the shortest path to destruction: a floral...

The Shortest Path to Destruction: A Floral

Shop

Operations Research I: Linear Programming Project

Kathedra Burton

Mari Juno

Casey Haley

December 6, 2004 “On my honor, as an Aggie, I have neither given nor received unauthorized aid on this academic work.”

__________________________

__________________________

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Signature of Each Group Member

1

Contents

1.0 Executive Summary ........................................................................................2

2.0 Problem Description........................................................................................4

2.1 Problem 1 Grummins Engine Truck Production ...................................4

2.2 Problem 2 Wheat Stock........................................................................5

2.3 Problem 3 Floral Dilemma....................................................................7

3.0 Computational Results....................................................................................9

3.1 Problem 1 Grummins Engine Truck Production ...................................9

3.2 Problem 2 Wheat Stock......................................................................12

3.3 Problem 3 Floral Dilemma..................................................................15

4.0 Conclusions & Recommendations ................................................................21

5.0 References....................................................................................................23

6.0 Appendix.......................................................................................................24

2

1 Executive SummaryThe purpose of this project was to take 3 situational problems and solve

each one using linear programming methods. The problems addressed are 1)

Multi-period Inventory, 2) Multi-period Work Scheduling and 3) Floral Buying and

Scheduling. All problems were solved using the LINDO optimization software.

In the Multi-period inventory problem, Grummins Engine must decide how

many trucks to produce in a three-year period to both maximize profit and satisfy

a government imposed pollution constraint. Grummins will produce according to

the following schedule and realize a profit of $3,600,000.

Table 1—Optimal Solution for Grummins

Year Truck Type Produced Sold Inventoried

1 1 100 100 0

2 200 200 0

2 1 180 180 0

2 100 100 0

3 1 170 170 0

2 150 150 0

Recommendations for Grummins:

1. Study customers to see if they will accept an increase in selling price.

2. Look into better manufacturing practices to reduce manufacturing cost.

In the multi-period work scheduling problem, a wheat warehouse must

decide on how much wheat to sell and purchase according to an initial monthly

inventory and a storage constraint of 20,000 bushels. The maximum profit is

$162. The optimal solution is shown below and is in 1000 bushels.

Table 2—Optimal Solution for Wheat Warehouse

Month Initial Stock Wheat Sold Wheat Bought

1 6 0 0

2 6 0 0

3 6 6 20

3

4 20 0 0

5 20 0 0

6 20 20 20

7 20 20 0

8 0 0 20

9 20 20 0

10 0 0 0

Recommendations:

1. Try to increase selling price.

2. Look for a different retailer to buy wheat at a lower price.

3. If 1 and 2 can be done, look into increasing storage space.

The third problem looks at a floral shop. The florist buys five different

colors of flowers between the months of February and September. She has to

decide on what color to buy each month and how many while minimizing her total

cost. LINDO provided an optimal solution with a total cost of $ 20,500.

Table 3—Optimal Solution for Floral Shop

Color Feb Mar Apr May June July Aug Sep

Red 500 500

White 500 500

Pink 500 500

Yellow 500

Purple 500

Recommendations:

1. Look for other suppliers.

2. Study customer demand to see if cheaper alternatives can be found such

as underused colors that can be bought at reduced prices.

4

2.1 Problem 1 : Grummins Engine Diesel Truck Production

Grummins Engine produces diesel trucks. New government emission

standards have dictated that the average pollution emissions of all trucks

produced in the next three years cannot exceed 10 grams per truck. Grummins

produces two types of trucks. Each type 1 truck sells for $20,000, cost $15,000

to manufacture, and emits 15 grams of pollution. Each type 2 sells for $17,000,

cost $14,000 to manufacture, and emits 5 grams of pollution. Production

capacity limits total truck production during each year to at most 320 trucks.

Grummins knows that the maximum number of ach truck type that can be sold

during each of the next years is given in Table 62. Demand can be met from

previous production or the current year’s production. It costs $2,000 to hold 1

truck (of any type) in inventory for one year [1, p117].

