 = 24
Profit = $55,200
ISM 6407 Fall 2009
7-20
A winner of the Texas Lotto has decided to invest $50,000 per year in the stock market. Under
consideration are stocks for a petrochemical firm and a public utility. Although a long-range goal
is to get the highest possible return, some consideration is given to the risk involved with the
stocks. A risk index on a scale of 110 (with 10 being the most risky) is assigned to each of the
two stocks. The total risk of the portfolio is found by multiplying the risk of each stock by the
dollars invested in that stock. The following table provides a summary of the return and risk:
Stock
Petrochemical
Utility
Estimated Return
12%
6%
Risk Index
9
4
The investor would like to maximize the return on the investment, but the average risk index of
the investment should not be higher than 6. How much should be invested in each stock? What is
the average risk for this investment? What is the estimated return for this investment?
Let X1 = the number of dollars invested in petrochemical stocks
X2 = the number of dollars invested in utility stocks
Maximize
Subject to:
.12X1 +
X1 +
3X1
.06X2
X2 50,000
2X2
0
X1, X2 0
(maximize return on investment)
(limit on total investment)
(average risk cannot exceed 6)
(non-negativity constraints)
Optimal Solution: X1 = $20,000 X2 = $30,000 Return = $4,200
The total risk is 300,000 (9 x $20,000 + 4 x $30,000), which yields an average risk of 6
(300,000/50,000 = 6).
7-21
Referring to the Texas Lotto situation in Problem 7-20, suppose the investor has changed his
attitude about the investment and wishes to give greater emphasis to the risk of the investment.
Now the investor wishes to minimize the risk of the investment as long as a return of at least 8%
is generated. Formulate this as an LP problem and find the optimal solution. How much should
be invested in each stock? What is the average risk for this investment? What is the estimated
return for this investment?
Let X1 = the number of dollars invested in petrochemical stocks
X2 = the number of dollars invested in utility stocks
Minimize
Subject to:
9X1 +
X1 +
.04X1
4X2
X2 50,000
.02X2
0
X1, X2 0
(minimize total risk)
(limit on total investment)
(average return must be at least 8%)
(non-negativity constraints)
Optimal Solution: X1 = $16,666.67 X2 = $33,333.33 Total risk = 283,333.33 (which
equates to an average risk of 283,333.33/50,000 = 5.67).
The total return would be $4000 (.12 x 16,666.67 + .09 x 33,333.33), which just happens to
be a return of exactly 8% ($4000/$50,000).
ISM 6407 Fall 2009
7-24
The stock brokerage firm of Blank, Leibowitz, and Weinberger has analyzed and recommended
two stocks to an investors club of college professors. The professors were interested in factors
such as short term growth, intermediate growth, and dividend rates. These data on each stock are
as follows:
Factor
Short term growth
potential, per dollar
invested
Intermediate
growth potential
(over next three
years), per dollar
invested
Dividend rate
potential
Stock
Louisiana Gas and
Trimex Insulation
Power
Company
.36
.24
1.67
1.5
4%
8%
Each member of the club has an investment goal of (1) an appreciation of no less than $720 in
the short term, (2) an appreciation of at least $5,000 in the next three years, and (3) a dividend
income of at least $200 per year. What is the smallest investment that a professor can make to
meet these three goals?
Let X1 = the number of dollars invested in Louisiana Gas and Power
X2 = the number of dollars invested in Trimex Insulation Co.
Minimize
Subject to:
X1
.36X1
1.67X1
.04X1
+
+
+
+
X2
.24X2
1.50X2
.08X2
X1, X2
720
5,000
200
0
Optimal Solution: X1 = $1,359
(minimize total investment)
(appreciation in the short term)
(appreciation in next three years)
(dividend income per year)
(non-negativity constraints)
X2 = $1,818.18 Total investment = $3,177.18
ISM 6407 Fall 2009
7-25
Woofer Pet Foods produces a low-calorie dog food for overweight dogs. This product is made
from beef products and grain. Each pound of beef costs $0.90, and each pound of grain costs
$0.60. A pound of the dog food must contain at least 9 units of Vitamin 1 and 10 units of
Vitamin 2. A pound of beef contains 10 units of Vitamin 1 and 12 units of Vitamin 2. A pound of
grain contains 6 units of Vitamin 1 and 9 units of Vitamin 2. Formulate this as an LP problem to
minimize the cost of the dog food. How many pounds of beef and grain should be included in
each pound of dog food? What is the cost and vitamin content of the final product?
