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海湾城市搬家公司的管理
海湾城市搬家公司是当地的一家公司,专门从事城际搬家。在提交给其赞助商的商业计划中,海湾城市的货运能力至少为36吨。公司正在更换所有的卡车，包括1吨接卡车和2.5吨搬运车类卡车。1吨的卡车将由一名工人负责,而大货车将有四个人员来完成大型的搬家行动。海湾城搬家公司雇佣48个工人，目前设施了40辆卡车。接卡车公司花费了24000美元,移动面包车耗资60000美元。公司希望在卡车的货运能力保持至少36吨的情况下供应最低的投资,并且不需要任何新员工或运输设施。
虽然连续性假设是不严谨的(因为每个购买卡车的数量必须是整数),但是使用一个线性规划模型来确定海湾城市搬家公司的最优卡车和货车购买量。你会发现选择最佳的解决方案是可行的。
准备一份报告,列出这些选项中的几个,并讨论各自的利弊。
Management at bay city movers
Bay City Movers is a local company that specializes in intercity moves. In the business plan submitted to its backers, Bay City has committed itself to a total trucking capacity of at least 36 tons. The company is in the process of replacing its entire fleet of trucks with 1 ton pick up trucks and 2.5 ton moving van type trucks. The 1 ton pick up trucks will be manned by one worker, whereas the large vans will utilize a total of four personnel for larger moves. Bay City Movers currently employs 48 workers and has facilities for 40 trucks. Pick up trucks cost the company $24,000 and the moving vans cost $60,000. The company wishes to make a minimum investment in trucks that will provide a trucking capacity of at least 36 tons while not requiring any new hires or trucking facilities.
Although the continuity assumption is violated (since the number of each truck purchased must be integer), use a linear programming model to determine the optimal purchase of pick up trucks and vans for Bay City Movers. You will find that alternative optimal solutions are possible.
Prepare a report detailing several of these options and discuss the pros and cons of each. Among the alternatives, you should present in your report are the following:
Purchasing only one type of truck.
Purchasing the same number of pick up trucks as moving vans.
Purchasing the minimum total number of trucks.
Also include in your report pertinent sensitivity information you feel would be interest to management at Bay City Movers.
Decision Variables
X1 = No. of Pick up Trucks
X2 = No. of Moving Vans
Constraints
In Addition to Non-negativity Constraint for the decision variables, there are three functional constraints:
No. of workers
Trucking Capacity.
Facilities
No. of Workers:
The pick up trucks uses only one worker, while the large vans utilize 4 workers. Currently Bay City movers has maximum of 48 workers.
X1 + 4X2 ≤ 48
Trucking Capacity:
Pick up Trucks has a Capacity of 1 ton while moving vans has capacity of 2.5 ton. Company wants its trucking Capacity to be atleast 36 tons.
X1 + 2.5X2 ≥ 36
Facilities:
Bay City movers have facilities of 40 trucks, including both pick up trucks and moving vans.
X1 + X2 ≤ 40
Non-negativity of Decision Variables:
No. of pick up trucks and moving vans cannot be negative and it’s impossible. Therefore,
X1, X2 ≥ 0
The Mathematical Model
Min 24000X1 + 60000X2 (Objective Function)
Subject to:
X1 + 4X2 ≤ 48 (No. of workers)
X1 + 2.5X2 ≥ 36 (Trucking Capacity)
X1 + X2 ≤ 40 (Facilities)
X1, X2 ≥ 0 (Non-Negativity)
Detailed Step by Step Description of the Problem and The Graphical Solution Algorithm:
For solving the Bay City Movers Problem, we first should have all the limitation’s understanding. Further, we should know how these limitations will affect the optimization. In this case, Company’s total trucking capacity and no. of workers are necessary for understanding the optimal solution. Then only it is possible to write a linear equation that represents these limitations.
Variable X1 will symbolize no. of Pick up Trucks and variable X2 will embody no. of Moving Vans. It is also seen that 1 ton pick up trucks will be manned by one worker. Therefore 1X1 represents the no. of workers required for 1 ton pick up truck. Similarly, 4X2 represents the no. of workers for large moving vans. The total no. of workers available is 48 and therefore total workers for both trucks and vans must be less than or equal to 48. The linear equation X1 + 4X2 ≤ 48 places right in all the conditions.