Assumptions:

1. There will be no trucks left over from year three.

2. The divisibility assumption is not valid.

3. All variables are nonnegative.

Table 62 [1]

Year Type 1 Type 2

1 100 200

2 200 100

3 300 150

Decision Variables:

Xij = number of trucks made

Yij = number of trucks sold

Zij = number of trucks in inventory and carried over

where i = year made from 1to 3 and j = type of truck 1 and 2

5

Objective Function:

Maximize 20000y11 + 20000y21 + 20000y31 + 17000y12 + 17000y22 + 17000y32 –

15000x11 – 15000x21 – 15000x31 – 14000x12 – 14000x22 – 14000x32 -

2000z11 – 2000z12 - 2000z21 - 2000z22

s.t.

y11 < 100 y12 < 200

x11+x12 < 320 x11-y11-z11 = 0

x12-y12-z12 = 0 x21+x22 < 320

z11+y21 < 200 z12+y22 < 100

x21+z11-y21-z21 = 0 x22+z12-y22-z22 = 0

x31+x32 < 320 z21+y31 < 300

z22+y32 < 150 x31+z21-y31 = 0

x32+z22-y32 = 0 5y11+5y21+5y31-5y12-5y22-5y32 < 0

2.2 Problem 2: Wheat Stock

You own a wheat warehouse with a capacity of 20,000 bushels. At the

beginning of month 1, you have 6,000 bushels of wheat. Each month, wheat can

be bought and sold at the price per 1000 bushels given in table 46. The

sequence of events during each month is as follows:

1. You observe your initial stock of wheat.

2. You can sell any amount of wheat up to your initial stock at the current

month’s selling price.

3. You can buy (at the current month’s buying price) as much wheat as you

want, subject to the warehouse size limitation.

*Everything shown in per thousand bushels [1, 112]

Assumptions:


Table 46 [1]

Month Selling Price ($) Purchase Price ($)

1 3 8

6

2 6 8

3 7 2

4 1 3

5 4 4

6 5 3

7 5 3

8 1 2

9 3 5

10 2 5

Decision Variables:

Xi = initial stock of wheat

Yi = amount of wheat sold

Zi = amount of wheat bought

Objective Function:

Maximize 3y1 + 6y2 + 7y3 + y4 + 4y5 + 5y6 + 5y7 + y8 + 3y9 + 2y10 - 8z1 - 8z2- 2z3 -

3z4 - 4z5 - 3z6 - 3z7 - 2z8 - 5z9 - 5z10

s.t.

x1 = 6 y1-x1 < 0

z1+x1-y1 < 20 x2-z1-x1+y1 = 0

y2-x2 < 0 z2+x2-y2 < 20

x3-z2-x2+y2 = 0 y3-x3 < 0

z3+x3-y3 < 20 x4-z3-x3+y3 = 0

y4-x4 < 0 z4+x4-y4 < 20

x5-z4-x4+y4 = 0 y5-x5 < 0

z5+x5-y5 < 20 x6-z5-x5+y5 = 0

y6-x6 < 0 z6+x6-y6 < 20

x7-z6-x6+y6 = 0 y7-x7 < 0

z7+x7-y7 < 20 x8-z7-x7+y7 = 0

7

y8-x8 < 0 z8+x8-y8 < 20

x9-z8-x8+y8 = 0 y9-x9 < 0

z9+x9-y9 < 20 x10-z9-x9+y9 = 0

y10-x10 < 0 z10+x10-y10 < 20

2.3 Problem 3: Florist Dilemma

A flower shop owner must buy bundles of flowers to make bouquets. She

buys five different colors of flowers between the months of February and

September. She must buy at least 500 bundles of flowers a month. The nursery

where she buys them can only provide her with 1000 of each color of flower

during these months. Below in Table 1 are the prices of each color flower for

each month.

