Inventory Control

Question 1
The demand of a certain part is 1000 parts/year and its cost is Rs. 1000/part. The orders are placed based on the economic order quantity (EOQ). The cost of ordering is Rs. 100/order and the lead time for receiving the orders is 5 days. If the holding cost is Rs. 20/part/year, the inventory level for placing the orders is ________ parts (round off to the nearest integer).
A
14
B
8
C
18
D
22
GATE ME 2022 SET-2   Industrial Engineering
Question 1 Explanation: 
Inventory control:
Annual demand (D) = 1000 units
Lead time (LT) = 5 days
Inventory level for placing the order = Re-order level (ROL)
Rate of consumption (d) =D/365
ROL=d \times LT=\frac{1000}{365} \times 5=13.69 \approx 14 \; units

Question 2
The product structure diagram shows the number of different components required at each level to produce one unit of the final product P. If there are 50 units of on-hand inventory of component A, the number of additional units of component A needed to produce 10 units of product P is _________ (in integer).

A
160
B
50
C
110
D
170
GATE ME 2022 SET-1   Industrial Engineering
Question 2 Explanation: 


To produce 10 units of 'P'
No. of units of 'A' required = (4x10)+(2x3x2x10) = 160 units
Net requirement of 'A' = 160 - 50 = 110 units
Question 3
Which one of the following is NOT a form of inventory?
A
Raw materials
B
Work-in-process materials
C
Finished goods
D
CNC Milling Machines
GATE ME 2022 SET-1   Industrial Engineering
Question 3 Explanation: 


CNC milling machines will not be treated as inventory.
Question 4
A factory produces m(i=1,2,...m) products, each of which requires processing on n(j=1,2,...n) workstations. Let a_{ij} be the amount of processing time that one unit of the i^{th} product requires on the j^{th} workstation. Let the revenue from selling one unit of the i^{th} product be r_i and h_i be the holding cost per unit per time period for the i^{th} product. The planning horizon consists of T \;(t=1,2,...T) time periods. The minimum demand that must be satisfied in time period t is d_{it}, and the capacity of the j^{th} workstation in time period t is c_{jt}. Consider the aggregate planning formulation below, with decision variables S_{it} (amount of product i sold in time period t ), X_{it} (amount of product i manufactured in time period t ) and I_{it} (amount of product i held in inventory at the end of time period t.

max\sum_{t=1}^{T}\sum_{i=1}^{m}(r_iS_{it}-h_iI_{it})
subject to
S_{it}\geq d_{it}\;\;\forall i,t
< capacity constraint >
< inventory balance constraint >
X_{it},S_{it}, I_{it} \geq 0;\; I_{i0}=0

The capacity constraints and inventory balance constraints for this formulation are
A
\sum_{j}^{m}a_{ij}X_{it}\leq c_{jt}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-d_{it}\; \forall \;i,t
B
\sum_{i}^{m}a_{ij}X_{it}\leq c_{jt}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-d_{it}\; \forall \;i,t
C
\sum_{i}^{m}a_{ij}X_{it}\leq d_{it}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-S_{it}\; \forall \;i,t
D
\sum_{i}^{m}a_{ij}X_{it}\leq d_{it}\;\forall \; i ,t\text{ and }I_{it}=I_{i,t-1}+S_{it}-X_{it}\; \forall \;i,t
GATE ME 2021 SET-2   Industrial Engineering
Question 4 Explanation: 
\begin{aligned} m & \rightarrow i \ldots m \leftarrow \text { product } \\ n & \rightarrow i \ldots n \leftarrow \text { workstation } \\ a_{\bar{j}} & \rightarrow \text { time } \end{aligned}
r_{i} \rightarrow selling price
h_{i} \rightarrow holding cost
T \rightarrow t=1,2, \ldots T
d_{i t} \rightarrow demand of product in time t
c_{j t} \rightarrow capacity of workstation in time t
S_{i t} \rightarrow Number of product sold in time t
x_{i t} \rightarrow Number of product produced in time t
I_{i t} \rightarrow Number of product i hold in inventory at end of period t
Capacity constraint
a_{i j} x_{i t} \leq c_{j t}
Inventory constraint
l_{i t}=I_{i, t-1}+x_{i t}-S_{i t}
Question 5
For a single item inventory system, the demand is continuous, which is 10000 per year. The replacement is instantaneous and backorders (S units) per cycle are allowed as shown in the figure.

