LINEAR PROGRAMMING AND SENSITIVITY ANALYSIS FOR SHOPPING

One problem when people usually meet in shopping is what people should buy in “clearance sale” with aim to sale again. Usually people confuse to buy between/among two types of products or more. The objective of this problem usually is maximize profit, but this problem will be more difficult because people should decide among different constraints such as sales demand, budget, and capacity. How do we solve this problem?

Solving of this problem will become simple with Linear Programming (LP). LP is a mathematical technique for optimization of linear objective function. In 1939, Leonid Kantorovich, a Russian mathematician, founds this technique. Then, George B. Datzig, John von Neumann, Leonid Khaciyan, and Narendra Karmarkar develop this technique. Clearance sale shirt and t-shirt example below will show that LP can be used to solve this problem. For example, a department store opens New Year clearance sale for shirt and t-shirt. One shirt prices $25 and $15 for a t-shirt. A shop looks this opportunity to make a $400. In addition, his inventory only can keep 100 clothes, whereas he has kept 53 shirts and 29 t-shirts. Then, he considers that he can sell shirt at least twice than t-shirt in one week. How much shirt and t-shirt should he buy in this clearance sale?

Consider that X₁ is for shirt and X₂ is for t-shirt. The objective of this problem is maximize profit from (40-25) X1 + (35-15) X₂ = 15 X₁ + 20 X₂. There are three constraint of this problem. First, sales demand constraint is X₁ – 2X₂ >= 0. Second, budget constraint is 25X₁ +15 X₂ <= 400. Third, capacity constraints is X₁ + X₂ <= (100-(53+29)) ~ 18. So the mathematical equation is:

Max Profit: z = 15X₁ + 20 X₂

Constraint: X₁ – 2X₂ >= 0

25X₁ +15 X₂ <= 400

X₁ + X₂ <= 18

There are many ways to solve this problem such as LP Graphic and Excel Solver. The graphic and the excel solver output will be showed below.

From graphics, there are three alternatives of optimum solutions. First, in point (18, 0) will make profit $270. Second, in point (13, 5) will make profit $295. Third, in point (12.3, 6.15) will make profit $307.5, but this answer is invisible because shirt and t-shirt cannot be decimal. Hence, the nearest options are (12, 6) with profit $300. Furthermore, He should buy 12 shirts and 6 t-shirts. It is very suitable with the answer from Excel Solver table below.

The owner of shop also can analyze the changing in the parameters of this LP problem. This analysis is called sensitivity analysis. In this case, Excel Solver is also used to define the sensitivity analysis.

The table above shows that the final answer will not be change if:

1. The profit of Shirt changes in range from 0 to 20 infinity

2. The profit of T-Shirt changes in range from 15 to infinity

3. The constraint of budget change in range from 290 to infinity

4. The constraint of capacity change in range from 0 to 18.461538462

5. The constraint of demand change in range from -36 to 3

Hence, have fun in shopping.

USING EOQ TO ESTIMATE THE OPTIMIZE YOUR FOOD AND BEVERAGE STORAGE IN HOME

In the big company, inventory management uses EOQ to calculate the most economic quantity of order. The purposes are to know the optimum order with the lowest cost parameter. There are three main basic components to calculate EOQ such as re-ordering cost, demand, and holding cost.

EOQ also can use in the daily activity to calculate of consumer good storage in home. In the daily life, everyone must keep something in home to store vegetables, meats, fruits, milk, book, bath equipments, and other consumer goods. To store everything, people usually don’t consider about the optimal time, size, and quantity when they buy something to stock in home. For example, how much mother should buy milk for her children in the market. This example will be calculate later.

The first component is re-ordering cost (RC). The definition of re-order cost is total expenses involved in repeating supply such as order preparation, communications, transportation, inspection, administration, and other costs. For example, the distance from house to market is 20 km. Assume that the car’s consuming is 10 km per litre and gasoline price is $1 per litre. Therefore, the re-ordering cost is 2 x 20 / 10 x $1 = $4 (the calculation is for 2 ways, go to the market and go home).

The second component is demand (D). Demand in EOQ is define as the estimate of consume or how much the product will be sold in a year. For example, the children can consume 8 cans of milk in a month. Hence the demand is 8 x 12 = 96 bottles annually.

The second component is holding cost or storage cost (HC). It can be defined as expenses involved in keeping and maintain a stock. It is including rent of space, equipment, materials, labour, insurance, security, interest, and other direct expenses. For example, a $200 locker can store 10 cans of milk. It estimates that have a reliability age 10 years. Therefore, we can calculate the holding cost as $50/10/10 = $2 per can annually.

The calculation of EOQ for this example is EOQ = (2xRCxD/HC)^0.5 = (2 x 4 x 96 /2)^0.5 = 19.6 ~ 20 cans. It means that each order or buy is 20 cans of milk. Moreover, the calculation of total annual inventory cost is VC = (RCxD/EOQ) + (HC*EOQ/2) = (4x96/19.6) + (2x19.6/2) = 19.6 + 19.6 = $39.2. In addition, the time between order is 360/5 = 72 days (around 2.4 months).

By this calculation, each family is able to calculate the most optimal storage of daily consumer goods. In this example if the mother buys 15 cans every purchase (below EOQ), the total annual inventory cost will be VC = (RCxD/Q) + HC*Q/2 = (4x96/15) + (2x15/2) = 25.6 + 15 = $40.6. It will be the same if the mother buys 25 cans every purchase (above EOQ), the total annual inventory cost will be VC = (RCxD/Q) + HC*Q/2 = (4x96/25) + (2x25/2) = 15.36 + 25 = $40.36.

For conclusion, each family can save money with EOQ calculation in daily life.