Question

WeHaul Trucking is planning its truck purchases for the coming year. It allocated $600,000 for the purchase of additional trucks, of which three sizes are available. A large truck costs $150,000 and will return the equivalent of $15,000 per year to profit. A medium-sized truck costs $90,000 and will return the equivalent of $12,000 per year. A small truck costs $50,000 and will return the equivalent of $9,000 per year. WeHaul has maintenance capacity to service either four large trucks, five medium-sized trucks, or eight small trucks, or some equivalent combination. WeHaul believes that it will be able to hire a maximum of seven new drivers for these added trucks. The company cannot spend more than one/half of the total funds it actually spends to purchase medium-sized trucks. (Hint: this is not necessarily one half of the total funds it has allocated for the purchase of additional trucks).

a) Formulate a linear programming model to be used for determining how many of each size of truck to purchase if the company wants to maximize its profit. Ignore the time value of money. Provide the linear programming variables, the objective function, and the constraints for the problem.

b) At optimality, how much profit will result and what is the optimal combination of trucks?

c) Using your sensitivity analysis output, provide two sensitivity analysis interpretations. One must be for the objective function and one must be for one of the constraints

d) Now suppose that there is a requirement that WeHaul must purchase at least one of each size truck. Also, the number of larger trucks cannot be greater than the number of medium trucks. Write the constraint(s) for this requirement.  

Answer 1

A.

The boxes in green, are decision variables
The boxes in blue are constrains
The boxes in orange is the objective function

The solution:

		large	Medium	Small
Dec Var	No of trucks	4	3	0	7	<=	7
	Each truck capacity	<=	<=	<=	Total driver
		4	5	8
	Cost	150000	90000	50000
	Profit	15000	12000	9000
	Total cost	600000	270000	0
			<=	less than 50% in Med
			435000
	Total profit	96000
	Objective func	Maximization

B

Total profit

96000

	large	Medium	Small
No of trucks	4	3	0

C.

Variable Cells
			Final	Reduced
	Cell	Name	Value	Gradient
	$C$3	No of trucks large	4	3000
	$D$3	No of trucks Medium	3	0
	$E$3	No of trucks Small	0	-3000
Constraints
			Final	Lagrange
	Cell	Name	Value	Multiplier
	$D$8	Total cost <=	270000	0
	$F$3	No of trucks	7	12000

Objective Cell (Max)
	Cell	Name	Original Value	Final Value
	$C$11	Total profit <=	0	96000
Variable Cells
	Cell	Name	Original Value	Final Value	Integer
	$C$3	No of trucks large	0	4	Contin
	$D$3	No of trucks Medium	0	3	Contin
	$E$3	No of trucks Small	0	0	Contin
Constraints
	Cell	Name	Cell Value	Formula	Status	Slack
	$D$8	Total cost <=	270000	$D$8<=$D$10	Not Binding	165000
	$F$3	No of trucks	7	$F$3<=$H$3	Binding	0
	$C$3	No of trucks large	4	$C$3<=$C$5	Binding	0
	$D$3	No of trucks Medium	3	$D$3<=$D$5	Not Binding	2
	$E$3	No of trucks Small	0	$E$3<=$E$5	Not Binding	8

	Objective
Cell	Name	Value
$C$11	Total profit <=	96000
	Variable			Lower	Objective		Upper	Objective
Cell	Name	Value		Limit	Result		Limit	Result
$C$3	No of trucks large	4		1.8	63000		1.8	63000
$D$3	No of trucks Medium	3		0	60000		3	96000
$E$3	No of trucks Small	0		0	96000		0	96000

D.

With new added constraints:

we add 2 constraints:

3	<=	3
Med > =Large

No of trucks	3.00000	3.00000	1.00000
Min trucks	>=	>=	>=
	1	1	1

THe solution:

		large	Medium	Small
Dec Var	No of trucks	3.00000	3.00000	1.00000	7	<=	7
	Min trucks	>=	>=	>=	Total driver
		1	1	1
	Each truck capacity	<=	<=	<=	3	<=	3
		4	5	8	Med > =Large
	Cost	150000	90000	50000
	Profit	15000	12000	9000
	Total cost	450000.0	270000	50000
			<=	less than 50% in Med
			385000
	Total profit	90000.00
	Objective func	Maximization