As soon as the quantity (Q units) ordered from the supplier is received, the backordered quantity is issued to the customers. The ordering cost is Rs. 300 per order. The carrying cost is Rs. 4 per unit per year. The cost of backordering is Rs. 25 per unit per year. Based on the total cost minimization criteria, the maximum inventory reached in the system is ________ (round off to nearest integer).
A
181.94
B
1319.09
C
1137.14
D
1802.16
GATE ME 2020 SET-2   Industrial Engineering
Question 5 Explanation: 
Given data: D=10000 items/year, C_{o}=Rs. 300/order,
C_{h}=Rs. 4/unit/year, C_{b}=Rs. 25/unit/year,
For minimum tool cost
\begin{aligned} \text { Quantity ordered, } Q=& \sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}+C_{h}}{C_{o}}\right)}\\ &=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25+4}{25}\right)} \\ Q &=1319.09 \text { unit } \end{aligned}
Now, for minimum cost, optimum units backordered
\begin{aligned} (Q-S) \times C_{h}&=(S) \times C_{b}\\ S\left(C_{b}+C_{h}\right) &=Q \times C_{h} \\ S &=\frac{1319.09 \times 4}{(25+4)}=181.94 \text { units } \end{aligned}
Maximum inventory in the system, Q_{\max }=Q-S
\begin{aligned} &=1319.09-181.94 \\ &=1137.15 \text { unit } \\ &\approx 1137 \text { units } \end{aligned}
Alternate:
Maximum inventory in system =\sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}}{C_{b}+C_{h}}\right)}
Q_{\max }=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25}{29}\right)}=1137.147 \text { units }
Question 6
Consider two cases as below.

Case 1: A company buys 1000 pieces per year of a certain part from vendor 'X'. The changeover time is 2 hours and the price is Rs. 10 per piece. The holding cost rate per part is 10% per year.

Case 2: For the same part, another vendor 'Y' offers a design where the changeover time is 6 minutes, with a price of Rs. 5 per piece, and a holding cost rate per part of 100% per year. The order size is 800 pieces per year from 'X' and 200 pieces per year from 'Y'.

