INEN 420 Semester Project Grummins Engine Company...

INEN 420 Semester Project

Grummins Engine Company Wheat Warehouse Power Generation

By

Class of ‘93

Javier Garcia Dinh Nguyen Rhoda Read

Class of ’93: Garcia, Nguyen, Read

CONTENTS

1.0 Executive Summary ………………………………………….……………………. Page 3

2.0 Problem Description ………………………………………………….……………...…….

2.1 Problem 1 (Grummins Engine Company) ………………...............……….. Page 3

2.2 Problem 2 (Wheat Warehouse) …………….……………………………… Page 5

2.3 Problem 3 (Power Generation) ………....………………………………….. Page 7

3.0 Computational Results ………………………………………………..……………………

3.1 Problem 1 (Grummins Engine Company) ………………...……..……….. Page 14

3.2 Problem 2 (Wheat Warehouse) ………………………….……..………… Page 17

3.3 Problem 3 (Power Distribution) ………………………………………….. 1

4.0 Conclusions and Recommendations ………………………..……………………. Page 29

5.0 References ……………………………………………………………………….. Page 32

Class of ’93: Garcia, Nguyen, Read Page 2

1.0 Executive Summary

This report focuses on three different possible scenarios. First, a diesel truck manufacturing

company wants to maximize profit, but is restricted by industry and government regulations. Its

model fits the profile of an optimal solution but upon testing the ranges produced in the

sensitivity analysis, it became clear that the original LP is degenerate. A similar case surfaced

after analyzing a model to maximize profit for a wheat farmer with a small warehouse and a rigid

selling/purchasing schedule. Again, we deduced that the original LP was degenerate. Lastly,

power generation in Puerto Rico was studied to determine how to minimize costs. The ranges

produced by its sensitivity analysis, again, indicate degeneracy. However, in all three cases the

LPs were functional. The interpretations and recommendations should be made with caution and

understanding of the effects of degeneracy is important.

2.1 Problem 1 (Grummins Engine Company)

The Grummins Engine Co. produces 2 types of diesel trucks that have different selling prices,

manufacturing costs, and pollution emission. We want to formulate an LP that can be used to

determine how to maximize profit during the next three years given the following conditions:

Sales Price, Manufacturing Costs and Pollution Emission:

Sales Price Manufacturing Costs Emissions

Type 1 $20,000.00 $15,000.00 15 grams

Type 2 $17,000.00 $14,000.00 5 grams

Maximum Demand for Trucks:

Year Type 1 Type 2

1 100 200

2 200 100

3 300 150


• Production capacity limits total truck production during each year to at most 300 trucks.

• It cost $2,000.00 to hold 1 truck (of any type) in inventory for one year.

• From the table above, at most 300 type 1 trucks can be sold in year 3. Demand may be

met from previous production or the current year’s production.

Grummins Engine Co. wants a plan to help them arrange their product in the next three years to

get the maximum profit. This means that there should be no trucks in stock at the end of the third

year.

Assumptions include that the company should not keep more trucks in inventory than demand

predicts, so production is regulated by the amount of trucks in inventory and the amount of

trucks that can be sold. Also, we assume that trucks are only produced and sold, not acquired by

any other method such as auctions, trading, etc.

We determined our decision variables to be the following:

Pij= Number of trucks (each type i) produced for each year j

Sij= Number of trucks (each type i) sold for each year j

Rij= Number of trucks (each type i) that remain in stock at the end of each year j

With i=1, 2; j=1, 2, 3.

The objective function for the maximum profit is as follows:

Max Z= 20 (S11+S12+S13) + 17 (S21+S22+S23) - 15 (P11+P12+P13) -14 (P21+P22+P23) - 2 (R11+R12+R21+R22) (in $ thousands)

s.t. P11+P21 <=320 (Production) P12+P22 <=320 P13+P23 <=320 S11 <=100 (Sale) S12 <=200 S13 <=300 S21 <=200 S22 <=100 S23 <=150 R11-P11+S11 =0 (Remain in stock) R21-P21+S21 =0 R12-P12+S12-R11 =0


R22-P22+S22-R21 =0 P13+R12-S13 =0 P23+R22-S23 =0 5P11+5P12+5P13-5P21-5P22-5P23<=0 (Emissions requirement) Pij, Sij, Rij >=0; i=1,2; j=1,2,3

2.2 Problem 2 (Wheat Warehouse)

As owner of a wheat warehouse with a capacity of 20,000 bushels, we want to formulate an LP

that can be used to determine how to maximize the profit earned over the next 10 months, given

the following conditions:

• We begin Month 1 with 6,000 bushels of wheat.

• Each month, wheat can be bought and sold at the price per 1000 bushels listed in the

following table: Month Selling Price ($) Purchase Price ($)

1 3 8

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

• Each month, the sequence of events is as follows:

o Observe the initial stock of wheat.

o Sell any amount of wheat up to the initial stock at the current month’s selling

price.

o Buy (at the current month’s buying price) as much wheat as wanted, subject to the

warehouse size limitation.

Assumptions include that wheat can only be sold or purchased (at the given rates). In this way,

ending inventory is simply determined by the following equation:


ending inventory = beginning inventory – amount sold + amount purchased

For Month 1, our beginning inventory is 6,000 bushels of wheat. We can then sell up to 6,000

bushels. Next, we can purchase any amount of wheat up to our capacity (20,000 bushels) minus

the beginning inventory plus the amount previously sold. Finally, our ending inventory is as

stated above (6,000 – amount sold + amount purchased). For Month 2, our beginning inventory

is the ending inventory from Month 1. We can then sell up to our ending inventory from Month

1. Next, we can purchase up to 20,000 minus the beginning inventory plus the amount

previously sold. Finally, our ending inventory for Month 2 is the ending inventory from Month 1

– amount sold during Month 2 + amount purchased during Month 2. The remaining months

follow the same process and are detailed in the constraints listed later.

We determined our decision variables to be the following:

si = amount of wheat (in thousands) sold during month i, i = 1,…,10 pi = amount of wheat (in thousands) purchased during month i, i=1,…,10 ej = # of bushels (in thousands) left at the end of month j, j=1,…,9 For our objective function, we want to maximize profit over the next 10 months, and determined

it to be the following equation:

max z = 3s1 – 8p1 + 6s2 – 8p2 + 7s3 – 2p3 + s4 – 3p4 + 4s5 – 4p5 + 5s6 – 3p6 + 5s7

– 3p7 + s8 – 2p8 + 3s9 – 5p9 + 2s10 – 5p10

Therefore, the LP formulation is as follows:

max z = 3s1 – 8p1 + 6s2 – 8p2 + 7s3 – 2p3 + s4 – 3p4 + 4s5 – 4p5 + 5s6 – 3p6 + 5s7

– 3p7 + s8 – 2p8 + 3s9 – 5p9 + 2s10 – 5p10 s.t. s1 <= 6 (selling restrictions – up to current inventory) s2 – e1 <= 0 s3 – e2 <= 0 s4 – e3 <= 0 s5 – e4 <= 0 s6 – e5 <= 0 s7 – e6 <= 0 s8 – e7 <= 0 s9 – e8 <= 0 s10 – e9 <= 0 p1 – s1 <= 14 (purchasing restrictions – up to warehouse capacity)


p2 – s2 + e1 <= 20 p3 – s3 + e2 <= 20 p4 – s4 + e3 <= 20 p5 – s5 + e4 <= 20 p6 – s6 + e5 <= 20 p7 – s7 + e6 <= 20 p8 – s8 + e7 <= 20 p9 – s9 + e8 <= 20 p10 – s10 + e9 <= 20 e1 + s1 – p1 = 6 (ending inventory for each month) e2 + s2 – p2 – e1 = 0 e3 + s3 – p3 – e2 = 0 e4 + s4 – p4 – e3 = 0 e5 + s5 – p5 – e4 = 0 e6 + s6 – p6 – e5 = 0 e7 + s7 – p7 – e6 = 0 e8 + s8 – p8 – e7 = 0 e9 + s9 – p9 – e8 = 0 si, pi >= 0, i = 1,…,10; ei >= 0, i = 1,…,9

2.3 Problem 3 (Power Generation)

GENERAL BACKGROUND

Puerto Rico enjoys a highly diversified economy, a strong tourist sector, and good trade relations with the United States, its largest trading partner. Despite mixed current economic indicators, Puerto Rico’s short-term economic outlook looks relatively positive given the strengthening of the U.S. economy, which is expected to grow over 4% this year. The island’s real gross domestic product (GDP) is expected to grow

3.3% in 2004 and 2.8% in 2005, and 2.4% over the medium term (2006–10). As consumers take advantage of low interest rates, private consumption has increased. Over the past year, high oil prices have had an adverse affect on Puerto Rico's economy and inflation, as the Commonwealth is heavily dependent on oil imports to meet its domestic energy needs, particularly for electricity generation.