Table 4: Flower Prices


Red 15 14 5 2 4 2 9 14

White 14 9 4 5 1 4 6 12

Pink 16 4 3 3 3 7 5 10

Yellow 13 10 7 1 2 6 7 16

Purple 12 11 8 4 6 5 15 13

Assumptions:


Decision Variables:

xij = number of bundles of flowers purchased each month

where i = color of flower (from 1 to 5)

and j = month purchased (from 1 to 8)

8

Objective Function:

Min z = 15x11 + 14x12 + 5x13 + 2x14 + 4x15 + 2x16 + 9x17 + 14x18 + 14x21 + 9x22 + 4x23

+ 5x24 + 1x25 + 4x26 + 6x27 + 12x28 + 16x31 + 4x32 + 3x33 +

3x34 + 3x35 + 7x36 + 5x37 + 10x38 + 13x41 + 10x42 + 7x43 +

1x44 + 2x45 + 6x46 + 7x47 + 16x48 + 12x51 + 11x52 + 8x53 +

4x54 + 6x55 + 5x56 + 15x57 + 13x58

s. t.

x11 + x21 + x31 + x41 + x51 >= 500

x12 + x22 + x32 + x42 + x52 >= 500

x13 + x23 + x33 + x43 + x53 >= 500

x14 + x24 + x34 + x44 + x54 >= 500

x15 + x25 + x35 + x45 + x55 >= 500

x16 + x26 + x36 + x46 + x56 >= 500

x17 + x27 + x37 + x47 + x57 >= 500

x18 + x28 + x38 + x48 + x58 >= 500

x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 <= 1000

x21 + x22 + x23 + x24 + x25 + x26 + x27 + x28 <= 1000

x31 + x32 + x33 + x34 + x35 + x36 + x37 + x38 <= 1000

x41 + x42 + x43 + x44 + x45 + x46 + x47 + x48 <= 1000

x51 + x52 + x53 + x54 + x55 + x56 + x57 + x58 <= 1000

9

3.1 Problem 1: Grummins Engine Diesel Truck Production

In the Grummins problem, LINDO found the optimal solution in 10

iterations with an objective function value of $3,600,000. Table 1 shows the

number of trucks produced, sold, and in inventory each year.

Table 5: Optimal Solution

Year Truck Type Number Produced

Number Sold Number Inventoried

1 1 100 100 0

2 200 200 0

2 1 180 180 0

2 100 100 0

3 1 170 170 0

2 150 150 0

The data shows that every truck Grummin produces in each year is sold

that same year so no storage costs are accrued.

The sensitivity analysis for this problem is shown below along with the

right-hand side analysis in Tables 6 and 7.

Table 6: Sensitivity Analysis

Variable Current Coefficient Range

X11 -15000 -15000 to -13000

X21 -15000 -17000 to -15000

X31 -15000 -15000 to -10000

X12 -14000 -22000 to -4000

X22 -14000 -19000 to -4000

X32 -14000 -19000 to ∞

Y11 20000 20000 to ∞

Y21 20000 0 to 20000

Y31 20000 20000 to 40000

10

Y12 17000 9000 to ∞

Y22 17000 9000 to ∞

Y32 17000 9000 to ∞

Z11 -2000 -∞ to 0

Z21 -2000 -∞ to 0

Z12 -2000 -∞ to 8000

Z22 -2000 -∞ to 8000

Table 7: Right Hand Side (RHS) Sensitivity Analysis

Row Current RHS Allowable Range

2 100 80 to 120

3 200 20 to 220

4 320 300 to ∞

5 0 -100 to 20

6 0 -200 to 20

7 320 280 to ∞

8 200 180 to ∞

9 100 0 to 120

10 0 -180 to 40

11 0 -100 to 40

12 320 300 to 450

13 300 170 to ∞

14 150 50 to 160

15 0 -130 to 20

16 0 -130 to 20

17 0 -900 to 100

The sensitivity report gives the range in which each variable can change

without altering the optimal solution. For values in the objective function, the

sensitivity analysis shows the selling price of type 1 (all sold at $20,000) trucks

produced in year 1 can increase to infinity but cannot decrease without changing

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the optimal solution. For the same type of trucks produced in year 2, the price