Assume the cost of downtime as Rs. 200 per hour. The percentage reduction in the annual cost for Case 2, as compared to Case 1 is___________ (round off to 2 decimal places).
A
5.32
B
8.19
C
5.82
D
6.22
GATE ME 2020 SET-1   Industrial Engineering
Question 6 Explanation: 
Given Data : 1000 pieces/year from 'X'
Changeover time =2 hrs.
cost of downtime = Rs.200/hour
So, Total cost of downtime
\begin{aligned} 2 \times 200 &=\text { Rs. } 400 / \text { downtime } \\ C &=\text { Rs. } 10 / \text { piece } \\ \text { Holding cost, } C_{h} &=10 \% \text { of Rs. } 10 \\ C_{h} &=\text { Rs. } 1 / \text { unit/year } \end{aligned}
So, total cost for Case I :
\begin{array}{l} =\text { Material cost }+\text { Downtime cost }+\text { Inventory Holding cost } \\ =1000 \times 10+(1 \times 400)+\frac{1000}{2} \times 1 \end{array}
Total cost for case I=Rs.10,900 /-
Case II : Order quantity 800 units from X and 200 units from Y.
For Y: Change overtime =6min.=0.1hour
Downtime cost =0.1 \times 200= Rs. 20 /-
Unit cost, C= Rs.5/piece
Holding cost, C_{h}=100% of unit cost = Rs. 5/-
So, total cost for case II :
\begin{aligned} &=\text { cost for }^{\prime} X+\text { cost for ' } Y \\ &=\left(800 \times 10+400+\frac{800}{2} \times 1\right)+\left(200 \times 5+20+\frac{200}{2} \times 5\right) \\ &=8000+800+1000+520 \end{aligned}
Total cost for case II=Rs.10,320 /-
So, percentage reduction in total cost of case II :
=\frac{10900-10320}{10900} \times 100=\frac{580}{10900} \times 100=5.32 \%
Question 7
For an assembly line, the production rate was 4 pieces per hour and the average processing time was 60 minutes. The WIP inventory was calculated. Now, the production rate is kept the same, and the average processing time is brought down by 30 percent. As a result of this change in the processing time, the WIP inventory
A
decreases by 25%
B
increases by 25%
C
decreases by 30%
D
increases by 30%
GATE ME 2020 SET-1   Industrial Engineering
Question 7 Explanation: 
WIP = Throughput rate x Processing time
(WIP)1 = 4 x 1 = 4 units
(Processing Time)2 = 1 x 0.7 = 0.7
Throughput rate = 4
(WIP)2 = 4 x 0.7 = 2.8
% decrease = 30%
Question 8
In the Critical Path Method (CPM), the cost-time slope of an activity is given by
A
\frac{Crash \; Cost -Normal \; Cost}{Crash \; Time}
B
\frac{Normal \; Cost}{Crash \; Time -Normal \; Time }
C
\frac{Crash \; Cost}{Crash \; Time -Normal \; Time }
D
\frac{Crash \; Cost -Normal \; Cost}{Normal \; Time -Crash \; Time}
GATE ME 2020 SET-1   Industrial Engineering
Question 9
The annual demand of valves per year in a company is 10,000 units. The current order quantity is 400 valves per order. The holding cost is Rs. 24 per valve per year and the ordering cost is Rs. 400 per order. If the current order quantity is changed to Economic Order Quantity, then the saving in the total cost of inventory per year will be Rs.______(round off to two decimal places).
A
254
B
563
C
478
D
941
GATE ME 2019 SET-2   Industrial Engineering
Question 9 Explanation: 
Annual demand (D) = 10000 units
Holding cost (C_{c}) = 24 /unit / year
Ordering cost (C_{0}) = 400 /order
Order quantity (Q)=400 units
\begin{aligned} \left.\mathrm{TC}\right|_{\mathrm{Q}=400}-\left.\mathrm{TC}\right|_{\mathrm{EOQ}}&=\left[\frac{\mathrm{Q}}{2} \times \mathrm{C}_{\mathrm{C}}+\frac{\mathrm{D}}{\mathrm{Q}} \times \mathrm{C}_{0}\right]\\ &-[\sqrt{2 \mathrm{DC}_{0} \mathrm{C}_{\mathrm{c}}}] \\ &=\left[\frac{400}{2} \times 24+\frac{10000}{400} \times 400\right] \\ &-[\sqrt{2 \times 10000 \times 400 \times 24}] \\ &=(4800+10000)-(13856) \\ &=943.60 \end{aligned}
Question 10
A local tyre distributor expects to sell approximately 9600 steel belted radial tyres next year. Annual carrying cost is Rs. 16 per tyre and ordering cost is Rs. 75. The economic order quantity of the tyres is
A
64
B
212
C
300
D
1200
GATE ME 2018 SET-2   Industrial Engineering
Question 10 Explanation: 
\begin{aligned} D &=9600 \\ C_{h} &=\text { Rs. } 16 / \text { year } \\ C_{0} &=\text { Rs. } 75 / \text { order } \\ E O Q &=\sqrt{\frac{2 D C}{C_{h}}}=\sqrt{\frac{2 \times 9600 \times 75}{16}} \\ &=\sqrt{1200 \times 75}=300 \end{aligned}
There are 10 questions to complete.

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