Let X1 = the number of pounds of beef in each pound of dog food
X2 = the number of pounds of grain in each pound of dog food
Minimize
Subject to:
.90X1
X1
10X1
12X1
+
+
+
+
.60X2
X2
6X2
9X2
X1, X2
=
0
(minimize cost per pound of dog food)
1 (total weight should be one pound)
9 (at least 9 units of vitamin 1 in a pound)
10 (at least 10 units of vitamin 2 in a pound)
(non-negativity constraints)
Optimal Solution: X1 = .75 X2 = .25 Cost = $.825
ISM 6407 Fall 2009
SOLUTIONS TO SELECT PROBLEMS FROM CHAPTER 8
8-1
(Production problem) Winkler Furniture manufactures two different types of china cabinets: a
French Provincial model and a Danish Modern model. Each cabinet produced must go through
three departments: carpentry, painting, and finishing. The table below contains all relevant
information concerning production times per cabinet produced and production capacities for
each operation per day, along with net revenue per unit produced. The firm has a contract with
an Indiana distributor to produce a minimum of 300 of each cabinet per week (or 60 cabinets per
day). Owner Bob Winkler would like to determine a product mix to maximize his daily revenue.
(a) Formulate as an LP problem.
(b) Solve using an LP software program or spreadsheet.
Cabinet Style
French Provincial
Danish Modern
Dept. capacity (hrs)
Carpentry
(Hours/Cabinet)
3
2
360
Painting
(Hours/Cabinet)
1.5
1
200
Finishing
(Hours/Cabinet)
.75
.75
125
Net Revenue
per Cabinet ($)
28
25
Let X1 = the number of French Provincial cabinets produced each day
X2 = the number of Danish Modern cabinets produced each day
Maximize
Subject to:
28X1
3X1
1.5X1
.75X1
X1
+
+
+
+
25X2
(maximize revenue)
2X2
360 (carpentry hours available)
X2
200 (painting hours available)
.75X2
125 (finishing hours available)
60 (contract requirement on F.P.
60 cabinets)
X2 (contract requirement on D.M. cabinets)
X1, X2 0
(non-negativity constraints)
Optimal Solution: X1 = 60
X2 = 90
Revenue = $3,930
ISM 6407 Fall 2009
8-2
(Investment decision problem) The Heinlein and Krarnpf Brokerage firm has just been instructed
by one of its clients to invest $250,000 for her money obtained recently through the sale of land
holdings in Ohio. The client has a good deal of trust in the investment house, but she also has her
own ideas about the distribution of the funds being invested. In particular, she requests that the
firm select whatever stocks and bonds they believe are well rated, but within the following
guidelines:
(a) Municipal bonds should constitute at least 20% of the investment.
(b) At least 40% of the funds should be placed in a combination of electronic firms, aerospace
firms, and drug manufacturers.
(c) No more than 50% of the amount invested in municipal bonds should be placed in a highrisk, high-yield nursing home stock.
Subject to these restraints, the clients goal is to maximize projected return on investments. The
analysts at Heinlein and Krampf, aware of these guidelines, prepare a list of high-quality stocks
and bonds and their corresponding rates of return.
Projected Rate
of Return (%)
5.3
6.8
4.9
8.4
11.8
Investment
Los Angeles municipal bonds
Thompson Electronics, Inc.
United Aerospace Corp.
Palmer Drugs
Happy Days Nursing Homes
(a) Formulate this portfolio selection problem using LP. (b) Solve this problem.