ENERGY OVERVIEW

Puerto Rico lacks domestic hydrocarbon reserves (including oil, natural gas, and coal) and relies

on imports to meet its energy needs. Imported oil, mainly from U.S. and Caribbean suppliers, is

the source of about 90% of Puerto Rico's power. Although consumption of natural gas has been

increasing over the past few years, Puerto Rico still relies overwhelmingly on oil. Many industry

analysts agree that gas- or coal-fired facilities are needed to supplement oil-burning power

plants. However, plans to widen and/or diversify the electric power supply through co-

generation and agreements with independent power producers have barely progressed due to

opposition from environmental groups and powerful labor unions.

In 2004, Puerto Rico generated an estimated 22.1 billion kilowatt hours (Bkwh) of electricity,

predominantly from five oil-fired generators, with a fraction coming from small hydroelectric

dams. Also, the country consumed about 223,000 barrels per day (bbl/d) of oil, all imported,

primarily for transportation and electric power generation. As of 2003, installed generation

capacity was 4.9 gigawatts. The five oil-fired plants are: the Costa Sur plant (1,090 MW); the

Aguirre plant (900 MW); the Palo Seco plant (602 MW); the San Juan plant (400 MW); and the

Arecibo plant (248 MW). The Puerto Rico Electric Power Authority (PREPA) accounts for a

majority of net electricity generation, and is the Commonwealth's sole distributor of electric

power. PREPA also purchases excess power generation from co-generators, primarily in the

cement industry, and from independent power producers. Early this year, Puerto Rico began

importing liquefied natural gas (LNG) to supply the EcoEléctrica facility, a 540-megawatt (MW)

natural gas-fired power plant that will supply power generation under a contract to the island at

the end of this year. Also, they begin importing coal to the new 454-MW coal-fired plant in

Guayama (ASE-PR).

With power consumption increasing more than 3% per year for more than a decade, both PREPA

and independent power producers have been investing in new capacity in order to meet growing

demand and to diversify energy sources. The following table provides the power plant capacity,

expected demand for each substation for next year and the cost of shipping power from a plant to

a substation.


Cost of Shipping to Substation # (As of 30 SEP 2004)

FROM 1 2 3 4 5 6 7 8 9 10 11 12

(Plant) Supply (MW)

San Juan 6 5 6 7 9 10 9 12 14 17 11 10 400

Palo Seco 6 6 5 8 10 11 10 11 13 16 10 9 602

Aguirre 10 10 11 12 11 13 9 7 8 10 12 13 900

Costa Sur 15 14 15 13 18 20 17 9 6 8 9 10 1090

Arecibo 9 8 9 10 12 15 11 11 9 10 6 4 248

ASE-PR 9 8 9 9 9 9 7 5 6 7 6 10 454

Eco

Electrica 14 13 14 13 16 17 14 7 5 8 7 8 507

Expected

Demand in

MW (2005) 190 175 175 195 165 190 200 190 200 175 145 190 2190

Last September, the tropical storm Jeanne hit the island. PREPA initial system-operations report

shows the island’s largest electricity-generating plants collapsed between noon and 1:00 p.m.

while operating at their average capacity, causing expensive damage to the equipment. Not only

did the thermoelectric plants collapse, but so did the island’s gas and steam turbines and the co-

generation plants. The failures meant the system was producing only some 607 megawatts

(MW) to power the entire island out of a maximum capacity of 3,240 MW, or less than 19%. If

a plant is operating at more than 50% of its peak capacity when objects such as broken trees and

flying debris strike or knock transmission poles down, the resultant shock to the power grid can

be destructive. "When the technicians went to inspect the systems on Thursday, the filed reports

indicated cracks in equipment, failed start-ups, a damaged generator in Costa Sur No. 3, and a

cracked boiler in Costa Sur No. 4," said the PREPA source. "Aguirre’s steam turbines alone had

an oil leak in the electrohydraulic system, a broken cooling fan, and a damaged rotor. The

damage to Puerto Rico’s power grid was estimated at $60 million.

PREPA wants to add two (2) new substations, ASE-PR and Eco Electrica, to the power grid

beginning next year and want them to provide at least 20% of the power demand during high

peak demand. Both ASE-PR and Eco Electrica are privately-owned and will be operational in

November 2004. Also, the Palo Seco plant will be able to supply only 50% of its maximum

output and Costa Sur only 70% of its maximum output during high peak demand beginning next


year due to an upgrade to their plants. The peak power demand occurs at the same time (around

0100 PM) on each substation. Assumptions include that all plants operate at 90% of their

maximum capacity and that all power supply to the substation is only being used by the intended

sources. Also, in the event of any bad weather, we will assume that at most two plants, Palo

Seco and Aguirre, will be disconnected. We want to formulate two LPs with different

conditions:

The first condition is to minimize the cost of meeting each substation’s peak power demand for

next year during high peak demand. The second condition is to minimize the cost of meeting

each substation’s peak power demand if Palo Seco and Aguirre are disconnected due to bad

weather.

EcoElectrica

ASE-PR

New Power Plant

Substations

12 11

10

9

8

7

6

5 4

3 2 1


FORMULATION (Both Conditions):

First, we defined our decision variables by determining how much power is sent to each

substation during high peak hour (0100 PM). We define seven plants (i = 1, 2, …., 7). Plant 1 is

San Juan, Plant 2 is Palo Seco, Plant 3 is Aguirre, Plant 4 is Costa Sur, Plant 5 is Arecibo and

Plants 6 and 7 are the two new facilities (ASE-PR and Eco Electrica). The twelve substations are

defined j = 1,2,3,4,…,12. See map for locations.

Xij = number of megawatts produced at plant i and sent to substations j

For our objective function, we want to minimize the cost of meeting each substation’s peak

power demand:

Min Z = 6X11 + 5X12 + 6X13 + 7X14 + 9X15 + 10X16 + 9X17 + 12X18 + 14X19 +

17X110 + 11X111 + 10X112 + 6X21 + 6X22 + 5X23 + 8X24 + 10X25 + 11X26 +

10X27 + 11X28 + 13X29 +16X210 + 10X211 + 9X212 + 10X31 + 10X32 + 11X33 +

12X34 + 11X35 + 10X36 + 9X37 + 6X38 + 7X39 + 9X310 + 12X311 + 13X312 +

15X41 + 14X42 + 15X43 + 13X44 + 18X45 + 20X46 + 17X47 + 9X48 + 5X49 + 8X410

+ 9X411 + 10x412 + 9X51 + 8X52 + 9X53 + 10X54 + 12X55 + 15X56 + 11X57 +

11X58 + 9X59 + 9X510 + 6X511 + 4X512 + 9X61 + 8X62 + 9X63 + 9X64 + 9X65 +

9X66 + 7X67 + 5X68 + 6X69 + 7X610 + 6X611 + 10X612 + 14X71 + 13X72 + 14X73

+ 13X74 + 16X75 + 17X76+ 14X77 + 7X78 + 5X79 + 8X710 + 7X711 + 8X712

PREPA faces three types of constraints. First, the total power supplied by each plant cannot

exceed the plant capacity. Examples are: amount of power sent from San Juan to the twelve

substations cannot exceed 360 megawatts (=90% MAX). Also, we have restrictions to Palo Seco

and Costa Sur power generation. Palo Seco can only supply 50% of its maximum capacity and

Costa Sur 70%. The formulation problem contains the following constraints for supply:

X11+X12+X13+X14+X15+X16+X17+X18+X19+X110+X111+X112 <= 360 X21+X22+X23+X24+X25+X26+X27+X28+X29+X210+X211+X212 <= 301 X31+X32+X33+X34+X35+X36+X37+X38+X39+X310+X311+X312 <= 810 X41+X42+X43+X44+X45+X46+X47+X48+X49+X410+X411+X412 <= 743 X51+X52+X53+X54+X55+X56+X57+X58+X59+X510+X511+X512 <= 224 X61+X62+X63+X64+X65+X66+X67+X68+X69+X610+X611+X612 <= 409 X71+X72+X73+X74+X75+X76+X77+X78+X79+X710+X711+X712 <= 451


The following constraint ensures the ASE-PR and Eco Electrica supply at least 20% of total

power demand (3,240 MW):

4X61 + 4X62+ 4X63 + 4X64 + 4X65 + 4X66 + 4X67 + 4X68 + 4X69 + 4X610 + 4X611 + 4X612+ 4X71 + 4X72 + 4X73 + 4X74 + 4X75 + 4X76 + 4X77 + 4X78 + 4X79 + 4X710 + 4X711 + 4X712 - X11 - X12 - X13 - X14 - X15 - X16 - X17 - X18 -X19 - X110 - X111 - X112 - X21 - X22 - X23 - X24 - X25 - X26 - X27 - X28 - X29 -X210 - X211 - X212 - X31 - X32 - X33 - X34 - X35 - X36 - X37 - X38 - X39 - X310 -X311 - X312 - X41 - X42 - X43 - X44 - X45 - X46 - X47 - X48 - X49 - X410 - X411 -X412 - X51 - X52 - X53 - X54 - X55 - X56 - X57 - X58 - X59 - X510 - X511 - X512 >= 0

Also, we need to ensure that each substation will receive sufficient power to meet its peak

demand. The demand constraint is as follows:

X11+X21+X31+X41+X51+X61+X71 >= 190 (Substation 1)












Sign Restrictions: Xij >= 0

Therefore, the LP formulation for condition 1 is as follows:

Min Z = 6X11 + 5X12 + 6X13 + 7X14 + 9X15 + 10X16 + 9X17 + 12X18 + 14X19 + 17X110 + 11X111 + 10X112 + 6X21 + 6X22 + 5X23 + 8X24 + 10X25 + 11X26 + 10X27 + 11X28 + 13X29 +16X210 + 10X211 + 9X212 + 10X31 + 10X32 + 11X33 + 12X34 + 11X35 + 10X36 + 9X37 + 6X38 + 7X39 + 9X310 + 12X311 + 13X312 + 15X41 + 14X42 + 15X43 + 13X44 + 18X45 + 20X46 + 17X47 + 9X48 + 5X49 + 8X410 + 9X411 + 10x412 + 9X51 + 8X52 + 9X53 + 10X54 + 12X55 + 15X56 + 11X57 + 11X58 + 9X59 + 9X510 + 6X511 + 4X512 + 9X61 + 8X62 + 9X63 + 9X64 + 9X65 + 9X66 + 7X67 + 5X68 + 6X69 + 7X610 + 6X611 + 10X612 + 14X71 + 13X72 + 14X73 + 13X74 + 16X75 + 17X76+ 14X77 + 7X78 + 5X79 + 8X710 + 7X711 + 8X712


s.t.

X11+X12+X13+X14+X15+X16+X17+X18+X19+X110+X111+X112 <= 360 X21+X22+X23+X24+X25+X26+X27+X28+X29+X210+X211+X212 <= 301 X31+X32+X33+X34+X35+X36+X37+X38+X39+X310+X311+X312 <= 810 X41+X42+X43+X44+X45+X46+X47+X48+X49+X410+X411+X412 <= 743 X51+X52+X53+X54+X55+X56+X57+X58+X59+X510+X511+X512 <= 224 X61+X62+X63+X64+X65+X66+X67+X68+X69+X610+X611+X612 <= 409 X71+X72+X73+X74+X75+X76+X77+X78+X79+X710+X711+X712 <= 451

4X61 + 4X62+ 4X63 + 4X64 + 4X65 + 4X66 + 4X67 + 4X68 + 4X69 + 4X610 + 4X611 + 4X612+ 4X71 + 4X72 + 4X73 + 4X74 + 4X75 + 4X76 + 4X77 + 4X78 + 4X79 + 4X710 + 4X711 + 4X712 - X11 - X12 - X13 - X14 - X15 - X16 - X17 - X18 -X19 - X110 - X111 - X112 - X21 - X22 - X23 - X24 - X25 - X26 - X27 - X28 - X29 -X210 - X211 - X212 - X31 - X32 - X33 - X34 - X35 - X36 - X37 - X38 - X39 - X310 -X311 - X312 - X41 - X42 - X43 - X44 - X45 - X46 - X47 - X48 - X49 - X410 - X411 -X412 - X51 - X52 - X53 - X54 - X55 - X56 - X57 - X58 - X59 - X510 - X511 - X512 >= 0

X11+X21+X31+X41+X51+X61+X71 >= 190 (Substation 1) X12+X22+X32+X42+X52+X62+X72 >= 175 (Substation 2) X13+X23+X33+X43+X53+X63+X73 >= 175 (Substation 3) X14+X24+X34+X44+X54+X64+X74 >= 195 (Substation 4) X15+X25+X35+X45+X55+X65+X75 >= 165 (Substation 5) X16+X26+X36+X46+X56+X66+X76 >= 190 (Substation 6) X17+X27+X37+X47+X57+X67+X77 >= 200 (Substation 7) X18+X28+X38+X48+X58+X63+X78 >= 190 (Substation 8) X19+X29+X39+X49+X59+X69+X79 >= 200 (Substation 9) X110+X210+X310+X410+X510+X610+X710 >= 175 (Substation 10) X111+X211+X311+X411+X511+X611+X711 >= 145 (Substation 11) X112+X212+X312+X412+X512+X612+X112 >= 200 (Substation 12) Xij >= 0, i=1,……..,7; j = 1,………..,12

The LP formulation for the condition 2 is as follows:

Min Z = 6X11 + 5X12 + 6X13 + 7X14 + 9X15 + 10X16 + 9X17 + 12X18 + 14X19 + 17X110 + 11X111 + 10X112 + 6X21 + 6X22 + 5X23 + 8X24 + 10X25 + 11X26 + 10X27 + 11X28 + 13X29 +16X210 + 10X211 + 9X212 + 10X31 + 10X32 + 11X33 + 12X34 + 11X35 + 10X36 + 9X37 + 6X38 + 7X39 + 9X310 + 12X311 + 13X312 + 15X41 + 14X42 + 15X43 + 13X44 + 18X45 + 20X46 + 17X47 + 9X48 + 5X49 + 8X410 + 9X411 + 10x412 + 9X51 + 8X52 + 9X53 + 10X54 + 12X55 + 15X56 + 11X57 + 11X58 + 9X59 + 9X510 + 6X511 + 4X512 + 9X61 + 8X62 + 9X63 + 9X64 + 9X65 + 9X66 + 7X67 + 5X68 + 6X69 + 7X610 + 6X611 + 10X612 + 14X71 + 13X72 + 14X73 + 13X74 + 16X75 + 17X76+ 14X77 + 7X78 + 5X79 + 8X710 + 7X711 + 8X712


s.t.

