set up an algebraic model which clearly states the solution problem objective and constraints

Thus, DuoCera only produces small bricks when it is possible to change their capacity at least to 50 hours for cutting, 116 hours for firing, 40 hours for glazing and have more than 4000

COLLEGE OF BUSINESS

SCHOOL OF INTERNATIONAL BUSINESS - MARKETING

FINAL INDIVIDUAL PROJECT

BUSINESS MODELING AND APPLICATIONS

LE THI HA VI

Ho Chi Minh City, December 19 2022th

UEH UNIVERSITY COLLEGE OF BUSINESS

UEH UNIVERSITY COLLEGE OF BUSINESS

Business Models & Applications

Final Individual Project

The name of the subject

Business Models & Applications

The class:

IBC05

Subject CODE:

22C1BUS50320202

Name of the student:

Le Thi Ha Vi

Name of Instructor:

Dr Ha Quang An

December, 2022, Ho Chi Minh City SCHOOL OF INTERNATIONAL BUSINESS – MARKETING

ENDORSEMENT 2

TEACHER COMMENT 3

MAIN CONTENTS 4

1 Linear programming 4

1.1 Set up an algebraic model - which clearly states the Solution, Problem Objective, and Constraints 4

1.2 Solve with SOLVER 5

1.3 Conclusion 6

1.4 Sensitivity analysis – what is the Shadow Price of this problem? 6

1.5 DuoCera sees small bricks selling well and doing well so they are trying to produce only small bricks So under what conditions should DuoCera produce ONLY SMALL Brick? 7

1.6 Because of some technical issues, the brick kiln had to be cut down by 3 hours/week – that is, only 37 hours a week instead of 40 hours like the original So has DuoCera's production plan changed? 9

2 Decision making 10

2.1 In the initial case, which brick kiln should DuoCera use, and why 10

2.2 Should DuoCera take a survey if it's free? And if there is a fee, how much does it cost DuoCera won't accept surveys? 11

3 Network Distribution 14

REFERENCES 16

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I declare that this thesis has been composed solely by myself and that it has not been submissed by any other organizations and individuals Except where states otherwise by reference or acknowledgment, the work presented is entirely my own If any fraud is discovered, I will accept full responsibility for my essay’s content

December 19, 2022 Signature

Le Thi Ha Vi

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1 Linear programming

1.1 Set up an algebraic model - which clearly states the Solution, Problem Objective, and Constraints

This is a product-mix problem Start by entering the data The data for this problem are Hour Used per Unit Product for cutting, firing, and glazing; clay used per unit product; hours and materials available per week; unit revenue of each type per batch The data are shown with blue, changing cells with yellow, and objective cells with orange

Algebraic symbols

L = number of the large type batches of bricks to be produced per week

S = number of the small type batches of bricks to be produced per week

R = total revenue per week from the two products, in VND

Relevant data

DuoCera Brick Product-Mix

Units

Revenue

Available

Clay Used per Unit Product Clay Available

The decisions to be made: the production rates for the two types of products The overall measure of performance for these decisions: the total revenue from the two products

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Algebraic model

Choose the values of L and S so as to maximize

R = 4700000L + 5900000S

The number of hours of production time available per week for the two products in the respective steps are 60, 105 and 40

subject to satisfying all the following constraints:

0,3L + 0,25S ≤ 60

0.27L + 0,58S ≤ 105

0,16L + 0,2S ≤ 40

In addition, because of the limit of materials:

32.8L + 20S ≤ 6000

1.2 Solve with SOLVER

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The optimal solution is to produce 56.7010309278351 batches of large type per week and 154.639175257732 batches of small type per week

DuoCera needs to produce quantity of brick as follows:

Numer of batch (Large type & Small Type ) * 100

• Large Type = 56.70 *100 = 5670 Bricks

• Small Type = 154.64*100 = 15464 Bricks

Thus, DuoCera needs to produce 5670 bricks of large type and 15464 bricks of small type

to fit the plan

1.4 Sensitivity analysis – what is the Shadow Price of this problem?

Microsoft Excel 16.0 Sensitivity

Report

Worksheet: [Book3]Sheet1

Report Created: 12/15/2022 1:42:22

PM

6

Variable Cells

Objecti ve Allowabl e Allowabl e

Coeffici

$C$

15

Units Produced

Large type

56.70103

470000

1953448 276

$D$

15

Units Produced

Small type

154.6391

590000 0

4196296

Constraints

Constra int Allowabl e Allowabl e

R.H.

$E$

12

Materials Clay

Used

4952.577

1047.422 68

$E$

7 Cutting Hour Used

55.67010

4.329896 907

$E$

103092.7

$E$

2920103

1.577464 789

3.793103 448 The shadow price remains (0; 0; 103092,7836; 29202030,93)

It means

The shadow price in the firing process = 103 092,7835 VND

So, if we change the constraint on the right side of the permissible range, the total revenue will change by 103 092,7835 VND for each modified in time used (hour) The shadow price in Glazing process = 29 201 030,93 VND

So, if we increase or decrease the constraint right-hand side in the allowable range, the total revenue will increase or decrease by 29 201 030,93 VND by each increased or decreased time used (hour)

1.5 DuoCera sees small bricks selling well and doing well so they are trying to produce only small bricks So under what conditions should DuoCera produce ONLY SMALL Brick?

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DuoCera should produce only small bricks when the new revenue is higher or equal to the old revenue (when producing types of bricks)

In addition, the new plan is to produce only small bricks so L = 0

T = New revenue

It means

T ≥ 1178865979

5900000S ≥ 1178865979

S ≥ 200

Thus, if DuoCere want to produce only small brics, they must produce at least 200 batches of small bricks

meet the following constrains:

Available Hour of Cutting process ≥ 0,25 x 200 ≈ 50

Available Hour of Firing process ≥ 0,58 x 200 ≈ 116

Available Hour of Glazing process ≥ 0,2 x 200 ≈ 40

Available Clay ≥ 32,8 x 200 ≈ 4000

8

Thus, DuoCera only produces small bricks when it is possible to change their capacity at least to 50 hours for cutting, 116 hours for firing, 40 hours for glazing and have more than 4000 kg clay

1.6 Because of some technical issues, the brick kiln had to be cut down by 3 hours/week – that is, only 37 hours a week instead of 40 hours like the original So has DuoCera's production plan changed?

To see what happens when the brick kiln had to be cut down by 3 hours/week, you need

to substitute the new number in cell G9 and run Solver again

Results

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So, DuoCera's production plan has changed to 1185 bricks of large type 17551 bricks of, small type and the total revenue changed to 1091 262 887 VND to fit the new capacity

2 Decision making

2.1 In the initial case, which brick kiln should DuoCera use, and why

The decision is whether the production of this type of brick is being done in Dong Nai (with a fixed cost to open a brick kiln of about 300,000,000 VND) or processed at the joint factory in Vinh Long (with a fixed cost of only 120,000,000 VND/month) This decision is represented in the decision tree by a decision node with two branches For each decision, there are two outcomes: strong growth (62% probability) or moderate growth (38% probability) This is represented in the decision tree by event nodes with two branches

Profit = [(Revenue of each batch – Cost of each batch) Number of batches of bricks] – Fixed cost

If the Dong Nai factory is used and growth is strong, the profit would be:

[(5,9 – 2,5)6000] – 300 = 20100 (milion VND)

If the Dong Nai factory is used and growth is moderate, the profit would be

[(5.9 – 3)3000] – 300 = 8400

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The expected payoff of the Dong Nai factory would be

20100 x 0,62 + 8400 x 0,38 = 15654

If the Vinh Long factory is used and growth is strong, the profit would be

[(5.9 – 2.3)6000] – 1440 = 20160

If the Vinh Long is used and growth is moderate, the profit would be

[(5,9 – 2,3)3000] – 1440 = 9360

The expected payoff of Vinh Long factory would be

20160 x 0,62 + 9360 x 0,38 = 16056

The resulting solved decision tree is shown below The decision to produce in Vinh Long has a higher expected profit of 16056 million VND

2.2 Should DuoCera take a survey if it's free? And if there is a fee, how much does it cost DuoCera won't accept surveys?

After the survey, businesses find 2 observations: customers will have a more favorable view of the product - that is, the number of customers will tend to buy more

- and customers will have a skeptical view - ie the number of customers will tend to buy less

Possible Findings from a Survey

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Favorable: customers will have a more favorable view of the product; the number of customers will tend to buy more

Skeptical: customers will have a skeptical view; the number of customers will tend to buy less

To use either finding to calculate the posterior probability of each state of nature, it is necessary to estimate the probability of obtaining this finding for each state of nature

P (finding | state) = Probability that the indicated finding will occur, given that the state of nature is the indicated one