cannot increase but can decrease to $18,000 while trucks produced in year 3

exhibit the exact opposite behavior. The price cannot decrease but will allow for

an increase of $2,000 to $ 22,000. All type 2 trucks (all sold at $17,000) show

the same behavior. The selling price can increase to infinity as well as decrease

to $9,000 meaning the company would take a $5,000 loss. The storage fee for

inventory ($2,000) on type 1 trucks can increase to $4000 and decrease to

infinity before the optimal solution is affected. The fee on type 2 trucks can only

increase to $3,000 but can decrease to infinity as well. The cost of production for

type 1 trucks ($15,000) produced in year 1 can increase to $17,000 but cannot

fall below $15,000. For trucks produced in year 2, the price cannot increase but

can decrease by $2,000 to $13,000. In year 3, the price of production can

increase to infinity but can only decrease to $10,000. For type 2 trucks, the cost

of production ($14,000) can increase to $24,000 and decrease to $6,000 for

trucks produced in year 1. In year 2, the price can also increase to $24,000, but

can only decrease to $9,000. In year 3, production price can increase to infinity

but can only decrease to $9,000.

The sensitivity analysis also studied the right-hand side (RHS) of the

constraints to determine how widely they could vary before a change occurred in

the optimal solution. The year 1 demand for type 1 trucks can vary from 80 to

120 and for type 2 it can vary from 20 to 220. For year 1, total truck production

can vary from 300 to infinity without changing the solution. Total truck production

for year 2 can vary from 280 to infinity and truck production in year 3 can vary

from 300 to 450. The inventory from year1 on type 1 and type 2 trucks can be in

the range of -100 to 20 and -200 to 20, respectively. The number of type 1 trucks

sold in year 2 can vary 160 to infinity and for type 2 it can vary from 0 to 120.

Type 1 trucks inventoried in year 2 have a range of -180 to 40 and type 2 trucks

have a range of -100 to 40. The number of type 1 trucks sold in year 3 can vary

from 170 to infinity and the number of type 2 trucks can be in the range of 60 to

160. The amount of trucks left over after sales in year 3 for type 1and type 2 can

12

both vary from -130 to 20 trucks. The average amount of pollution can vary from

-900 to 100 grams.

3.2 Problem 2: Wheat Stock

LINDO found the optimal solution to this problem in 20 iterations. The

wheat warehouse will realize a profit of $162. The table below shows how much

wheat you initially have, how much is sold and how much is bought each month

(all shown in 1000s of bushels).

Table 7: Optimal Solution


1 6 0 0

2 6 0 0

3 6 6 20

4 20 0 0

5 20 0 0

6 20 20 20

7 20 20 0

8 0 0 20

9 20 20 0

10 0 0 0

The sensitivity analysis for the problem is shown below in Table 8.


Variable Current Coefficient Allowable Range

X1 0 -∞ to ∞

X2 0 -4 to 1

X3 0 -1 to 1

X4 0 -1 to 1

X5 0 -2 to ∞

X6 0 -1 to ∞

13

X7 0 -2 to ∞

X8 0 -∞ to 1

X9 0 -1 to ∞

X10 0 -∞ to 1

Y1 3 -∞ to 7

Y2 6 -∞ to 7

Y3 7 6 to 8

Y4 1 -∞ to 2

Y5 4 2 to 4

Y6 5 4 to ∞

Y7 5 3 to ∞

Y8 1 -∞ to 2

Y9 3 2 to 5

Y10 2 0 to 3

Z1 -8 -221e to -7

Z2 -8 -221e to -7

Z3 -2 -3 to -1

Z4 -3 -∞ to -2

Z5 -4 -∞ to -4

Z6 -3 -5 to ∞

Z7 -3 -5 to -2

Z8 -2 -3 to -1

Z9 -5 -∞ to -3

Z10 -5 -∞ to 0

The SA shows that the selling price of wheat in month 1 and 2 can vary

from negative infinity to $7 per 1000 bushels without altering the optimal solution.

