# Inventory Control

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 5
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 6
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 6 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 7
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 7 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}
 Question 8
A food processing company uses 25,000 kg of corn flour every year. The quantity-discount price of corn flour is provided in the table below: The order processing charges are Rs. 500/order. The handling plus carry-over charge on an annual basis is 20% of the purchase price of the corn flour per kg. The optimal order quantity (in kg) is _____________
 A 1500 B 1600 C 1800 D 2000
GATE ME 2016 SET-2   Industrial Engineering
Question 8 Explanation:
The optimal order quantity (in kg) is 1500.
 Question 9
The annual demand for an item is 10,000 units. The unit cost is Rs. 100 and inventory carrying charges are 14.4% of the unit cost per annum. The cost of one procurement is Rs. 2000. The time between two consecutive orders to meet the above demand is _______ month(s).
 A 2 months B 3 months C 5 months D 6 months
GATE ME 2016 SET-1   Industrial Engineering
Question 9 Explanation:
\begin{aligned} Q &=\sqrt{\frac{2 D C_{0}}{C_{h}}} \\ &=\sqrt{\frac{2 \times 10000 \times 2000}{14.4}} \\ &=1666.67 \end{aligned}
(N)No. of orders per year $=\frac{D}{Q}$
\begin{aligned} &=\frac{1}{N} \times 12=\frac{12}{6} \\ &=\text{2 months} \end{aligned}
 Question 10
The annual requirement of rivets at a ship manufacturing company is 2000 kg. The rivets are supplied in units of 1 kg costing Rs. 25 each. If it costs Rs. 100 to place an order and the annual cost of carrying one unit is 9% of its purchase cost, the cycle length of the order (in days) will be ________
 A 92 B 76 C 98 D 88
GATE ME 2015 SET-3   Industrial Engineering
Question 10 Explanation:
\begin{aligned} Q &=\sqrt{\frac{2 D C_{0}}{C_{h}}} \\ Q &=\sqrt{\frac{2 \times 2000 \times 100}{0.09 \times 25}} \\ &=421.637 \\ N &=\frac{D}{Q}=4.7434 \\ T &=\frac{1}{N}=0.2108 \text { year } \\ &=76.948 \text { days } \end{aligned}
There are 10 questions to complete. 