P (favorable I strong growth) = 0,6

P (skeptical I strong growth) = 0,4

P (favorable I moderate growth) = 0,2

P (skeptical I moderate growth) = 0,8

Recall that the prior probabilities are

P (strong growth) = 0,62

P (moderate growth) = 0.38

The joint probability is determined by the following formula

P(state and finding) = P(state) P(finding | state)

P(Strong and Favorable) = 0,62 x 0,6 = 0,372

P(Strong and Skeptical) = 0,62 x 0.4 = 0,248

P(Moderate and Favorable) = 0,38 x 0,2 = 0,076

P(Moderate anf Skeptical) = 0,38 x 0.8 = 0,304

Find each probability of just a particular finding without specifying the state of nature Since any finding can be obtained with any state of nature, the formula for calculating the probability of just a particular finding is

P(finding) = P(State 1 and finding) + P(State 2 and finding)

P(Favorable) = 0,372 + 0,076 = 0,440

P(Skeptical) = 0,248 + 0,304 = 0,552

Thus, there is a 44,8% chance that the survey will indicate a favorable attitude and a 55,2% chance that the survey will indicate a skeptical attitude toward the product

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Finally, we now are ready to calculate each posterior probability of a particular state of nature given a particular finding from the survey The formula involves combining the joint with the unconditional probabilities underneath the diagram as follows

P (state I finding)=P(state finding∧ )

P(finding) P(Strong I Favorable) = 0,83

P(Strong I Skeptical) = 0,45

P(Moderate I Favorable) = 0,16

P(Moderate I Skeptical) = 0,55

Given a favorable attitude, the probability of strong growth increases to 83%, and moderate growth to 17% Given a skeptical attitude, the probability of strong growth sinks to 45%, and moderate growth to 55%

The revised decision tree is shown below It begins with an event node with two branches for the two possible outcomes of the survey After the survey results are known, there is a decision of whether to open a new factory in Dong Nai, or process at the joint factory in Vinh Long, represented by a pair of branches

Finally, after the decision is made, there will either be strong or moderate growth This is represented by event nodes with two branches and uses the posterior probabilities given the results of the survey

If the survey indicates a favorable attitude toward the product, they should process

it at the joint factory in Vinh Long

If the survey indicates a skeptical attitude toward the product, they also should process it at the joint factory in Vinh Long

The expected profit is 16.058.690.000 VND

The expected value of experimentation is the expected payoff with the information minus the expected payoff without the information

With the survey information, the expected payoff is 16058,59 million VND Without the survey information, the expected payoff is 16056 million VND Thus, the expected value of experimentation in part b is 2,59 million VND

So, DuoCare should take a survey if it’s free

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survey before it would no longer be worthwhile to conduct.

3 Network Distribution

This is a shortest-path problem To set up a spreadsheet model, first list all of the arcs as shown in B4:C18, along with their distance (F4:F19) Then list all of the nodes as shown

in H4:H15 along with each node's supply or demand (K4:J15) We are sending one unit from Ho Nai, Dong Nai to Phan Rang (Start), and Ninh Thuan (End), so the supply is 1 at Start and the demand is 1 at End Every other node has demand 0 because if you enter the node, you must also leave it

The changing cells are whether or not to include an arc on the route These are shown in OnRoute (D4:D18) below If one unit is shipped through an arc, it must mean they traveled along that route

For each node, calculate the net flow as a function of the changing cells This can be done using the SUMIF function In each case, the first SUMIF function calculates the flow leaving the node and the second one calculates the flow entering the node For example, consider the Start node (H4) SUMIF(From,H4,OnRoute)sums each individual entry in OnRoute (the changing cells in D4:D18) if that entry is in a row where the entry

in From (B4:B18) is the same as in H4 are rows 4 and 5, the sum in the OnRoute column

is only over these same rows, so this sum is D4+D5

The goal is to minimize the total distance of the route The cost is the SUMPRODUCT of the Distance with OnRoute, or Total distance = SUMPRODUCT(OnRoute,Distance) This formula is entered into Total_Distance (E20)

The Solver information and solved spreadsheet are shown below

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OnRoute (D4:D18) indicates whether that arc should be included in the route The optimal route is to go B – E – F – H – J or Dong Nai – Bau Ham – Tan Nghia – Tan Hai – Suoi

Da – Ca Na – Ninh Thuan

The minimum Total_Distance (E20) is 220

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1 Frederick S Hillier and Mark S Hillier (2019 – sixth edition) Introduction to Management Science A Modeling and Case Studies Approach with Spreadsheets 2.Dr Ha Quang An (2022) Lecture slides of Business Models & Applications Subject

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