NGA-SCE Decision Analysis & Modeling Assignment Solution - June 2019

This assignment solution addresses key concepts in Decision Analysis and Modeling, providing step-by-step solutions to three problems. The first question involves linear programming, utilizing both primal and dual models to determine optimal resource allocation for a salad composition. The second question focuses on developing and interpreting a multiple linear regression model, analyzing the impact of various factors on profit generation, and suggesting an alternative exponential regression model. The final question explores transportation problems, employing the Vogel's Approximation method to find an initial solution and then using the Modified Distribution method to optimize the solution, minimizing the total transportation cost. The solution includes detailed calculations, graphical representations, and interpretations of the results, demonstrating a thorough understanding of the concepts and techniques involved.

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DECISION ANALYSIS AND MODELLING
Student’s Name
Code + Name of Course
Professor’s Name
University
City (State)
Date

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Question 1
Primal model
Primal form of linear programming is the original form of linear programming. In this case lets
lettuce be x and tomato be y. The linear model will be
Min z=100 x +20 y
Subject to
2 x+ y ≥ 8
2 x+2 y ≥10
x , y ≥ 0
Using graphical technique, we draw the graph of the model as presented in the diagram below.
From the graphical solution of linear program designed above the optimal point will be at x = 0
and y = 8. The salad should thus e composed of only 8 grams of tomato. This will cost 160Rs
which is the minimal cost possible under the given constraints.
Dual form
This is another linear program problem that is related to the primal model. It will be designed as
follows.
Max ; Z=8 x +10 y

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Subject to
2 x+2 y +w ≤ 100
x +2 y + z ≤20
Solving the dual model gives the solution presented by the table below.
From this solution, it’s evident that the cost will be 160Rs. This is similar to the optimal solution
of the primal model hence the linear model is proved to be the dual of the initial model.
Question 2
Developing a multiple linear regression model

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Looking at the output of the multiple
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.992281243
R Square 0.984622065
Adjusted R Square 0.976341639
Standard Error 924.7666897
Observations 21
ANOVA
df SS MS F Significance F
Regression 7 711834866.4 101690695.2 118.9095842 9.25393E-11
Residual 13 11117514.6 855193.4304
Total 20 722952381
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 42608.49397 18419.79052 2.313191017 0.03772355 2814.955873 82402.03206 2814.955873 82402.03206
Raw materials consumed 0.420050819 0.206673114 2.032440564 0.063059743 -0.026439297 0.866540936 -0.026439297 0.866540936
Number of factory workers -1671.600624 822.1880417 -2.033112304 0.062983585 -3447.829899 104.6286512 -3447.829899 104.6286512
Number of sales personnel -318.8303339 3379.535328 -0.094341471 0.926276513 -7619.87253 6982.211862 -7619.87253 6982.211862
Capacity utilization 2671.774972 6148.000447 0.434576249 0.670996445 -10610.17249 15953.72244 -10610.17249 15953.72244
Sale price 1203.207774 412.0216291 2.920253912 0.011934685 313.0891609 2093.326388 313.0891609 2093.326388
Offi ce expenses -0.639034144 3.12265515 -0.204644482 0.841019127 -7.385120456 6.107052167 -7.385120456 6.107052167
Electricity 1.584361104 1.392194658 1.138031305 0.275646976 -1.423292599 4.592014807 -1.423292599 4.592014807
Looking at the p values of the coefficient of the independent variables, we can observe that raw
materials, number of factory workers, sales personnel, capacity utilization, office expenses and
electricity have p values greater than 0.05. it can thus be concluded that the variables have no
significant influence on the value of the profits. Only the sale price has an impact on the profit
generated.
Alternate regression model
The most preferable model to accommodate the high profit values will ne the exponential
regression model.
Question 3
a. The Vogel’s Approximation method
Destination 1 Destination 2 Destination 3 Supply
Origin 1 20 17 4 120
Origin 2 35 10 5 60
Demand 40 30 110 180

