How to Choose the Best Supplier for the Class in the School

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Introduction of the Problem

Description of the Selected Real-world Scenario

The real-world scenario this task has taken into consideration in the case of a teacher, who would like to choose the best supplier for purchasing the supplies for his class in the school. There are two companies, which offer the required supplies to the teacher, out of which he has to choose one company. The options before the teacher are company A, which offers a 10% discount on every dollar purchase the teacher, makes without any restrictions. On the other hand, the second option is company B, which offers the teacher a 20% discount on all purchases; but company B has a condition to be fulfilled by the teacher for getting the discount of 20%. The condition is that the discount is available in respect of those purchases made by the teacher above the limit of $ 20 and the discount is calculated on the value of purchases exceeding $ 20. If the purchase is, 20 or less than 20 there will be no discount. This effectively means that the teacher has to make in each article a purchase of at least $20 to avail the offer of a discount from company B.

Explanation on the Needs of the Hypothetical Customer

The hypothetical customer here is the teacher. With his intention to purchase the supplies required for his class, he has the objective of making the purchases at the best possible prices available in the market. The objective of the teacher is to save as much as possible in the amount he is going to spend on the purchases of the school supplies. There are two options present before him and it is for him to evaluate the options mathematically to decide on his supply source.

Discussion on the Two Cost Options under Consideration

Cost Option 1

In the first cost option, there is company A, which deals in the supplies required by the teacher. This company A follows a discount policy that it will offer a discount of 10% on all the purchases made by customers like the teacher irrespective of the total amount of purchases made by the customer. This implies that every dollar purchase by the customer becomes automatically entitled to a 10% discount on the total purchases.

Cost Option 2

The second cost option is that the teacher can purchase the supplies from company B. According to the discount policy of company B, the customer will become entitled to a discount of 20% on the dollar purchases, when the customer makes a purchase of more than $ 20 in each item of purchase. Thus, company B places a condition or restriction on the availability of the discount. This implies that the customer can claim the discount of 20% only when on purchases exceeding $ 20 in value, in respect of an individual item of purchase. The discount will be applied on the amount of purchase exceeding $20 and if the value is less than or equal to $ 20 no discount will be offered by company B.

Evaluation of the cost Options (Algebraic Method)

Development of a Linear Algebraic Equation

The following process describes the development of a linear algebraic equation, which comprises clearly defined variables representing the cost of the two different cost options enumerated above.

Option 1: Selection of Company A as the source for the supplies

Concerning the selection of company A as the source of supply, the algebraic equation is developed by assuming x as the cost before the discount is applied and y representing the cost after applying the discount. With this assumption, the following relationship is represented by the equation

y = 0.90 x

Option 2: Selection of Company B as the source for the supplies

In the case of company B, the algebraic equation is developed assuming x as the cost before applying the discount and z as the cost after applying the discount rate of 20%. We have to incorporate the condition of minimum purchase value of $ 20 in the equation. With these assumptions, the algebraic equation takes the following form

z = x; if x is < or x is less than or equal to 20 and

z = 0.80 x + 4; if x is greater than or equal to 20.

Reasoning behind the Way of constructing Each Equation

Under the first option of choosing company A we assumed x be the original cost before any discount is applied. Since the company A offers a discount of 10%, the discount is to be applied on the original cost of x. 10% of x is calculated by multiplying the x with the fraction of 10/100 representing the value of 10%. This gives a result of 0.10x, which represents the discount on the purchase cost of x. For obtaining the cost after discount, we have to subtract the discount from the original cost. Therefore the cost after discount y is calculated as x – 0.10x = (1-0.10) x = 0.90x

The following expression explains the relationship between original cost (x) and cost after discount of 10% (y), which is

y = 0.90 x

For arriving at the algebraic equation under the second option of using Company B for the supplies, the following assumptions and calculations were made. Here also we assume that the original cost be x before any discount is applied and z is the cost after discount. The condition for offering the discount is that the purchase value should be more than $ 20. Therefore, when the cost x is less than or equal to 20 there will be no discount offered by the company. If the purchase value is, less than 20 the cost after discount (z) will remain the same as the original cost (x) and therefore we can algebraically state that

z = x if x is less than or equal to 20 ——– (1)

When the original cost (x) is greater than 20, company B offers a discount of 20%. The discount is applied on the amount of purchase exceeding 20. We calculate the amount exceeding 20 by deducting 20 from the original cost of 20 and this is expressed as x-20. The amount of discount is 20% of (x-20). The discount is calculated by using the expression

(x – 20)x(20/100)=(x-20) x 0.20 = 0.20x – 4.

Since 20% is represented by the fraction 20/200. When multiplying (x – 20) by 20/100 we get (x – 20) x 0.20 which gives the resulting figure of 0.20 x – 4.

Now the cost after discount z is calculated

Z = cost before applying discount – the amount of discount

= x – (0.20x – 4)
= x – 0.20x + 4
= 0.80 x + 4 ——- (2)

From the above equations (1) and (2) we can associate the cost before discount (x) and cost after discount (y) as follows

z=x if x ≤ 20

z = 0.80x + 4 if x>20

Solving the System of Linear Equations

The linear equations arrived at in the previous section can be solved algebraically. We can use either substitution method or the elimination method for determining the point at which the two cost options are likely to yield same results. The following sections show the relevant work including the steps used to solve the system of linear equations. The workings also exhibit the algebraic process employed to find the solution to the set of equations. The workings include all mathematical calculations leading to the identified solutions. Finally, the workings show the solution of the system of equations.

