Merton truck company (case study)
Merton Truck Company has been experiencing difficulties related to their financial
performance, and they do not know which the optimal product mix is to maximize profits
and performance. Group 2 was asked to make recommendations and analyze the given
situation to eliminate the difficulties and come up with the right product mix and the optimal
solutions considering different alternatives and scenarios.?
Model of linear programming
Solution
LP is solved by using excel solver the output of the case study is given below
Sensitivity report
Variable
Cell
$G$
6
$H$
6
Final
Valu
e
Name
weekly profit model
101
weekly profit model
102
Reduce
d
Objective
Coefficie
nt
Cost
Allowabl
e
Allowabl
e
Increase
Decrease
2000
0
3000
2000
500
1000
0
5000
1000
2000
Fina
l
Valu
e
Shado
w
Constrai
nt
Allowab
le
Allowabl
e
Price
R.H. Side
Increase
Decrease
engine assembly total
4000
2000
4000
500
500
metal stamping total
model 101 assembly
total
model 102 assembly
total
6000
500
6000
500
1000
4000
0
5000
1E+30
1000
3000
0
4500
1E+30
1500
Constraints
Cell
$I$1
1
$I$1
2
$I$1
3
$I$1
4
Name
Activity report
Objective cell
Cell
$I$1
0
Name
Original
Value
profit total
Final
Value
0
11000000
Variable cell
Cell
$G$
6
$H$
6
Name
Original
Value
Final Value
Integer
weekly profit model 101
0
2000 Contin
weekly profit model 102
0
1000 Contin
Constraint
Cell
$I$1
1
$I$1
2
$I$1
3
$I$1
4
Name
Cell Value
Engine assembly total
4000
Metal stamping total
Model 101 assembly
total
Model 102 assembly
total
6000
4000
3000
Formula
$I$11<=$K$1
1
$I$12<=$K$1
2
$I$13<=$K$1
3
$I$14<=$K$1
4
Status
Slac
k
Binding
0
Binding
Not
Binding
Not
Binding
0
BY USING R
> library (lpSolve)
> library (ROI)
> plugins <- ROI_available_solvers()[,c("Package", "Repository")]
> plugins
> lp <- OP (objective = L_objective (c (3000,5000), c ("x", "y")),
+
constraints = L_constraint(L = rbind(c(1,2), c(2,2), c(2,0),c(0,3)),
+
dir = c ("<=", "<=", "<=","<="),
+
rhs = c (4000,6000,5000,4500)),
+
maximum = TRUE)
> sol <- ROI_solve(lp, solver = "lpsolve")
> sol
Optimal solution found.
The objective value is: 1.100000e+07
> sol$solution
x y
2000 1000
1000
1500
Question 1 (a)
Best product mix for Merton is 2000 units of model 101 and 1000 units of model 102.
(b)
If we increase one unit then the best product mix is 1999 units of model 101 and 1001 units of
model 102. Extra unit of engine assembly worth of 2000 dollars.
Cell
$G$
6
$H$
6
Cell
$I$
9
Name
Valu
e
Weekly profit Model 101
1999
Weekly profit Model 102
1001
Original
Value
Name
Profit Total
Final Value
11000000
11002000
(c)
If the capacity is increase 4100
hours, then the extra unit of
assembly worth of 2000*100 =
$ 200,000. It is 100 times of part (b).
(d)
Allowable increase in the engine assembly capacity is 500 units. The profit at 4000 and 4500 is
same there will be no change in production decision.
Question # 2
The shadow price of the engine assembly is 2000$. This is the price which the manager willing to
pay to the third party. The maximum number of machine hours is 500 hrs. So by these value there
is no effect on the profit
Question # 3 (a)
From the data Merton should not produce Model 103. This is because the contribution of $2000
is far below contributions of $3000 and $5000 derived from trucks 101 and 102 and thus not
worthwhile to deploy resources into its production compared to other truck models.
(b)
As the reduced cost of the model 103 is 350. If we increase 351 in the present contribution means
the contribution is 2351 them it will make worthy.
Model
101
Model
102
Model
103
Weekly profit
0
857.1428
6
2857.142
9
By increasing the contribution, the production of model 103 is 2857.149 and model 102 is 857.14
and no model of 101 produce.
Question # 4
The best product mix give the profit of 2.4 million per month. If Merton wants to assemble engine
overtime which increase the overhead value about 0.75 million. With the increase of the overhead
the profit is decrease. The new profit is 1.65 million, which is less than the previous so Merton
cannot produce engine on overtime.
Question # 5
The optimal mix is 2500 unit of model 101 and 500 unit of model 102.
The objective function for this problem is 9000X1 + 5000X2
Weekly profit
Model
Model
101
102
2500
500
The profit is maximizing $25million by increasing the model 101 three times then the model 102.