5.2 Explain how the two advances in technology identified in AC 5.1 have impacte

Question

5.2 Explain how the two advances in technology identified in AC 5.1 have impacted upon supply chain operations (800 - 1,000 words)

(5.1Describe the key elements and functions of two advances in technology within the supply chain (800 - 1,000 words) )

ANS:

Two significant advances supply chain technology are described below: Data Integration: Supply chain optimization and efficiency is a target every logistics provider is aiming for where they reduce overhead and ultimately speed up customer delivery. Distributing information throughout the supply chain is one of the biggest challenges 3PL providers face because there are often multiple software platforms, applications and partnerships in place for each level of the supply chain, from procurement to manufacturing to distribution and everything in between. These systems are often proprietary and don’t always circulate data easily with one another. Hence data may have to be extracted manually and feed into the system. This leads to a delay in generating output and even erroneous data feed into the system. Integration is the key to an efficient supply chain because it breaks down the walls between software applications. The power of optimization provided by integration is why logistics providers are looking to assimilate their enterprise resource planning, transportation management, warehouse management and partner systems together. The TMS (Transport Management System) is designed to facilitate integration and synthesize data from ERPs, WMS, and partner systems. Customized technology solutions such as SaaS, support integration, resulting in reduced warehousing costs, better inventory management, and quicker shipping and delivery. Key elements include- data from different systems and the software used to gather, interpret and present data in an acceptable format. There are two elements to providing the data and data integration services: Establish a service-oriented architecture (SOA) to enable access to services throughout the enterprise and outside its firewalls with customers, suppliers and partners if desired. Service-oriented architecture can be implemented with Web-services This is done to make the functional building-blocks accessible over standard Internet protocols that are independent of platforms and programming languages. These services can represent either new applications or just wrappers around existing legacy systems to make them network-enabled. Implementers commonly build SOAs using web services standards Enable SOA data and data integration services you need to have the underlying data integration processes working. Processes include ETL (extract, transform and load), ELT (extract, load and transform), EAI (enterprise application integration), EII (enterprise information integration), data cleansing, data profiling, data auditing, metadata management. Malware protection system: This is used because a lot of data has to be used for the Analytics part where the customer buying pattern, and other relevant information are analyzed for business expansion. Unfortunately these set of data also contains valuable customer 2. Radio Technology: One of the greatest problems for any supply chain manager is the increase in anomalies pertaining to an order when it is in transit. This not only leads to losses but eventually also has a negative impact on a brand. If a product is lost during transit, the supplier bears all the costs and moreover, they have to bear with the ensuing cost of disruption. By adoption of Radio Frequency Identification (RFID) technology, a company can effectively monitor every product both at the production line and in the supply line. RFID chips are placed on all items which helps employees to quickly detect any anomalies in an order. The product or order can be tracked at any point in time and any issues related to the transit or product can be discovered through online systems. This has been in market since a long time but hasn't been used efficiently because of the cost involved. Key elements for this system are chips that are attached to the product are: RFID Tag: It is a transponder mounted on a substrate that is programmed with unique information. Tags are activated when they pass through a radio frequency field produced by the antenna of the reader. Reader/Scanner: A reader (also known as transceiver or interrogator) handles radio communications through the antennas. A reader can transmit signals to a tag, synchronize a tag with the reader, and interrogates all or part of the tag's contents. Thus, the main purpose of the reader is to transmit and collect information. Antenna: An antenna consists of a coil with a winding and a matching network. The primary purpose of an antenna is to radiate the electromagnetic waves produced by the reader, and in the same manner, receives radio frequency signals from the transponder. Reader Interface Layer: The reader interface layer is used as a conduit between the readers and the hardware elements such as computers.

Explanation / Answer

enter the following program in matlab to get the tone mary had a little lamb

fsamp=44100;

% 44100 samples per sec

% 261.63 Hz => Nper = 44100/261.63

song = zeros(1,fsamp/4);

A = [1.0000 -2.4584 1.8539 -0.2387 -0.0466 -0.1614 -0.1004 0.1980 -0.0420];

fc=261.63;

fd=293.66;

ff=349.23;

fg=392.00;

Nfc=ceil(fc);

Nfd=ceil(fd);

Nff=ceil(ff);

Nfg=ceil(fg);

Npc = ceil(44100/fc);% period of note c

Npd = ceil(44100/fd);% period of note d

Npf = ceil(44100/ff);% period of note f

Npg = ceil(44100/fg);% period of note g

b=0.1;

Nper=Npc;

M = Nfc/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1 yout1];

Nper=Npd;

M = Nfd/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1];

Nper=Npc;

M = Nfc/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1];

Nper=Npg;

M = Nfg/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1];

Nper=Npg;

M = Nfg/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1];

Nper=Npf;

M = Nff/2;

xin = zeros(1,M*Nper);

for ii=1:M

xin(1+(Nper-1)*ii) = 1;

end

yout=filter(b,A,xin);

Ny=size(yout,2);

u=[0:Ny-1];

wt=sin(pi*u/Ny);

wt=wt.*wt;

yout1=yout.*wt;

song=[song yout1];

plot(song)

Fs=44100;

bits=16;

soundsc(song,fsamp);

audiowrite('hb.wav', song, Fs, 'BitsPerSample', bits);

amp=10;

audioplayer(amp*song,fsamp)

the tone is as follows

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.