In month 3, it can vary from $6 to $8 per 1000 bushels. The selling price can

vary from negative infinity to $2 per 1000 bushels in month 4. Month 5 can only

decrease to $2 with no increase over the current $4 per 1000 bushels price. The

price can rise to an infinite leveling months 6 and 7 but the lower limit is $4 and

14

$3, respectively. In month 8, the amount can fluctuate from negative infinity to

$2. In months 9 and 10, the amount can be in the ranges $2-$5 and $0-$3,

respectively.

In months1and 2, the price wheat is bought at can vary from negative

infinity to $9 without affecting the solution. The price in months 3 and 4 can be in

the ranges of $1-$3 and infinity-$4, respectively. In month 5, the price can be

anywhere from negative infinity to $4. In months 6 and 7, the range is $1-infinity

and $1-$4, respectively. The ranges for months 8, 9, and 10 are $1-$3, negative

infinity-$7, and negative infinity-$10. Once again, as long as the prices stay in

the allowable ranges a new solution will not have to be found.

The initial stock (xi) has no price on it so it does not appear in the objective

function, but there is still an allowable range for it. The stock in months 1 and 2

can be in the ranges infinity and -4000 to 1000 bushels, respectively. In months

3 and 4, the range is -1000 to 1000 bushels. The ranges for bushels in months

5, 6, and 7 are -2000 to infinity, -1000 to infinity, and -2000 to infinity. In months

8, 9, and 10, the ranges are infinity to 1000 bushels, -1000 bushels to infinity,

and infinity to 1000 bushels.

As far as the right-hand side analysis (RHS), SA provided the following

information presented in Table 9. As long as the constraints’ RHS stay within the

lower and upper limit, the optimal solution will stay optimal.

Table 9: RHS Ranges

Constraint Current (bushels) Lower Limit Upper Limit

Initial Stock 6000 0 20000

First Month’s Sell 0 0 Infinity

First Month’s Buy 20000 6000 Infinity

Initial Stock (Month 2) 0 -6000 14000

Second Month’s Sell 0 -6000 Infinity

Second Month’s Buy 20000 6000 Infinity

Initial Stock (Month 3) 0 -6000 Infinity

Third Month’s Sell 0 -6000 Infinity

Third Month’s Buy 20000 20000 Infinity

15

Initial Stock (Month 4) 0 0 20000

Fourth Month’s Sell 0 -20000 Infinity

Fourth Month’s Buy 20000 20000 20000

Initial Stock (Month 5) 0 0 Infinity

Fifth Month’s Sell 0 -20000 Infinity

Fifth Month’s Buy 20000 0 20000


Sixth Month’s Sell 0 -20000 Infinity

Sixth Month’s Buy 20000 0 Infinity


Seventh Month’s Sell 0 0 Infinity

Seventh Month’s Buy 20000 0 Infinity


Eighth Month’s Sell 0 0 Infinity

Eighth Month’s Buy 20000 0 Infinity


Ninth Month’s Sell 0 0 Infinity

Ninth Month’s Buy 20000 0 Infinity


Tenth Month’s Sell 0 0 Infinity

Tenth Month’s Buy 0 0 Infinity

3.3 Problem 3: Florist Dilemma

The optimal solution was found in 16 iterations. The minimum cost is

$20,500. The table below shows how the florist should purchase the flowers.

Table 10: Optimal Solution for Floral Shop


Red 500 500

White 500 500

Pink 500 500

16

Yellow 500

Purple 500

The sensitivity analysis and right-hand side analysis is shown below in

Tables 11 and 12.


