Project 1: Application of LP in Manufacturing and Project Management

This project report applies the principles of Linear Programming (LP) to solve optimization problems in manufacturing and project management. The assignment focuses on maximizing profit by determining optimal production quantities for various products, considering constraints such as resource availability, man-hours, and demand. The student employs both Excel Solver and MATLAB to find solutions, providing detailed formulations of the objective function and constraints. The report includes sensitivity analysis and a similar problem to further demonstrate the application of LP in different scenarios, such as maximizing the lifespan of products purchased by a company. The report also includes a detailed analysis of the constraints and the optimal solutions obtained using both Excel and MATLAB, showcasing the practical application of LP in decision-making for resource allocation and production planning.

Running head: Application of LP in Manufacturing and Project Management
Project 1
Name of the Student
Name of the University
Author Note

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
Question 9:
Given problem is
Maximize, Profit P = 10a + 15b + 22c + 17d
Where, a, b, c and d are the quantities of product 1, 2, 3 and 4.
The quantities must be less than the maximum demand of each product in the coming week.
Hence, a <= 50, b <= 60, c <= 85 and d <= 70.
The man-hours needed in three stages of 4 products are given by the following table.
20% of the man-hours which is previously employed in stage B can be employed in stage A
and 20% of the man-hours which is previously employed in stage C can be employed in stage
A.
The nominal time available till next week for every stage A, B, C is 160, 200 and 80 man-
hours respectively.
Hence,
2*a + 2*b + 3*c + 4*c + 0.2*(2a + 2b + 4c + 2d) + 0.3(a + 3b + 3c + d) <= 160
2a + 2b + 4c + 2d <= 200

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
a + 3b + 3c + d <= 80
Product 1 units assembled / product 4 units assembled must be in between 0.9 and 1.15.
0.9 <= (a/d) <= 1.15
Or, a/d => 0.9 Or, a >= 0.9d Or, a – 0.9d >= 0
Or, a/d <= 1.15 Or, a <= 1.15d Or, a -1.15d <= 0
Hence, the linear programming problem is the following.
Objective function:
Max 10a + 15b + 22c + 17d
Subjected to Constraints:
a <= 50, b <= 60, c <= 85 and d <= 70
2*a + 2*b + 3*c + 4*d + 0.2*(2a + 2b + 4c + 2d) + 0.3(a + 3b + 3c + d) <= 160
2a + 2b + 4c + 2d <= 200
a + 3b + 3c + d <= 80
a – 0.9d >= 0
a -1.15d <= 0
Excel solver solution:

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
Product 1 quantity Product 2 quantity Product 3 quantityProduct 4 qua
8 0 21 8
Profit(£) 678
Constraints
2*a + 2*b + 3*c + 4*c + 0.2*(2a + 2b + 4c + 2d) + 0.3(a + 3b + 3c + d) <= 160 157.9 <= 160
2a + 2b + 4c + 2d 116 <= 200
a + 3b + 3c + d 79 <= 80
a – 0.9d 0.8 >= 0
a -1.15d -1.2 <= 0
a >= 50 8 <= 50
b >= 60 0 <= 60
c >= 85 21 <= 85
d >= 70 8 <= 70
Sensitivity report:
Cell Name Origi
nal
Value
Final
Value
$B$7 Profit(£) Product 2 quantity 0 678
Cell Name Origi
nal
Value
Final
Value
Integ
er
$A$3 Product 1 quantity 0 8 Integ
er
$B$3 Product 2 quantity 0 0 Integ
er

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
$C$3 Product 3 quantity 0 21 Integ
er
$D$3 Product 4 quantity 0 8 Integ
er
Cell Name Cell
Value
Formul
a
Stat
us
Sl
ac
k
$B$11 2*a + 2*b + 3*c + 4*c + 0.2*(2a + 2b + 4c +
2d) + 0.3(a + 3b + 3c + d) <= 160 Product 2
quantity
157.9 $B$11<
=$D$11
Not
Bindi
ng
2.1
$B$12 2a + 2b + 4c + 2d Product 2 quantity 116 $B$12<
=$D$12
Not
Bindi
ng
84
$B$13 a + 3b + 3c + d Product 2 quantity 79 $B$13<
=$D$13
Not
Bindi
ng
1
$B$14 a – 0.9d Product 2 quantity 0.8 $B$14>
=0
Not
Bindi
ng
0.8
$B$15 a -1.15d Product 2 quantity -1.2 $B$15<
=0
Not
Bindi
ng
1.2
$B$16 a >= 50 Product 2 quantity 8 $B$16< Not 42

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
=$D$16 Bindi
ng
$B$17 b >= 60 Product 2 quantity 0 $B$17<
=$D$17
Not
Bindi
ng
60
$B$18 c >= 85 Product 2 quantity 21 $B$18<
=$D$18
Not
Bindi
ng
64
$B$19 d >= 70 Product 2 quantity 8 $B$19<
=$D$19
Not
Bindi
ng
62
$A$3=
Integer
$B$3=I
nteger
$C$3=I
nteger
$D$3=
Integer
Hence, the optimum solution for which the maximum profit is generated is 8 units of product
1 quantity, 0 units of product 2 quantity, 8 units of product 3 quantity and 21 units of product
4 quantity. The maximum profit is £ 678.
Solution by MATLAB:

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
In MATLAB for solving maximization problem the problem is first converted into
minimization problem and then solved.
Hence, the objective function becomes
Minimize -(10a + 15b + 22c + 17d)
Such that,
a <= 50, b <= 60, c <= 85 and d <= 70
2*a + 2*b + 3*c + 4*d + 0.2*(2a + 2b + 4c + 2d) + 0.3(a + 3b + 3c + d) <= 160
Or, 2.7a + 3.3b + 4.7c + 4.7d <= 160
2a + 2b + 4c + 2d <= 200
a + 3b + 3c + d <= 80
a – 0.9d >= 0 Or, -a + 0.9d <= 0
a -1.15d <= 0
a, b, c and d are all integers.
MATLAB code:
obj = [-10;-15;-22;-17]; % specifying objective function
intvars = [1 2 3 4]; % specifying all the quantities are positive integers
A = [2.7,3.3,4.7,4.7;2,2,4,2;1,3,3,1;-1,0,0,0.9;1,0,0,-1.15]; % specifying left hand side of
constraints. All the inequalities are converted to A <= b form
b = [160;200;80;0;0]; % specifying RHS of constraints. All the inequalities are converted to
A <= b form

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
lbound = [0;0;0;0]; % specifying lowest bound of quanities of four types of product
ubound = [50;60;85;70]; % specifying highest bound of quanities of four types of product
opt = optimoptions('intlinprog','RootLPAlgorithm','primal-simplex'); % selecting algorithm to
solve LP problem
prod_quantity = intlinprog(obj,intvars,A,b,[],[],lbound,ubound,[],opt); % Obtaining solution
in vector form
Profit = -sum(obj.*(prod_quantity));% obtaining maximum profit
sprintf('The optimum product quanties for which maximum profit of £%g can be generated
are \n product 1= %i \n product 2 = %g \n product 3 = %i \n product 4 =
%i',Profit,prod_quantity(1),prod_quantity(2),prod_quantity(3),prod_quantity(4))
Output:
optimizationmatlab
LP: Optimal objective value is -687.383628.
Cut Generation: Applied 3 Gomory cuts.
Lower bound is -684.648716.
Heuristics: Found 1 solution using rounding.
Upper bound is -661.000000.
Relative gap is 3.47%.

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
Branch and Bound:
nodes total num int integer relative
explored time (s) solution fval gap (%)
6 0.02 2 -6.780000e+02 7.363770e-01
8 0.02 3 -6.810000e+02 2.932551e-01
9 0.03 3 -6.810000e+02 0.000000e+00
Optimal solution found.
Intlinprog stopped because the objective value is within a gap tolerance of the optimal value,
options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within
tolerance,
options.IntegerTolerance = 1e-05 (the default value).
ans =
'The optimum product quantities for which maximum profit of £681 can be generated are
product 1= 9

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
product 2 = 1
product 3 = 20
product 4 = 8'
Hence, by the primal-simplex method the solution of the linear programming problem with
integer number of quantities generate a maximum profit of £681 when 9 product 1, 1 product
2, 20 product 3 and 8 product 4 quantities are produced.

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
A similar problem like problem 9 is formulated below.
Similar problem:
The four similar-purpose used products produced by a factory namely P1, P2, P3 and P4 has
lifetimes of 5, 10, 15 and 25 years respectively. A company XYZ uses every products of the
factory and want to have buy products such that maximum lifespan can be utilized by the
company. The company has limited space available and can only keep maximum of 500
products in inventory.
The price of the four products are $150, $250, $300 and $ 500 for each unit of product P1,
P2, P3 and P4 respectively. Company can only bear up to $500000 for buying products from
factory.
The company places orders at the start of month and receives the products from factory at the
end of the month. The factory only produces products after receiving orders. Hence, the
factory has one month two produce the orders. There are three stages in making a product and
the man-hours needed per day to produce unit product is given below.
P1 P2 P3 P4
Stage 1 2 3 4 5
Stage 2 3 5 6 8
Stage 3 1 2 2 3
The maximum man-hours that can be employed in a day in stage 1 is 30, in stage 2 it is 40,
and in stage 3 it is 50.
There is some minimum demand of the products in the company as per it employees. The
employees need at least 3 P1, at least 5 P2, no demand for P3 and at least 8 P4 products.

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APPLICATION OF LP IN MANUFACTURING AND PROJECT MANAGEMENT
LP formulation:
The linear programming problem can be formulated as given below.
Variables: Quantities of P1, P2, P3 and P4 indicated by a, b, c and d as bought by the
company.
Objective function:
Maximize total lifespan.
Total lifespan = 5P1 + 10P2 + 15P3 + 25P4
Subjected to constraints:
a + b + c+ d <= 500 (limited space constraint)
150a + 250b + 300c + 500d <= 500000 (limited price constraint)
(2*a + 3*b + 4*c + 5*d)/30 <= 30 (stage 1 man-hours constraint)
(3*a + 5*b + 6*c+ 8*d)/30 <= 40 (stage 1 man-hours constraint)
(a + 2*b + 2*c + 3*d)/30 <= 50 (stage 1 man-hours constraint)
Lower bound constraints:
a >= 3
b >= 5
c >= 0
d >= 8
Additionally, the number of product quantities are needed to be positive integers.
MATLAB code: