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SOLUTIONARY - FORMULATION OF LINEAR PROGRAMS

1. Diet Problem – OZARK FARMS .

OZARK FARMS uses at least 800 pounds of specialty feed daily. The special
food is a mixture of corn and soybean seeds, with the following compositions:
Pound for Pound of Livestock Feed

Livestock Feed Proteins Fiber Cost (/pound)

Corn 0.09 0.02 0.30
Soybean 0.60 0.06 0.90

The daily dietary requirements for the special food stipulate at least 30%
protein and at most 5% fiber. OZARK FARMS wants to determine the minimum
daily cost of the feed mix. Apply a PL model.

SOLUTION:

Livestock Feed Proteins Fiber Cost (/pound) Production

Corn 0.09 0.02 0.30 1
Soybean 0.60 0.06 0.90 1
Available 0.30 (mix) 0.05 (mix) - 800

a) Definition of variables
X 1 : Quantity of pounds of corn to use in the daily mix.
X 2 : Quantity to use in pounds of soybean seed in the daily mixture.

b) Objective Function: minimize

Min Z = 0.30X 1 + 0.90X 2

c) Restrictions:

For the amount of food available: X 1 + X 2 ≥ 800

For the daily dietary requirements of the food:

Proteins: 0.09X 1 + 0.60X 2 ≥ 0.30 (X 1 + X 2 )

Fiber: 0.02X 1 + 0.06X 2  0.05 (X 1 + X 2 )

d) Non-negativity constraints:
X 1  0, X 2  0

e) The PL model is:

Min Z = 0.30X 1 + 0.90X 2

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Subject to

X 1 + X 2 ≥ 800
0.09X 1 + 0.60X 2 ≥ 0.30 (X 1 + X 2 )
0.02X 1 + 0.06X 2  0.05 (X 1 + X 2 )

X1,X20

2. XYZ Company produces screws and nails. The raw material for the screws costs
S/.2.00 per unit, while the material for each nail costs S/. 2.50. A nail requires two
hours of labor in department No. 1 and three hours in department No. 2, while a screw
requires 4 hours in department No. 1 and 2 hours in department No. 2, the hourly
wage in both departments is S/.2.00. If both products are sold at S/.18.00 and the
number of hours of labor available per week in the departments is 160 and 180
respectively, express the proposed problem as a linear program, such that profits are
maximized.

SOLUTION:

a) Definition of variables

X 1 : Quantity of screws to be produced per week.

X 2 : Quantity of nails to be produced per week.

Hours MO – Hours MO – Raw Hourly Sale

DEPARTME DEPARTME Material wage price
NT 1 NT 2 Cost payment
Screws 4 2 S/. 2 S/. 2 S/. 18
Nails 2 3 S/. 2.5 S/. 2 S/. 18
Available hours 160 180 - - -

PROFIT = SALES PRICE – COST

COST = (hourly wage payment x total hours DPTO1 and DPTO2) + (Raw Material
Cost)

Cost of Screws = (S/.2 x 6 hours) + (S/. 2)

= S/. 14

UTILITY – SCREWS = S/.18 – S/.14 = S/. 4

Cost of Nails = (S/.2 x 5 hours) + (S/. 2.50)

= S/. 12.50

PROFIT – NAILS = S/.18 – S/.12.50 = S/. 5.5

b) Objective Function: maximize profits

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Max Z = 4X 1 + 5.50X 2

c) Restrictions

M.Obra - DEPARTMENT 1: 4X 1 + 2X 2  160

M.Obra - DEPARTMENT 2: 2X 1 + 3X 2  180

d) Non-negativity constraints
X10,X20

e) The PL model is:

Max Z = 4X 1 + 5.50X 2

Subject to
4X 1 + 2X 2  160
2X 1 + 3X 2  180

X1,X20

3. A young mathematician was asked to entertain a visitor to his company for 90

minutes. He thought it would be a great idea to get the guest drunk. The
mathematician was given S/.50.00.
The young man knew that the visitor liked to mix his drinks, but that he always
drank less than 8 glasses of beer, 10 gins, 12 whiskeys and 24 martinis. The
time he spent drinking was 15 minutes per glass of beer, 6 minutes per glass
of gin, 7 minutes per glass of whiskey and 4 minutes per glass of martini. The
prices of the drinks were: Beer S/.1.00 per glass, gin S/.2.00 per glass,
whiskey S/.2.00 per glass and martini S/.4.00 per glass. The mathematician
assumes that the object is to maximize alcohol consumption during the 90
minutes he had to entertain his guest. He got a chemist friend to give him the
alcohol content of the drinks quantitatively, the alcoholic units for a glass of
beer being 17, for a glass of gin 15, for a glass of whiskey 16 and for a glass of
martini 7. The visitor always drank a minimum of 2 whiskeys.