Variable Current Coefficient Allowable Range

X11 15 12 to ∞

X12 14 7 to ∞

X13 5 0 to 5

X14 2 1 to ∞

X15 4 2 to ∞

X16 2 0 to 5

X17 9 7 to ∞

X18 14 13 to ∞

X21 14 11 to ∞

X22 9 6 to ∞

X23 4 4 to ∞

X24 5 0 to ∞

X25 1 -1 to 1

X26 4 1 to ∞

X27 6 -1 to 6

X28 12 12

X31 16 9 to ∞

X32 4 -3 to 7

X33 3 2 to ∞

X34 3 -2 to ∞

X35 3 -1 to ∞

X36 7 -1 to ∞

X37 5 4 to ∞

17

X38 10 7 to11

X41 13 12 to ∞

X42 10 7 to ∞

X43 7 5 to ∞

X44 1 0 to 2

X45 2 2 to ∞

X46 6 2 to ∞

X47 7 7 to ∞

X48 16 13 to ∞

X51 12 0 to 13

X52 11 7 to ∞

X53 8 5 to ∞

X54 4 1 to ∞

X55 6 2 to ∞

X56 5 2 to ∞

X57 15 7 to ∞

X58 13 13

Table 12: Right-Hand Side (RHS) Sensitivity Analysis

Row Current RHS Allowable Range

2 500 0 to 1000

3 500 500 to 1000

4 500 0 to 500

5 500 0 to 1000

6 500 500

7 500 0 to 500

8 500 500

9 500 500 to 1000

10 1000 1000 to ∞

11 1000 1000

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12 1000 500 to 1000

13 1000 500 to ∞

14 1000 500 to ∞

By looking at the sensitivity analysis we can see that the price of red

flowers bought in February can increase to infinity or decrease by 3 from $15 to

$12 before the optimal solution changes. The price of red flowers bought in

March can increase to infinity or decrease by 7 from $14 to $7 and the price of

red flowers bought in April can’t increase at all, but can decrease by 5 from $5 to

$0 before the optimal solution changes. The price of red flowers bought in May

can increase to infinity or decrease by 1 from $2 to $1 and the price of red

flowers bought in June can increase to infinity or decrease by 2 from $4 to $2

before the optimal solution changes. The price of red flowers bought in July can

increase by 3 from $2 to $5 or can decrease by 2 from $2 to $0 before the

optimal solution changes. The price of red flowers bought in August can

increase to infinity or decrease by 2 from $9 to $7 and the price of red flowers

bought in September can increase to infinity or decrease by 1 from $14 to $13

before the optimal solution changes.

By looking at the sensitivity analysis we can see that the price of white


before the optimal solution changes. The price of white flowers bought in March

can increase to infinity or decrease by 3 from $9 to $6 and the price of white

flowers bought in April can increase to infinity, but can’t decrease by any before

the optimal solution changes. The price of white flowers bought in May can

increase to infinity or decrease by 5 from $5 to $0 and the price of white flowers

bought in June can’t increase at all, but can decrease by 2 before the optimal

solution changes. The price of white flowers bought in July can increase to

infinity or decrease by 3 from $4 to $1 before the optimal solution changes. The

price of white flowers bought in August can’t increase at all, but can decrease by

7, while the price of white flowers bought in September can’t increase at all and

can’t decrease at all before the optimal solution changes.

19

By looking at the sensitivity analysis we can see that the price of pink


before the optimal solution changes. The price of pink flowers bought in March

can increase by 3 from $4 to $7 or can decrease by 7 and the price of pink

flowers bought in April can increase to infinity but can decrease by 1 from $3 to

$2 before the optimal solution changes. The price of pink flowers bought in May,

June July, and August can all increase to infinity before the optimal solution

changes. The price of pink flowers bought in May, June, and July can all

decrease to zero, while the price of pink flowers bought in August can decrease

by 1 from $5 to $4 before the optimal solution changes. The price of pink flowers

bought in September can increase by 1 from $10 to $11 or decrease by 3 from

$10 to $7 before the optimal solution changes.

By looking at the sensitivity analysis we can see that the price of yellow


before the optimal solution changes. The price of yellow flowers bought in March

can increase to infinity or decrease by 3 from $10 to $7, the price of yellow

flowers bought in April can increase to infinity or can decrease by 2 from $7 to $5

before the optimal solution changes. The price of yellow flowers bought in May

can increase by 1 from $1 to $2 or decrease by 1 from $1 to $0, while the price of

yellow flowers bought in June can increase to infinity, but can’t decrease at all

before the optimal solution changes. The price of yellow flowers bought in July

can increase to infinity or can decrease by 4 from $6 to $2 before the optimal

solution changes. The price of yellow flowers bought in August can increase to

infinity, but can’t decrease at all, while the price of yellow flowers bought in

September can increase to infinity or decrease by 3 from $16 to $13 before the

optimal solution changes.

By looking at the sensitivity analysis we can see that the price of purple

flowers bought in June can increase by 1 from $12 to $13 or decrease by 12 from

$12 to $0 before the optimal solution changes. The price of purple flowers

bought in March can increase to infinity or decrease by 4 from $11 to $7, while

the price of purple flowers bought in April can increase to infinity or decrease by

20

3 from $8 to $5 before the optimal solution changes. The price of purple flowers

bought in May can increase to infinity or decrease by 3 from $4 to $1, while the

price of purple flowers bought in June can increase to infinity or decrease by 4

from $6 to $2 before the optimal solution changes. The price of purple flowers

bought in July can increase to infinity or decrease by 3 from $5 to $2 before the

optimal solution changes. The price of purple flowers bought in August can

increase to infinity or decrease by 8 from $15 to $7, while the price of purple

flowers bought in September cannot increase or decrease at all before the

optimal solution changes.

The right-hand side ranges are the ranges for which the current basis

remains optimal. From the analysis we see that for the months of February and

May the number of flower bundles purchased can increase by 500 from 500 to

1000 or can decrease by 500 from 500 to 0 before the current basis is no longer

optimal. For the months March and September the number of flower bundles

purchased can increase by 500 from 500 to 1000, but cannot decrease at all

before the current basis is no longer optimal. For the months of April and July

the number of flower bundles purchased cannot increase at all, but can decrease

by 500 from 500 to 0 before the current basis is no longer optimal. For the

months of June and August the number of flower bundles purchased cannot

increase or decrease at all before the current basis is no longer optimal.

From the right-hand side analysis we can also see that the number of red

flower bundles purchased can increase to infinity, but cannot decrease before the

current basis is no longer optimal. The number of white flower bundles

purchased cannot increase or decrease before the current basis is no longer

optimal. The number of pink flower bundles purchased cannot increase, but can

decrease by 500 from 1000 to 500 before the current basis is no longer optimal.

The number of yellow and purple flower bundles purchased can each increase to

infinity or decrease by 500 from 1000 to 500 before the current basis is no longer

optimal.

21

4 Conclusions & Recommendations Grummins Engine will have a profit of $3,600,000. The optimal solution

provided by LINDO is shown below.

Year Truck Type Produced Sold Inventoried

1 1 100 100 0

2 200 200 0

2 1 180 180 0

2 100 100 0

3 1 170 170 0

2 150 150 0

Grummins should take the following recommendations under consideration:

1. Study customers to see if they will accept an increase in selling price.

2. Look into better manufacturing practices to reduce manufacturing cost.

Wheat warehouse will earn a profit of $162. The optimal solution is shown in

the table below.


1 6 0 0

2 6 0 0

3 6 6 20

4 20 0 0

5 20 0 0

6 20 20 20

7 20 20 0

8 0 0 20

9 20 20 0

10 0 0 0

The wheat warehouse is given the following recommendations:

1. Try to increase selling price.

2. Look for a different retailer to buy wheat at a lower price.

3. If 1 and 2 can be done, look into increasing storage space.

22

The floral shop’s minimum cost is $20,500. The optimal solution is below.


Red 500 500

White 500 500

Pink 500 500

Yellow 500

Purple 500

The floral shop is presented with the recommendations below:

1. Look for other suppliers.

2. Study customer demand to see if cheaper alternatives can be found such

as underused colors that can be bought at reduced prices.

23

5 References

[1] Winston, W.L. and M. Venkataramanan, Introduction to Mathematical Programming, 4th Edition, Duxbury Press, Belmont, CA, 2003.

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APPENDIX

the shortest path to destruction: a floral...

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