X11+X12+X13+X14+X15+X16+X17+X18+X19+X110+X111+X112 <= 360 X21+X22+X23+X24+X25+X26+X27+X28+X29+X210+X211+X212 <= 0 X31+X32+X33+X34+X35+X36+X37+X38+X39+X310+X311+X312 <= 0 X41+X42+X43+X44+X45+X46+X47+X48+X49+X410+X411+X412 <= 743 X51+X52+X53+X54+X55+X56+X57+X58+X59+X510+X511+X512 <= 224 X61+X62+X63+X64+X65+X66+X67+X68+X69+X610+X611+X612 <= 409 X71+X72+X73+X74+X75+X76+X77+X78+X79+X710+X711+X712 <= 451 4X61 + 4X62+ 4X63 + 4X64 + 4X65 + 4X66 + 4X67 + 4X68 + 4X69 + 4X610 + 4X611 + 4X612+ 4X71 + 4X72 + 4X73 + 4X74 + 4X75 + 4X76 + 4X77 + 4X78 + 4X79 + 4X710 + 4X711 + 4X712 - X11 - X12 - X13 - X14 - X15 - X16 - X17 - X18 -X19 - X110 - X111 - X112 - X21 - X22 - X23 - X24 - X25 - X26 - X27 - X28 - X29 -X210 - X211 - X212 - X31 - X32 - X33 - X34 - X35 - X36 - X37 - X38 - X39 - X310 -X311 - X312 - X41 - X42 - X43 - X44 - X45 - X46 - X47 - X48 - X49 - X410 - X411 -X412 - X51 - X52 - X53 - X54 - X55 - X56 - X57 - X58 - X59 - X510 - X511 - X512 >= 0 X11+X21+X31+X41+X51+X61+X71 >= 190 (Substation 1) X12+X22+X32+X42+X52+X62+X72 >= 175 (Substation 2) X13+X23+X33+X43+X53+X63+X73 >= 175 (Substation 3) X14+X24+X34+X44+X54+X64+X74 >= 195 (Substation 4) X15+X25+X35+X45+X55+X65+X75 >= 165 (Substation 5) X16+X26+X36+X46+X56+X66+X76 >= 190 (Substation 6) X17+X27+X37+X47+X57+X67+X77 >= 200 (Substation 7) X18+X28+X38+X48+X58+X63+X78 >= 190 (Substation 8) X19+X29+X39+X49+X59+X69+X79 >= 200 (Substation 9) X110+X210+X310+X410+X510+X610+X710 >= 175 (Substation 10) X111+X211+X311+X411+X511+X611+X711 >= 145 (Substation 11) X112+X212+X312+X412+X512+X612+X112 >= 200 (Substation 12) Xij >= 0, i=1,……..,7; j = 1,………..,12

3.1 Problem 1 (Grummins Engine Company)

The optimal solution obtained through Lindo is as follows:

Z = 3,600.00 (in $ thousands) S11 =100 S12 =200

S13 =150 S21 =200 S22 =100 S23 =150 P11 =100 P12 =200


P13 =150 P21 =200 P22 =100 P23 =150 R11 =0 R12 =0 R21 =0 R22 =0

Lindo determined that 10 iterations were necessary to find this optimal solution. The solution

indicates that the best way to get the maximum profit is to not have remaining inventory at the

end of each year. Therefore, the company should sell every truck they make each year, for both

types of trucks.

After examining the sensitivity analysis report, we determined the ranges for our decision

variables could be as follows, and still remain the optimal solution:

Decision

Variables

Current Sale &

Production

($ thousands)

Range of Increase/Decrease Objective Function Coefficient

Range of Decision Variables

S11 20 0<=∆<= +∞ 20<=C11<= +∞

S12 20 0<=∆<= +∞ 20<=C12<= +∞

S13 20 -5<=∆<= 0 15<=C13<= 20

S21 17 -8<=∆<= +∞ 9<=C21<= +∞

S22 17 -8<=∆<= +∞ 9<=C22<= +∞

S23 17 -8<=∆<= +∞ 9<=C23<= +∞

P11 15 -2<=∆<=0 13<=C14<=15

P12 15 -2<=∆<=0 13<=C15<=15

P13 15 0<=∆<=2 15<=C16<=17

P21 14 -2<=∆<=8 12<=P24<=22

P22 14 -2<=∆<=2 12<=P25<=16

P23 14 -∞<=∆<= 2 -∞<=P26<= 16

R11 2 -2 <=∆<=+∞ 0 <=C1<=+∞

R12 2 -2 <=∆<=+∞ 0 <=C2<=+∞

R21 2 -2 <=∆<=+∞ 0 <=C3<=+∞

R22 2 -2 <=∆<=+∞ 0 <=C4<=+∞


The selling price for a type 1 truck (which emits 15 gm of pollution) can vary from $20,000 up to

+∞ for the first two years and the manufacturing price can also increase up to $20,000 more. In

other words, the company can increase the selling price C11 but due to the constraint in truck

sale S11 <= 100, it will not affect the current basis. By increasing C11, the current basis remains

optimal and the only change will be the optimal Z (Cbv B-1b). However, the company has to

reduce the price for the type 1 truck at the third year down to $15,000 to guarantee that all type 1

trucks will be sold. It is similar for the manufacturing costs for type 1 trucks. The company can

make up to $2,000.00 for each type 1 truck for the first two years, but then they have to reduce

its price on the third year. When we change the value of S11 or S12 of the objective function

coefficient out of range, that variable will leave the basis and change the optimal solution, thus

decreasing the profit for Grummins Engine Company. Similarly, when we increase the value of

S13 out of range (above 20), the optimal solution will include less trucks sold in year 2 (plus

leftover inventory) in order to sell as many trucks as possible during year 3, when the selling

price is higher.

For type 2 trucks, the selling price can vary from $9,000.00 up to +∞ for all three years. The

manufacturing costs are also the same; their price can vary from $6,000.00 to $16,000.00 for the

first year and produce more profit for each year later. But we have the same situation as we

explain in the prior paragraph (Constraint in selling trucks will not affect the current basis, only

the optimal Z). Also as type 1 trucks, if we change the objective function coefficient (S21, S22,

S23) out of the range, that variable also will leave the basis and change the optimal solution, thus

decreasing the company’s profit.

Similarly, Grummins Engine Company could reduce the manufacturing prices (P11, P12, P13,

P21, P22, P23) out of range to less than the minimum value and the maximum profit will

increase. Otherwise, the maximum profit will decrease when manufacturing costs increase above

the range maximum. However, upon closer inspection of the variable P22, the optimal solution

changes if P22 = 16, which is in the allowable range. Similarly, if P21 = 12, which is within the

allowable range, the optimal solution changes. These particular occurrences indicate that our

original LP is degenerate.

The ranges for the right-hand side variables are listed in the table below:


Decision Variable

Current Right-Hand Side Range of Increase/Decrease Ranges of Right-Hand Sides

b1 320 -20 <= ∆ <= ∞ 300 <= b1 <= ∞

b2 320 -20 <= ∆ <= ∞ 300 <= b2 <= ∞

b3 320 -20 <= ∆ <= ∞ 300 <= b3 <= ∞

b4 100 -20 <= ∆ <= 20 80 <= b4 <= 120

b5 200 -20 <= ∆ <= 20 180 <= b5 <= 220

b6 300 -150 <= ∆ <= ∞ 150 <= b6 <= ∞

b7 200 -150 <= ∆ <= 20 50 <= b7 <= 220

b8 100 -100 <= ∆ <= 20 0 <= b8 <= 120

b9 150 -150 <= ∆ <= 10 0 <= b9 <= 160

b10 0 -20 <= ∆ <= 20 -20 <= b10 <= 20

b11 0 -20 <= ∆ <= 150 -20 <= b11 <= 150

b12 0 -20 <= ∆ <= 20 -20 <= b12 <= 20

b13 0 -20 <= ∆ <= 100 -20 <= b13 <= 100

b14 0 -150 <= ∆ <= 150 -150 <= b14 <= 150

b15 0 -150 <= ∆ <= 10 -150 <= b15 <= 10

b16 0 -750 <= ∆ <= 100 -750 <= b16 <= 100

According to this analysis, the range for b1 could be increased to any amount & the solution

would remain optimal. That is, our limit for total truck production could be any amount over

300 trucks per year. Also, the range of optimality for the amount of each truck we can sell each

year is given by the ranges of b4 – b9. For instance, the ranges indicate that if the right-hand side

of the constraint for the amount of type 1 trucks sold in year 1 (b4) is less than 80 or more than

120, we should receive a new optimal solution. However, upon testing this range, we see that

the optimal value stays the same for any value chosen greater than 120. This, too, seems to

indicate that our original LP is degenerate.