Next constraint we look at is the trucking capacity. By looking at the problem, we see that Bay City has committed itself to a total trucking capacity of at least 36. Trucks have the capacity to carry 1 ton only while moving vans have capacity of 2.5 tons. So the total capacity of trucks and moving vans must be more than or equal to 36 tons and therefore the linear equation, X1 + 2.5X2 ≥ 36 satisfies the conditions.
Considering the next constraint, are the facilities of the company. Bay City Movers has total facilities of 40 trucks. Company has a limit of 40 trucks including only pick up trucks and moving vans. Therefore Total no. of trucks must be less than or equal to total trucking facility i.e., 40. The linear equation that represents these constraints is X1 + X2 ≤ 40.
The final constraint and most important one is the non- negative constraint. There cannot be negative no. of trucks and X1 and X2 must be greater than zero.
After understanding the constraints, we must remember the goal. In this case it is minimization in the investments in trucks. We want to minimize the investment in trucks with minimum capacity of 36 tons. We can see that the Cost of Pick up truck is $24000 and the cost of moving vans is $60000. We combine these costs to the variables that symbolize pick up trucks and moving vans. This can be seen in the equation Min 24000X1 + 60000X2. This is known as the Objective Function of the problem.
While making a graph of the problem, the horizontal axis represents No. of trucks and the vertical axis represents no. of vans. First we graph the no. of workers constraint. According to the constraint equation, if we put X2 equal to 0, than X1 would be 48. So we draw a point of X-axis that represents X1. Similarly, we put X1 equal to 0, then X2 value comes out to be 12. So we plot the point at Y-Axis which represents X2. When a line is then drawn across the graph joining these two points, it will represent the time constraint.
Now we draw the trucking capacity constraint. This is determined by finding the point in the X-axis that represents the case of Pick up trucks capacity. This point is 36. Now we draw a point on the Y-axis that represents moving vans capacity. This point is 14.4. Joining these two dots with a line, we get the Trucking capacity constraint line.
Finally we look at the Facility constraint. This is represented by the point in X axis, where X2 is 0 and X1 is 40, and the point in Y axis, where X1 is 0 and X2 is 40. Line joining these two points will represent the facility constraint equation.
All the constraints will be graphed just till the limit of those constraints. Such as, No. of workers is limited to 48, trucking capacity has a minimum limit of 36 and facilities are limited to 40. When all of these lines are drawn on graph and the areas excluded are removed, including the negative areas, the area that is left is called the feasible region. Any of the points of the feasible region is represented as solution to the given constraints. In order to minimize investment, a point where two constraint lines cross must be taken into consideration.
As we know, a Formulation of Bay City Movers problem is:
X1 = the number of Pick up trucks
X2 = the number of Moving vans
Max Objective Function 24000X1+60000X2
Subject to:
C1 = X1 + 4X2 ≤ 48 (constraint of No. of workers)#p#分页标题#e#
C2 = X1 + 2.5X2 ≥ 36 (constraint of Trucking Capacity)
C3 = X1 + X2 ≤ 40 (constrain of Facilities)
C4 = X1, X2 ≥ 0 (Non-Negativity)
1. Drawing the graph to find the feasible region.
There are three such points in this problem. After analysing the graph, optimal solution has the following optimal values of X1 and X2.
X1 (No. of pick up trucks) = 16
X2 (No. of moving vans) = 8
2. Binding and Non-Binding Constraints.
C1: X1 + 4X2 ≤ 48; (Binding Constraints)
C2: X1 + 2.5X2 ≥ 36; (Binding Constraints)
C3: X1 + X2 ≤ 40; (Redundant Constraints)
C4: X1, X2 ≥ 0; (Non-Binding Constraints)
C1: X1 + 4X2 ≤ 48; (Binding Constraints)
C2: X1 + 2.5X2 ≥ 36; (Binding Constraints)
C3: X1 + X2 ≤ 40; (Non-Binding Constraints)
C4: X1, X2 ≥ 0; (Redundant Constraints)
3. Applying to constraints.
According to the Optimal Values, Bay City movers are considering 16 no. of trucks and 8 no. of moving vans to minimize the investment.