Scenario 1

In the first scenario, we consider the cost before discount is not more than $ 20.

This is expressed as. 0 ≤ x ≤ 20.

Here the system of equations to be solved:

y = 0.90 x ————— (1)
z = x ———————- (2)
y – z ———————- (3)

We can substitute equation (1) for the value of y and equation (2) for the value of z in equation (3). By doing these substitutions we get

0.90 x = x

Subtracting 0.90 x from both the sides of the equation

0 = x – 0.90 x = (1 – 0.90 x) = 0.10 x

That is

0 = 0.10 x
Or
0.10 x = 0

We divide both sides by 0.10 to solve the equation. We get

x=0/0.10=0

Now we substitute the value of x obtained above in equations (1) and (2) we get

y = 0.90 x = 0.90 x 0 = 0

z = x = 0

Therefore the solution to the system of equations is

x = 0 y = 0 and z = 0.

This implies that when the value of purchase by the teacher does not exceed a total value of $ 20, at a point where no purchase is made, the offer from both the companies remain the same. When the amount of purchases increase from the 0 point up to $ 20, it is advisable to purchase from company A, as the company A gives an offer of 10% discount and company B does not offer any discount until the point the purchase value exceeds $ 20.

Scenario (2)

In the second scenario, we consider the case in which the cost before discount is more than $20. That is x > 20.

In this scenario the system of equations that need to be solved is

y = 0.90 x ——————– (4)
z = 0.80 x + 4 —————- (5)
y = z —————————– (6)

We substitute equation (4) for the value of y and equation (5) for the value of z in equation (6). The resulting expression is

0.90 x = 0.80 x + 4

Now we subtract 0.80 x from both sides of the equation. This results in

0.10 x = 4

We divide both sides by 0.10

x=4/0.10=40

We substitute the value of x in equations (4) and (5) we get

y = 0.90 x = 0.90 x 40 = 36
z = 0.80 x + 4 = 32 + 4 = 36

Therefore, the solution to the system of equation is given by

X = 40; y = 36; z = 36.

This implies that when the purchase value exceeds $ 20, the offers of discount from both companies A and B remain equivalent until the purchase value reaches $ 40.

1. When the value of purchase exceeds $ 20 but less than $ 40, it is advisable to purchase from company A to minimize the cost.
2. When the value of purchase is equal to $ 40 the offers of discounts from both the companies are same and purchases can be made from either of the companies
3. When the value of purchases exceeds $ 40, it is advisable to purchase the supplies from company B to minimize the cost.

Graphical Representation of the Scenarios

The graphical representation of the different scenarios include the depiction of each option with clear labeling of the cost to the understanding of the hypothetical customer in respect of each of the options on the graph. The graph also shows the point of intersection of the cost options.

Discussion on Recommendation to the Customer

The first section discusses the graphical representation of the cost solutions explaining the interpretation of the graph for reaching a cost solution.

The first option available to the customer is a discount of 10% offered by Company A in respect of all purchases irrespective of any limits on the value of purchase. In this case, if the customer decides to purchase the supplies with company A, he has to pay 90% of the value of purchases. In the graph, the actual amount that the customer needs to pay on his purchases from company A is depicted by the curve in blue color. Any point intersecting x and y-axis in the blue curve will indicate the price; he has to pay when he opts to make the purchases from company A.

In the case of purchases from company B, the customer will not get any discount until such point where the purchase value exceeds $ 20. For all purchases exceeding $ 20, company B offers a discount of 20%. This scenario is depicted by the red curve in the graph. Any point along this red curve will indicate the price the customer has to pay to make purchases from B taking in to account the discount of 20% offered by company B.

From the graph it can be observed that when the value of purchase (x) is less than 40, the cost curve in color blue is below the cost curve of company B represented by the red curve. This indicates that company A offers better discount when the purchase value reaches the total of $ 40.

When the value of purchases (x) is greater than 40, the cost curve of company B in color red is below the cost curve of company A depicted in blue. This indicates that for purchase value exceeding $ 40 Company B offers better price by allowing a discount of 20%.

It can also be seen from the graph that at the point where x = 40, both the red and blue curves intersect each other, implying that at this point purchasing from company A or Company B makes no different.

From the above discussion, it may be inferred that the best decision for the customer to purchase from any source depends on the amount, he is going to spend. If it is less than $ 40, the teacher can choose company A and when it is more than $ 40, he can choose company B.

The decision of the customer can be arrived using the algebraic equations also. The following discussion explains this.

From the calculations we made earlier, it can be seen that the cost function of company A is represented by

y = 0.90 x

The cost function of Company B is represented by the equation

z = x if x is less than or equal to 20 and
z = 0.80 x + 4 if x is greater than 20

This leads the customer to the following conclusions.

When the purchase value (x) is less than 40, we have y is less than z. Therefore, in this situation company A offers better price when the purchase value is less than $ 40.

When the purchase value (x) is greater than 40, we have z is greater than y. This implies that company B offers better prices when the purchase value is greater than $ 40.

When the purchase value (x) is $ 40 we have y =z. This means that at the point where the purchase value is just $ 40 the prices of both the companies are the same and the customer is free to purchase from either company A or company B.

Thus other things remaining equal, the customer will benefit from making small purchases from company A and large purchases from company B.

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