3.2 Problem 2 (Wheat Warehouse)

The optimal solution obtained through Lindo is as follows:

z = 162 (in thousands $) s3 = 6 (in thousands) p3 = 20 ” s5 = 0 ” p5 = 0 ” s6 = 20 ” p6 = 20 ”


s7 = 20 ” p7 = 0 ” p8 = 20 ” s9 = 20 ” s10 = 0 ”

Lindo determined that 20 iterations were necessary to find this optimal solution. The solution

indicates that we should hold our initial stock of wheat (6,000 bushels) until Month 3, when we

should sell the entire stock at $7 per bushel and then purchase 20,000 bushels at $2 per bushel.

Again, this solution suggests that we should hold this stock until Month 6, when we should sell

all of it at $5 per bushel and purchase 20,000 more bushels at $3 per bushel. Then, in Month 7,

we should sell all 20,000 bushels at $5 per bushel and then have zero bushels in inventory until

we could purchase 20,000 bushels in Month 8 at $2 per bushel. Finally, the solution indicates we

should sell the entire stock in Month 9 at $3 per bushel, leaving the wheat warehouse empty

through Month 10. This solution makes sense because it encourages us to sell the wheat at a

higher rate than we can purchase it, thus maximizing profit.

After examining the sensitivity analysis report, we determined that the ranges for our decision

variables could be as follows, and still maintain the optimal solution:

Decision Variable

Current Selling/Purchase Price

Range of Increase/Decrease Objective Function

Coefficient Ranges of Decision Variables

s1 $3 -∞ <= ∆ <= 4 -∞ <= c1 <=7

p1 $8 -1 <= ∆ <= ∞ 7 <= c11 <= ∞

s2 $6 -∞ <= ∆ <= 1 -∞ <= c2 <= 7

p2 $8 -1 <= ∆ <= ∞ 7 <= c21 <= ∞

s3 $7 -1 <= ∆ <= 1 6 <= c3 <= 8

p3 $2 -1 <= ∆ <= 1 1 <= c31 <= 3

s4 $1 -∞ <= ∆ <=1 -∞ <= c4 <= 2

p4 $3 -1 <= ∆ <= ∞ 2 <= c41 <= ∞

s5 $4 -2 <= ∆ <=0 2 <= c5 <= 4

p5 $4 0 <= ∆ <= ∞ 4 <= c51 <= ∞

s6 $5 -1 <= ∆ <= ∞ 4 <= c6 <= ∞

p6 $3 -∞ <= ∆ <= 2 -∞ <= c61 <= 5

s7 $5 -2 <= ∆ <= ∞ 3 <= c7 <= ∞

p7 $3 -1 <= ∆ <= 2 2 <= c71 <= 5

s8 $1 -∞ <= ∆ <= 1 -∞ <= c8 <= 2

p8 $2 -1 <= ∆ <= 1 1 <= c81 <= 3

s9 $3 -1 <= ∆ <= 2 2 <= c9 <= 5


Decision Variable

Current Selling/Purchase Price

Range of Increase/Decrease Objective Function

Coefficient Ranges of Decision Variables

p9 $5 -2 <= ∆ <= ∞ 3 <= c91 <= ∞

s10 $2 -2 <= ∆ <= 1 0 <= c10 <= 3

p10 $5 -5 <= ∆ <= ∞ 0 <= c11 <= ∞

That is, the selling price of wheat c1 during Month 1 can vary from –∞ (we pay someone to take

our wheat – which is not profitable) to $7 (up $4 from the given selling price). In the same way,

the purchase price of wheat during Month 1 can vary from $7 (down $1 from the given purchase

price) to ∞ (again, this is not profitable), etc. We see that for the basic variables in the optimal

solution, the range of possible increases/decreases is much smaller, and therefore, more sensitive

to change(s). When we increase/decrease the objective function coefficients outside of the

range, that variable will enter/leave the basis & change the optimal solution. We tested this by

changing the coefficient of c1 to $8 & solving the LP again. As expected, s1 entered the basis,

was part of the optimal solution, and increased the z-value of the LP. By changing the range for

c31, the optimal solution changed as expected when c31 = 4, which is out of the allowable range.

Our new basis doesn’t include p3 as basic variable, and the new optimal Z = 142,000. Also, the

allowable range for c9 was found to be between 2 and 5, but setting c9 = 7, the current basis

changed but remains optimal and the only change will be the optimal Z (Cbv B-1b) with Z =

222,000. This happen is because a constraint on the selling quantity s8– e8 <= 0. Event if we

increased the selling price to a value greater than $5, we can only sell what we have on inventory

(20,000 bushels). Similarly, picking values outside the ranges of c3 and c9 will only affect the Z-

value, but will not affect the current basis. However by changing c5 , the basis will remain the

same but s5 will be equal to 20,000. Originally s5 was a basic variable equal to 0. In other

words, this leads us to believe that the original LP is degenerate, especially since there are at

least one basic variable in the optimal solution equal to zero.

In looking at the sensitivity ranges for the right-hand side, we see that the only major affectation

is to increase/decrease the storage capacity of the warehouse. Then, we could sell more wheat

and purchase more as determined by a new optimal solution. The ranges for the right-hand side

are listed in the table below:


Decision Variable

Current Right-Hand Side Range of Increase/Decrease Ranges of Right-Hand Sides

b1 6 -6 <= ∆ <= ∞ 0 <= b1 <= ∞

b2 0 -6 <= ∆ <= ∞ -6 <= b2 <= ∞

b3 0 -6 <= ∆ <= ∞ -6 <= b3 <= ∞

b4 0 -20 <= ∆ <= ∞ -20 <= b4 <= ∞

b5 0 -20 <= ∆ <= ∞ -20 <= b5 <= ∞

b6 0 -20 <= ∆ <= ∞ -20 <= b6 <= ∞

b7 0 0 <= ∆ <= ∞ 0 <= b7 <= ∞

b8 0 0 <= ∆ <= ∞ 0 <= b8 <= ∞

b9 0 0 <= ∆ <= ∞ 0 <= b9 <= ∞

b10 0 0 <= ∆ <= ∞ 0 <= b10 <= ∞

b11 14 -14 <= ∆ <= ∞ 0 <= b11 <= ∞

b12 20 -14 <= ∆ <= ∞ 6 <= b12 <= ∞

b13 20 0 <= ∆ <= ∞ 20 <= b13 <= ∞

b14 20 0 <= ∆ <= 0 20 <= b14 <= 20

b15 20 -20 <= ∆ <= 0 0 <= b15 <= 20

b16 20 -20 <= ∆ <= ∞ 0 <= b16 <= ∞

b17 20 -20 <= ∆ <= ∞ 0 <= b17 <= ∞

b18 20 -20 <= ∆ <= ∞ 0 <= b18 <= ∞

b19 20 -20 <= ∆ <= ∞ 0 <= b19 <= ∞

b20 20 -20 <= ∆ <= ∞ 0 <= b20 <= ∞

b21 6 -6 <= ∆ <= 14 0 <= b21 <= 20

b22 0 -6 <= ∆ <= ∞ -6 <= b22 <= ∞

b23 0 0 <= ∆ <= 20 0 <= b23 <= 20

b24 0 0 <= ∆ <= ∞ 0 <= b24 <= ∞

b25 0 -20 <= ∆ <= ∞ -20 <= b25 <= ∞

b26 0 -20 <= ∆ <= ∞ -20 <= b26 <= ∞

b27 0 -20 <= ∆ <= 0 -20 <= b27 <= 0

b28 0 -20 <= ∆ <= ∞ -20 <= b28 <= ∞

b29 0 -20 <= ∆ <= 0 -20 <= b29 <= 0

According to this analysis, it is interesting to note that the optimal solution will not change

regardless of how large our warehouse storage capacity may be, as seen in the range for b1. On

the other hand, the range for b15 indicates that the original warehouse capacity is the maximum

value that will keep the current solution optimal. We tested this in Lindo & any increase in the

right-hand side of constraint #16 changes the optimal solution where the z-value is greater and p6

increases, as expected.