On applying to constraints,
C1: X1 + 4X2 ≤ 48; 48 ≤ 48 (constraint of No. of workers Satisfied)
C2: X1 + 2.5X2 ≥ 36; 36 ≤ 36 (constraint of Trucking Capacity Satisfied)
C3: X1 + X2 ≤ 40; 24 ≤ 40 (constrain of Facilities Satisfied)
C4: X1, X2 ≥ 0; 16, 8 ≤ 0 (Non-Negativity Satisfied)
This also shows that Bay City Movers have extra slack of 16 facilities.
4. Applying to Objective function.
Bay City movers are seeking to minimize the investment on trucks. According to the optimal solution, the minimum no. of trucks is 16 and minimum no. of moving vans is 8. Therefore,
24000(16) + 60000(8) = 864000
So the minimum investment would be $864,000.
Algebraic Solution Algorithm:
The mathematical model
X1 + 4X2 ≤ 48 (No. of workers)
X1 + 2.5X2 ≥ 36 (Trucking Capacity)
X1 + X2 ≤ 40 (Facilities)
X1, X2 ≥ 0 (Non-Negativity)
Algebraic Method
Considering all the equations as binding equation, that is all with = sign.
X1 + 4X2 = 48
X1 + 2.5X2 = 36
X1 + X2 = 40
X1 = 0
X2 = 0
Here are 5 equations and 2 unknowns. There can be at most 5! [2! (5-2)!] = 10 basic solutions. Solving each pair of equations randomly and then checking the solutions of constraints for feasibility. We find that there are 6 non feasible solutions and 4 feasible solutions. We also get more than one alternative optimal values.
Therefore, from the above table, we conclude that optimal value to be decided is X1 = 16 and X2 = 8, which give the optimal solution of $864,000. We do not consider the other optimal value because, there the no. of moving vans is 0 and as per the problem, we want to invest in both the trucks.
Computer Implementation:
The computation is done by using WINQSB software. I have used Linear and Integer Programming for solving the above problem. On opening a new model, Variables and Constraints numbers and names are added. Also minimization factor is added. Values are also entered as the table below shows:
After all the values are added, Solve problem icon is clicked and following message appears:
This has confirmed that the problem is solved and has got an optimal solution. After this we proceed towards the Problem Table which seems as below:
Decision Variable: Identification of the unknown variable to be determined and representation these variables in term of algebraic symbols is defined as Decision variable (Sivarethinamohan, 2008). It is usually designated as X1, X2, X3, etc. In this case, No. of trucks and No. of Vans are the decision variables.
Solution Value: A solution value is feasible solution that optimizes the Objective Function. Minimizing or maximizing the solution value gives us the objective function. For Bay City Movers, Solution values for No. of trucks is 16 and No. of vans is 8.
Unit Cost or Profit: Unit cost or Profit is the coefficient of decision variable in the objective function. In this problem, Unit cost for No. of trucks is 24,000 and for no. of Vans in 60,000.
Total Contribution: Total contribution of a variable to the objective function is the multiplication of solution values and the unit cost. For Bay City Movers, Total contribution for No. of trucks is $384,000 and for No. of vans is $480,000.
Reduced Cost: Reduced cost is the cost for any decision variable which is not currently in the optimal solution indicates how much better that coefficient must be before that variable enters at a positive level. The reduced cost for any variable already in the optimal solution is automatically 0 (Winston, 2009). In this case, both the decision variables have 0 reduced costs.
Basis Status: This represents whether the variable has a basic status or not i.e., the variable has a value of 0 or a positive value with respect to constraints. In Bay City movers, both have basic status.
Allowable Minimum/ Maximum: It is the range of values of objective function coefficient, in which the optimal solution values does not change. This is also known as the Range of Optimality. Within this range, objective function’s value can change but, the values of optimal solution remain unchanged. For this problem, Range of optimality for unit cost of No. of trucks is 24,000 to maximum and range of optimality for unit cost of No. of Vans is from minimum to 60,000.