Let X1 = dollars invested in Los Angeles municipal bonds
X2 = dollars invested in Thompson Electronics
X3 = dollars invested in United Aerospace
X4 = dollars invested in Palmer Drugs
X5 = dollars invested in Happy Days Nursing Homes
Maximize
.053X1
+
.068X2
+
.049X3
+
.084X4
+
.118X5
Subject to:
X1
+
X2
+
X3
+
X4
+
X5
250,000
.8X1
-
.2X2
-
.2X3
-
.2X4
-
.2X5
0
(municipal bond restriction)
-.4X1
+
.6X2
+
.6X3
+
.6X4
-
.4X5
0
(electronics, aerospace, drugs combo)
+
X5
0
(nursing home as a percent of bonds)
X1, X2, X3, X4, X5
-.5X1
(maximize return on investment)
0
(total funds available)
(non-negativity constraints)
Optimal Solution: X1 = $50,000 X2 = $0 X3 = $0 X4 = $175,000 X5 = $25,000 ROI = $20,300
ISM 6407 Fall 2009
8-3
(Restaurant work scheduling problem). The famous Y. S. Chang Restaurant is open 24 hours a
day. Waiters and busboys report for duty at 3AM., 7 AM., 11 AM., 3 P.M., 7 P.M., or 11 P.M.,
and each works an 8-hour shift. The following table shows the minimum number of workers
needed during the six periods into which the day is divided. Changs scheduling problem is to
determine how many waiters and busboys should report for work at the start of each time period
to minimize the total staff required for one days operation. (Hint: Let Xi equal the number of
waiters and busboys beginning work in time period i, where i = 1, 2,3,4,5,6.)
Period
1
2
3
4
5
6
Time
3 A.M7 A.M.
7 A.M11 A.M.
11 A.M3 P.M.
3 P.M7 P.M.
7 P.M11 P.M.
11 P.M3 A.M.
Number of Waiters and
Busboys Required
3
12
16
9
11
4
Let Xi = the number workers beginning work at the start of time period i (i=1,2,3,4,5,6)
Minimize
Subject to:
X1
+
X2
+
X3
+
X4
+
X5
(min. staff size)
3
(period 1)
X2
12
(period 2)
X2
16
(period 3)
X4
9
(period 4)
X4
X5
11
(period 5)
X5
X6
4
(period 6)
X1, X2, X3, X4, X5, X6
+
+
X6
X1
X1
+
+
X6
X3
X3
+
+
+
0
(non-negativity)
ISM 6407 Fall 2009
8-4
(Animal feed mix problem) The Battery Park Stable feeds and houses the horses used to pull
tourist-filled carriages through the streets of Charlestons historic waterfront area. The stable
owner, an ex-racehorse trainer, recognizes the need to set a nutritional diet for the horses in his
care. At the same time, he would like to keep the overall daily cost of feed to a minimum.
The feed mixes available for the horses diet are an oat product, a highly enriched grain,
and a mineral product. Each of these mixes contains a certain amount of five ingredients needed
daily to keep the average horse healthy. The table below shows these minimum requirements,
units of each ingredient per pound of feed mix, and costs for the three mixes.
In addition, the stable owner is aware that an overfed horse is a sluggish worker.
Consequently, he determines that 6 pounds of feed per day are the most that any horse needs to
function properly. Formulate this problem and solve for the optimal daily mix of the three feeds.
Diet
Requirement
(Ingredients)
A
B
C
D
E
Cost/lb
Let
Oat
Product
(units/lb)
2
.5
3
1
.5
$0.09
Feed Mix
Enriched
Grain
(units/lb)
3
1
5
1.5
.5
$0.14
Mineral
Product
(units/lb)
1
.5
6
2
1.5
$0.17
Minimum Daily
Requirement
(units)
6
2
9
8
5
X1 = the number pounds of oat product per horse each day
X2 = the number pounds of enriched grain per horse each day
X3 = the number pounds of mineral product per horse each day
Minimize
s.t.
.09X1
2X1
.5X1
3X1
X1
.5X1
X1
+
+
+
+
+
+
+
.14X2
3X2
X2
5X2
1.5X2
.5X2
X2
+
+
+
+
+
+
+
.17X3
X3
.5X3
6X3
2X3
1.5X3
X3
X1, X2, X3
6
2
9
8
5
6
0
(minimize cost)
(ingredient A)
(ingredient B)
(ingredient C)
(ingredient D)
(ingredient E)
(maximum feed per day)
(non-negativity constraints)
ISM 6407 Fall 2009
8-6
(Media selection problem) The advertising director for Diversey Paint and Supply, a chain of
four retail stores on Chicagos North Side, is considering two media possibilities. One plan is for
a series of half- page ads in the Sunday Chicago Tribune newspaper, and the other is for
advertising time on Chicago TV. The stores are expanding their lines of do-it-yourself tools, and
the advertising director is interested in an exposure level of at least 40% within the citys
neighborhoods and 60% in northwest suburban areas.