3.3 Problem 3 (Power Generation)

The optimal solution obtained through Lindo for condition 1 is as follows:

Z = 14192 X12 = 175 X14 = 185 X15 = 0 X21 = 190 X22 = 0 X24 = 10 X25 = 101 X35 = 30 X36 = 190 X37 = 0 X38 = 15 X49 = 200 X410 = 175 X511 = 24 X512 = 200 X63 = 175 X65 = 34 X67 = 200 X79 = 0 X710 = 0 X711 = 121

Lindo determined that 26 iterations were necessary to find the optimal solution. The solution

also highlighted the power supply from each plant. For example the San Juan (1) plant will

supply power to the substations 2 and 4, total 360 MW. Palo Seco (2) supplies power to

substations 1, 4 and 5, total 301 MW (100% capacity). ASE-PR (6) and Eco Electrica (7) supply

530 MW (24% of the total high peak demand). Also Costa Sur (4) supplies 375 MW (less than

the 70% max capacity of the plant). The optimal solution holds the constraints.

After examining the sensitivity analysis, we determined that the ranges for our decision variables

could be as follows, and still maintain the optimal solution: Decision Variables

Cost of Shipping Power ($)

Range of Increase/Decrease Objective Function Coefficient Ranges of Decision Variables

X11 6 -1 <= ∆ <= ∞ 5 <= C11<= ∞

X12 5 -7 <= ∆ <= 0 -2 <= C12 <= 5


Decision Variables



X13 6 -3 <= ∆ <= ∞ 3 <= C13 <= ∞

X15 9 0 <= ∆ <= ∞ 9 <= C15 <= ∞

X16 10 -2 <= ∆ <= ∞ 8 <= C16 <= ∞

X17 9 -2 <= ∆ <= ∞ 7 <= C17 <= ∞

X18 12 -8 <= ∆ <= ∞ 4 <= C18 <= ∞

X19 14 -11 <= ∆ <= ∞ 3 <= C19 <= ∞

X110 17 -11 <= ∆ <= ∞ 6 <= C110 <= ∞

X111 11 -6 <= ∆ <= ∞ 5 <= C111 <= ∞

X112 10 -2 <= ∆ <= ∞ 8 <= C112 <= ∞

X21 6 -7 <= ∆ <= ∞ 1 <= C21 <= ∞

X22 6 0 <= ∆ <= ∞ 6 <= C22 <= ∞

X23 5 -1 <= ∆ <= ∞ 4 <= C23 <= ∞

X25 10 -2 <= ∆ <= 0 8 <= C25 <= 10

X26 11 -2 <= ∆ <= ∞ 9 <= C26 <= ∞

X27 10 -2 <= ∆ <= ∞ 8 <= C27 <= ∞

X28 11 -6 <= ∆ <= ∞ 5 <= C28 <= ∞

X29 13 -9 <= ∆ <= ∞ 4 <= C29 <= ∞

X210 16 -9 <= ∆ <= ∞ 7 <= C210 <= ∞

X211 10 -4 <= ∆ <= ∞ 6 <= C211 <= ∞

X212 9 -5 <= ∆ <= ∞ 4 <= C212 <= ∞

X31 10 -3 <= ∆ <= ∞ 7 <= C31 <= ∞

X32 10 -3 <= ∆ <= ∞ 7 <= C32 <= ∞

X33 11 -6 <= ∆ <= ∞ 5 <= C33 <= ∞

X34 12 -3 <= ∆ <= ∞ 9 <= C34<= ∞

X35 11 -1 <= ∆ <= 0 10 <= C35 <= 11

X36 10 -10 <= ∆ <= 1 0 <= C36 <= 11

X39 8 -2 <= ∆ <= ∞ 6 <= C39 <= ∞

X310 10 -1 <= ∆ <= ∞ 9 <= C310 <= ∞

X311 12 -5 <= ∆ <= ∞ 7 <= C311 <= ∞

X312 13 -8 <= ∆ <= ∞ 5 <= C312 <= ∞

X41 15 -8 <= ∆ <= ∞ 7 <= C41 <= ∞

X42 14 -7 <= ∆ <= ∞ 7 <= C42 <= ∞

X43 15 -10 <= ∆ <= ∞ 5 <= C43 <= ∞

X44 13 -4 <= ∆ <= ∞ 9 <= C44 <= ∞

X45 18 -7 <= ∆ <= ∞ 11 <= C45 <= ∞

X46 20 -10 <= ∆ <= ∞ 10 <= C46 <= ∞

X47 17 -8 <= ∆ <= ∞ 9 <= C47 <= ∞

X48 9 -3 <= ∆ <= ∞ 6 <= C48 <= ∞


Decision Variables



X49 6 -5 <= ∆ <= 0 1 <= C49 <= 6

X410 8 -8 <= ∆ <= 0 0 <= C410 <= 8

X411 9 -2 <= ∆ <= ∞ 7 <= C411 <= ∞

X412 10 -5 <= ∆ <= ∞ 5 <= C412 <= ∞

X51 9 -3 <= ∆ <= ∞ 6 <= C51 <= ∞

X52 8 -2 <= ∆ <= ∞ 6 <= C52 <= ∞

X53 9 -5 <= ∆ <= ∞ 4 <= C53 <= ∞

X54 10 -2 <= ∆ <= ∞ 8 <= C54 <= ∞

X55 12 -2 <= ∆ <= ∞ 10 <= C55 <= ∞

X56 15 -6 <= ∆ <= ∞ 9 <= C56 <= ∞

X57 11 -3 <= ∆ <= ∞ 8 <= C57 <= ∞

X58 11 -6 <= ∆ <= ∞ 5 <= C58 <= ∞

X59 9 -5 <= ∆ <= ∞ 4 <= C59 <= ∞

X510 10 -2 <= ∆ <= ∞ 8 <= C510 <= ∞

X511 6 -1 <= ∆ <= 1 5 <= C511 <= 7

X512 4 -5 <= ∆ <= 1 -1 <= C512 <= 5

X61 9 -4 <= ∆ <= ∞ 5 <= C61 <= ∞

X62 8 -3 <= ∆ <= ∞ 5 <= C62 <= ∞

X63 9 -5 <= ∆ <= 1 4 <= C63 <= 10

X64 9 -2 <= ∆ <= ∞ 7 <= C64 <= ∞

X65 9 0 <= ∆ <= 1 9 <= C65 <= 10

X66 9 -1 <= ∆ <= ∞ 8 <= C66 <= ∞

X67 7 -9 <= ∆ <= 0 -2 <= C67 <= 7

X68 5 -7 <= ∆ <= ∞ -2 <= C68 <= ∞

X69 6 -3 <= ∆ <= ∞ 3 <= C69 <= ∞

X610 7 -1 <= ∆ <= ∞ 6 <= C610 <= ∞

X611 6 -1 <= ∆ <= ∞ 5 <= C611 <= ∞

X612 10 -7 <= ∆ <= ∞ 3 <= C612 <= ∞

X71 14 -7 <= ∆ <= ∞ 7 <= C71 <= ∞

X72 13 -6 <= ∆ <= ∞ 7 <= C72 <= ∞

X73 14 -9 <= ∆ <= ∞ 5 <= C73 <= ∞

X74 13 -4 <= ∆ <= ∞ 9 <= C74 <= ∞

X75 16 -5 <= ∆ <= ∞ 11 <= C75 <= ∞

X76 17 -7 <= ∆ <= ∞ 10 <= C76 <= ∞

X77 14 -5 <= ∆ <= ∞ 9 <= C77 <= ∞

X78 7 -1 <= ∆ <= ∞ 6 <= C78 <= ∞

X711 7 -1 <= ∆ <= 1 6 <= C711 <= 8

X712 8 -8 <= ∆ <= ∞ 0 <= C712 <= ∞


Looking at Plant 1, the ranges for X11 (NBV) are 5 <= C11<= ∞. If we decrease the cost of

shipping in X11 to $4, then X11 will enter the basis with C11 = 190 and a new optimal solution

Z = 14,002. But, by increasing the shipping cost to any value greater than $5, the current basis

will remain optimal and the values of the decision variables will remain the same. Another

example is the basic variable X12. X12 holds for -2 <= C12 <= 5. But if we increase the

shipping cost C12 to more than $6, the solution is no longer optimal and we will get another

optimal solution (Z = $14,256 IAW Lindo). Also, X15 (BV = 0) holds for 9 <= C15 <= ∞. But

if we increase it to $10, X15 will exit the basis, but the solution will remain unchanged because

of the degenerate LP. Because we are dealing with a real-life problem, it’s unrealistic to bring

the shipping price to zero ($0). By increasing the shipping price on any BV above the range will

help to get another solution and a NBV to enter the basis. Any variation on the shipping cost on

any of the NBV (decreasing) or BV (increasing) will change by increasing or decreasing the

shipping cost, and we can get a new optimal solution.