Objective function: “The objective function is a mathematical formulation of the criterion by which all decisions should be evaluated” (Boyer, 2010). The goal is to either minimize or maximize the objective function. It is the expression that a Company is seeking to optimize. The Objective function is attained by adding the total contributions. In Bay City Movers Problem, Objective function is 864,000.
Constraints: “Constraints are the restriction placed on the decision scenario” (Boyer, 2010). It is the restraining condition that can change the optimal values of the objective function. In this problem, there are total 4 constraints, namely, No. of workers, capacity, facilities and non-negativity.
Left hand Side: When the problem is solved, the left hand side value of a constraint equalizes the combination of the decision variable value times it co-efficient in the constraint. For example, constraint (C1) for Bay City Movers was X1 + 4X2 ≤ 48. After the problem was solved, 1(16) + 4(8) = 48; 48 = 48.
Direction: It denotes the relation between RHS and the LHS of the linear problem.
Right Hand Side: The RHS of a constraint is a constant value that represents the requirement of maximum (<, =) or minimum (>).
Slack: It represents the amount by which the available amount of resource exceeds its usage by the activities (Sivarethinamohan, 2008). “The slack of any activity is the amount of the time the activity can be delayed beyond its earliest” (Winston, 2009). It is associated with the “<” constraints.
Surplus: Surplus is the amount representing the excess of the LHS over the minimum requirement (Sivarethinamohan, 2008). It Associated with “>” constraints.
Shadow Price: Shadow Price is the price hidden in the perception of buyer’s mind unknown to anyone else and always in the shadow. This is the value that buyer is willing to pay for an extra unit of constrained resources cause this resource will generate some extra profit. For example, if Bay City Movers suddenly had to increase the capacity by 37, the objective function value would increase by $24000 (the shadow price) to $888,000.
Allowable Minimum / Maximum RHS: The range of RHS values of a resource in which the shadow price remains constant. It is also known as the range of feasibility and represents whether the solution of set of binding constraint equations is a feasible solution or not. For example, Bay City Movers may have 30 to 44 trucking capacity and the shadow price will remain $24,000 for a one-unit increase.
Sensitivity Analysis:
Sensitivity Analysis is done after the optimum solution is obtained by solving the linear programming. It determines whether the change in the given model’s coefficients will leave the current solution unchanged or not and if not how a new optimum can be obtained affectively. It refers to the solution’s sensitivity towards the changes in resources, changes in profit composition and addition of new constraints (Sivarethinamohan, 2008).
1. Cost Coefficients Sensitivity Analysis:
Bay City Mover’s optimal solution was verified to be 16 No. of trucks and 8 No. of Vans. By this, Bay City Movers was able to minimize its investment to $864,000. After the optimal solution is determined, sensitivity of the coefficients of the objective function is seen and their effect on the optimal point. Sensitivity Analysis creates a set of values, having no effect on the optimal solution, which is applied to the coefficients of the objective function. However, optimal solution is un-affected; the objective function value will change as the coefficient changes.
The table above provides the range of optimality, which determines the allowable minimum and allowable maximum of the coefficient at which optimal solution do not change. The range of the unit cost for X1 is between 24000 and maximum Infinity, while the range for the unit cost of X2 is between minimum Infinity and 60000. The use of this method is better because it allows the values to be visible.
2. Right Hand Side Sensitivity Analysis:
The RHS analysis emphasizes on the effect on the optimal value by unit change in the total number of a binding constraint or profits. The single unit change is also known as the shadow price. For Bay City Movers, it is clear that its second binding constraint, X1 + 2.5X2 ≥ 36 has 24000 shadow price. The shadow price will be constant as long as the constraint (capacity) remains within the range of 30 and 44. These values are also known as the lower and upper boundaries. The use of this method is better because it allows the values to be visible but a small problem is there that the value may not always be an integer or rounded off. For Example, as seen in range of feasibility of first constraint (No. of workers), upper boundary is 57.6. This number makes it difficult to graph. The graph is just for giving a general visual of the points that would lie if they were graphed. Any ways, by looking at the workings, it is simple to visualize that when trucking capacity increases/decreases by 1 unit, the total profit also increases/decreases by $24,000. This is where the shadow price is ascertained to be 24,000. In Bay City Mover’s situation, once the ranges are calculated, they will come to know all of the different combinations of trucking capacity and the No. of workers that can be used to still calculate the optimal value.