SOLUTION:

DRINKS TIME COSTS MIXTURES ALCOHOLIC

CONSUMPTI
ON
Beer 15 S/. 1 8 17
Geneva 6 S/. 2 10 15
Whiskey 7 S/. 2 2 – 12 16
Martini 4 S/. 4 24 7
AVAILABLE 90 S/. 50 - -

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a) Definition of Variables:

x 1 = Amount to drink from glasses of beer.

x 2 = Amount to drink from glasses of gin.
x 3 = Amount to drink from glasses of whiskey.
x 4 = Amount to drink from martini glasses.

b) Objective Function: maximize alcohol consumption.

Max Z = 17x 1 + 15x 2 +16x 3 + 7x 4

c) Restrictions :

 Restriction 1: Cost of drinks.

1x 1 + 2x 2 + 2x 3 + 4x 4 ≤ 50

 Restriction 2: Limit number of beer glasses.

x1<8

 Restriction 3: Limit number of gin glasses.

x2 < 10

 Restriction 4: Limit number of whiskey glasses.

2 ≤ x 3 < 12

 Restriction 5: Limit number of martini glasses.

X 4 < 24

 Restriction 6: The time you spend drinking.

15x 1 + 6x 2 + 7x 3 + 4x 4 ≤ 90

d) No Negativity Restrictions:

x1 ≥ 0, x2 ≥ 0 , x3 ≥ 0, x4 ≥ 0

e) The PL model is:

Max Z = 17x 1 + 15x 2 +16x 3 + 7x 4

Subject to
1x 1 + 2x 2 + 2x 3 + 4x 4 ≤ 50
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x1<8
x2 < 10
2 ≤ x 3 < 12
X 4 < 24
15x 1 + 6x 2 + 7x 3 + 4x 4 ≤ 90

x1 , x2, x3 , x4 ≥ 0

4. LAN-PERU is considering the probability of acquiring passenger aircraft in the

world market: USA, England or Russia. The cost of the plane – USA (A) is $6.7
million, the plane – English (B) is $5 million and the plane – Russian (C) is $3.5
million.
The board of directors of said company has authorized the purchase of aircraft
worth 150 million. LAN-PERU economists have calculated that whichever type
A with the highest capacity will provide a net profit of $420 thousand per year,
plane B will provide a net profit of $300 thousand and plane C a net profit of
$230 thousand per year.
On the other hand, it is known that the Peruvian Air Force could only provide
30 properly trained pilots. If only smaller aircraft are acquired, the repair and
service services available to LAN-PERÚ will only be able to keep a maximum
of 40 units in operation. Furthermore, it is known that maintaining airplane B
requires 1 1/3 more than airplane C, and that airplane A requires 1 2/3 more
than airplane C. Determine a LP model for the purchase of aircraft, obtaining
maximum profits.

SOLUTION:

PLANES UTILITY COST PILOTS MAINTENANCE

(millions)
USA (A) 420 6.7 1 2
1
3
England (B) 300 5 1 1
1
3
Russia (C) 230 3.5 1 1
AVAILABLE - 150 30 40

a) Definition of variables:
X 1 : Quantity to purchase of aircraft A.
X 2 : Quantity to purchase of aircraft B.
X 3 : Quantity to purchase of aircraft C.

b) Objective Function: maximize profits.

Max Z = 420X 1 + 300X 2 + 230X 3

c) Restrictions:

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 For the cost of each plane: 6.7X 1 + 5X 2 + 3.5X 3  150 (millions)

 For the number of pilots available: X 1 + X 2 + X 3  30

2 1
 Aircraft maintenance capacity:1 x 1 +1 X 2 + X 3  40
3 3

d) Non-negativity constraints:

X 1  0, X 2  0, X 3  0

e) The PL model is:

Max Z = 420X 1 + 300X 2 + 230X 3

Subject to

6.7X 1 + 5X 2 + 3.5X 3  150

X 1 + X 2 + X 3  30
2 1
1 x 1 +1 X 2 + X 3  40
3 3

X1,X2,X30

5. A seller is in charge of two products A and B. You want to set up a calling

schedule for the coming months. He expects to be able to sell at most 20 units
of product A and at most 78 units of product B.
He must sell at least 48 units of product B, to satisfy his minimum sales quota,
he receives a 10% commission on the total sale he makes. But he must pay
his own costs (which are estimated at 30 soles per hour for making calls) from
his commission. He is willing to spend no more than 160 hours per month
calling his clients. The following data is available in the following table:

PRODUCT SALE PRICE TIME USED PROBABILITY OF A

Soles/Unit Time/call CALL SALE
TO 3 000 3 0.5
b 1 400 1 0.6

Formulate the problem in a way that maximizes the amount of profit the seller
expects.