The TV viewing time under consideration has an exposure rating per spot of 5% in city
homes and 3% in the northwest suburbs. The Sunday newspaper has corresponding exposure
rates of 4% and 3% per ad. The cost of a half-page Tribune advertisement is $925; a television
spot costs $2,000.
Diversey Paint would like to select the least costly advertising strategy that would meet
desired exposure levels.
(a) Formulate using LP.
(b) Solve the problem.
Let
X1 = the number of newspaper ads placed
X2 = the number of TV spots purchased
Minimize
Subject to:
925X1 + 2,000X2
.04X1 +
.05X2
.03X1 +
.03X2
X1, X2 0
(minimize cost)
.4 (city exposure)
.6 (suburb exposure)
(non-negativity constraints)
ISM 6407 Fall 2009
8-11
(College meal selection problem) Kathy Roniger, campus dietician for a small Idaho college, is
responsible for formulating a nutritious meal plan for students. For an evening meal, she feels
that the following five meal-content requirements should be met: (1) between 900 and 1,500
calories; (2) at least 4 milligrams of iron; (3) no more than 50 grams of fat; (4) at least 26 grams
of protein; and (5) no more than 50 grams of carbohydrates. On a particular day, Ronigers food
stock includes seven items that can be prepared and served for supper to meet these
requirements. The cost per pound for each food item and the contribution to each of the five
nutritional requirements are given in the accompanying table:
Calories/
Pound
295
1216
394
358
128
118
279
Food Item
Milk
Ground Meat
Chicken
Fish
Beans
Spinach
Potatoes
Iron
(mg/lb)
0.2
0.2
4.3
3.2
3.2
14.1
2.2
Table of Food Values and Costs
Fat
Protein
Carbs.
(gm/lb)
(gm/lb)
(gm/lb)
16
16
22
96
81
0
9
74
0
0.5
83
0
0.8
7
28
1.4
14
19
0.5
8
63
Cost/
Pound ($)
0.60
2.35
1.15
2.25
0.58
1.17
0.33
What combination and amounts of food items will provide the nutrition Roniger requires at the
least total food cost?
Let X1 = the number of pounds of milk per student in the evening meal
X2 = the number of pounds of ground meat per student in the evening meal
Etc., down to X7 = the number of pounds of potatoes per student in the evening meal
Minimize
S.T. (Cal.)
.6X1
+
2.35X2
+
1.15X3
+
2.25X4
+
.58X5
+
1.17X6
+
295X1
+
1216X2
+
394X3
+
358X4
+
128X5
+
118X6
+
.33X7
279X7
900
1500
(Cal.)
295X1
+
1216X2
+
394X3
+
358X4
+
128X5
+
118X6
+
279X7
(Iron)
.2X1
+
.2X2
+
4.3X3
+
3.2X4
+
3.2X5
+
14.1X6
+
2.2X7
4
(Fat)
16X1
+
96X2
+
9X3
+
.5X4
+
.8X5
+
1.4X6
+
.5X7
50
26
50
(Protein)
16X1
(Carbs.)
22X1
+
81X2
+
74X3
+
83X4
+
7X5
+
14X6
+
8X7
+
28X5
+
19X6
+
63X7
X1, X2, X3, X4, X5, X6, X7
0
ISM 6407 Fall 2009
8-12
(High tech production problem) Quitmeyer Electronics Incorporated manufactures the following
six microcomputer peripheral devices: internal modems, external modems, graphics circuit
boards, CD drives, hard disk drives, and memory expansion boards. Each of these technical
products requires time, in minutes, on three types of electronic testing equipment, as shown in
the table the following table:
Internal
Modem
Test device 1
7
Test device 2
2
Test device 3
5
External
Modem
3
5
1
Circuit
Board
12
3
3
CD
Drive
6
2
2
Hard
Drive
18
15
9
Memory
Board
17
17
2
The first two test devices are available 120 hours per week. The third (device 3) requires more
preventive maintenance and may be used only 100 hours each week. The market for all six
computer components is vast, and Quitmeyer Electronics believes that it can sell as many units
of each product as it can manufacture. The table that follows summarizes the revenues and
material costs for each product:
Device
Internal modem
External modem
Graphics circuit board
CD drive
Hard disk drive
Memory expansion board
Revenue Per
Unit Sold ($)
200
120
180
130
430
260
Material Cost
Per Unit ($)
35
25
40
45
170
60
In addition, variable labor costs are $15 per hour for test device 1, $12 per hour for test device 2.
and $18 per hour for test device 3. Quitmeyer Electronics wants to maximize its profits.