Alternatives Discussion and Managerial Implications:
1. Purchasing only one type of truck:
For analysing this situation, we consider one truck at a time.
Only pick up trucks:
In this case, we consider only Pick Up trucks and not the Moving Vans. We edit WINQSB by removing one variable of No. of vans (X2).
When we click to solve it, we get the solution to be optimal and feasible.
After clicking Ok tab, we proceed to WINQSB table to analyse the situation.
Here, Decision variable is only one that is No. of truck, having a solution value of 36. Its unit cost is $24,000 which gives a total contribution of $864,000. This is also the Objective function of the Problem. Its reduced cost is 0. Its range of optimality is ranging from 0 to maximum infinity.
By removing one type of truck, No. of worker’s LHS is reduced to 36 and having a slack of 12 and its range of feasibility has changed to 36 to infinity. In capacity constraints, LHS is saturated with 0 surplus. Its shadow price remains unchanged and range of feasibility changes to 0 to 48. The last constraint is of Facility. Its LHS is fully untouched and has slack of 40. It has 0 Shadow price and Range of feasibility is from 0 to infinity.
Equations for this Alternative:
Decision Variables: X1 = No. of Trucks
Objective Function: Min 24000X1
Subject to:
X1 ≤ 48 (No. of workers)
X1 ≥ 36 (Trucking Capacity)
X1 ≤ 40 (Facilities)
X1 ≥ 0 (Non-Negativity)
Only Moving Vans:
In this case, we consider only Moving vans and not the pick up trucks. We edit WINQSB by removing one variable of No. of trucks (X1).
When we click to solve it, we get the unfeasible solution.
2. Purchasing the same no. of pick up trucks as moving vans:
On putting the same no. of pick up trucks as the no. of moving vans, we get the solution that is infeasible.
3. Purchasing the minimum total no. of trucks:
Here, in this alternative, we have minimized the unit cost to 1 for both the trucks. This gives us different range of optimalities and different objective function. For no. of trucks, Range of optimality ranges from 0.40 to maximum infinity, and for no. of vans, it ranges from minimum infinity to 2.5. The objective function attained is 24.
Among constraints, Workers and Capacity constraints have 0 Slack or Surplus and have shadow price of -1 and 2 respectively. Worker’s range of feasibility ranges from 36 to 57.6 and capacity’s ranges from 30 to 48. As for the case of facility, there is a slack of 25 and its range of feasibility ranges from 24 to infinity.
Equations for this Alternative:
Decision Variables: X1 = No. of Trucks
X2 = No. of Vans
Objective Function: Min X1 + X2
Subject to:
X1 + 4X2 ≤ 48 (No. of workers)
X1 + 2.5X2 ≥ 36 (Trucking Capacity)
X1 + X2 ≤ 40 (Facilities)
X1, X2 ≥ 0 (Non-Negativity)
Managerial Implication:
X1 = the number of Pick up trucks
X2 = the number of Moving vans
Max Objective Function 24000X1+60000X2
Subject to:
C1 = X1 + 4X2 ≤ 48 (constraint of No. of workers)
C2 = X1 + 2.5X2 ≥ 36 (constraint of Trucking Capacity)
C3 = X1 + X2 ≤ 40 (constrain of Facilities)
C4 = X1, X2 ≥ 0 (Non-Negativity)
The impact on the optimal value:
If we increasing the right-hand side of constraint # 3 by 50, the impact is none because the shadow price is zero and the change is within the sensitivity limits.
The impact of deleting the second constraint on the optimal solution:
Because this constraint is a binding constraint, the impact of deleting the second constraint is a better optimal solution.
The impact of adding the new constraint on the optimal solution:
The impact of adding the new constraint on the optimal solution is none since the current solution satisfies this added condition.
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