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.”
__________________________
__________________________
___________________________
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 Summary
The 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
1
2
3
Truck Type
Produced
Sold
Inventoried
1
100
100
0
2
200
200
0
1
180
180
0
2
100
100
0
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
500
Red
July
Aug
500
500
White
500
500
Pink
Sep
500
500
Yellow
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 $2,000 to hold 1
truck (of any type) in inventory for one year [1, p117].
Assumptions:
1. There will be no trucks left over from year three.
2. The divisibility assumption is not valid.
3. All variables are nonnegative.
Table 62 [1]
Year
Type 1
Type 2
1
100
200
2
200
100
3
300
150
Decision Variables:
Xij = number of trucks made
Yij = number of trucks sold
Zij = number of trucks in inventory and carried over
where i = year made from 1to 3 and j = type of truck 1 and 2
5
Objective Function:
Maximize 20000y11 + 20000y21 + 20000y31 + 17000y12 + 17000y22 + 17000y32 –
15000x11 – 15000x21 – 15000x31 – 14000x12 – 14000x22 – 14000x32 2000z11 – 2000z12 - 2000z21 - 2000z22
s.t.
y11 < 100
y12 < 200
x11+x12 < 320
x11-y11-z11 = 0
x12-y12-z12 = 0
x21+x22 < 320
z11+y21 < 200
z12+y22 < 100
x21+z11-y21-z21 = 0
x22+z12-y22-z22 = 0
x31+x32 < 320
z21+y31 < 300
z22+y32 < 150
x31+z21-y31 = 0
x32+z22-y32 = 0
5y11+5y21+5y31-5y12-5y22-5y32 < 0
2.2 Problem 2: Wheat Stock
You own a wheat warehouse with a capacity of 20,000 bushels. At the
beginning of month 1, you have 6,000 bushels of wheat. Each month, wheat can
be bought and sold at the price per 1000 bushels given in table 46.
The
sequence of events during each month is as follows:
1. You observe your initial stock of wheat.
2. You can sell any amount of wheat up to your initial stock at the current
month’s selling price.
3. You can buy (at the current month’s buying price) as much wheat as you
want, subject to the warehouse size limitation.
*Everything shown in per thousand bushels [1, 112]
Assumptions:
1. All variables are nonnegative.
Table 46 [1]
Month
Selling Price ($)
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
Color
Feb
Mar
Apr
May
June
July
Aug
Sep
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:
1. All variables are nonnegative.
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
Number Sold
Number
Produced
1
2
3
Inventoried
1
100
100
0
2
200
200
0
1
180
180
0
2
100
100
0
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
11
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
Month
Initial Stock
Wheat Sold
Wheat Bought
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.
Table 8: Sensitivity Analysis
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
Initial Stock (Month 6)
0
-20000
Infinity
Sixth Month’s Sell
0
-20000
Infinity
Sixth Month’s Buy
20000
0
Infinity
Initial Stock (Month 7)
0
-20000
Infinity
Seventh Month’s Sell
0
0
Infinity
Seventh Month’s Buy
20000
0
Infinity
Initial Stock (Month 8)
0
-20000
0
Eighth Month’s Sell
0
0
Infinity
Eighth Month’s Buy
20000
0
Infinity
Initial Stock (Month 9)
0
-200000
Infinity
Ninth Month’s Sell
0
0
Infinity
Ninth Month’s Buy
20000
0
Infinity
Initial Stock (Month 10)
0
-20000
0
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
Color
Feb
Mar
May
June
500
Red
500
July
Aug
Sep
500
500
White
Pink
Apr
500
500
16
500
Yellow
Purple 500
The sensitivity analysis and right-hand side analysis is shown below in
Tables 11 and 12.
Table 11: Sensitivity Analysis
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
18
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
flowers bought in June can increase to infinity or decrease by 3 from $14 to $11
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
flowers bought in June can increase to infinity or decrease by 7 from $16 to $9
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
flowers bought in June can increase to infinity or decrease by 1 from $13 to $12
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
1
2
3
Truck Type
Produced
Sold
Inventoried
1
100
100
0
2
200
200
0
1
180
180
0
2
100
100
0
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.
Month
Initial Stock
Wheat Sold
Wheat Bought
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.
Color
Feb
Mar
Apr
May
June
500
Red
July
Aug
500
500
White
500
500
Pink
Yellow
Sep
500
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.
24
APPENDIX