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This is a balancing problem that dictated that demand should be equal to the supply
The first step is to calculate the penalty for the rows and the columns.
For the rows this is the difference between the lowest value and the second lowest value in the
row
Also, for the column we take the difference between the lowest value and the second lowest
value. The results are indicated in the table below.
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 20 17 4 120 13
Origin 2 35 10 5 60 5
Demand 40 30 110 180
Penalties 15 7 1
The maximum penalty is 15 which is in column 1, taking the column we allocate the maximum
units to the place with the minimum cost. In this case position (1, 1) has the minimum cost so we
have
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 17 4 80 13
Origin 2 35 10 5 60 5
Demand 0 30 110 140
Penalties 15 7 1
Now we remove the fist column form the calculate the penalties for the reaming columns.
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 17 4 80 13
Origin 2 10 5 60 5
Demand 30 110 140
Penalties 7 1

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We again take the highest penalty which is 13 for this case, the minimum cost is 4, we hence
allocate the maximum value possible to the cell and have.
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 (17) 80 0 13
Origin 2 (10) (5) 60 5
Demand 30 30 60
Penalties 7 1
Calculating the new penalties give
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 80
Origin 2 (10) (5) 60 5
Demand 30 30 60
Penalties 10 5
We pick column 2 and allocate the values at the minimal cost which is 10
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 80
Origin 2 30 (5) 30 5
Demand 0 30 30
Penalties 10 5
The new penalties calculation will thus be
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 80
Origin 2 30 (5) 30 5
Demand 30 30
Penalties 5

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The maximum penalty is thus 5 and we allocate the units at this column to give
Destination 1 Destination 2 Destination 3 Supply Penalties
Origin 1 40 80
Origin 2 30 30
Demand
Penalties
The final table of demand and supply will be as follows where the values in bracket represent the
cost.
Destination 1 Destination 2 Destination 3 Supply
Origin 1 40 (20) 80 (4) 120
Origin 2 30 (10) 30 (5) 60
Demand 40 30 110 180
The total cost of supplying the three destinations will thus be
( 40∗20 )+ (30∗10 ) + ( 80∗4 ) + ( 30∗5 )=1570
b. Modified Distribution method
The initial transportation problem derived by the Vogel’s Approximation method is as
indicated in the table below.
Destination 1 Destination 2 Destination 3 Supply ui
Origin 1 40 (20) 80 (4) 120 80
Origin 2 30 (10) 30 (5) 60 30
Demand 40 30 110 180
vj -40 0 0
The fist step in obtaining the optimal solution using the modified distribution method is
to check whether the solution obtained by Vogel approximation is optimal.
The formulas for ui and vj are given by
ui=cij−vj
vj=cij−ui

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After obtaining all the values of ui and vj, we calculate the opportunity cost for all the
empty cells
This is calculated using the formula
dij=cij−(ui+ vj)
Destination 1 Destination 2 Destination 3 Supply ui
Origin 1 40 (20) -63 30 50 (4) (-) 120 80
Origin 2 35 0(10) (-) 60 (5) (+) 60 30
Demand 40 30 110 180
vj -40 0 0
Being that the opportunity cost of destination 2 origin 1 is -63, we conclude that the
solution obtained above is not an optimal solution. We hence modify the allocation of the
cells with the help of a closed loop
Destination 1 Destination 2 Destination 3 Supply ui
Origin 1 40 (20) 30 50 (4) 120 80
Origin 2 (35) (10) 60 (5) 60 30
Demand 40 30 110 180
vj -40 0 0
Checking if the new allocations are optimal by obtaining the opportunity cost
Destination 1 Destination 2 Destination 3 Supply ui
Origin 1 40 (20) 30 50 (4) 120 4
Origin 2 14 0 60 (5) 60 5
Demand 40 30 110 180
vj 16 5 0
The values of the opportunity cost are all >= 0, the solution is thus optimal
The final optimal solution will thus be summarized by the table below.
Destination 1 Destination 2 Destination 3 Supply
Origin 1 40 30 50 120
Origin 2 60 60
Demand 40 30 110 180
This gives an optimal cost of ( 40∗20 )+ (30∗17 )+ ( 50∗4 ) + ( 60∗5 )=1810

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