For the right hand side of the constraints:

RHS Current Value Range of Increase/Decrease Objective Function Coefficient Ranges of Decision Variables

b1 0 -∞ <= ∆ <=625 -∞ <= b1< = 625

b2 360 -101 <= ∆ <=10 259 <= b2 <= 370

b3 301 -101 <= ∆ <= 30 200 <= b3 <= 331

b4 810 -575 <= ∆ <= ∞ 235 <= b4 <= ∞

b5 743 -368 <= ∆ <= ∞ 375 <= b5 <= ∞

b9 190 -30 <= ∆ <= 101 160 <= b9 <= 291

b11 175 -30 <= ∆ <= 15 145 <= b11 <= 190

b20 200 -121 <= ∆ <= 24 79 <= b20 <= 224

In this part, we examined the supply and demand constraints, which are listed in the table above.

For example, if Plant 3 (b3) decreases its output by less than 101 MW or increases by more than

30 MW, then we get a new solution with a new constraints and a new optimal solution. For

example, if we decrease its output to 190 MW, the new Z-value we obtain is 14323. Likewise,

for a demand constraint, if substation 1 (b9) increases his demand over 291 MW, the new

solution is 14472. Even if the current basis remains optimal (between the ranges) the values of

the decision variables and Z change.


The optimal solution obtained through Lindo for condition 2 is as follows:

Z = 17237.00 X11 = 174 X12 = 175 X15 = 0 X112 = 11 X21 = 0 X36 = 0 X44 = 195 X49 = 200 X410 = 175 X51 = 16 X52 = 0 X55 = 30 X512 = 178 X63 = 175 X65 = 44 X66 = 190 X67 = 0 X75 = 91 X77 = 200 X78 = 15 X711 = 145

Lindo determined that 32 iterations were necessary to find the optimal solution. The power

supply by ASE-PR and Eco Electrica increases dramatically due to the shutdown of Palo Seco

and Aguirre Plant. ASE-PR (6) supplies power to substations 3, 5, and 6, for a total of 314 MW

(70% capacity). Eco Electrica (7) supplies 451 MW (100 % capacity). The solution holds the

constraints.

After examining the sensitivity analysis, we determined that the ranges for our decision variables

could be as follows, and still maintain the optimal solution: Decision Variables



X11 6 0 <= ∆ <= 0 X11 will not change

X12 5 -13 <= ∆ <= 0 -8 <= C12 <= 5

X13 6 -5 <= ∆ <= ∞ 1 <= C13 <= ∞

X15 9 0 <= ∆ <= ∞ 9 <= C15 <= ∞

X16 10 -1 <= ∆ <= ∞ 9 <= C16 <= ∞

X17 9 -2 <= ∆ <= ∞ 7 <= C17 <= ∞


Decision Variables



X18 12 -12 <= ∆ <= ∞ 0 <= C18 <= ∞

X19 14 -17 <= ∆ <= ∞ -3 <= C19 <= ∞

X110 17 -17 <= ∆ <= ∞ 0 <= C110 <= ∞

X111 11 -11 <= ∆ <= ∞ 0 <= C111 <= ∞

X112 10 -1 <= ∆ <= 1 9 <= C112 <= 11

X21 6 ∞ <= ∆ <= 1 ∞ <= C21 <= 7

X22 6 -1 <= ∆ <= ∞ 5 <= C22 <= ∞

X23 5 -4 <= ∆ <= ∞ 1 <= C23 <= ∞

X25 10 -1 <= ∆ <= ∞ 9 <= C25 <= ∞

X26 11 -2 <= ∆ <= ∞ 9 <= C26 <= ∞

X27 10 -3 <= ∆ <= ∞ 7 <= C27 <= ∞

X28 11 -11 <= ∆ <= ∞ 0 <= C28 <= ∞

X29 13 -16 <= ∆ <= ∞ -3 <= C29 <= ∞

X210 16 -16 <= ∆ <= ∞ 0 <= C210 <= ∞

X211 10 -10 <= ∆ <= ∞ 0 <= C211 <= ∞

X212 9 -8 <= ∆ <= ∞ 1 <= C212 <= ∞

X31 10 -3 <= ∆ <= ∞ 7 <= C31 <= ∞

X32 10 -4 <= ∆ <= ∞ 6 <= C32 <= ∞

X33 11 -9 <= ∆ <= ∞ 2 <= C33 <= ∞

X34 12 -6 <= ∆ <= ∞ 6 <= C34<= ∞

X35 11 -1 <= ∆ <= ∞ 10 <= C35 <= ∞

X36 13 -∞ <= ∆ <= 1 -∞ <= C36 <= 14

X39 7 -9 <= ∆ <= ∞ -2 <= C39 <= ∞

X310 9 -8 <= ∆ <= ∞ 1 <= C310 <= ∞

X311 12 -11 <= ∆ <= ∞ 1 <= C311 <= ∞

X312 13 -11 <= ∆ <= ∞ 2 <= C312 <= ∞

X41 15 -1 <= ∆ <= ∞ 14 <= C41 <= ∞

X42 14 -1 <= ∆ <= ∞ 13 <= C42 <= ∞

X43 15 -6 <= ∆ <= ∞ 9 <= C43 <= ∞

X44 13 -13 <= ∆ <= 1 0 <= C44 <= 14

X45 18 -1 <= ∆ <= ∞ 17 <= C45 <= ∞

X46 20 -3 <= ∆ <= ∞ 17 <= C46 <= ∞

X47 17 -2 <= ∆ <= ∞ 15 <= C47 <= ∞

X48 9 -1 <= ∆ <= ∞ 8 <= C48 <= ∞

X49 6 -5 <= ∆ <= 1 1 <= C49 <= 7

X410 8 -8 <= ∆ <= 1 0 <= C410 <= 9

X411 9 -1 <= ∆ <= ∞ 8 <= C411 <= ∞

X412 10 -1 <= ∆ <= ∞ 9 <= C412 <= ∞


Decision Variables



X51 9 0 <= ∆ <= 0 9 <= C51 <= 9

X52 8 0 <= ∆ <= ∞ 8 <= C52 <= ∞

X53 9 -5 <= ∆ <= ∞ 4 <= C53 <= ∞

X54 10 -2 <= ∆ <= ∞ 8 <= C54 <= ∞

X55 12 -1 <= ∆ <= 0 11 <= C55 <= 12

X56 15 -3 <= ∆ <= ∞ 12 <= C56 <= ∞

X57 11 -1 <= ∆ <= ∞ 10 <= C57 <= ∞

X58 11 -8 <= ∆ <= ∞ 3 <= C58 <= ∞

X59 9 -9 <= ∆ <= ∞ 0 <= C59 <= ∞

X510 10 -4 <= ∆ <= ∞ 4 <= C510 <= ∞

X511 6 -3 <= ∆ <= ∞ 3 <= C511 <= ∞

X512 4 -0.5 <= ∆ <= 0.5 3.5 <= C512 <= 4.5

X61 9 -3 <= ∆ <= ∞ 6 <= C61 <= ∞

X62 8 -3 <= ∆ <= ∞ 5 <= C62 <= ∞

X63 9 -9 <= ∆ <= 4 0 <= C63 <= 13

X64 9 -4 <= ∆ <= ∞ 5 <= C64 <= ∞

X65 9 -1 <= ∆ <= 0 8 <= C65 <= 9

X66 9 -1 <= ∆ <= 1 7 <= C66 <= 8

X67 7 0 <= ∆ <= ∞ 7 <= C67 <= ∞

X68 5 -13 <= ∆ <= ∞ -8 <= C68 <= ∞

X69 6 -9 <= ∆ <= ∞ -3 <= C69 <= ∞

X610 7 -7 <= ∆ <= ∞ 0 <= C610 <= ∞

X611 6 -6 <= ∆ <= ∞ 0 <= C611 <= ∞

X612 10 -9 <= ∆ <= ∞ 1 <= C612 <= ∞

X71 14 -1 <= ∆ <= ∞ 13 <= C71 <= ∞

X72 13 -1 <= ∆ <= ∞ 12 <= C72 <= ∞

X73 14 -6 <= ∆ <= ∞ 8 <= C73 <= ∞

X74 13 -1 <= ∆ <= ∞ 12 <= C74 <= ∞

X75 16 0 <= ∆ <= 1 15 <= C75 <= 17

X76 17 -1 <= ∆ <= ∞ 16 <= C76 <= ∞

X77 14 -15 <= ∆ <= 0 -1 <= C77 <= 14

X78 7 -4 <= ∆ <= 1 3 <= C78 <= 8

X711 7 -8 <= ∆ <= 1 -1 <= C711 <= 8

X712 8 -9 <= ∆ <= ∞ -1 <= C712 <= ∞

Looking at Plant 7, if we decrease the cost of shipping in C71 (NBV) to less than $13, then X71