SOLUTION:

a) Definition of variables:

X 1 : Number of calls to sell product A.

X 2 : Number of calls to sell product B.
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b) Objective Function: Maximize the seller's profit.

PROFIT-SELLER = % COMMISSION (PROBAB. (SALES PRICE)-COST

Max Z = 0.1 (3000 (0.5)X 1 + 1400 (0.6)X 2 ) – 30 (3 X 1 + X 2 )

Max Z = 0.1 (1500 X 1 + 840 X 2 ) – 90 X 1 - 30 X 2

Maz Z = 150 X 1 + 84 X 2 – 90 X 1 - 30 X 2

Maz Z = 60 X 1 + 54 X 2

c) Restrictions

 Quantity of products sold:

Product A: 0.5 X 1  20

Product B: 0.6 X 2  78

48 ≤ X 2

 Time spent making calls: 3 X 1 + X 2  160

d) Non-negativity constraints:
X 1  0, X 2  0

e) The PL model is:

Maz Z = 60 X 1 + 54 X 2

Subject to
0.5 X 1  20
0.6 X 2  78
48 ≤ X 2
3 X 1 + X 2  160

X1,X20

6. A contractor is considering a proposal for paving a road; the specifications

require a minimum thickness of 12 inches and a maximum of 48 inches. The
path must be paved in concrete, asphalt or gravel, or any combination of the
three.
However, specifications require a final consistency equal to or greater than that
of a 54-inch-thick concrete surface. The contractor has determined that 3

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inches of his asphalt is as strong as 1 inch of concrete and 6 inches of gravel is

as strong as 1 inch of concrete. Each inch of thickness costs S/.100 for
concrete, S/.380 for asphalt, and S/.150 for gravel. Determine the combination
of materials he should use to minimize his cost.

SOLUTION:

PAVING COSTS MINIMAL MAXIMUM CONSISTENCY

THICKNE THICKNE
SS SS
Concrete 100 1 1 1
Asphalt 380 1 1 1
3
Gravel 150 1 1 1
6
AVAILABLE - 12 48 54

a) Definition of variables

X 1 : Amount of concrete to be used.

X 2 : Amount of asphalt to be used.
X 3 : Amount of gravel to be used.

b) Objective Function: minimize costs.

Min Z = 100X 1 + 380X 2 + 150X 3

c) Restrictions:

 Minimum thickness: X 1 + X 2 + X 3 ≥ 12

 Maximum thickness: X 1 + X 2 + X 3  48

 Surface consistency compared to concrete:

1 1
x 1 + x 2 + X 3 ≥ 54
3 6

d) Non-negativity constraints

X 1  0, X 2  0, X 3  0

e) The PL model is:

Min Z = 100X 1 + 380X 2 + 150X 3

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Subject to

X 1 + X 2 + X 3 ≥ 12
X 1 + X 2 + X 3  48
1 1
x 1 + x 2 + X 3  54
3 6

X 1  0, X 2  0, X 3  0

7. A farmer can raise sheep, pigs, and cattle.

It has space for 30 sheep or 50 pigs or 20 head of cattle or any combination of
these.
The benefits (utilities) given per animal are S/.500, S/.500 and S/.100 for
sheep, pigs and cows respectively. The farmer must raise, by law, at least as
many pigs as sheep and cows combined .

SOLUTION:

a) Definition of variables

X 1 : Number of sheep to raise.

X 2 : Number of pigs to raise.
X 3 : Number of cows to breed.

b) Objective Function: maximize profits.

Max Z = 500X 1 + 500X 2 + 100X 3

c) Restrictions:

 Sheep space limit: X 1 ≤ 30

 Pig space limit: X 2  50

 Cow space limit: X 3  20

 By law: X2≥X1+X3

d) Non-negativity constraints:
X 1  0, X 2  0, X 3  0

e) The PL model is:

Max Z = 500X 1 + 500X 2 + 100X 3

Subject to

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X 1 ≤ 30
X 2  50
X 3  20
X2≥X1+X3

X1,X2,X30

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