(a) Formulate this problem as an LP model.
(b) Solve the problem by computer. What is the best product mix?
(c) What is the value of an additional minute of time per week on test device 1? Test device 2?
Test device 3? Should Quitmeyer Electronics add more test device time? If so, on which
equipment?
Let X1 = the number of internal modems scheduled for manufacture each week
X2 = the number of external modems scheduled for manufacture each week
Etc., down to X6 = the number of mem. expansion boards scheduled for mfg. each week
Maximize
161.35X 1
+
92.95X2
+
135.50X3
+
82.50X4
+
249.80X5
+
191.75X6
S.T.
7X1
+
3X2
+
12X3
+
6X4
+
18X5
+
17X6
7200
2X1
+
5X2
+
3X3
+
2X4
+
15X5
+
17X6
7200
5X1
+
1X2
+
3X3
+
2X4
+
9X5
+
2X6
6000
X1, X2, X3, X4, X5, X6
0
ISM 6407 Fall 2009
8-15
(Material blending problem) Amalgamated Products has just received a contract to construct
steel body frames for automobiles that are to be produced at the new Japanese factory in
Tennessee. The Japanese auto manufacturer has strict quality control standards for all of its
component subcontractors and has informed Amalgamated that each frame must have the
following steel content:
Material
Manganese
Silicon
Carbon
Minimum Percent
2.1
4.3
5.05
Maximum Percent
2.3
4.6
5.35
Amalgamated mixes batches of eight different available materials to produce one ton of steel
used in the body frames. The table below details these materials. Formulate and solve the LP
model that will indicate how much of each of the eight materials should be blended into a 1-ton
load of steel so that Amalgamated meets its requirements while minimizing cost.
Material
Available
Alloy 1
Alloy 2
Alloy 3
Iron 1
Iron 2
Carbide 1
Carbide 2
Carbide 3
Manganese
(%)
70.0
55.0
12.0
1.0
5.0
0
0
0
Silicon
(%)
15.0
30.0
26.0
10.0
2.5
24.0
25.0
23.00
Carbon
(%)
3.0
1.0
0
3.0
0
18.0
20.0
25.0
Pounds
Available
No limit
300
No limit
No limit
No limit
50
200
100
Cost Per
Pound ($)
0.12
0.13
0.15
0.09
0.07
0.10
0.12
0.09
Let X1 = the number of pounds of alloy 1 in one ton of steel
X2 = the number of pounds of alloy 2 in one ton of steel
Etc., down to X8 = the number of pounds of carbide 3 in one ton of steel
Minimize
S.T. (Mnmin)
(Mn-max)
.12X1
+
.13X2
+
.15X3
+
.09X4
+
.07X5
.7X1
+
.55X2
+
.12X3
+
.01X4
+
.05X5
42
.7X1
+
.55X2
+
.12X3
+
.01X4
+
.05X5
46
(Si-min)
.15X1
+
.30X2
+
.26X3
+
.10X4
+
.025X5
.23X8
86
.23X8
92
101
(Si-max)
.15X1
+
.30X2
+
.26X3
+
.10X4
+
.025X5
+
+
+
.10X6
.24X6
.24X6
+
+
+
.12X7
.25X7
.25X7
+
+
+
.09X8
(C-min)
.03X1
+
.01X2
+
.03X4
+
.18X6
+
.20X7
+
.25X8
(C-max)
.03X1
+
.01X2
+
.03X4
+
.18X6
+
.20X7
+
.25X8
107
300
50
200
Alloy 2 lim.
X2
Carbide 1 lim.
X6
Carbide 2 lim.
X7
X8
Carbide 3 lim.
Weighs 1 ton
X1
+
X2
+
X3
+
X4
+
X5
+
X6
+
X1, X2, X3, X4, X5, X6, X7, X8
X7
0
+
100
X8
=
2000