(TIỂU LUẬN) final assignment erp

UNIVERSITY OF ECONOMICS HO CHI MINH CITY
SCHOOL OF INTERNATIONAL BUSINESS - MARKETING

FINAL ASSIGNMENT
ERP

Lecturer:
Group:
Class:

DH44IBC01

YEAR 2020


A. Case 1 – Production Scheduling of the Boeing Company
I.

Overview:
Denotation: Xi as the Production of each month with i={1,2,3,4}
Xj as the Installation of each month with j={1,2,3,4,5}
The Supply for the production (maximum production) is divided into two categories:
Regular time and Overtime.
The cost incurred for the production process is well stated on the basis of Excel
Solver. An additional storage cost of $15,000 has been counted into the cost per unit section.
Since the Supply and Demand for production is not equal to each other, the
company’s capability is not well balanced. Moreover, the production for the previous months
will be counted in the following month. Therefore, we created a Fifth month (Dummy),
where the unused capacity for the production will be stored, thus, the Total Demand and Total
Supply of the problem will be balanced.
II.

Solved by Solver:
1. Step 1: Input data
In the Data array (D5:H12):
- Data in D5, D6, E7, E8, F9, F10, G11, and G12 is the Unit Cost of Production in
either Regular time and Overtime Production of each month. Therefore, the costs will
be equal to the Unit Cost of Production given in the case, respectively.
- As is mentioned in the case, there will be a storage cost of 0.015 ($ million) per
engine stored until it's scheduled installation. Thus, the cost of an engine produced in
Month A and stored to be used in the following months will need to add 0.015 to it’s
Unit Production Cost.
+ The cost of an engine produced in Month 1 then stored in the next 3 months
are described in array (E5:G6).
+ The cost of an engine produced in Month 2 then stored in the next 2 months
are described in array (F7:G8).
+ The cost of an engine produced in Month 3 then stored in the 4th month is
described in array (G9:G10).
- The other data cells do not have value (-) because i>j (refer to the Denotation). For
example, the Production of Month 2 cannot produce engines for installing in Month 1
(which i=2>j=1).


2.

-

Ste
p 2:
Add

constraints
Constraint 1: Total Demand must be equal Total Supply.
Constraint 2: Total Demand of each month cannot less than the Scheduled Installation
of that month.
Constraint 3: Total Supply of each month cannot exceed the Maximum Production of
that month.


-

Constraint 4, 5, and 6: These cells cannot have value because engines cannot be
produced in an “i>j” situation.

3. Step 3: Solve


III.

Solved by QM for Windows:
We convert all the data and constraints from Excel spreadsheet and Solver to QM for
Windows.
1. Step 1 : Setting Variables and Constraints
We choose a Linear Programming Module to solve the problem.

2. Step 2: Adding the inputs data



3. Step 3: Solve

IV.

Conclusions:
In the first month of production, the company should install 10 units from the regular
time of the first month.
In the second month, the company should install 10 units from the regular work time
of the first month and 5 units from regular work time of the second month
For the third month, the company should install 10 units from regular work time of
the third month and 10 units from overtime of the third month
In the fourth month, they will be better off if they install 10 units from regular work
time of the third month and 5 units from the regular work time of the fourth month
The total cost of the recommended solution is 77.4 million dollars, being
minimized so that Boeing can utilize cost savings as much as possible.


B. Case 2 – Crash landing on Inventory Control

a)
Based on Jeong-hyuk’s current inventory policy (which is 50 units of hammer), the
unit holding cost is:

√

√

2 KD
2∗75∗600
=

= 50 => h = $36
h
h
So, the unit holding cost is $36.
The unit acquisition cost is $20.
The unit of holding cost as a percentage of the unit acquisition cost:
36/20 = 180%.

Q=

b)
The unit holding cost: 20%*20 = $4
Based on the unit holding cost, which is equal 20 percent of the unit acquisition cost,
the optimal order quantity is:
2 KD
2∗75∗600
Q=
==
= 150 units of hammer.
h
4
So, the value of TVC is: Annual setup cost + Annual holding cost
600
150
D
Q
+ h*
= 75*
+ 4*
= $600.

= K*
2
150
2
Q

√

√

 SOLVER
Step 1: Insert data into Excel and add the formula:
- Annual Setup Cost: =K*(D/Q)
- Annual Holding Cost: =h*(Q/2)
- Order quantity (Q): =SQRT(2*D*K/h)
- Total Variable Cost: =Annual Setup cost + Annual Holding Cost
(Give a temporary value for Q > 0)


Step 2: Click Data -> Solver. The above box appears.
Step 3: Set Objective (G7), Variable(C11). Select “GRG Nonlinear” Solving Method and tick
“Make Unconstrained Variables Non-negative”.

Step 4: Click Solve -> OK. The results we need appear in below table.

 QM
Step 1: - Click Module -> Inventory.
- Click New -> Economic Order Quantity (EOQ) Model. The above box will appear.
- Add title, choose No reorder point -> OK.

Step 2: Add data according to the information we have in the table below

Step 3: Click Solve and the results we need will appear as the below table.


Based on current inventory policy (Q=50, h=36), the value of TVC is:
600
50
D
Q
TVC = K*
+ h*
= 75*
+ 36*
= $1800
50
2
2
Q

c)
Because the wholesaler typically delivers an order of hammers in 5 working days (out
of 25 working days in average month), the lead time is 5 days and the working days is
300 days (which equals 25*12).
5
L
= 600*
= 10 units of hammer.
So, the reorder point should be: D*

300
WD
 SOLVER
Step 1: Insert data into Excel and add the formula:
- Annual Setup Cost: =K*(D/Q)
- Annual Holding Cost: =h*(Q/2)
- Order quantity (Q): =SQRT(2*D*K/h)
- Total Variable Cost: =Annual Setup cost + Annual Holding Cost
(Give a temporary value for Q > 0)


Step 2: Click Data -> Solver. The above box appears.
Step 3: Set Objective (G7), Variable(C11). Select “GRG Nonlinear” Solving Method and tick
“Make Unconstrained Variables Non-negative”.

Step 3: Click Solve and the results we need will appear as the below table.

 QM
Step 1: - Click Module -> Inventory.
- Click New -> Economic Order Quantity (EOQ) Model. The above box will appear.
- Add title, choose No reorder point -> OK.


Step 2: Add data according to the information we have in the table below

Step 3: Click Solve and the results we need will appear as the below table.


d)
o As calculated by QM, the reorder point after Jeong-hyuk adds a safety stock of 5

hammers is 15 hammers.
- New reorder point = (Average Daily Usage * Average Lead Time Days) + Safety
Stock
50
* 5) + 5 = 15 hammers.
=(
25
This is the result we have when using the unit holding cost in b):

In this case, the annual holding cost for safety stock is $20.
o And we have the different result when using the unit holding cost in a):

But in this case, the annual holding cost for safety stock is much higher ($180).
e)
Using the current inventory policy, Jeong-hyuk has to order 12 times to get the
demand rate (which is 600 hammers/year) leading to the higher setup cost (about
$900) 3 times than the new policy (which is 4 orders with the $300 of setup cost). So
does the holding cost (the current policy: $900, the new policy: $300). Totally,
applying new inventory policy helps him save $1200 per year ($1800-$600 = $1200).


Moreover, using the new policy helps him reduce annual holding cost for safety stock
also (from $180 per year to $20 per year, it is 9 times less).
So, Jeong-hyuk should use the new inventory policy for his store to save $1360
per year.