# Decision-Making Models

LINEAR PROGRAMMING 5

LINEAR PROGRAMMING

Problem 1

This is a decision problem with various alternatives available based on choice taken at any point based on the available options. By considering a number decision variables and the possible outcomes, the model can be represented as an integer programming problem with either Yes/No decisions (1/0). The following is a formulation of the scenario.

Salvaging cost of \$60000 — Govt Agency Approval

Conditions:

Approval Evaluated at a substantial value – withheld

» » » » low/moderate – Granted

Evaluations:

Submission of a bid:

Bid too low (< \$30000) – No evaluation (Replica build @ \$135 000)

Bid >= \$30000 – Conduct evaluation (Cost \$100 000) – Salvaging not approved

Let us assume that the probability of bid are:

Low (< \$30000) = 0.5

High \$30000 <= Bid <= \$60000) = 0.5

Network Diagram Assumptions

1. At no point in time are more than one decision acceptable

2. The probability of making both low and acceptable bids are 0.5 0.5

3. On the acceptance of the bid and approval, salvage is automatically not considered, on contrary, it is considered

Problem 2

This problem requires a formulation of the solution to help West Digital Company meet the monthly demand of HDDs but at the minimum possible cost. The Company has two assembly points and a number of points under which the DVDs ass before they are finally distributed to the market. The understanding of the problem is that, the formulation should give the number of units that each point should have at any given time to ensure maximum production at minimum cost.

Solution Formulation

NB: Values in 1000’s

Let the HDDs getting to Assembly point 1 be F1

» » » » » point 2 be F2

Let the HDDs leaving the Packaging point 10 be P10

» » » » » point 11 be P11

But F1+F2 = P10 + P11 or (F1+F2 ) — (P10 + P11) = 0

Let the decision variable xij be the number of HDDs in every station

Minimize:

8X1 + 10X2 + 1.6X3+ 1.8X4+ 1.4X5+ 1.7X6+ 0.8X7+ 1.0X8+ 1.2X9+ 0.6X10+ 0.5X11

X13 + X14 + X15+ X23+ X24+ X25 — X36+ X37+ X38+ X39+ X46+ X47+ X48+ X49+ X56+ X57+ X58+ X59 = 0

X36+ X37+ X38+ X39+ X46+ X47+ X48+ X49+ X56+ X57+ X58+ X59 — X610 + X611+ X710+ X711+ X810+ X811+ X910+ X911 = 0

X1 + X2+ X3+ X4+ X5+ X6+ X7+ X8+ X9+ X10+ X11 >= 4000

X9 <= 0.95

X10 <= 2.1

X11 <= 2.3

c) Solution interpretation:

The solution to the above problem shows the number of items to be moved from one point of the processing process to the other to obtain the highest possible amount of distributed HDDs at the meeting the demand requirement of at least 4000 HDDs in a month. In the excel attached (quiz 2), the column ‘flow’ shows the number of units moved from one unit to another every time.

Problem 3

This problem requires a formulation of a solution that allows the scheduling of the staffs to ensure that the daily demands of customer services are met. The call center have estimated the number of staffs required for each duration based on the average number calls. Having set the number of staffs to 70, the problem in this case is scheduling 70 or less staffs to meet every duration’s needs of the customers.

To solve this problem, some assumptions must be made. We must assume a given time when the shifts are expected to start. In this case, we are assuming the shifts to start at midnight. Given that each staff works for 8 hours, and each shift goes for 4 hours, it only implies that each staff will go exactly two shifts. This implies that, if the number of staffs beginning work at 4.00p.m are more than or equal to the number of staffs required in the last shift (8.00p.m to midnight), then no new staff will be assigned in during this time.

Number of staff=70]o sol

Number of shifts=3 @8hrs

 Time period Duration Av. no of calls 0.00-4.00am 4.00am-8.00am 8.00am-noon Noon-4.00pm 4.00pm-8.00pm 8.00pm-midnight

Given that each staff works 8hrs

No new staff required from 8.00pm since staff reporting at 4.00pm serves up to midnight

Let Xt be the number of staff starting at period X

The total workforce to minimize is SXt<=70

During time period:

X2+X3>=26

X3+X4>=30

X4+X5>=20

1. Let say 13 staffs reported at midnight: This restricts the number of those staffs beginning work to be 13 instead of 8. This lowers the number of staffs assigned for the next shift.

The 13 staffs will work for two shifts

That is from midnight to 8.00am

This implies that

13<=X1>=8

13<=X1+X2>=12

At period 3 new staffs will reports

Therefore:

Max. SXt<=57

X3+X4>=30

X4+X5>=20

1. Given that between period 3 and 4 the number of staffs should be <=30: At this point, the call center has restricted the number staffs to 30 staffs or less during the period between 8a.m – 4p.m. This changes the formulation as follows, since during this specific time, the number of staffs should not exceed 30 at the center.

Max. SXt<=70

X1+X2>=12

X2+X3>=26

X3+X4<=30(the new condition)

X4+X5>=20

Problem 4

The main problem here is to help General Motors company identify the most economical warehouses and close the once which are not so economical. The decision should be arrived at based on the cost of distributing the vehicles to the market through the warehouses to meet the customer demands.

Let the number of 1000’s of vehicles be Xij

X11 + X12 + X13+X14 <= 12

X21+X22+X23+X24 <= 15

X31+X32+X33+X34 <= 13

X41+X42+X43+X44 <= 10

X51+X52+X53+X54 <= 16

The Cost:

52X11+20X12+30X13+62X14 <= 2000

16X21+45X22+11X23+32X24 <= 900

10X31+70X32+35X33+48X34 <= 1500

28X41+22X42+58X43+25X44 <= 1000

42X51+48X52+24X53+30X54 <= 850

Annual demand for W-1,, W— 2, W— 3 and W— 4

From the solutions in excel, the warehouses with values of 1 are the once to be considered, those with values of 0 are the once to be closed. It implies that, when the company closes these specific warehouses, they’ll reducing the operation cost by greatest percentage.