will enter the basis and the solution is no longer optimal. Now that the shipping cost is less, the


new solution will include X71 (Plant 7 supplies 135 MW to Substation 1) and the new optimal

solution is Z = 17,102. For X711 (BV), if the shipping cost (C711) increases to more than $9,

the solution is no longer optimal and X67 enters the basis with 44 MW. Also the optimal Z-

value will change ($17,516 IAW Lindo). Any variation on the shipping cost of any of the NBV

(decreasing) or BV (increasing) will change by increasing or decreasing the shipping cost, and

we will get a new optimal solution.

For the right hand side of the constraints:

RHS Current Value Range of Increase/Decrease Objective Function Coefficient Ranges of Decision Variables

b1 0 -∞ <= ∆ <=2286 -∞ <= b1 <= 2286

b2 360 -87 <= ∆ <= 8 273 <= b2 <= 368

b3 0 0 <= ∆ <= 8 0 <= b3 <= 8

b4 0 0 <= ∆ <= 11 0 <= b4 <=11

b5 743 -173 <= ∆ <= ∞ 570 <= b5 <= ∞

b7 409 -44 <= ∆ <= 11 365 <= b7 <= 420

b10 175 -8 <= ∆ <= 87 167 <= b10 <= 262

b12 195 -195 <= ∆ <= 173 0 <= b12 <= 368

b15 200 -11 <= ∆ <= 89 189 <= b15 <= 289

b20 200 -11 <= ∆ <= 174 189 <= b20 <= 374

In this part, we examined the supply and demand constraints, which are listed in the table above.

For example, if Plant 7 (b7) decreases his power supply by less than 44 MW or increases by

more than 11 MW, then we get a new optimal solution with new constraints and new optimal

solution. For example, if we decrease its power supply to 360 MW, the new Z-value obtained is

17,634. Likewise, if the power demand on the substation 20 (b20) increases over 374 MW, the

solution is no longer optimal and the new Z = 18,812 MW. Even if the current basis remains

optimal (between the ranges) the values of the decision variables and Z change. Finally, for

both conditions, due to the fact that there are more than one basic variable equal to zero, we

conclude that this LP is also degenerate.

In all three scenarios, Lindo provided adequate ranges. Even with the sensitivity analysis that

Lindo provided, some of the ranges do not seem feasible to apply to real-world situations. For

instance, we would never pay consumers for using electricity and market demands would not


allow us to increase the selling price to infinity, as suggested. By having a constraint that affect

a basic variable, even if the ranges indicated that the solution will be suboptimal if we leave the

ranges, the constraint will force the basis to remain the same and the only change will be the Z-

optimal. We observed this situation in all three scenarios. Therefore, it is important to consider

all aspects of the situation before changing the variables.

4.0 Conclusions and Recommendations

In this project, we attempted to maximize or minimize three different situations. The solutions

for all three problems using LINDO show that the companies can gain substantial profit or

minimize their expenses by implementing recommendations developed by each model.

First, our recommendation to Cummings Engine Company is to produce and sell trucks is as

follows: Produce Sell

Truck Type 1st Year 2nd Year 3rd Year 1st Year 2nd Year 3rd Year

Type 1 100 200 150 100 200 150

Type 2 200 100 150 200 100 150

During the formulation, we began by defining the decision variables that would describe each

situation and decision to be made (e.g., determine the number of type 1 trucks to produce during

the first year). The formulation presented a challenge for the group due to a set of conditions

that later became constraints. For example, Grummins Engine Company produces two different

types of trucks. The production of each truck was based on how many trucks can be sold every

year during a three year period. Also, if a truck is not sold, the company incurs an additional

charge for inventory. Because they want to maximize its profit, we have to make sure that our

solution takes into consideration all related costs (production, overhaul, etc). During our

presentation of the model to the company we must state that during the formulation phase, we

assumed that Grummins will sell all trucks produced in that year. Also the government imposes

certain restrictions and specifications that we must follow. These restrictions can cost millions

of dollars to company if they are violated.


For the inventory problem, we determined the following:

Month Selling Quantity Purchase

Quantity

1 0 0

2 0 0

3 6 20

4 0 0

5 0 0

6 20 20

7 20 0

8 0 20

9 20 0

10 0 0

The above table represented the optimal solution. During the formulation phase, we needed to

determine what real-life facts can affect the situation. The capacity of the warehouse, initial

inventory at the beginning of each month, and the monthly selling limitation are some examples

of a real-life situation. Sometimes, we must determine all constraints and limitations before we

begin the formulation phase. If a limitation is ignored during the formulation phase, the optimal

solution can be completely wrong. The above table clearly confirmed the inventory procedures

for the next 10 months. We can recommend as an option to increase the capacity of the

warehouse in order to increase profits.

In the power distribution case, we were asked to develop an LP for two different conditions. The

first condition was to minimize the cost of meeting each substation’s peak power demand for

next year during high peak demand and the second condition was to minimize the cost of

meeting each substation’s peak power demand if two plants were disconnected due to bad

weather. This model doesn’t include the operating limits of the generators, loads and the

transmission line network. The only two types of critical points that we identified were the

transmission line flow limits and generator capability limits (plants). Our final solution was as

follows:


Condition 1 Condition 2

Plant i to

Substation j

Power

Transmitted

Plant i to

Substation j

Power

Transmitted

X12 175 X11 174

X14 185 X12 175

X21 190 X112 11

X24 10 X44 195

X25 101 X49 200

X35 30 X410 175

X36 190 X51 16

X38 15 X55 30

X49 200 X512 178

X410 175 X63 175

X511 24 X65 44

X512 200 X66 190

X63 175 X75 91

X65 34 X77 200

X67 200 X78 15

X711 121 X711 145

Because generating companies and power systems have the problem of deciding how best to

meet the varying demand for electricity, which has a daily and weekly cycle, we must develop a

model that will support the high peak demand. Electricity cannot be stored; it is necessary to

start-up and shut-down a number of generating units at various power stations each day. During

the second condition, the problem is to decide when and which generating units to start-up and

shut-down, in order to minimize the total fuel cost or to maximize the total profit, over a study

period of typically a day, subject to a large number of difficult constraints that must be satisfied.

By assuming that two plants will be disconnected at any time, we can develop an optimal

solution. The most important constraint is that the total generation must equal the forecast

demands for electricity.

Also, most of these transportation problems must take into consideration the supply of imported

power. In the United States, the supply of imported power is price responsive. The quantity of

imported power can increase in the face of higher power prices, dampening market power. In

our case, the market is owned by one source, so they can control any changes on the shipping


cost of the decision variables. The source can use our formulation to predict the worst case

scenario and to determine any backup plan to sustain the demand if anything should happen.

Furthermore, the source can save millions of dollars by improving the transmission line networks

and also by obtaining more fuel cells.

5.0 References

Winston, W.L. and M.Venkataramanan, Introduction to Mathematical Programming, 4th Edition,

Duxbury Press, Belmont, CA, 2003.

Gonen, Turan, Electric Power Distribution System Engineering, McGraw-Hill Publishing

Company, Oklahoma City, OK, 1986.

Puerto Rico Electric Power Authority, Tarifas para Servicio de Electricidad, November 1989.

Autoridad de Energia Electrica, www.prepa.com, 2002.


INEN 420 Semester Project Grummins Engine Company...

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